Claim: \(f(x) = \frac 1 x\) is uniformly continuous on \((c, \infty)\) for any \(c > 0\).

• Note that \begin{align*} {\left\lvert {x} \right\rvert}, {\left\lvert {y} \right\rvert} > c > 0 \implies {\left\lvert {xy} \right\rvert} = {\left\lvert {x} \right\rvert}{\left\lvert {y} \right\rvert} > c^2 \implies \frac{1}{{\left\lvert {xy} \right\rvert}} < \frac 1 {c^{2}} .\end{align*}

• Letting \(\varepsilon\) be arbitrary, choose \(\delta < \varepsilon c^2\).

  • Note that \(\delta\) does not depend on \(x, y\).

• Then \begin{align*} {\left\lvert {f(x) - f(y)} \right\rvert} &= {\left\lvert {\frac 1 x - \frac 1 y} \right\rvert} \\ &= \frac{{\left\lvert {x-y} \right\rvert}}{xy} \\ &\leq \frac{\delta}{xy} \\ &< \frac{\delta}{c^2} \\ &< \varepsilon ,\end{align*} which shows uniform continuity.

Claim: \(f\) is not uniformly continuous when \(c=0\).

• Toward a contradiction, let \(\varepsilon < 1\).
• Let \(x_n = \frac 1 n\) for \(n\geq 1\).
• Choose \(n\) large enough such that \({\left\lvert {x_n - x_{n+1}} \right\rvert} = \frac 1 n - \frac 1 {n+1} < \delta\).
  • Why this can be done: by the archimedean property of \({\mathbb{R}}\), choose \(n\) such that \({1\over n} < \varepsilon\).
  • Then \begin{align*} {1 \over n} - {1\over n+1} = {1 \over n(n+1)} \leq {1\over n} < \varepsilon\quad\text{since }n+1 > 1 .\end{align*}
• Note \(f(x_n) = n\) and thus \begin{align*} {\left\lvert {f(x_{n+1}) - f(x_{n})} \right\rvert} = (n+1) - n = 1 > \varepsilon ,\end{align*} a contradiction.

• Set \(f_N(x) = \sum_{n=1}^N {x^n \over n!}\).

  • Then by definition, \(f_N(x) \to f(x)\) pointwise on \({\mathbb{R}}\).

• For any compact interval \([-M, M]\), we have \begin{align*} {\left\lVert {f_N(x) - f(x)} \right\rVert}_\infty &= \sup_{-M \leq x \leq M} ~{\left\lvert {\sum_{n=N+1}^\infty {x^n \over {n!}} } \right\rvert} \\ &\leq \sup_{-M\leq x \leq M} ~\sum_{n=N+1}^\infty {\left\lvert { {x^n \over {n!}} } \right\rvert} \\ &\leq \sum_{n=N+1}^\infty {M^n \over n!} \\ &\leq \sum_{n=0}^\infty {M^n \over {n!} } \quad\text{since all additional terms are positive} \\ &= e^M \\ &<\infty ,\end{align*} so \(f_N \to f\) uniformly on \([-M, M]\) by the M-test.

  Here we’ve used that \(e^x\) is equal to its power series expansion.

• Thus \(f\) converges on any bounded interval, since any bounded interval is contained in some larger compact interval.

Claim: \(f\) does not converge on \({\mathbb{R}}\).

• If \(\sum_{n=0}^\infty g_n(x)\) converges uniformly on a set \(A\), then \(\sup_{x\in A} {\left\lvert {f_n(x)} \right\rvert} \to 0\).
• But taking \(A = {\mathbb{R}}\) and \(g_n(x) = {x^n \over n!}\), we have \begin{align*} \sup_{x\in {\mathbb{R}}} {\left\lvert {g_n(x)} \right\rvert} = \sup_{x\in {\mathbb{R}}} {x^n \over n!} = \infty .\end{align*}

Claim: If \(F_N\to F\) uniformly with each \(F_N\) continuous, then \(F\) is continuous.

• Follows from an \(\varepsilon/3\) argument: \begin{align*} {\left\lvert {F(x) - F(y} \right\rvert} \leq {\left\lvert {F(x) - F_N(x)} \right\rvert} + {\left\lvert {F_N(x) - F_N(y)} \right\rvert} + {\left\lvert {F_N(y) - F(y)} \right\rvert} \leq \varepsilon\to 0 .\end{align*}

  • The first and last \(\varepsilon/3\) come from uniform convergence of \(F_N\to F\).
  • The middle \(\varepsilon/3\) comes from continuity of each \(F_N\).

• Now setting \(F_N\mathrel{\vcenter{:}}=\sum_{n=1}^N f_n\) yields a finite sum of continuous functions, which is continuous.

• Each \(F_N\) is continuous and \(F_N\to F\) uniformly, so applying the claim yields the desired result.

Switching to polar coordinates and integrating over one quarter of the unit disc \(D \subseteq I^2\), we have \begin{align*} \int_{I^2} f \, dA &\geq \int_D f \, dA \\ &\geq \int_0^{\pi/2} \int_0^1 \frac{\cos(\theta)\sin(\theta)}{r^4} ~r~dr~d\theta \\ &= \int_0^{\pi/2} \cos(\theta)\sin(\theta) \int_0^1 {1 \over r^3} ~dr~d\theta \\ &= \qty{\int_0^1 {1\over r^3}\,dr} \qty{\int_0^{\pi/2} \cos(\theta)\sin(\theta)\,d\theta }\\ &= \qty{\int_0^1 {1\over r^3}\,dr} \qty{-{1\over 2}\cos^2(\theta)\Big|_0^{\pi/2}} \\ &= -{1\over 2r^2}\Big|_0^1 \qty{1\over 2} \\ &= \qty{1\over 4}\qty{ -1 + \lim_{r\to 0} {1\over r^2} } \\ &= \infty ,\end{align*}

so \(f\) is not integrable.

\(1\implies 2\):

• Let \(\left\{{x_n}\right\} \overset{n\to\infty}\to x_0\) be arbitrary; we want to show \(\left\{{f(x_n)}\right\}\overset{n\to\infty}\to f(x_0)\).
  • We thus want to show that for every \(\varepsilon>0\), there exists an \(N(\varepsilon)\) such that \begin{align*}n\geq N(\varepsilon) \implies \rho(f(x_n), f(x_0)) < \varepsilon.\end{align*}
• Let \(\varepsilon>0\) be arbitrary, then by (1) choose \(\delta\) such that \(\rho(f(x), f(x_0)) < \varepsilon\) when \(d(x, x_0) < \delta\).
• Since \(x_n\to x\), there is some \(N\) such that \(n\geq N \implies d(x_n, x_0) < \delta\)
• Then for \(n\geq N\), \(d(x_n, x_0) < \delta\) and thus \(\rho(f(x_n), f(x_0)) < \varepsilon\), so \(f(x_n)\to f(x_0)\) as desired.

\(2\implies 1\):

Note that we need a \(\delta\) for every sequence, so picking a sequence for the forward implication is not a good idea here.

• By contrapositive, we will show that \(\not 1\implies \not 2\).
• Thus we need to show that if \(f\) is not \(\varepsilon{\hbox{-}}\delta\) continuous at \(x_0\), then there exists a sequence \(x_n\to x_0\) where \(f(x_n)\not\to f(x_0)\).
• Negating \(1\), we have that there exists an \(\varepsilon>0\) such that for all \(\delta\), there exists an \(x\) with \(d(x, x_0) < \delta\) but \(\rho(f(x), f(x_0))>\varepsilon\)
• So take a sequence of deltas \(\delta_n = {1\over n}\), apply this to produce a sequence \(x_n\) with \(d(x_n, x_0) < {1\over n}\) and \(\rho(f(x_n), f(x_0)) > \varepsilon\).
• Then \(x_n \to x_0\) but \(f(x_n) \not\to f(x_0)\).

2.6.3 1

• Set \(S \mathrel{\vcenter{:}}=\sum_{i\in I} \alpha(i)\), we will show that \(S<\infty \implies I\) is countable.

• Write \(I = {\coprod}_{n\in {\mathbb{N}}} S_n\) where \(S_n \mathrel{\vcenter{:}}=\left\{{i\in I {~\mathrel{\Big|}~}\alpha(i) \geq {1\over n}}\right\}\).

• We now have the inequality \begin{align*} S = \sum_{i\in I} \alpha(i) \geq \sum_{i\in S_n} \alpha(i) \geq \sum_{i\in S_n} {1\over n} = {1\over n} \sum_{i\in S_n} 1 = \qty{1\over n} {\left\lvert {S_n} \right\rvert} \\ \\ \infty > \implies n S \geq {\left\lvert {S_n} \right\rvert} ,\end{align*} so \(S_n\) is a countable set.

• But then \(I\) is a countable union of countable sets and thus countable.

2.6.4 2

\todo \begin{align*}inline\end{align*} {Not sure.}

• Define a sequence of operators \begin{align*} T_N: \ell^2 &\to \ell^1\\ \left\{{b_n}\right\} &\mapsto \sum_{n=1}^N a_n b_n .\end{align*}

• By assumption, these are well defined: the image is \(\ell^1\) since \({\left\lvert {T_N(\left\{{b_n}\right\})} \right\rvert} < \infty\) for all \(N\) and all \(\left\{{b_n}\right\} \in \ell^2\).

• So each \(T_N \in \qty{\ell^2}^\vee\) is a linear functional on \(\ell^2\).

• For each \(x\in \ell^2\), we have \({\left\lVert {T_N(x)} \right\rVert}_{{\mathbb{R}}} = \sum_{n=1}^N a_n b_n < \infty\) by assumption, so each \(T_N\) is pointwise bounded.

• By the Uniform Boundedness Principle, \(\sup_N {\left\lVert {T_N} \right\rVert}_{\text{op}} < \infty\).

• Define \(T = \lim_{N \to\infty } T_N\), then \({\left\lVert {T} \right\rVert}_{\text{op}} < \infty\).

• By the Riesz Representation theorem, \begin{align*} \sqrt{\sum a_n^2} \mathrel{\vcenter{:}}={\left\lVert {\left\{{a_n}\right\}} \right\rVert}_{\ell^2} = {\left\lVert {T} \right\rVert}_{\qty{\ell^2}^\vee} = {\left\lVert {T} \right\rVert}_{\text{op}} < \infty .\end{align*}

• So \(\sum a_n^2 < \infty\).

• Suppose \(p\) is a polynomial, then \begin{align*} \lim_{k\to\infty} \int_0^1 kx^{k-1} p(x) \, dx &= \lim_{k\to\infty} \int_0^1 \qty{ {\frac{\partial }{\partial x}\,}x^k } p(x) \, dx \\ &= \lim_{k\to\infty} \left[ x^k p(x) \Big|_0^1 - \int_0^1 x^k \qty{{\frac{\partial p}{\partial x}\,}(x) } \, dx \right] \quad\text{integrating by parts}\\ &= p(1) - \lim_{k\to\infty} \int_0^1 x^k \qty{{\frac{\partial p}{\partial x}\,}(x) } \, dx ,\end{align*}

• Thus it suffices to show that \begin{align*} \lim_{k\to\infty} \int_0^1 x^k \qty{{\frac{\partial p}{\partial x}\,} (x) } \, dx = 0 .\end{align*}

• Integrating by parts a second time yields \begin{align*} \lim_{k\to\infty} \int_0^1 x^k \qty{{\frac{\partial p}{\partial x}\,}(x) } \, dx &= \lim_{k\to\infty} {x^{k+1} \over k+1} {\frac{\partial p}{\partial x}\,}(x) \Big|_0^1 - \int_0^1 {x^{k+1} \over k+1} \qty{ {\frac{\partial ^2 p}{\partial x^2}\,}(x)} \, dx \\ &= \lim_{k\to\infty} {p'(1) \over k+1} - \lim_{k\to\infty} \int_0^1 {x^{k+1} \over k+1} \qty{ {\frac{\partial ^2p}{\partial x^2}\,}(x)} \, dx \\ &= - \lim_{k\to\infty} \int_0^1 {x^{k+1} \over k+1} \qty{ {\frac{\partial ^2p}{\partial x^2}\,}(x)} \, dx \\ &= - \int_0^1 \lim_{k\to\infty} {x^{k+1} \over k+1} \qty{ {\frac{\partial ^2p}{\partial x^2}\,}(x)} \, dx \quad\text{by DCT} \\ &= - \int_0^1 0 \qty{ {\frac{\partial ^2p}{\partial x^2}\,}(x)} \, dx \\ &= 0 .\end{align*}

  • The DCT can be applied here because polynomials are smooth and \([0, 1]\) is compact, so \({\frac{\partial ^2 p}{\partial x^2}\,}\) is bounded on \([0, 1]\) by some constant \(M\) and \begin{align*} \int_0^1 {\left\lvert {x^k {\frac{\partial ^2 p}{\partial x^2}\,} (x)} \right\rvert} \leq \int_0^1 1\cdot M = M < \infty.\end{align*}

• So the result holds when \(f\) is a polynomial.

• Now use the Weierstrass approximation theorem:

  • If \(f: [a, b] \to {\mathbb{R}}\) is continuous, then for every \(\varepsilon>0\) there exists a polynomial \(p_\varepsilon(x)\) such that \({\left\lVert {f - p_\varepsilon} \right\rVert}_\infty < \varepsilon\).

• Thus \begin{align*} {\left\lvert { \int_0^1 kx^{k-1} p_\varepsilon(x)\,dx - \int_0^1 kx^{k-1}f(x)\,dx } \right\rvert} &= {\left\lvert { \int_0^1 kx^{k-1} \qty{p_\varepsilon(x) - f(x)} \,dx } \right\rvert} \\ &\leq {\left\lvert { \int_0^1 kx^{k-1} {\left\lVert {p_\varepsilon-f} \right\rVert}_\infty \,dx } \right\rvert} \\ &= {\left\lVert {p_\varepsilon-f} \right\rVert}_\infty \cdot {\left\lvert { \int_0^1 kx^{k-1} \,dx } \right\rvert} \\ &= {\left\lVert {p_\varepsilon-f} \right\rVert}_\infty \cdot x^k \Big|_0^1 \\ &= {\left\lVert {p_\varepsilon-f} \right\rVert}_\infty \\ \\ &\overset{\varepsilon\to 0}\to 0 \end{align*}

  and the integrals are equal.

• By the first argument, \begin{align*}\int_0^1 kx^{k-1} p_\varepsilon(x) \,dx = p_\varepsilon(1) \text{ for each } \varepsilon\end{align*}

• Since uniform convergence implies pointwise convergence, \(p_\varepsilon(1) \overset{\varepsilon\to 0}\to f(1)\).

3.2.3 a

• Prove a stronger result: \begin{align*} a_k \to S \implies S_N\mathrel{\vcenter{:}}=\frac 1 N \sum_{k=1}^N a_k \to S .\end{align*}

• For any \(\varepsilon> 0\), use convergence \(a_k \to S\): choose (and fix) \(M = M(\varepsilon)\) large enough such that \begin{align*} k\geq M+1 \implies {\left\lvert {a_k - S} \right\rvert} < \varepsilon .\end{align*}

• With \(M\) fixed, choose \(N = N(M, \varepsilon)\) large enough so that \({1\over N} \sum_{k=1}^{M} {\left\lvert {a_k - S} \right\rvert} < \varepsilon\).

• Then \begin{align*} \left|\left(\frac{1}{N} \sum_{k=1}^{N} a_{k}\right)-S\right| &= {1\over N} {\left\lvert { \qty{\sum_{k=1}^N a_k} - NS } \right\rvert} \\ &= {1\over N} {\left\lvert { \qty{\sum_{k=1}^N a_k} - \sum_{k=1}^N S } \right\rvert} \\ &=\frac{1}{N}\left|\sum_{k=1}^{N}\left(a_{k}-S\right)\right| \\ &\leq \frac{1}{N} \sum_{k=1}^{N}\left|a_{k}-S\right| \\ &= {1\over N} \sum_{k=1}^{M} {\left\lvert {a_k - S} \right\rvert} + \sum_{k=M+1}^N {\left\lvert {a_k - S} \right\rvert} \\ &\leq {1\over N} \sum_{k=1}^{M} {\left\lvert {a_k - S} \right\rvert} + \sum_{k=M+1}^N {\varepsilon} \quad \text{since } a_k \to S\\ &= {1\over N} \sum_{k=1}^{M} {\left\lvert {a_k - S} \right\rvert} + (N - M){\varepsilon} \\ &\leq \varepsilon+ (N(M, \varepsilon) - M(\varepsilon))\varepsilon .\end{align*}

3.2.4 b

• Define \begin{align*} \Gamma_n \mathrel{\vcenter{:}}=\sum_{k=n}^\infty \frac{a_k}{k} .\end{align*}

• \(\Gamma_1 = \sum_{k=1}^n \frac{ a_k } k\) is the original series and each \(\Gamma_n\) is a tail of \(\Gamma_1\), so by assumption \(\Gamma_n \overset{n\to\infty}\to 0\).

• Compute \begin{align*} \frac 1 n \sum_{k=1}^n a_k &= \frac 1 n (\Gamma_1 + \Gamma_2 + \cdots + \Gamma_{n} \mathbf{- \Gamma_{n+1}}) \\ .\end{align*}

• This comes from consider the following summation:

  

• Use part (a): since \(\Gamma_n \overset{n\to\infty}\to 0\), we have \({1\over n} \sum_{k=1}^n \Gamma_k \overset{n\to\infty}\to 0\).

• Also a minor check: \(\Gamma_n \to 0 \implies {1\over n}\Gamma_n \to 0\).

• Then \begin{align*} \frac 1 n \sum_{k=1}^n a_k &= \frac 1 n (\Gamma_1 + \Gamma_2 + \cdots + \Gamma_{n} \mathbf{- \Gamma_{n+1}}) \\ &= \qty{ {1\over n } \sum_{k=0}^n \Gamma_k } - \qty{{1\over n}\Gamma_{n+1} } \\ &\overset{n\to\infty}\to 0 .\end{align*}

Case of a characteristic function of an interval \([a, b]\):

• First suppose \(f(x) = \chi_{[a, b]}(x)\).

• Note that \(\sin(nx)\) has a period of \(2\pi/n\), and thus \({\left\lfloor (b-a) \over (2\pi / n) \right\rfloor} = {\left\lfloor n(b-a)\over 2\pi \right\rfloor}\) full periods in \([a, b]\).

• Taking the absolute value yields a new function with half the period

  • So \({\left\lvert {\sin(nx)} \right\rvert}\) has a period of \(\pi/n\) with \({\left\lfloor n(b-a) \over \pi \right\rfloor}\) full periods in \([a, b]\).

• We can compute the integral over one full period (which is independent of which period is chosen)

  • We can use translation invariance of the integral to compute this over the period \(0\) to \(\pi/n\).
  • Since \(\sin(nx)\) is positive, it equals \({\left\lvert {\sin(nx)} \right\rvert}\) on its first period, so we have \begin{align*} \int_{\text{One Period}} {\left\lvert {\sin(nx)} \right\rvert} \, dx &= \int_0^{\pi/n} \sin(nx)\,dx \\ &= {1\over n} \int_0^\pi \sin(u) \,du \quad u = nx \\ &= {1\over n} \qty{-\cos(u)\mathrel{\Big|}_0^\pi} \\ &= {2 \over n} .\end{align*}

• Then break the integral up into integrals over full periods \(P_1, P_2, \cdots, P_N\) where \(N \mathrel{\vcenter{:}}={\left\lfloor n(b-a)/\pi \right\rfloor}\)

• Noting that each period is of length \(\pi\over n\), so letting \(L_n\) be the regions falling outside of a full period, we have

• Thus \begin{align*} \int_a^b {\left\lvert {\sin(nx)} \right\rvert} \, dx &= \qty{ \sum_{j=1}^{N} \int_{P_j} {\left\lvert {\sin(nx)} \right\rvert} \, dx } + \int_{L_n} {\left\lvert {\sin(nx)} \right\rvert}\,dx \\ &= \qty{ \sum_{j=1}^{N} {2\over n} } + \int_{L_n} {\left\lvert {\sin(nx)} \right\rvert}\,dx \\ &= N \qty{2\over n} + \int_{L_n} {\left\lvert {\sin(nx)} \right\rvert}\,dx \\ &\mathrel{\vcenter{:}}={\left\lfloor (b-a) n \over \pi \right\rfloor} {2\over n} + R_n \\ &\mathrel{\vcenter{:}}=(b-a)C_n + R_n \end{align*} where (claim) \(C_n \overset{n\to\infty}\to {2\over \pi}\) and \(R(n) \overset{n\to\infty}\to 0\).

• \(C_n \to {2\over \pi}\): \begin{align*} {n-1 \over n} \qty{2\over \pi} = {n-1 \over \pi} \qty{2\over n} \leq {\left\lfloor n\over \pi \right\rfloor}\qty{2\over n} \leq {n \over \pi}\qty{2\over n} = {2 \over \pi} ,\end{align*} then use the fact that \({n-1 \over n} \to 1\).

  • Then equality follows by the Squeeze theorem.

• \(R_n \to 0\):

  • We use the fact that \(m(L_n) \to 0\), then \(\int_{L_n} {\left\lvert {\sin(nx)} \right\rvert} \leq \int_{L_n} 1 = m(L_n) \to 0\).
  • This follows from the fact that \(L_n\) is the complement of \(\cup_j P_j\), the set of full periods, so \begin{align*} m(L_n) &= m(b-a) - \sum m(P_j) \\ &= \qty{b-a} - {\left\lfloor n(b-a) \over \pi \right\rfloor}\qty{\pi \over n} \\ &\overset{n\to \infty}\to (b-a) - (b-a) \\ &= 0 .\end{align*} where we’ve used the fact that \begin{align*} \qty{\pi \over n} \qty{(b-a)n-1 \over \pi} &\leq {\left\lfloor n(b-a) \over \pi \right\rfloor}\qty{\pi \over n} \\ &\leq \qty{\pi \over n} \qty{(b-a)n\over \pi} \\ &= (b-a) ,\end{align*} then taking \(n\to \infty\) sends the LHS to \(b-a\), forcing the middle term to be \(b-a\) by the Squeeze theorem.

General case:

• By linearity of the integral, the result holds for simple functions:

  • If \(f = \sum c_j \chi_{E_j}\) where \(E_j = [a_j, b_j]\), we have \begin{align*} \int_0^1 f(x) {\left\lvert {\sin(nx)} \right\rvert}\,dx &= \int_0^1 \sum c_j \chi_{E_j}(x) {\left\lvert {\sin(nx)} \right\rvert}\,dx \\ &= \sum c_j \int_0^1 \chi_{E_j}(x) {\left\lvert {\sin(nx)} \right\rvert}\,dx \\ &= \sum c_j (b_j - a_j) {2\over \pi} \\ &= {2\over \pi} \sum c_j (b_j - a_j) \\ &= {2\over \pi} \sum c_j m(E_j) \\ &\mathrel{\vcenter{:}}={2\over \pi} \int_0^1 f .\end{align*}

• Since \(f\in L^1\), where simple functions are dense, choose \(s_n\nearrow f\) where \({\left\lVert {s_N - f} \right\rVert}_1 < \varepsilon\), then \begin{align*} {\left\lvert { \int_0^1 f(x) {\left\lvert {\sin(nx)} \right\rvert} \,dx - \int_0^1 s_N(x) {\left\lvert {\sin(nx)} \right\rvert}\,dx } \right\rvert} &= {\left\lvert { \int_0^1 \qty{f(x) - s_N(x)} {\left\lvert {\sin(nx)} \right\rvert} \,dx } \right\rvert} \\ &\leq \int_0^1 {\left\lvert { f(x) - s_N(x)} \right\rvert} {\left\lvert {\sin(nx)} \right\rvert} \,dx \\ &= {\left\lVert { \qty{f - s_N} {\left\lvert {\sin(nx)} \right\rvert} } \right\rVert}_1 \\ &\leq {\left\lVert {f-s_N} \right\rVert}_1 \cdot {\left\lVert {{\left\lvert {\sin(nx)} \right\rvert}} \right\rVert}_\infty \quad\text{by Holder}\\ &\leq \varepsilon\cdot 1 ,\end{align*}

• So the integrals involving \(s_N\) converge to the integral involving \(f\), and \begin{align*} \lim_{n\to\infty} \int f(x){\left\lvert {\sin(nx)} \right\rvert} &= \lim_{n\to\infty} \lim_{N\to\infty} \int s_N(x) {\left\lvert {\sin(nx)} \right\rvert} \\ &= \lim_{N\to\infty} \lim_{n\to\infty} \int s_N(x) {\left\lvert {\sin(nx)} \right\rvert} \quad\text{because ?}\\ &= \lim_{N\to \infty} {2\over \pi} \int s_N(x) \\ &= {2\over \pi} \int f ,\end{align*} which is the desired result.

3.4.1 a

\(f_n\to 0\) pointwise:

• Finding the maximum: can check that \({\frac{\partial f_n}{\partial x}\,} = x(1-x)^{n-1} \qty{1 + (n^2-1)x}\)
• This has critical points \(x=0, 1, {-1 \over n^2 + 1}\), and the latter is a global max on \([0, 1]\).
• Set \(x_n \mathrel{\vcenter{:}}={-1 \over n^2 + 1}\)
• Compute \begin{align*} \lim f_n(x_n) = \lim_{n\to \infty } {-n \over n^2 + 1} \qty{1 + x_n}^n = 0\cdot 1 = 0 .\end{align*}
• So \(\sup f_n \to 0\), forcing \(f_n \to 0\) pointwise.

The convergence is not uniform:

• Let \(x_n = \frac 1 n\) and \(\varepsilon > e^{-1}\), then \begin{align*} {\left\lVert {nx(1-x)^n - 0} \right\rVert}_\infty &\geq {\left\lvert {nx_n (1-x_n)^n} \right\rvert} \\ &= {\left\lvert {\left( 1 - \frac 1 n\right)^n} \right\rvert} \\ &> e^{-1} \\ &> \varepsilon .\end{align*}

  • Here we’ve used that \((1 + {x\over n})^n \leq e^x\) for all \(x\in {\mathbb{R}}\) and all \(n\).
  • Follows from \(1+y \leq e^y\) applied to \(y = x/n\).

• Thus \({\left\lVert {f_n - 0} \right\rVert}_\infty = {\left\lVert {f_n} \right\rVert}_\infty > e^{-1} > 0\).

3.4.2 b

• Noting that \(\sin(x) \leq 1\), we have \begin{align*} {\left\lvert {\int_0^1 n(1-x)^{n} \sin(x)} \right\rvert} &\leq \int_0^1 {\left\lvert {n(1-x)^n \sin(x)} \right\rvert} \\ &\leq \int_0^1 {\left\lvert {n (1-x)^n} \right\rvert} \\ &= n\int_0^1 (1-x)^n \\ &= -\frac{n(1-x)^{n+1}}{n+1} \\ &\overset{n\to\infty}\longrightarrow 0 .\end{align*}

3.5.3 a

• \(f_n\) has a root: \begin{align*} ae^{-nax} = be^{-nbx} &\iff {1\over n} = e^{-nbx} e^{nax} = e^{n(b-a)x} \iff x = {\ln\qty{a\over b} \over n(a-b)} \mathrel{\vcenter{:}}= x_n .\end{align*}

• Thus \(f_n\) only changes sign at \(x_n\), and is strictly positive on one side of \(x_n\).

• Then \begin{align*} \int_{\mathbb{R}}\sum_n {\left\lvert {f_n(x)} \right\rvert}\,dx &= \sum_n \int_{\mathbb{R}}{\left\lvert {f_n(x)} \right\rvert} \,dx \\ &\geq \sum_n \int_{x_n}^\infty f_n(x) \, dx \\ &= \sum_n {1\over n} \qty{ e^{-bnx} - e^{-anx}\Big|_{x_n}^\infty } \\ &= \sum_n {1\over n} \qty{ e^{-bnx_n} - e^{-anx_n} } .\end{align*}

3.5.4 b

?

• Set \(f_N(x) \mathrel{\vcenter{:}}=\sum_{n=1}^N n^{-x}\), so \(f(x) = \lim_{N\to\infty} f_N(x)\).

• If an interchange of limits is justified, we have \begin{align*} {\frac{\partial }{\partial x}\,} \lim_{N\to\infty} \sum_{n=1}^N n^{-x} &= \lim_{h\to 0} \lim_{N\to\infty} {1\over h} \left[ \qty{\sum_{n=1}^N n^{-x}} - \qty{\sum_{n=1}^N n^{-(x+h)} }\right] \\ &\mathop{\mathrm{=}}_{?} \lim_{N\to\infty} \lim_{h\to 0} {1\over h} \left[ \qty{\sum_{n=1}^N n^{-x}} - \qty{\sum_{n=1}^N n^{-(x+h)} }\right] \\ &= \lim_{N\to\infty} \lim_{h\to 0} {1\over h} \left[ {\sum_{n=1}^N n^{-x}} - {n^{-(x+h)} }\right] \quad\text{(1)} \\ &= \lim_{N\to\infty} \sum_{n=1}^N \lim_{h\to 0} {1\over h} \left[ n^{-x} - n^{-(x+h)} \right] \quad\text{since this is a finite sum} \\ &\mathrel{\vcenter{:}}=\lim_{N\to\infty} \sum_{n=1}^N {\frac{\partial }{\partial x}\,}\qty{1 \over n^x} \\ &= \lim_{N\to\infty} \sum_{n=1}^N -{\ln(n) \over n^x} ,\end{align*} where the combining of sums in (1) is valid because \(\sum n^{-x}\) is absolutely convergent for \(x>1\) by the \(p{\hbox{-}}\)test.

• Thus it suffices to justify the interchange of limits and show that the last sum converges on \((1, \infty)\).

• Claim: \(\sum n^{-x}\ln(n)\) converges.

  • Use the fact that for any fixed \(\varepsilon>0\), \begin{align*} \lim_{n\to\infty} {\ln(n) \over n^\varepsilon} \mathop{\mathrm{=}}^{L.H.} \lim_{n\to\infty}{1/n \over \varepsilon n^{\varepsilon-1}} = \lim_{n\to\infty} {1\over \varepsilon n^\varepsilon} = 0 ,\end{align*}

  • This implies that for a fixed \(\varepsilon>0\) and for any constant \(c>0\) there exists an \(N\) large enough such that \(n\geq N\) implies \(\ln(n)/n^\varepsilon< c\), i.e. \(\ln(n) < c n^{\varepsilon}\).

  • Taking \(c=1\), we have \(n\geq N \implies \ln(n) < n^\varepsilon\)

  • We thus break up the sum: \begin{align*} \sum_{n\in {\mathbb{N}}} {\ln(n) \over n^x} &= \sum_{n=1}^{N-1} { \ln(n) \over n^x} + \sum_{n=N}^\infty {\ln(n) \over n^x} \\ &\leq \sum_{n=1}^{N-1} { \ln(n) \over n^x} + \sum_{n=N}^\infty {n^\varepsilon\over n^x} \\ &\mathrel{\vcenter{:}}= C_\varepsilon+ \sum_{n=N}^\infty {n^\varepsilon\over n^x} \quad \text{with $C_\varepsilon<\infty$ a constant}\\ &= C_\varepsilon+ \sum_{n=N}^\infty {1 \over n^{x-\varepsilon}} ,\end{align*} where the last term converges by the \(p{\hbox{-}}\)test if \(x-\varepsilon> 1\).

  • But \(\varepsilon\) can depend on \(x\), and if \(x\in (1, \infty)\) is fixed we can choose \(\varepsilon< {\left\lvert {x-1} \right\rvert}\) to ensure this.

• Claim: the interchange of limits is justified.

Let \(L\) be the LHS and \(R\) be the RHS.

Claim: \(L\leq R\). - Since \({\left\lvert {\phi } \right\rvert}\leq {\left\lVert {\phi} \right\rVert}_\infty\) a.e., we can write \begin{align*} L^{1\over n} &\mathrel{\vcenter{:}}=\int_{\mathbb{R}}{ {\left\lvert {\phi(x)} \right\rvert}^n \over 1+ x^2} \\ &\leq \int_{\mathbb{R}}{ {\left\lVert {\phi} \right\rVert}_\infty^n \over 1+ x^2} \\ &= {\left\lVert {\phi} \right\rVert}_\infty^n \int_{\mathbb{R}}{1\over 1 + x^2} \\ &= {\left\lVert {\phi} \right\rVert}_\infty^n \arctan(x)\Big|_{-\infty}^{\infty} \\ &= {\left\lVert {\phi} \right\rVert}_\infty^n \qty{{\pi \over 2} - {-\pi \over 2} } \\ &= \pi {\left\lVert {\phi} \right\rVert}_\infty^n \\ \\ \implies L^{1\over n} &\leq \sqrt[n]{\pi {\left\lVert {\phi} \right\rVert}_\infty^n} \\ \implies L &\leq \pi^{1\over n} {\left\lVert {\phi} \right\rVert}_\infty \\ &\overset{n\to \infty }\to {\left\lVert {\phi} \right\rVert}_\infty ,\end{align*} where we’ve used the fact that \(c^{1\over n} \overset{n\to\infty}\to 1\) for any constant \(c\).

Claim: \(R\leq L\).

• We will show that \(R\leq L + \varepsilon\) for every \(\varepsilon>0\).
• Set \begin{align*} S_\varepsilon\mathrel{\vcenter{:}}=\left\{{x\in {\mathbb{R}}^n{~\mathrel{\Big|}~}{\left\lvert {\phi(x)} \right\rvert} \geq {\left\lVert {\phi} \right\rVert}_\infty - \varepsilon}\right\} .\end{align*}
• Then we have \begin{align*} \int_{\mathbb{R}}{{\left\lvert {\phi(x)} \right\rvert}^n \over 1 +x^2}\,dx &\geq \int_{S_\varepsilon} {{\left\lvert {\phi(x)} \right\rvert}^n \over 1 +x^2}\,dx \quad S_\varepsilon\subset {\mathbb{R}}\\ &\geq \int_{S_\varepsilon} { \qty{{\left\lVert {\phi} \right\rVert}_\infty - \varepsilon}^n \over 1 +x^2}\,dx \qquad\text{by definition of }S_\varepsilon\\ &= \qty{{\left\lVert {\phi} \right\rVert}_\infty - \varepsilon}^n \int_{S_\varepsilon} { 1 \over 1 +x^2}\,dx \\ &= \qty{{\left\lVert {\phi} \right\rVert}_\infty - \varepsilon}^n C_\varepsilon\qquad\text{where $C_\varepsilon$ is some constant} \\ \\ \implies \qty{ \int_{\mathbb{R}}{{\left\lvert {\phi(x)} \right\rvert}^n \over 1 +x^2}\,dx }^{1\over n} &\geq \qty{{\left\lVert {\phi} \right\rVert}_\infty - \varepsilon} C_\varepsilon^{1 \over n} \\ &\overset{n\to\infty}\to \qty{{\left\lVert {\phi} \right\rVert}_\infty - \varepsilon} \cdot 1 \\ &\overset{\varepsilon\to 0}\to {\left\lVert {\phi} \right\rVert}_\infty ,\end{align*} where we’ve again used the fact that \(c^{1\over n} \to 1\) for any constant.

• Use the fact that \(L^p\) has small tails: if \(h\in L^2({\mathbb{R}})\), then for any \(\varepsilon> 0\), \begin{align*} \forall \varepsilon,\, \exists N\in {\mathbb{N}}{\quad \operatorname{such that} \quad}\int_{{\left\lvert {x} \right\rvert} \geq {N}} {\left\lvert {h(x)} \right\rvert}^2 \,dx < \varepsilon .\end{align*}

• So choose \(N\) large enough so that \begin{align*} \int_{{\left\lVert {x} \right\rVert} \geq N}{\left\lvert {g(x)} \right\rvert}^2 < \varepsilon\\ \int_{{\left\lVert {x} \right\rVert} \geq N}{\left\lvert {f(x)} \right\rvert}^2 < \varepsilon\\ .\end{align*}

• Then write \begin{align*} \int_{{\mathbb{R}}^d} f(x) g(x+n) \,dx = \int_{{\left\lVert {x} \right\rVert} \leq N} f(x)g(x+n)\,dx + \int_{{\left\lVert {x} \right\rVert} \geq N} f(x) g(x+n)\,dx .\end{align*}

• Bounding the second term: apply Cauchy-Schwarz \begin{align*} \int_{{\left\lVert {x} \right\rVert} \geq N} f(x) g(x+n)\,dx \leq \qty{ \int_{{\left\lVert {x} \right\rVert} \geq N} {\left\lvert {f(x)} \right\rvert}^2}^{1\over 2} \cdot \qty{ \int_{{\left\lVert {x} \right\rVert} \geq N} {\left\lvert {g(x)} \right\rvert}^2}^{1\over 2} \leq \varepsilon^{1\over 2} \cdot {\left\lVert {g} \right\rVert}_2 .\end{align*}

• Bounding the first term: also Cauchy-Schwarz, after variable changes \begin{align*} \int_{{\left\lVert {x} \right\rVert} \leq N} f(x) g(x+n)\,dx &= \int_{-N}^N f(x) g(x+n)\,dx \\ &= \int_{-N+n}^{N+n} f(x-n) g(x)\,dx \\ &\leq \int_{-N+n}^{\infty} f(x-n) g(x)\,dx \\ &\leq \qty{\int_{-N+n}^{\infty} {\left\lvert {f(x-n)} \right\rvert}^2}^{1\over 2}\cdot \qty{\int_{-N+n}^{\infty} {\left\lvert {g(x)} \right\rvert}^2}^{1\over 2} \\ &\leq {\left\lVert {f} \right\rVert}_2 \cdot \varepsilon^{1\over 2} .\end{align*}

• Then as long as \(n\geq 2N\), we have \begin{align*} \int {\left\lvert {f(x) g(x+n)} \right\rvert} \leq \qty{{\left\lVert {f} \right\rVert}_2 + {\left\lVert {g} \right\rVert}_2} \cdot \varepsilon^{1\over 2} .\end{align*}

4.1.3 a

• If \(m_*(E) = \infty\), then take \(B = {\mathbb{R}}^n\) since \(m({\mathbb{R}}^n) = \infty\).

• Suppose \(N \mathrel{\vcenter{:}}= m_*(E) < \infty\).

• Since \(m_*(E)\) is an infimum, by definition, for every \(\varepsilon> 0\) there exists a covering by closed cubes \(\left\{{Q_i(\varepsilon)}\right\}_{i=1}^\infty \rightrightarrows E\) depending on \(\varepsilon\) such that \begin{align*} \sum_{i=1}^\infty {\left\lvert {Q_i(\varepsilon)} \right\rvert} < N + \varepsilon .\end{align*}

• For each fixed \(n\), set \(\varepsilon_n = {1\over n}\) to produce such a covering \(\left\{{Q_i(\varepsilon_n)}\right\}_{i=1}^\infty\) and set \(B_n \mathrel{\vcenter{:}}=\cup_{i=1}^\infty Q_i(\varepsilon_n)\).

• The outer measure of cubes is equal to the sum of their volumes, so \begin{align*} m_*(B_n) = \sum_{i=1}^\infty {\left\lvert {Q_i(\varepsilon_n)} \right\rvert} < N + \varepsilon_n = N + {1\over n} .\end{align*}

• Now set \(B \mathrel{\vcenter{:}}=\cap_{n=1}^\infty B_n\).

  • Since \(E\subseteq B_n\) for every \(n\), \(E\subseteq B\)
  • Since \(B\) is a countable intersection of countable unions of closed sets, \(B\) is Borel.
  • Since \(B_n \subseteq B\) for every \(n\), we can apply subadditivity to obtain the inequality \begin{align*} E \subseteq B \subseteq B_n \implies N \leq m_*(B) \leq m_*(B_n) < N + {1\over n} {\quad \operatorname{for all} \quad} n\in {\mathbb{Z}}^{\geq 1} .\end{align*}

• This forces \(m_*(E) = m_*(B)\).

4.1.4 b

Suppose \(m_*(E) < \infty\).

• By (a), find a Borel set \(B\supseteq E\) such that \(m_*(B) = m_*(E)\)
• Note that \(E\subseteq B \implies B\cap E = E\) and \(B\cap E^c = B\setminus E\).
• By assumption, \begin{align*} m_*(B) &= m_*(B\cap E) + m_*(B\cap E^c) \\ m_*(E) &= m_*(E) + m_*(B\setminus E) \\ m_*(E) - m_*(E) &= m_*(B\setminus E) \qquad\qquad\text{since } m_*(E) < \infty \\ \implies m_*(B\setminus E) &= 0 .\end{align*}
• So take \(N = B\setminus E\); this shows \(m_*(N) = 0\) and \(E = B\setminus (B\setminus E) = B\setminus N\).

If \(m_*(E) = \infty\):

• Apply result to \(E_R\mathrel{\vcenter{:}}= E \cap[R, R+1)^n \subset {\mathbb{R}}^n\) for \(R\in {\mathbb{Z}}\), so \(E = {\coprod}_R E_R\)
• Obtain \(B_R, N_R\) such that \(E_R = B_R \setminus N_R\), \(m_*(E_R) = m_*(B_R)\), and \(m_*(N_R) = 0\).
• Note that
  • \(B\mathrel{\vcenter{:}}=\cup_R B_R\) is a union of Borel sets and thus still Borel
  • \(E = \cup_R E_R\)
  • \(N\mathrel{\vcenter{:}}= B\setminus E\)
  • \(N' \mathrel{\vcenter{:}}=\cup_R N_R\) is a union of null sets and thus still null
• Since \(E_R \subset B_R\) for every \(R\), we have \(E\subset B\)
• We can compute \begin{align*} N = B\setminus E = \qty{ \cup_R B_R } \setminus \qty{\cup_R E_R } \subseteq \cup_R \qty{B_R\setminus E_R} = \cup_R N_R \mathrel{\vcenter{:}}= N' \end{align*} where \(m_*(N') = 0\) since \(N'\) is null, and thus subadditivity forces \(m_*(N) = 0\).

4.2.1 a

• The Borel \(\sigma{\hbox{-}}\)algebra is closed under countable unions/intersections/complements,
• \(B = \limsup_n B_n\) is an intersection of unions of measurable sets.

4.2.2 b

• Tails of convergent sums go to zero, so \(\sum_{n\geq M} \mu(B_n) \xrightarrow{M\to\infty} 0\),
• \(B_M \mathrel{\vcenter{:}}=\cap_{m = 1}^M \cup_{n\geq m} B_n \searrow B\).

\begin{align*} \mu(B_M) &= \mu\left(\cap_{m\in {\mathbb{N}}} \cup_{n\geq m} B_n\right) \\ &\leq \mu\left( \cup_{n\geq m} B_n \right) \quad \text{for all } m\in {\mathbb{N}}\text{ by countable subadditivity} \\ &\to 0 ,\end{align*}

• The result follows by continuity of measure.

4.2.3 c

• To show \(\mu(B) = 1\), we’ll show \(\mu(B^c) = 0\).

• Let \(B_k = \cap_{m=1}^\infty \cup_{n = m}^K B_n\). Then \begin{align*} \mu(B_K^c) &= \mu \left(\cup_{m=1}^\infty \cap_{n=m}^K B_n^c\right) \\ &\leq \sum_{m=1}^\infty \mu\left( \cap_{n=m}^K B_n^c \right) \quad\text{ by subadditivity} \\ &= \sum_{m=1}^\infty \prod_{n=m}^K \qty{1 - \mu(B_n)} \quad \text{by assumption} \\ &\leq \sum_{m=1}^\infty \prod_{n=m}^K e^{-\mu(B_n^c)} \quad\text{by hint} \\ &= \sum_{m=1}^\infty \exp{-\sum_{n=m}^K \mu(B_n^c)} \\ &\overset{K\to\infty}\to 0 \end{align*} since \(\displaystyle\sum_{n=m}^K \mu(B_n^c) \overset{K\to\infty}\to \infty\) by assumption

• We can apply continuity of measure since \(B_K^c \xrightarrow{K\to\infty} B^c\).

4.3.3 a

See Folland p.26

• Lemma 1: \(\mu({\coprod}_{k=1}^\infty E_k) = \lim_{N\to\infty} \sum_{k=1}^N \mu(E_k)\).

• Suppose \(F_0 \supseteq F_1 \supseteq \cdots\).

• Let \(A_k = F_k \setminus F_{k+1}\), since the \(F_k\) are nested the \(A_k\) are disjoint

• Set \(A \mathrel{\vcenter{:}}={\coprod}_{k=1}^\infty A_k\) and \(F \mathrel{\vcenter{:}}=\cap_{k=1}^\infty F_k\).

• Note \(X = X\setminus Y ~{\coprod}~ X\cap Y\) for any two sets (just write \(X\setminus Y \mathrel{\vcenter{:}}= X\cap Y^c\))

• Note that \(A\) contains anything that was removed from \(F_0\) when passing from any \(F_j\) to \(F_{j+1}\), while \(F\) contains everything that is never removed at any stage, and these are disjoint possibilities.

• Thus \(F_0 = F {\coprod}A\), so \begin{align*} \mu(F_0) &= \mu(F) + \mu(A) \\ &= \mu(F) + \mu({\coprod}_{k=1}^\infty A_k) \\ &= \mu(F) + \lim_{n\to\infty} \sum_{k=0}^n \mu(A_k) \quad \text{by countable additivity}\\ &= \mu(F) + \lim_{n\to\infty} \sum_{k=0}^n \mu(F_k) - \mu(F_{k+1}) \\ &= \mu(F) + \lim_{n\to\infty} \left( \mu(F_1) - \mu(F_n) \right) \quad\text{(Telescoping)} \\ &=\mu(F) + \mu(F_1) - \lim_{N\to\infty} \mu(F_n) ,\end{align*}

• Since \(\mu\) is a finite measure, \(\mu(F_1) < \infty\) and can be subtracted, yielding \begin{align*} \mu(F_1) &= \mu(F) + \mu(F_1) - \lim_{n\to\infty} \mu(F_n) \\ \implies \mu(F) &= \lim_{n\to\infty} \mu(F_n) \\ \implies \mu\qty{\cap_{k=1}^\infty F_k} &= \lim_{n\to\infty} \mu(F_n) .\end{align*}

4.3.4 b

• Toward a contradiction, negate the implication: suppose there exists an \(\varepsilon>0\) such that for all \(\delta\), we have \(m(E) < \delta\) but \(\mu(E) > \varepsilon\).

• The sequence \(\left\{{\delta_n \mathrel{\vcenter{:}}={1\over 2^n}}\right\}_{n\in {\mathbb{N}}}\) and produce sets \(A_n\in {\mathcal{B}}\) such \(m(A_n) < {1\over 2^n}\) but \(\mu(A_n) > \varepsilon\).

• Define \begin{align*} F_n &\mathrel{\vcenter{:}}=\cup_{j\geq n} A_j \\ C_m &\mathrel{\vcenter{:}}=\cap_{k=1}^m F_k \\ A &\mathrel{\vcenter{:}}= C_\infty \mathrel{\vcenter{:}}=\cap_{k=1}^\infty F_k .\end{align*}

• Note that \(F_1 \supseteq F_2 \supseteq \cdots\), since each increase in index unions fewer sets.

• By continuity for the Lebesgue measure, \begin{align*} m(A) = m \qty{\cap_{k=1}^\infty F_k } = \lim_{k\to \infty} m (F_k) = \lim_{k\to\infty} m\qty{\cup_{j\geq k} A_j } \leq \lim_{k\to\infty} \sum_{j\geq k} m(A_j) = \lim_{k\to\infty} \sum_{j\geq k} {1\over 2^n} = 0 ,\end{align*} which follows because this is the tail of a convergent sum

• Thus \(m(A) = 0\) and by assumption, this implies \(\mu(A) = 0\).

• However, by part (a), \begin{align*} \mu(A) = \lim_n \mu\left( \cup_{k=n}^\infty A_k \right) \geq \lim_n \mu(A_n) = \lim_n \varepsilon = \varepsilon > 0 .\end{align*}

4.4.1 Indirect Proof

• Since \(E\) is measurable, \(E^c\) is measurable.
• Since \(E^c\) is measurable exists an open \(O\supseteq E^c\) such that \(m(O\setminus E^c) = 0\).
• Set \(B \mathrel{\vcenter{:}}= O^c\), then \(O\supseteq E^c \iff {\mathcal{O}}^c \subseteq E \iff B\subseteq E\).
• Computing measures yields \begin{align*} E\setminus B \mathrel{\vcenter{:}}= E\setminus {\mathcal{O}}^c \mathrel{\vcenter{:}}= E\cap({\mathcal{O}}^c)^c = E\cap{\mathcal{O}}= {\mathcal{O}}\cap(E^c)^c \mathrel{\vcenter{:}}={\mathcal{O}}\setminus E^c ,\end{align*} thus \(m(E\setminus B) = m({\mathcal{O}}\setminus E^c) = 0\).
• Since \({\mathcal{O}}\) is open, \(B\) is closed and thus Borel.

4.4.2 Direct Proof (Todo)

• Strategy: Borel-Cantelli.

• We’ll show that \(m(E) \cap[n, n+1] = 0\) for all \(n\in {\mathbb{Z}}\); then the result follows from \begin{align*} m(E) = m \qty{\cup_{n\in {\mathbb{Z}}} E \cap[n, n+1]} \leq \sum_{n=1}^\infty m(E \cap[n, n+1]) = 0 .\end{align*}

• By translation invariance of measure, it suffices to show \(m(E \cap[0, 1]) = 0\).

  • So WLOG, replace \(E\) with \(E\cap[0, 1]\).

• Define \begin{align*} E_j \mathrel{\vcenter{:}}=\left\{{x\in [0, 1] {~\mathrel{\Big|}~}\ \exists p\in {\mathbb{Z}}^{\geq 0} \text{ s.t. } {\left\lvert {x - \frac{p}{j} } \right\rvert} < \frac 1 {j^3}}\right\} .\end{align*}

  • Note that \(E_j \subseteq {\coprod}_{p\in {\mathbb{Z}}^{\geq 0}} B_{j^{-3}}\qty{p\over j}\), i.e. a union over integers \(p\) of intervals of radius \(1/j^3\) around the points \(p/j\). Since \(1/j^3 < 1/j\), this union is in fact disjoint.

• Importantly, note that \begin{align*} \limsup_{j\to\infty} E_j \mathrel{\vcenter{:}}=\cap_{n=1}^\infty \cup_{j=n}^\infty E_j = E \end{align*}

  since

  \begin{align*} x \in \limsup_j E_j &\iff x \in E_j \text{ for infinitely many } j \\ &\iff \text{ there are infinitely many $j$ for which there exist a $p$ such that } {\left\lvert {x - {p\over j}} \right\rvert} < j^{-3} \\ &\iff \text{ there are infinitely many such pairs $p, j$} \\ &\iff x\in E .\end{align*}

• Intersecting with \([0, 1]\), we can write \(E_j\) as a union of intervals: \begin{align*} E_j =& \qty{0, {j^{-3}}} \quad {\coprod}\quad B_{j^{-3}}\qty{1\over j} {\coprod} B_{j^{-3}}\qty{2\over j} {\coprod} \cdots {\coprod} B_{j^{-3}}\qty{j-1\over j} \quad {\coprod}\quad (1 - {j^{-3}}, 1) ,\end{align*} where we’ve separated out the “boundary” terms to emphasize that they are balls about \(0\) and \(1\) intersected with \([0, 1]\).

• Since \(E_j\) is a union of open sets, it is Borel and thus Lebesgue measurable.

• Computing the measure of \(E_j\):

  • For a fixed \(j\), there are exactly \(j+1\) possible choices for a numerator (\(0, 1, \cdots, j\)), thus there are exactly \(j+1\) sets appearing in the above decomposition.

  • The first and last intervals are length \(1 \over j^3\)

  • The remaining \((j+1)-2 = j-1\) intervals are twice this length, \(2 \over j^3\)

  • Thus \begin{align*} m(E_j) = 2 \qty{1 \over j^3} + (j-1) \qty{2 \over j^3} = {2 \over j^2} \end{align*}

• Note that \begin{align*} \sum_{j\in {\mathbb{N}}} m(E_j) = 2\sum_{j\in {\mathbb{N}}} \frac 1 {j^2} < \infty ,\end{align*} which converges by the \(p{\hbox{-}}\)test for sums.

• But then \begin{align*} m(E) &= m(\limsup_j E_j) \\ &= m(\cap_{n\in {\mathbb{N}}} \cup_{j\geq n} E_j) \\ &\leq m(\cup_{j\geq N} E_j) \quad\text{for every } N \\ &\leq \sum_{j\geq N} m(E_j) \\ &\overset{N\to\infty}\to 0 \quad\text{} .\end{align*}

• Thus \(E\) is measurable as a subset of a null set and \(m(E) = 0\).

4.6.1 a

It suffices to consider the bounded case, i.e. \(E \subseteq B_M(0)\) for some \(M\). Then write \(E_n = B_n(0) \cap E\) and apply the theorem to \(E_n\), and by subadditivity, \(m^*(E) = m^*(\cup_n E_n) \leq \sum_n m^*(E_n) = 0\).

Lemma: \(f(x) = x^2, f^{-1}(x) = \sqrt{x}\) are Lipschitz on any compact subset of \([0, \infty)\).

Proof: Let \(g = f\) or \(f^{-1}\). Then \(g\in C^1([0, M])\) for any \(M\), so \(g\) is differentiable and \(g'\) is continuous. Since \(g'\) is continuous on a compact interval, it is bounded, so \({\left\lvert {g'(x)} \right\rvert} \leq L\) for all \(x\). Applying the MVT, \begin{align*} {\left\lvert {f(x) - f(y)} \right\rvert} = f'(c) {\left\lvert {x-y} \right\rvert} \leq L {\left\lvert {x-y} \right\rvert} .\end{align*}

Lemma: If \(g\) is Lipschitz on \({\mathbb{R}}^n\), then \(m(E) = 0 \implies m(g(E)) = 0\).

Proof: If \(g\) is Lipschitz, then \begin{align*} g(B_r(x)) \subseteq B_{Lr}(x) ,\end{align*} which is a dilated ball/cube, and so \begin{align*} m^*(B_{Lr}(x)) \leq L^n \cdot m^*(B_{r}(x)) .\end{align*}

Now choose \(\left\{{Q_j}\right\} \rightrightarrows E\); then \(\left\{{g(Q_j)}\right\} \rightrightarrows g(E)\).

By the above observation, \begin{align*} {\left\lvert {g(Q_j)} \right\rvert} \leq L^n {\left\lvert {Q_j} \right\rvert} ,\end{align*}

and so \begin{align*} m^*(g(E)) \leq \sum_j {\left\lvert {g(Q_j)} \right\rvert} \leq \sum_j L^n {\left\lvert {Q_j} \right\rvert} = L^n \sum_j {\left\lvert {Q_j} \right\rvert} \to 0 .\end{align*}

Now just take \(g(x) = x^2\) for one direction, and \(g(x) = f^{-1}(x) = \sqrt{x}\) for the other. \(\hfill\blacksquare\)

4.6.2 b

Lemma: \(E\) is measurable iff \(E = K {\coprod}N\) for some \(K\) compact, \(N\) null.

Write \(E = K {\coprod}N\) where \(K\) is compact and \(N\) is null.

Then \(\phi^{-1}(E) = \phi^{-1}(K {\coprod}N) = \phi^{-1}(K) {\coprod}\phi^{-1}(N)\).

Since \(\phi^{-1}(N)\) is null by part (a) and \(\phi^{-1}(K)\) is the preimage of a compact set under a continuous map and thus compact, \(\phi^{-1}(E) = K' {\coprod}N'\) where \(K'\) is compact and \(N'\) is null, so \(\phi^{-1}(E)\) is measurable.

So \(\phi\) is a measurable function, and thus yields a well-defined map \(\mathcal L({\mathbb{R}}) \to \mathcal L({\mathbb{R}})\) since it preserves measurable sets. Restricting to \([0, \infty)\), \(f\) is bijection, and thus so is \(\phi\).

4.7.3 a

• Strategy: use approximation by simple functions to show absolute continuity and apply Radon-Nikodym

• Claim: \(\lambda \ll \mu\), i.e. \(\mu(E) = 0 \implies \lambda(E) = 0\).

  • Note that if this holds, by Radon-Nikodym, \(f = {\frac{\partial \lambda}{\partial \mu}\,} \implies d\lambda = f d\mu\), which would yield \begin{align*} \int g ~d\lambda = \int g f ~d\mu .\end{align*}

• So let \(E\) be measurable and suppose \(\mu(E) = 0\).

• Then \begin{align*} \lambda(E) \mathrel{\vcenter{:}}=\int_E f ~d\mu &= \lim_{n\to\infty} \left\{{\int_E s_n \,d\mu {~\mathrel{\Big|}~}s_n \mathrel{\vcenter{:}}=\sum_{j=1}^\infty c_j \mu(E_j),\, s_n \nearrow f}\right\} \end{align*} where we take a sequence of simple functions increasing to \(f\).

• But since each \(E_j \subseteq E\), we must have \(\mu(E_j) = 0\) for any such \(E_j\), so every such \(s_n\) must be zero and thus \(\lambda(E) = 0\).

4.7.4 b

• Set \(g(x) = x^2\), note that \(g\) is positive and measurable.

• By part (a), there exists a positive \(f\) such that for any \(E\subseteq {\mathbb{R}}\), \begin{align*} \int_E g ~dm = \int_E gf ~d\mu \end{align*}

  • The LHS is zero by assumption and thus so is the RHS.

  • \(m \ll \mu\) by construction.

  • Note that \(gf\) is positive.

• Define \(A_k = \left\{{x\in X {~\mathrel{\Big|}~}gf \cdot \chi_E > {1 \over k} }\right\}\), for \(k\in {\mathbb{Z}}^{\geq 0}\)

• Then by Chebyshev with \(p=1\), for every \(k\) we have

\begin{align*} \mu(A_k) \leq k \int_E gf ~d\mu = 0 \end{align*}

• Then noting that \(A_k \searrow A \mathrel{\vcenter{:}}=\left\{{x\in X {~\mathrel{\Big|}~}gf\cdot \chi_E(x) > 0}\right\}\), we have \(\mu(A) = 0\).

• Since \(gf\) is positive, we have \begin{align*} x\in E \iff gf\chi_E(x) > 0 \iff x\in A \end{align*} so \(E = A\) and \(\mu(E) = \mu(A)\).

• But \(m \ll \mu\) and \(\mu(E) = 0\), so we can conclude that \(m(E) = 0\).

• Claim: \(G\in {\mathcal{M}}\).

  • Claim: \begin{align*} G = \qty{ \cap_{N=1}^\infty \cup_{n=N}^\infty E_n}^c = \cup_{N=1}^\infty \cap_{n=N}^\infty E_n^c .\end{align*}

    • This follows because \(x\) is in the RHS \(\iff\) \(x\in E_n^c\) for all but finitely many \(n\) \(\iff\) \(x\in E_n\) for at most finitely many \(n\).

  • But \({\mathcal{M}}\) is a \(\sigma{\hbox{-}}\)algebra, and this shows \(G\) is obtained by countable unions/intersections/complements of measurable sets, so \(G\in {\mathcal{M}}\).

• Claim: \(\mu(G) = 0\).

  • We have \begin{align*} \mu(G) &= \mu\qty{\cup_{N=1}^\infty \cap_{n=N}^\infty E_n^c} \\ &\leq \sum_{N=1}^\infty \mu \qty{\cap_{n=N}^\infty E_n^c} \\ &\leq \sum_{N=1}^\infty \mu(E_M^c) \\ &\mathrel{\vcenter{:}}=\sum_{N=1}^\infty \mu(X\setminus E_N) \\ &\overset{N\to\infty}\to 0 .\end{align*}

Claim: \(K\) is compact.

• It suffices to show that \(K^c \mathrel{\vcenter{:}}=[0, 1]\setminus K\) is open; Then \(K\) will be a closed and bounded subset of \({\mathbb{R}}\) and thus compact by Heine-Borel.

• Strategy: write \(K^c\) as the union of open balls (since these form a basis for the Euclidean topology on \({\mathbb{R}}\)).

  • Do this by showing every point \(x\in K^c\) is an interior point, i.e. \(x\) admits a neighborhood \(N_x\) such that \(N_x \subseteq K^c\).

• Identify \(K^c\) as the set of real numbers in \([0, 1]\) whose decimal expansion does contain a 4.

  • We will show that there exists a neighborhood small enough such that all points in it contain a \(4\) in their decimal expansions.

• Let \(x\in K^c\), suppose a 4 occurs as the \(k\)th digit, and write \begin{align*} x = 0.d_1 d_2 \cdots d_{k-1}~ 4 ~d_{k+1}\cdots = \qty{\sum_{j=1}^k d_j 10^{-j}} + \qty{4\cdot 10^{-k}} + \qty{\sum_{j=k+1}^\infty d_j 10^{-j}} .\end{align*}

• Set \(r_x < 10^{-k}\) and let \(y \in [0, 1] \cap B_{r_x}(x)\) be arbitrary and write \begin{align*} y = \sum_{j=1}^\infty c_j 10^{-j} .\end{align*}

• Thus \({\left\lvert {x-y} \right\rvert} < r_x < 10^{-k}\), and the first \(k\) digits of \(x\) and \(y\) must agree:

  • We first compute the difference: \begin{align*} x - y &= \sum_{i=1}^\infty d_j 10^{-j} - \sum_{i=1}^\infty c_j 10^{-j} = \sum_{i=1}^\infty \qty{d_j - c_j} 10^{-j} \\ \end{align*}
  • Thus (claim) \begin{align*} {\left\lvert {x-y} \right\rvert} &\leq \sum_{j=1}^\infty {\left\lvert {d_j - c_j} \right\rvert} 10^j < 10^{-k} \iff {\left\lvert {d_j - c_j} \right\rvert} = 0 \quad \forall j\leq k .\end{align*}
  • Otherwise we can note that any term \({\left\lvert {d_j - c_j} \right\rvert}\geq 1\) and there is a contribution to \({\left\lvert {x-y} \right\rvert}\) of at least \(1\cdot 10^{-j}\) for some \(j < k\), whereas \begin{align*} j < k \iff 10^{-j} > 10^{-k} ,\end{align*} a contradiction.

• This means that for all \(j \leq k\) we have \(d_j = c_j\), and in particular \(d_k = 4 = c_k\), so \(y\) has a 4 in its decimal expansion.

• But then \(K^c = \cup_x B_{r_x}(x)\) is a union of open sets and thus open.

Claim: \(K\) is nowhere dense and \(m(K) = 0\):

• Strategy: Show \(\qty{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}^\circ = \emptyset\).

• Since \(K\) is closed, \(\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu = K\), so it suffices to show that \(K\) does not properly contain any interval.

• It suffices to show \(m(K^c) = 1\), since this implies \(m(K) = 0\) and since any interval has strictly positive measure, this will mean \(K\) can not contain an interval.

• As in the construction of the Cantor set, let

  • \(K_0\) denote \([0, 1]\) with 1 interval \(\left({4 \over 10}, {5 \over 10} \right)\) of length \(1 \over 10\) deleted, so \begin{align*}m(K_0^c) = {1\over 10}.\end{align*}
  • \(K_1\) denote \(K_0\) with 9 intervals \(\left({1 \over 100}, {5\over 100}\right), ~\left({14 \over 100}, {15 \over 100}\right), \cdots \left({94\over 100}, {95 \over 100}\right)\) of length \({1 \over 100}\) deleted, so \begin{align*}m(K_1^c) = {1\over 10} + {9 \over 100}.\end{align*}
  • \(K_n\) denote \(K_{n-1}\) with \(9^{n}\) such intervals of length \(1 \over 10^{n+1}\) deleted, so \begin{align*}m(K_n^c) = {1\over 10} + {9 \over 100} + \cdots + {9^{n} \over 10^{n+1}}.\end{align*}

• Then compute \begin{align*} m(K^c) = \sum_{j=0}^\infty {9^n \over 10^{n+1} } = {1\over 10} \sum_{j=0}^\infty \qty{9\over 10}^n = {1 \over 10} \qty{ {1 \over 1 - {9 \over 10 } } } = 1. \end{align*}

Claim: \(K\) has no isolated points:

• A point \(x\in K\) is isolated iff there there is an open ball \(B_r(x)\) containing \(x\) such that \(B_r(x) \subsetneq K^c\).

  • So every point in this ball should have a 4 in its decimal expansion.

• Strategy: show that if \(x\in K\), every neighborhood of \(x\) intersects \(K\).

• Note that \(m(K_n) = \left( \frac 9 {10} \right)^n \overset{n\to\infty}\to 0\)

• Also note that we deleted open intervals, and the endpoints of these intervals are never deleted.

  • Thus endpoints of deleted intervals are elements of \(K\).

• Fix \(x\). Then for every \(\varepsilon\), by the Archimedean property of \({\mathbb{R}}\), choose \(n\) such that \(\left( \frac 9 {10} \right)^n < \varepsilon\).

• Then there is an endpoint \(x_n\) of some deleted interval \(I_n\) satisfying \begin{align*}{\left\lvert {x - x_n} \right\rvert} \leq \left( \frac 9 {10} \right)^n < \varepsilon.\end{align*}

• So every ball containing \(x\) contains some endpoint of a removed interval, and thus an element of \(K\).

• Suppose it is not the case that \(f=g\) almost everywhere; then letting \(A\mathrel{\vcenter{:}}=\left\{{x\in [a,b] {~\mathrel{\Big|}~}f(x) \neq g(x)}\right\}\), we have \(m(A) > 0\).

• Write \begin{align*} A = A_1{\coprod}A_2 \mathrel{\vcenter{:}}=\left\{{f>g}\right\} {\coprod}\left\{{f then \(m(A_1) > 0\) or \(m(A_2) > 0\) (or both).

• Wlog (by relabeling \(f, g\) if necessary), suppose \(m(A_1) > 0\), and take \(E\mathrel{\vcenter{:}}= A_1\).

• Then on \(E\), we have \(f(x)>g(x)\) pointwise. This is preserved by monotonicity of the integral, thus \begin{align*} f(x) > g(x) \text{ on } E \implies \int_{E} f(x)\,dx > \int_{E} g(x)\, dx .\end{align*}

• By induction on the number of limits we can pass through the integral.

• For \(n=1\) we first pass one derivative into the integral: let \(x_n \to x\) be any sequence converging to \(x\), then \begin{align*} {\frac{\partial }{\partial x}\,} {\sin(x) \over x} &= {\frac{\partial }{\partial x}\,} \int_0^1 \cos(tx)\,dt \\ &= \lim_{x_n\to x} {1\over x_n - x} \qty{ \int_0^1 \cos(t x_n)\,dt - \int_0^1 \cos(tx) \,dt} \\ &= \lim_{x_n\to x} \qty{ \int_0^1 { \cos(tx_n) - \cos(tx) \over x_n - x} \,dt} \\ &= \lim_{x_n\to x} \qty{ \int_0^1 \qty{t\sin(tx)\mathrel{\Big|}_{x=\xi_n}} \,dt} {\quad \operatorname{where} \quad} \xi_n \in [x_n, x] \text{ by MVT}, \xi_n\to x \\ &= \lim_{\xi_n\to x} \qty{ \int_0^1 t \sin(t \xi_n) \,dt} \\ &=_{\text{DCT}} \int_0^1 \lim_{\xi_n \to x} t \sin(t \xi_n) \,dt \\ &= \int_0^1 t\sin(tx) \,dt \\ .\end{align*}

• Taking absolute values we obtain an upper bound \begin{align*} {\left\lvert { {\frac{\partial }{\partial x}\,} {\sin(x) \over x} } \right\rvert} &= {\left\lvert { \int_0^1 t\sin(tx) \,dt } \right\rvert} \\ &\leq \int_0^1 {\left\lvert {t\sin(tx)} \right\rvert} \,dt \\ &\leq \int_0^1 1 \, dt = 1 ,\end{align*} since \(t\in [0, 1] \implies {\left\lvert {t} \right\rvert} < 1\), and \({\left\lvert {\sin(xt)} \right\rvert} \leq 1\) for any \(x\) and \(t\).

• Note that this bound also justifies the DCT, since the functions \(f_n(t) = t\sin(t \xi_n )\) are uniformly dominated by \(g(t) = 1\) on \(L^1([0, 1])\).

Note: integrating by parts here yields the actual formula: \begin{align*} \int_0^1 t\sin(tx) \,dt &=_{\text{IBP}} \qty{-t\cos(tx) \over x}\mathrel{\Big|}_{t=0}^{t=1} - \int_0^1 {\cos(tx) \over x} \,dt \\ &= {-\cos(x) \over x} - {\sin(x) \over x^2} \\ &= {x\cos(x) - \sin(x) \over x^2} .\end{align*}

• For the inductive step, we assume that we can pass \(n-1\) limits through the integral and show we can pass the \(n\)th through as well. \begin{align*} {\frac{\partial ^n}{\partial x^n}\,} {\sin(x) \over x} &= {\frac{\partial ^n}{\partial x^n}\,} \int_0^1 \cos(tx)\,dt \\ &= {\frac{\partial }{\partial x}\,} \int_0^1 {\frac{\partial ^{n-1}}{\partial x^{n-1}}\,} \cos(tx)\,dt \\ &= {\frac{\partial }{\partial x}\,} \int_0^1 t^{n-1} f_{n-1}(x, t) \,dt \end{align*}
  • Note that \(f_n(x, t) = \pm \sin(tx)\) when \(n\) is odd and \(f_n(x, t) = \pm \cos(tx)\) when \(n\) is even, and a constant factor of \(t\) is multiplied when each derivative is taken.
• We continue as in the base case: \begin{align*} {\frac{\partial }{\partial x}\,} \int_0^1 t^{n-1} f_{n-1}(x, t) \,dt &= \lim_{x_k\to x} \int_0^1 t^{n-1} \qty{ f_{n-1}(x_n, t) - f_{n-1}(x, t) \over x_n - x} \,dt \\ &=_{\text{IVT}} \lim_{x_k\to x} \int_0^1 t^{n-1} {\frac{\partial f_{n-1}}{\partial x}\,}(\xi_k, t) \,dt \quad\text{where } \xi_k\in [x_k, x],\, \xi_k \to x \\ &=_{\text{DCT}} \int_0^1 \lim_{x_k\to x} t^{n-1} {\frac{\partial f_{n-1}}{\partial x}\,}(\xi_k, t) \,dt \\ &\mathrel{\vcenter{:}}=\int_0^1 \lim_{x_k\to x} t^{n} f_n(\xi_k, t) \,dt \\ &\mathrel{\vcenter{:}}=\int_0^1 t^{n} f_n(x, t) \,dt .\end{align*}
  • We’ve used the fact that \(f_0(x) = \cos(tx)\) is smooth as a function of \(x\), and in particular continuous
  • The DCT is justified because the functions \(h_{n, k}(x, t) = t^n f_n(\xi_k, t)\) are again uniformly (in \(k\)) bounded by 1 since \(t\leq 1 \implies t^n \leq 1\) and each \(f_n\) is a sin or cosine.
• Now take absolute values \begin{align*} {\left\lvert {{\frac{\partial ^n}{\partial x^n}\,} {\sin(x) \over x} } \right\rvert} &= {\left\lvert { \int_0^1 -t^n f_n(x, t) ~dt } \right\rvert} \\ &\leq \int_0^1 {\left\lvert {t^n f_n(x, t)} \right\rvert} ~dt \\ &\leq \int_0^1 {\left\lvert {t^n} \right\rvert} {\left\lvert {f_n(x, t)} \right\rvert} ~dt \\ &\leq \int_0^1 {\left\lvert { t^n} \right\rvert} \cdot 1 ~dt \\ &\leq \int_0^1 t^n ~dt \quad\text{since $t$ is positive} \\ &= \frac{1}{n+1} \\ &< \frac{1}{n} .\end{align*}
  • We’ve again used the fact that \(f_n(x, t)\) is of the form \(\pm \cos(tx)\) or \(\pm \sin(tx)\), both of which are bounded by 1.

Note that \begin{align*} \lim_{n} \qty{1 + {x^2 \over n}}^{-(n+1)} &= {1 \over \lim_{n} \qty{1 + {x^2 \over n}}^1 \qty{1 + {x^2 \over n}}^n } \\ &= {1 \over 1 \cdot e^{x^2}} \\ &= e^{-x^2} .\end{align*}

If passing the limit through the integral is justified, we will have \begin{align*} \lim_{n\to\infty} \int_0^n \qty{ 1 + {x^2\over n}}^{-(n+1)}\, dx &= \lim_{n\to\infty} \int_{\mathbb{R}}\chi_{[0, n]} \qty{ 1 + {x^2\over n}}^{-(n+1)} \, dx \\ &= \int_{\mathbb{R}}\lim_{n\to\infty} \chi_{[0, n]} \qty{ 1 + {x^2\over n}}^{-(n+1)} \, dx {\quad \operatorname{by the DCT} \quad} \\ &= \int_{\mathbb{R}}\lim_{n\to\infty} \qty{ 1 + {x^2\over n}}^{-(n+1)} \, dx \\ &= \int_0^\infty e^{-x^2} \\ &= {\sqrt \pi \over 2} .\end{align*}

Computing the last integral:

\begin{align*} \qty{\int_{\mathbb{R}}e^{-x^2}\, dx}^2 &= \qty{\int_{\mathbb{R}}e^{-x^2}\,dx} \qty{\int_{\mathbb{R}}e^{-y^2}\,dx} \\ &= \int_{\mathbb{R}}\int_{\mathbb{R}}e^{-(x+y)^2}\, dx \\ &= \int_0^{2\pi} \int_0^\infty e^{-r^2} r\, dr \, d\theta \qquad u=r^2 \\ &= {1\over 2} \int_0^{2\pi } \int_0^\infty e^{-u}\, du \, d\theta \\ &= {1\over 2} \int_0^{2\pi} 1 \\ &= \pi ,\end{align*} and now use the fact that the function is even so \(\int_0^\infty f = {1\over 2} \int_{\mathbb{R}}f\).

Justifying the DCT:

• Apply Bernoulli’s inequality: \begin{align*} {1 + {x^2\over n}}^{n+1} \geq {1 + {x^2\over n}}\qty{1 + x^2} \geq {1 + x^2} ,\end{align*} where the last inequality follows from the fact that \(1 + {x^2 \over n} \geq 1\)

\(L^2\) bound:

• Since \(f_k \to f\) almost everywhere, \(\liminf_n f_n(x) = f(x)\) a.e.
• \({\left\lVert {f_n} \right\rVert}_2 < \infty\) implies each \(f_n\) is measurable and thus \({\left\lvert {f_n} \right\rvert}^2 \in L^+\), so we can apply Fatou:

\begin{align*} {\left\lVert {f} \right\rVert}_2^2 &= \int {\left\lvert {f(x)} \right\rvert}^2 \\ &= \int \liminf_n {\left\lvert {f_n(x)} \right\rvert}^2 \\ &\underset{\text{Fatou}}\leq\liminf_n \int {\left\lvert {f_n(x)} \right\rvert}^2 \\ &\leq \liminf_n M \\ &= M .\end{align*}

• Thus \({\left\lVert {f} \right\rVert}_2 \leq \sqrt{M} < \infty\) implying \(f\in L^2\).

Equality of Integrals:

• Take the sequence \(\varepsilon_n = {1\over n}\)

• Apply Egorov’s theorem: obtain a set \(F_\varepsilon\) such that \(f_n \to f\) uniformly on \(F_\varepsilon\) and \(m(I\setminus F_\varepsilon) < \varepsilon\). \begin{align*} \lim_{n\to \infty} {\left\lvert { \int_0^1 f_n - f } \right\rvert} &\leq \lim_{n\to\infty} \int_0^1 {\left\lvert {f_n - f} \right\rvert} \\ &= \lim_{n\to\infty} \qty{ \int_{F_\varepsilon} {\left\lvert {f_n - f} \right\rvert} + \int_{I\setminus F_\varepsilon} {\left\lvert {f_n - f} \right\rvert} } \\ &= \int_{F_\varepsilon} \lim_{n\to\infty} {\left\lvert {f_n - f} \right\rvert} + \lim_{n\to\infty} \int_{I\setminus F_\varepsilon} {\left\lvert {f_n - f} \right\rvert} \quad\text{by uniform convergence} \\ &= 0 + \lim_{n\to\infty} \int_{I\setminus F_\varepsilon} {\left\lvert {f_n - f} \right\rvert} ,\end{align*}

  so it suffices to show \(\int_{I\setminus F_\varepsilon} {\left\lvert {f_n - f} \right\rvert} \overset{n\to\infty}\to 0\).

• We can obtain a bound using Holder’s inequality with \(p=q=2\): \begin{align*} \int_{I\setminus F_\varepsilon} {\left\lvert {f_n - f} \right\rvert} &\leq \qty{ \int_{I\setminus F_\varepsilon} {\left\lvert {f_n - f} \right\rvert}^2 } \qty{ \int_{I\setminus F_\varepsilon} {\left\lvert {1} \right\rvert}^2 } \\ &= \qty{ \int_{I\setminus F_\varepsilon} {\left\lvert {f_n - f} \right\rvert}^2 } \mu(F_\varepsilon) \\ &\leq {\left\lVert {f_n - f} \right\rVert}_2 \mu(F_\varepsilon) \\ &\leq \qty{ {\left\lVert {f_n} \right\rVert}_2 + {\left\lVert {f} \right\rVert}_2 } \mu(F_\varepsilon) \\ &\leq 2M \cdot \mu(F_\varepsilon) \end{align*} where \(M\) is now a constant not depending on \(\varepsilon\) or \(n\).

• Now take a nested sequence of sets \(F_{\varepsilon}\) with \(\mu(F_\varepsilon) \to 0\) and applying continuity of measure yields the desired statement.

• Note that \(x^{1\over n} \overset{n\to\infty}\to 1\) for any \(0 < x < \infty\).
• Thus the integrand converges to \({1\over e^x}\), which is integrable on \((0, \infty)\) and integrates to 1.
• Break the integrand up: \begin{align*} \int_0^\infty {1 \over \qty{ 1 + {x\over n} }^n x^{1\over n}} \,dx = \int_0^1 {1 \over \qty{ 1 + {x\over n} }^n x^{1\over n}} \,dx = \int_1^\infty {1 \over \qty{ 1 + {x\over n} }^n x^{1\over n}} \,dx .\end{align*}

\begin{align*} {\frac{\partial }{\partial t}\,} F(t) &= {\frac{\partial }{\partial t}\,} \int_{\mathbb{R}}f(x) \cos(xt) ~dx \\ &\overset{DCT}= \int_{\mathbb{R}}f(x) {\frac{\partial }{\partial t}\,} \cos(xt) ~dx \\ &= \int_{\mathbb{R}}xf(x) \cos(xt)~dx ,\end{align*} so it only remains to justify the DCT.

• Fix \(t\), then let \(t_n \to t\) be arbitrary.

• Define \begin{align*} h_n(x, t) = f(x) \left(\frac{\cos(tx) - \cos(t_n x)}{t_n - t}\right) \overset{n\to\infty}\to {\frac{\partial }{\partial t}\,} \qty{f(x) \cos(xt)} \end{align*} since \(\cos(tx)\) is differentiable in \(t\) and this is the limit definition of differentiability.

• Note that \begin{align*} {\frac{\partial }{\partial t}\,} \cos(tx) &\mathrel{\vcenter{:}}=\lim_{t_n \to t} \frac{\cos(tx) - \cos(t_n x)}{t_n - t} \\ &\overset{MVT} = {\frac{\partial }{\partial t}\,}\cos(tx)\mathrel{\Big|}_{t = \xi_n} \hspace{6em} \text{for some } \xi_n \in [t, t_n] \text{ or } [t_n, t] \\ &= x\sin(\xi_n x) \end{align*} where \(\xi_n \overset{n\to\infty}\to t\) since wlog \(t_n \leq \xi_n \leq t\) and \(t_n \nearrow t\).

• We then have \begin{align*}{\left\lvert {h_n(x)} \right\rvert} = {\left\lvert {f(x) x\sin(\xi_n x)} \right\rvert} \leq {\left\lvert {xf(x)} \right\rvert}\quad\text{since } {\left\lvert {\sin(\xi_n x)} \right\rvert} \leq 1\end{align*} for every \(x\) and every \(n\).

• Since \(xf(x) \in L^1({\mathbb{R}})\) by assumption, the DCT applies.

• Since \(\int {\left\lvert {f_n} \right\rvert} \overset{n\to\infty}\to \int {\left\lvert {f} \right\rvert}\), define \begin{align*} h_n &= {\left\lvert {f_n - f} \right\rvert} &\overset{n\to\infty}\to 0 ~a.e.\\ g_n &= {\left\lvert {f_n} \right\rvert} + {\left\lvert {f} \right\rvert} &\overset{n\to\infty}\to 2{\left\lvert {f} \right\rvert} ~a.e. \end{align*}

  • Note that \(g_n - h_n \overset{n\to\infty}\to 2{\left\lvert {f} \right\rvert} - 0 = 2{\left\lvert {f} \right\rvert}\).

• Then \begin{align*} \int 2 {\left\lvert {f} \right\rvert} &= \int \liminf_n (g_n - h_n) \\ &= \int \liminf_n(g_n) + \int \liminf_n(-h_n) \\ &= \int \liminf_n(g_n) - \int \limsup_n(h_n) \\ &= \int 2 {\left\lvert {f} \right\rvert} - \int \limsup_n(h_n) \\ &\leq \int 2{\left\lvert {f} \right\rvert} - \limsup_n \int h_n \quad\text{by Fatou} ,\end{align*}

• Since \(f\in L^1\), \(\int 2{\left\lvert {f} \right\rvert} = 2{\left\lVert {f} \right\rVert}_1 < \infty\) and it makes sense to subtract it from both sides, thus \begin{align*} 0 &\leq - \limsup_n \int h_n \\ &\mathrel{\vcenter{:}}=- \limsup_n \int {\left\lvert {f_n - f} \right\rvert} .\end{align*} which forces \(\limsup_n \int {\left\lvert {f_n -f} \right\rvert} = 0\), since

  • The integral of a nonnegative function is nonnegative, so \(\int {\left\lvert {f_n - f} \right\rvert} \geq 0\).
  • So \(\qty{ -\int {\left\lvert {f_n - f} \right\rvert} } \leq 0\).
  • But the above inequality shows \(\qty{ -\int {\left\lvert {f_n - f} \right\rvert} } \geq 0\) as well.

• Since \(\liminf_n \int h_n \leq \limsup_n \int h_n = 0\), \(\lim_n \int h_n\) exists and is equal to zero.

• But then \begin{align*} {\left\lvert {\int f_n - \int f} \right\rvert} &= {\left\lvert {\int f_n -f} \right\rvert} \leq \int {\left\lvert {f_n - f} \right\rvert} ,\end{align*} and taking \(\lim_{n\to\infty}\) on both sides yields \begin{align*} \lim_{n\to\infty} {\left\lvert {\int f_n - \int f} \right\rvert} \leq \lim_{n\to\infty} \int {\left\lvert {f_n - f} \right\rvert} = 0 ,\end{align*} so \(\lim_{n\to\infty} \int f_n = \int f\).

6.7.1 a

Claim: \(f_n\) does not converge uniformly to its limit.

• Note each \(f_n(x)\) is clearly continuous on \((0, \infty)\), since it is a quotient of continuous functions where the denominator is never zero.

• Note \begin{align*} x < 1 \implies x^n \overset{n\to\infty}\to 0{\quad \operatorname{and} \quad} x>1 \implies x^n \overset{n\to\infty}\to \infty .\end{align*}

• Thus \begin{align*} f_n(x) = \frac{x}{1+ x^n}\overset{n\to\infty}\longrightarrow f(x) \mathrel{\vcenter{:}}= \begin{cases} x, & 0 \leq x < 1 \\ \frac 1 2, & x = 1 \\ 0, & x > 1 \\ \end{cases} .\end{align*}

• If \(f_n \to f\) uniformly on \([0, \infty)\), it would converge uniformly on every subset and thus uniformly on \((0, \infty)\).

  • Then \(f\) would be a uniform limit of continuous functions on \((0, \infty)\) and thus continuous on \((0, \infty)\).
  • By uniqueness of limits, \(f_n\) would converge to the pointwise limit \(f\) above, which is not continuous at \(x=1\), a contradiction.

6.7.2 b

• If the DCT applies, interchange the limit and integral: \begin{align*} \lim_{n\to\infty }\int_0^\infty f_n(x)\, dx &= \int_0^\infty \lim_{n\to\infty}f_n(x) \, dx \quad\text{DCT}\\ &=\int_0^\infty f(x) \,dx \\ &= \int_0^1 x \,dx + \int_1^\infty 0\, dx \\ &= {1\over 2}x^2 \Big|_0^1 \\ &= {1\over 2} .\end{align*}

• To justify the DCT, write \begin{align*} \int_0^\infty f_n(x) = \int_0^1 f_n(x) + \int_1^\infty f_n(x) .\end{align*}

• \(f_n\) restricted to \((0, 1)\) is uniformly bounded by \(g_0(x) = 1\) in the first integral, since \begin{align*} x \in [0, 1] \implies \frac{x}{1+x^n} < \frac{1}{1+x^n} < 1 \mathrel{\vcenter{:}}= g(x) \end{align*} so \begin{align*} \int_0^1 f_n(x)\,dx \leq \int_0^1 1 \,dx = 1 < \infty .\end{align*} Also note that \(g_0\cdot \chi_{(0, 1)} \in L^1((0, \infty))\).

• The \(f_n\) restricted to \((1, \infty)\) are uniformly bounded by \(g_1(x) = {1\over x^{2}}\) on \([1, \infty)\), since \begin{align*} x \in (1, \infty) \implies \frac{x}{1+x^n} \leq {x \over x^n} = {1 \over x^{n-1}} \leq {1\over x^2}\in L^1([1, \infty) \text{ when } n\geq 3 ,\end{align*} by the \(p{\hbox{-}}\)test for integrals.

• So set \begin{align*}g \mathrel{\vcenter{:}}= g_0 \cdot \chi_{(0, 1)} + g_1 \cdot \chi_{[1, \infty)},\end{align*} then by the above arguments \(g \in L^1((0, \infty))\) and \(f_n \leq g\) everywhere, so the DCT applies.

• Fixing notation, set \(\tau_x f(y) \mathrel{\vcenter{:}}= f(y-x)\); we then want to show \begin{align*} {\left\lVert {\tau_x f -f} \right\rVert}_{L^1} \overset{x\to 0}\to 0 .\end{align*}
• Claim: by an \(\varepsilon/3\) argument, it suffices to show this for compactly supported functions:
  • Since \(f\in L^1\), choose \(g_n\subset C_c^\infty({\mathbb{R}}^1)\) smooth and compactly supported so that \begin{align*}{\left\lVert {f-g} \right\rVert}_{L^1} < \varepsilon.\end{align*}
  • Claim: \({\left\lVert {\tau_x f - \tau_x g} \right\rVert} < \varepsilon\).
    • Proof 1: translation invariance of the integral.
    • Proof 2: Apply a change of variables: \begin{align*} {\left\lVert {\tau_x f - \tau_x g} \right\rVert}_1 &\mathrel{\vcenter{:}}=\int_{\mathbb{R}}{\left\lvert {\tau_x f(y) - \tau_x g(y)} \right\rvert}\, dy \\ &= \int_{\mathbb{R}}{\left\lvert {f(y-x) - g(y-x)} \right\rvert}\, dy \\ &= \int_{\mathbb{R}}{\left\lvert {f(u) - g(u)} \right\rvert}\, du \qquad (u=y-x,\, du=dy) \\ &= {\left\lVert {f-g} \right\rVert}_1 \\ &< \varepsilon .\end{align*}
  • Then \begin{align*} {\left\lVert {\tau_x f - f} \right\rVert}_1 &= {\left\lVert {\tau_x f - \tau_x g + \tau_x g - g +g - f} \right\rVert}_{1} \\ &\leq {\left\lVert {\tau_x f - \tau_x g} \right\rVert}_1 + {\left\lVert {\tau_x g - g} \right\rVert}_1 + {\left\lVert {g - f} \right\rVert}_{1} \\ &\leq 2\varepsilon+ {\left\lVert {\tau_x g - g} \right\rVert}_1 .\end{align*}
• To show this for compactly supported functions:

  • Let \(g\in C_c^\infty({\mathbb{R}}^1)\), let \(E = {\operatorname{supp}}(g)\), and write \begin{align*} {\left\lVert {\tau_x g - g} \right\rVert}_1 &= \int_{\mathbb{R}}{\left\lvert {g(y-x) - g(y)} \right\rvert}\,dy \\ &= \int_E {\left\lvert {g(y-x) - g(y)} \right\rvert} \,dy + \int_{E^c} {\left\lvert {g(y-x) - g(y)} \right\rvert} \,dy\\ &= \int_E {\left\lvert {g(y-x) - g(y)} \right\rvert} \,dy .\end{align*}

  • But \(g\) is smooth and compactly supported on \(E\), and thus uniformly continuous on \(E\), so \begin{align*} \lim_{x\to 0} \int_E {\left\lvert {g(y-x) - g(y)} \right\rvert} \,dy &= \int_E \lim_{x\to 0} {\left\lvert {g(y-x) - g(y)} \right\rvert} \,dy \\ &= \int_E 0 \,dy \\ &= 0 .\end{align*}

• \({\left\lVert {f} \right\rVert}_p \leq {\left\lVert {f} \right\rVert}_\infty\):

  • Note \({\left\lvert {f(x)} \right\rvert} \leq {\left\lVert {f} \right\rVert}_\infty\) almost everywhere and taking \(p\)th powers preserves this inequality.

  • Thus \begin{align*} {\left\lvert {f(x)} \right\rvert} &\leq {\left\lVert {f} \right\rVert}_\infty \quad\text{a.e. by definition} \\ \implies {\left\lvert {f(x)} \right\rvert}^p &\leq {\left\lVert {f} \right\rVert}_\infty^p \quad\text{for } p\geq 0 \\ \implies {\left\lVert {f} \right\rVert}_p^p &= \int_X {\left\lvert {f(x)} \right\rvert}^p ~dx \\ &\leq \int_X {\left\lVert {f} \right\rVert}_\infty^p ~dx \\ &= {\left\lVert {f} \right\rVert}_\infty^p \int_X 1\,dx \\ &= {\left\lVert {f} \right\rVert}_\infty^p \cdot m(X) \quad\text{since the norm doesn't depend on }x \\ &= {\left\lVert {f} \right\rVert}_\infty^p \qquad \text{since } m(X) = 1 .\end{align*}

    • Thus \({\left\lVert {f} \right\rVert}_p \leq {\left\lVert {f} \right\rVert}_\infty\) for all \(p\) and taking \(\lim_{p\to\infty}\) preserves this inequality.
• \({\left\lVert {f} \right\rVert}_p \geq {\left\lVert {f} \right\rVert}_\infty\):

  • Fix \(\varepsilon > 0\).

  • Define \begin{align*} S_\varepsilon \mathrel{\vcenter{:}}=\left\{{x\in {\mathbb{R}}^n {~\mathrel{\Big|}~}{\left\lvert {f(x)} \right\rvert} \geq {\left\lVert {f} \right\rVert}_\infty - \varepsilon}\right\} .\end{align*}

    • Note that \(m(S_\varepsilon) > 0\); otherwise if \(m(S_\varepsilon) = 0\), then \(t\mathrel{\vcenter{:}}={\left\lVert {f} \right\rVert}_\infty - \varepsilon< {\left\lVert {f} \right\rVert}_\varepsilon\). But this produces a smaller upper bound almost everywhere than \({\left\lVert {f} \right\rVert}_\varepsilon\), contradicting the definition of \({\left\lVert {f} \right\rVert}_\varepsilon\) as an infimum over such bounds.

  • Then \begin{align*} {\left\lVert {f} \right\rVert}_p^p &= \int_X {\left\lvert {f(x)} \right\rvert}^p ~dx \\ &\geq \int_{S_\varepsilon} {\left\lvert {f(x)} \right\rvert}^p ~dx \quad\text{since } S_\varepsilon\subseteq X \\ &\geq \int_{S_\varepsilon} {\left\lvert {{\left\lVert {f} \right\rVert}_\infty - \varepsilon} \right\rvert}^p ~dx \quad\text{since on } S_\varepsilon, {\left\lvert {f} \right\rvert} \geq {\left\lVert {f} \right\rVert}_\infty - \varepsilon\\ &= {\left\lvert {{\left\lVert {f} \right\rVert}_\infty - \varepsilon} \right\rvert}^p \cdot m(S_\varepsilon) \quad\text{since the integrand is independent of }x \\ & \geq 0 \quad\text{since } m(S_\varepsilon) > 0 \end{align*}

  • Taking \(p\)th roots for \(p\geq 1\) preserves the inequality, so \begin{align*} \implies {\left\lVert {f} \right\rVert}_p &\geq {\left\lvert {{\left\lVert {f} \right\rVert}_\infty - \varepsilon} \right\rvert} \cdot m(S_\varepsilon)^{\frac 1 p} \overset{p\to\infty}\to {\left\lvert {{\left\lVert {f} \right\rVert}_\infty - \varepsilon} \right\rvert} \overset{\varepsilon \to 0}\to {\left\lVert {f} \right\rVert}_\infty \end{align*} where we’ve used the fact that above arguments work

  • Thus \({\left\lVert {f} \right\rVert}_p \geq {\left\lVert {f} \right\rVert}_\infty\).

7.2.1 Proof 1: Using Fourier Transforms

• Fix \(k \in {\mathbb{Z}}\).

• Since \(e^{2\pi i k x}\) is continuous on the compact interval \([0, 1]\), it is uniformly continuous.

• Thus there is a sequence of polynomials \(P_\ell\) such that \begin{align*} P_{\ell, k} \overset{\ell\to\infty}\to e^{2\pi i k x} \text{ uniformly on } [0,1] .\end{align*}

• Note applying linearity to the assumption \(\int f(x) \, x^n\), we have \begin{align*} \int f(x) x^n \,dx = 0 ~\forall n \implies \int f(x) p(x) \,dx = 0 \end{align*} for any polynomial \(p(x)\), and in particular for \(P_{\ell, k}(x)\) for every \(\ell\) and every \(k\).

• But then
  \begin{align*} {\left\langle {f},~{e_k} \right\rangle} &= \int_0^1 f(x) e^{-2\pi i k x} ~dx \\ &= \int_0^1 f(x) \lim_{\ell \to \infty} P_\ell(x) \\ &= \lim_{\ell \to \infty} \int_0^1 f(x) P_\ell(x) \quad\quad \text{by uniform convergence on a compact interval} \\ &= \lim_{\ell \to \infty} 0 \quad\text{by assumption}\\ &= 0 \quad \forall k\in {\mathbb{Z}} ,\end{align*} so \(f\) is orthogonal to every \(e_k\).

• Thus \(f\in S^\perp \mathrel{\vcenter{:}}={\operatorname{span}}_{\mathbb{C}}\left\{{e_k}\right\}_{k\in {\mathbb{Z}}}^\perp \subseteq L^2([0, 1])\), but since this is a basis, \(S\) is dense and thus \(S^\perp = \left\{{0}\right\}\) in \(L^2([0, 1])\).

• Thus \(f\equiv 0\) in \(L^2([0, 1])\), which implies that \(f\) is zero almost everywhere.

\(\hfill\blacksquare\)

7.2.2 Alternative Proof

• By density of polynomials, for \(f\in L^2([0, 1])\) choose \(p_\varepsilon(x)\) such that \({\left\lVert {f - p_\varepsilon} \right\rVert} < \varepsilon\) by Weierstrass approximation.

• Then on one hand, \begin{align*} {\left\lVert {f(f-p_\varepsilon)} \right\rVert}_1 &= {\left\lVert {f^2} \right\rVert}_1 - {\left\lVert {f\cdot p_\varepsilon} \right\rVert}_1 \\ &= {\left\lVert {f^2} \right\rVert}_1 - 0 \quad\text{by assumption} \\ &= {\left\lVert {f} \right\rVert}_2^2 .\end{align*}

  • Where we’ve used that \({\left\lVert {f^2} \right\rVert}_1 = \int {\left\lvert {f^2} \right\rvert} = \int {\left\lvert {f} \right\rvert}^2 = {\left\lVert {f} \right\rVert}_2^2\).

• On the other hand \begin{align*} {\left\lVert {f(f-p_\varepsilon)} \right\rVert} &\leq {\left\lVert {f} \right\rVert}_1 {\left\lVert {f-p_\varepsilon} \right\rVert}_\infty \quad\text{by Holder} \\ &\leq \varepsilon{\left\lVert {f} \right\rVert}_1 \\ &\leq \varepsilon{\left\lVert {f} \right\rVert}_2 \sqrt{m(X)} \\ &= \varepsilon{\left\lVert {f} \right\rVert}_2 \quad\text{since } m(X)= 1 .\end{align*}

  • Where we’ve used that \({\left\lVert {fg} \right\rVert}_1 = \int {\left\lvert {fg} \right\rvert} = \int {\left\lvert {f} \right\rvert}{\left\lvert {g} \right\rvert} \leq \int {\left\lVert {f} \right\rVert}_\infty {\left\lvert {g} \right\rvert} = {\left\lVert {f} \right\rVert}_\infty {\left\lVert {g} \right\rVert}_1\).

• Combining these, \begin{align*} {\left\lVert {f} \right\rVert}_2^2 \leq {\left\lVert {f} \right\rVert}_2 \varepsilon\implies {\left\lVert {f} \right\rVert}_2 < \varepsilon\to 0, .\end{align*} so \({\left\lVert {f} \right\rVert}_2 = 0\), which implies \(f=0\) almost everywhere.

8.1.1 a

Stated integral equality:

• Let \(\varepsilon> 0\)
• \(C_c({\mathbb{R}}^n) \hookrightarrow L^1({\mathbb{R}}^n)\) is dense so choose \(\left\{{f_n}\right\} \to f\) with \({\left\lVert {f_n - f} \right\rVert}_1 \to 0\).
• Since \(\left\{{f_n}\right\}\) are compactly supported, choose \(N_0\gg 1\) such that \(f_n\) is zero outside of \(B_{N_0}(\mathbf{0})\).
• Then \begin{align*} N\geq N_0 \implies \int_{{\left\lvert {x} \right\rvert} > N} {\left\lvert {f} \right\rvert} &= \int_{{\left\lvert {x} \right\rvert} > N} {\left\lvert {f - f_n + f_n} \right\rvert} \\ &\leq \int_{{\left\lvert {x} \right\rvert} > N} {\left\lvert {f-f_n} \right\rvert} + \int_{{\left\lvert {x} \right\rvert} > N} {\left\lvert {f_n} \right\rvert} \\ &= \int_{{\left\lvert {x} \right\rvert} > N} {\left\lvert {f-f_n} \right\rvert} \\ &\leq \int_{{\left\lvert {x} \right\rvert} > N} {\left\lVert {f-f_n} \right\rVert}_1 \\ &= {\left\lVert {f_n-f} \right\rVert}_1 \qty{\int_{{\left\lvert {x} \right\rvert} > N} 1} \\ &\overset{n\to\infty}\to 0 \qty{\int_{{\left\lvert {x} \right\rvert} > N} 1} \\ &= 0\\ &\overset{N\to\infty}\to 0 .\end{align*}

To see that this doesn’t force \(f(x)\to 0\) as \({\left\lvert {x} \right\rvert} \to \infty\):

• Take \(f(x)\) to be a train of rectangles of height 1 and area \(1/2^j\) centered on even integers.
• Then \begin{align*}\int_{{\left\lvert {x} \right\rvert} > N} {\left\lvert {f} \right\rvert} = \sum_{j=N}^\infty 1/2^j \overset{N\to\infty}\to 0\end{align*} as the tail of a convergent sum.
• However \(f(x) = 1\) for infinitely many even integers \(x > N\), so \(f(x) \not\to 0\) as \({\left\lvert {x} \right\rvert}\to\infty\).

8.1.2 b

8.1.2.1 Solution 1 (“Trick”)

• Since \(f\) is decreasing on \([1, \infty)\), for any \(t\in [x-n, x]\) we have \begin{align*} x-n \leq t \leq x \implies f(x) \leq f(t) \leq f(x-n) .\end{align*}

• Integrate over \([x, 2x]\), using monotonicity of the integral: \begin{align*} \int_x^{2x} f(x) \,dt \leq \int_x^{2x} f(t) \,dt \leq \int_x^{2x} f(x-n) \,dt \\ \implies f(x) \int_x^{2x} \,dt \leq \int_x^{2x} f(t) \,dt \leq f(x-n) \int_x^{2x} \,dt \\ \implies xf(x) \leq \int_x^{2x} f(t) \, dt \leq xf(x-n) .\end{align*}

• By the Cauchy Criterion for integrals, \(\lim_{x\to \infty} \int_x^{2x} f(t)~dt = 0\).

• So the LHS term \(xf(x) \overset{x\to\infty}\to 0\).

• Since \(x>1\), \({\left\lvert {f(x)} \right\rvert} \leq {\left\lvert {xf(x)} \right\rvert}\)

• Thus \(f(x) \overset{x\to\infty}\to 0\) as well.

8.1.2.2 Solution 2 (Variation on the Trick)

• Use mean value theorem for integrals: \begin{align*} \int_x^{2x} f(t)\, dt = xf(c_x) \quad\text{for some $c_x \in [x, 2x]$ depending on $x$} .\end{align*}

• Since \(f\) is decreasing, \begin{align*} x\leq c_x \leq 2x &\implies f(2x)\leq f(c_x) \leq f(x) \\ &\implies 2xf(2x)\leq 2xf(c_x) \leq 2xf(x) \\ &\implies 2xf(2x)\leq 2x\int_x^{2x} f(t)\, dt \leq 2xf(x) \\ .\end{align*}

• By Cauchy Criterion, \(\int_x^{2x} f \to 0\).

• So \(2x f(2x) \to 0\), which by a change of variables gives \(uf(u) \to 0\).

• Since \(u\geq 1\), \(f(u) \leq uf(u)\) so \(f(u) \to 0\) as well.

8.1.2.3 Solution 3 (Contradiction)

Just showing \(f(x) \overset{x\to \infty}\to 0\):

• Toward a contradiction, suppose not.

• Since \(f\) is decreasing, it can not diverge to \(+\infty\)

• If \(f(x) \to -\infty\), then \(f\not\in L^1({\mathbb{R}})\): choose \(x_0 \gg 1\) so that \(t\geq x_0 \implies f(t) < -1\), then

• Then \(t\geq x_0 \implies {\left\lvert {f(t)} \right\rvert} \geq 1\), so \begin{align*} \int_1^\infty {\left\lvert {f} \right\rvert} \geq \int_{x_0}^\infty {\left\lvert {f(t) } \right\rvert} \, dt \geq \int_{x_0}^\infty 1 =\infty .\end{align*}

• Otherwise \(f(x) \to L\neq 0\), some finite limit.

• If \(L>0\):

  • Fix \(\varepsilon>0\), choose \(x_0\gg 1\) such that \(t\geq x_0 \implies L-\varepsilon\leq f(t) \leq L\)
  • Then \begin{align*}\int_1^\infty f \geq \int_{x_0}^\infty f \geq \int_{x_0}^\infty \qty{L-\varepsilon}\,dt = \infty\end{align*}

• If \(L<0\):

  • Fix \(\varepsilon> 0\), choose \(x_0\gg 1\) such that \(t\geq x_0 \implies L \leq f(t) \leq L + \varepsilon\).
  • Then \begin{align*}\int_1^\infty f \geq \int_{x_0}^\infty f \geq \int_{x_0}^\infty \qty{L}\,dt = \infty\end{align*}

Showing \(xf(x) \overset{x\to \infty}\to 0\).

• Toward a contradiction, suppose not.
• (How to show that \(xf(x) \not\to + \infty\)?)
• If \(xf(x)\to -\infty\)
  • Choose a sequence \(\Gamma = \left\{{\widehat{x}_i}\right\}\) such that \(x_i \to \infty\) and \(x_i f(x_i) \to -\infty\).
  • Choose a subsequence \(\Gamma' = \left\{{x_i}\right\}\) such that \(x_if(x_i) \leq -1\) for all \(i\) and \(x_i \leq x_{i+1}\).
  • Choose a further subsequence \(S = \left\{{x_i \in \Gamma' {~\mathrel{\Big|}~}2x_i < x_{i+1}}\right\}\).
  • Then since \(f\) is always decreasing, for \(t\geq x_0\), \({\left\lvert {f} \right\rvert}\) is increasing, and \({\left\lvert {f(x_i)} \right\rvert} \leq {\left\lvert {f(2x_i)} \right\rvert}\), so \begin{align*} \int_1^{\infty} {\left\lvert {f} \right\rvert} \geq \int_{x_0}^\infty {\left\lvert {f} \right\rvert} \geq \sum_{x_i \in S} \int_{x_i}^{2x_i} {\left\lvert {f(t)} \right\rvert} \, dt \geq \sum_{x_i \in S} \int_{x_i}^{2x_i} {\left\lvert {f(x_i)} \right\rvert} &= \sum_{x_i \in S} x_i f(x_i) \to \infty .\end{align*}
• If \(xf(x) \to L \neq 0\) for \(0 < L< \infty\):
  • Fix \(\varepsilon> 0\), choose an infinite sequence \(\left\{{x_i}\right\}\) such that \(L-\varepsilon\leq x_i f(x_i) \leq L\) for all \(i\). \begin{align*} \int_1^\infty {\left\lvert {f} \right\rvert} \geq \sum_S \int_{x_i}^{2x_i} {\left\lvert {f(t)} \right\rvert}\,dt \geq \sum_S \int_{x_i}^{2x_i} f(x_i) \,dt = \sum_S x_i f(x_i) \geq \sum_S \qty{L-\varepsilon} \to \infty .\end{align*}
• If \(xf(x) \to L \neq 0\) for \(-\infty < L < 0\):
  • Fix \(\varepsilon> 0\), choose an infinite sequence \(\left\{{x_i}\right\}\) such that \(L \leq x_i f(x_i) \leq L + \varepsilon\) for all \(i\). \begin{align*} \int_1^\infty {\left\lvert {f} \right\rvert} \geq \sum_S \int_{x_i}^{2x_i} {\left\lvert {f(t)} \right\rvert}\,dt \geq \sum_S \int_{x_i}^{2x_i} f(x_i) \,dt = \sum_S x_i f(x_i) \geq \sum_S \qty{L} \to \infty .\end{align*}

8.1.2.4 Solution 4 (Akos’s Suggestion)

For \(x\geq 1\), \begin{align*} {\left\lvert {xf(x)} \right\rvert} = {\left\lvert { \int_x^{2x} f(x) \, dt } \right\rvert} \leq \int_x^{2x} {\left\lvert {f(x)} \right\rvert} \, dt \leq \int_x^{2x} {\left\lvert {f(t)} \right\rvert}\, dt \leq \int_x^{\infty} {\left\lvert {f(t)} \right\rvert} \,dt \overset{x\to\infty}\to 0 \end{align*} where we’ve used

• Since \(f\) is decreasing and \(\lim_{x\to\infty}f(x) =0\) from part (a), \(f\) is non-negative.
• Since \(f\) is positive and decreasing, for every \(t\in[a, b]\) we have \({\left\lvert {f(a)} \right\rvert} \leq {\left\lvert {f(t)} \right\rvert}\).
• By part (a), the last integral goes to zero.

8.1.2.5 Solution 5 (Peter’s)

• Toward a contradiction, produce a sequence \(x_i\to\infty\) with \(x_i f(x_i) \to \infty\) and \(x_if(x_i) > \varepsilon> 0\), then \begin{align*} \int f(x) \, dx &\geq \sum_{i=1}^\infty \int_{x_i}^{x_{i+1}} f(x) \, dx \\ &\geq \sum_{i=1}^\infty \int_{x_i}^{x_{i+1}} f(x_{i+1}) \, dx \\ &= \sum_{i=1}^\infty f(x_{i+1}) \int_{x_i}^{x_{i+1}} \, dx \\ &\geq \sum_{i=1}^\infty (x_{i+1} - x_i) f(x_{i+1}) \\ &\geq \sum_{i=1}^\infty (x_{i+1} - x_i) {\varepsilon\over x_{i+1}} \\ &= \varepsilon\sum_{i=1}^\infty \qty{ 1 - {x_{i-1} \over x_i}} \to \infty \end{align*} which can be ensured by passing to a subsequence where \(\sum {x_{i-1} \over x_i} < \infty\).

8.1.3 c

• No: take \(f(x) = {1\over x\ln x}\)
• Then by a \(u{\hbox{-}}\)substitution, \begin{align*} \int_0^x f = \ln\qty{\ln (x)} \overset{x\to\infty}\to \infty \end{align*} is unbounded, so \(f\not\in L^1([1, \infty))\).
• But \begin{align*} xf(x) = { 1 \over \ln(x)} \overset{x\to\infty} \to 0 .\end{align*}

8.2.3 a

Choose \(g\in C_c^0\) such that \({\left\lVert {f- g} \right\rVert}_1 \to 0\).

By translation invariance, \({\left\lVert {\tau_h f - \tau_h g} \right\rVert}_1 \to 0\).

Write \begin{align*} {\left\lVert {\tau f - f} \right\rVert}_1 &= {\left\lVert {\tau_h f - g + g - \tau_h g + \tau_h g - f} \right\rVert}_1 \\ &\leq {\left\lVert {\tau_h f - \tau_h g} \right\rVert} + {\left\lVert {g - f} \right\rVert} + {\left\lVert {\tau_h g - g} \right\rVert} \\ &\to {\left\lVert {\tau_h g - g} \right\rVert} ,\end{align*}

so it suffices to show that \({\left\lVert {\tau_h g - g} \right\rVert} \to 0\) for \(g\in C_c^0\).

Fix \(\varepsilon > 0\). Enlarge the support of \(g\) to \(K\) such that \begin{align*} {\left\lvert {h} \right\rvert} \leq 1 \text{ and } x \in K^c \implies {\left\lvert {g(x-h) - g(x)} \right\rvert} = 0 .\end{align*}

By uniform continuity of \(g\), pick \(\delta \leq 1\) small enough such that \begin{align*} x\in K, ~{\left\lvert {h} \right\rvert} \leq \delta \implies {\left\lvert {g(x-h) -g(x)} \right\rvert} < \varepsilon ,\end{align*}

then \begin{align*} \int_K {\left\lvert {g(x-h) - g(x)} \right\rvert} \leq \int_K \varepsilon = \varepsilon \cdot m(K) \to 0. \end{align*}

8.2.4 b

We have \begin{align*} \int_{\mathbb{R}}{\left\lvert {A_h(f)(x)} \right\rvert} ~dx &= \int_{\mathbb{R}}{\left\lvert {\frac{1}{2h} \int_{x-h}^{x+h} f(y)~dy} \right\rvert} ~dx \\ &\leq \frac{1}{2h} \int_{\mathbb{R}}\int_{x-h}^{x+h} {\left\lvert {f(y)} \right\rvert} ~dy ~dx \\ &=_{FT} \frac{1}{2h} \int_{\mathbb{R}}\int_{y-h}^{y+h} {\left\lvert {f(y)} \right\rvert} ~\mathbf{dx} ~\mathbf{dy} \\ &= \int_{\mathbb{R}}{\left\lvert {f(y)} \right\rvert} ~{dy} \\ &= {\left\lVert {f} \right\rVert}_1 ,\end{align*}

and (rough sketch)

\begin{align*} \int_{\mathbb{R}}{\left\lvert {A_h(f)(x) - f(x)} \right\rvert} ~dx &= \int_{\mathbb{R}}{\left\lvert { \left(\frac{1}{2h} \int_{B(h, x)} f(y)~dy\right) - f(x)} \right\rvert}~dx \\ &= \int_{\mathbb{R}}{\left\lvert { \left(\frac{1}{2h} \int_{B(h, x)} f(y)~dy\right) - \frac{1}{2h}\int_{B(h, x)} f(x) ~dy} \right\rvert}~dx \\ &\leq_{FT} \frac{1}{2h} \int_{\mathbb{R}}\int_{B(h, x)}{\left\lvert { f(y-x) - f(x)} \right\rvert} ~\mathbf{dx} ~\mathbf{dy} \\ &\leq \frac 1 {2h} \int_{\mathbb{R}}{\left\lVert {\tau_x f - f} \right\rVert}_1 ~dy \\ &\to 0 \quad\text{by (a)} .\end{align*}

It suffices to show that \(S\) is dense in simple functions, and since simple functions are finite linear combinations of characteristic functions, it suffices to show this for \(\chi_A\) for \(A\) a measurable set.

Let \(s = \chi_{A}\). By regularity of the Lebesgue measure, choose an open set \(O \supseteq A\) such that \(m(O\setminus A) < \varepsilon\).

\(O\) is an open subset of \({\mathbb{R}}\), and thus \(O = {\coprod}_{j\in {\mathbb{N}}} I_j\) is a disjoint union of countably many open intervals.

Now choose \(N\) large enough such that \(m(O \Delta I_{N, n}) < \varepsilon = \frac 1 n\) where we define \(I_{N, n} \mathrel{\vcenter{:}}={\coprod}_{j=1}^N I_j\).

Now define \(f_n = \chi_{I_{N, n}}\), then \begin{align*} {\left\lVert {s - f_n} \right\rVert}_1 = \int {\left\lvert {\chi_A - \chi_{I_{N, n}}} \right\rvert} = m(A \Delta I_{N, n}) \overset{n\to\infty}\longrightarrow 0 .\end{align*}

Since any simple function is a finite linear combination of \(\chi_{A_i}\), we can do this for each \(i\) to extend this result to all simple functions. But simple functions are dense in \(L^1\), so \(S\) is dense in \(L^1\).

\begin{align*} {\left\lVert {H(x)} \right\rVert}_1 &= \int _{\mathbb{R}}{\left\lvert {H(x, y)} \right\rvert} \, dx \\ &= \int _{\mathbb{R}}{\left\lvert { \int_{\mathbb{R}}f(y)g(x-y) \,dy } \right\rvert} \, dx \\ &\leq \int _{\mathbb{R}}\qty{ \int_{\mathbb{R}}{\left\lvert {f(y)g(x-y)} \right\rvert} \, dy } \, dx \\ &= \int _{\mathbb{R}}\qty{ \int_{\mathbb{R}}{\left\lvert {f(y)g(x-y)} \right\rvert} \, dx} \, dy \quad\text{by Tonelli} \\ &= \int _{\mathbb{R}}\qty{ \int_{\mathbb{R}}{\left\lvert {f(y)g(t)} \right\rvert} \, dt} \, dy \quad\text{setting } t=x-y, \,dt = - dx \\ &= \int _{\mathbb{R}}\qty{ \int_{\mathbb{R}}{\left\lvert {f(y)} \right\rvert}\cdot {\left\lvert {g(t)} \right\rvert} \, dt}\, dy \\ &= \int _{\mathbb{R}}{\left\lvert {f(y)} \right\rvert} \cdot \qty{ \int_{\mathbb{R}}{\left\lvert {g(t)} \right\rvert} \, dt}\, dy \\ &\mathrel{\vcenter{:}}=\int _{\mathbb{R}}{\left\lvert {f(y)} \right\rvert} \cdot {\left\lVert {g} \right\rVert}_1 \,dy \\ &= {\left\lVert {g} \right\rVert}_1 \int _{\mathbb{R}}{\left\lvert {f(y)} \right\rvert} \,dy \\ &\mathrel{\vcenter{:}}={\left\lVert {g} \right\rVert}_1 {\left\lVert {f} \right\rVert}_1 \\ &< \infty {\quad \operatorname{by assumption} \quad} .\end{align*}

• \(H\) is measurable on \({\mathbb{R}}^2\):
  • If we can show \(\tilde f(x, y) \mathrel{\vcenter{:}}= f(y)\) and \(\tilde g(x, y) \mathrel{\vcenter{:}}= g(x-y)\) are both measurable on \({\mathbb{R}}^2\), then \(H = \tilde f \cdot \tilde g\) is a product of measurable functions and thus measurable.
  • \(f \in L^1\), and \(L^1\) functions are measurable by definition.
  • The function \((x, y) \mapsto g(x-y)\) is measurable on \({\mathbb{R}}^2\):
    • Let \(g\) be measurable on \({\mathbb{R}}\), then the cylinder function \(G(x, y) = g(x)\) on \({\mathbb{R}}^2\) is always measurable
    • Define a linear transformation \(T \mathrel{\vcenter{:}}={\left[ {1, -1; 0, 1} \right]}\) which sends \((x,y) \to (x-y, y)\), then \(T\in \operatorname{GL}(2, {\mathbb{R}})\) is linear and thus measurable.
    • Then \((G\circ T)(x, y) = G(x-y, y) = \tilde g(x-y)\), so \(\tilde g\) is a composition of measurable functions and thus measurable.
• Apply Tonelli to \({\left\lvert {H} \right\rvert}\)
  • \(H\) measurable implies \({\left\lvert {H} \right\rvert}\) is measurable
  • \({\left\lvert {H} \right\rvert}\) is non-negative
  • So the iterated integrals are equal in the extended sense
  • The calculation shows the iterated integral is finite, to \(\int {\left\lvert {H} \right\rvert}\) is finite and \(H\) is thus integrable on \({\mathbb{R}}^2\).

Note: Fubini is not needed, since we’re not calculating the actual integral, just showing \(H\) is integrable.

9.2.1 a

\(\implies\):

• Suppose \(f\) is a measurable function.
• Note that \(\mathcal{A} = \left\{{f(x) - t \geq 0}\right\} \cap\left\{{t \geq 0}\right\}\).
• Define \(F(x, t) = f(x)\), \(G(x, t) = t\), which are cylinders on measurable functions and thus measurable.
• Define \(H(x, y) = F(x, t) - G(x, t)\), which are linear combinations of measurable functions and thus measurable.
• Then \(\mathcal{A} = \left\{{H \geq 0}\right\} \cap\left\{{G \geq 0}\right\}\) as a countable intersection of measurable sets, which is again measurable.

\(\impliedby\):

• Suppose \({\mathcal{A}}\) is a measurable set.

• Then FT on \(\chi_{{\mathcal{A}}}\) implies that for almost every \(x\in {\mathbb{R}}^n\), the \(x{\hbox{-}}\)slices \({\mathcal{A}}_x\) are measurable and \begin{align*} \mathcal{A}_x \mathrel{\vcenter{:}}=\left\{{t\in {\mathbb{R}}{~\mathrel{\Big|}~}(x, t) \in \mathcal{A}}\right\} = [0, f(x)] \implies m(\mathcal A_x) = f(x) - 0 = f(x) \end{align*}

• But \(x \mapsto m(\mathcal A_x)\) is a measurable function, and is exactly the function \(x \mapsto f(x)\), so \(f\) is measurable.

9.2.2 b

• Note \begin{align*} \mathcal{A} &= \left\{{(x, t) \in {\mathbb{R}}^n\times{\mathbb{R}}{~\mathrel{\Big|}~}0 \leq t \leq f(x)}\right\} \\ \mathcal{A}_t &= \left\{{x \in {\mathbb{R}}^n {~\mathrel{\Big|}~}t\leq f(x) }\right\} .\end{align*}

• Then \begin{align*} \int_{{\mathbb{R}}^n} f(x) ~dx &= \int_{{\mathbb{R}}^n} \int_0^{f(x)} 1 ~dt~dx \\ &= \int_{{\mathbb{R}}^n} \int_{0}^\infty \chi_\mathcal{A} ~dt~dx \\ &\overset{F.T.}= \int_{0}^\infty \int_{{\mathbb{R}}^n} \chi_\mathcal{A} ~dx~dt\\ &= \int_0^\infty m(\mathcal{A}_t) ~dt ,\end{align*} where we just use that \(\int \int \chi_\mathcal{A} = m(\mathcal{A})\)

• By Tonelli, all of these integrals are equal.

  • This is justified because \(f\) was assumed measurable on \({\mathbb{R}}^n\), thus by (a) \(\mathcal{A}\) is a measurable set and thus \(\chi_A\) is a measurable function on \({\mathbb{R}}^n\times{\mathbb{R}}\).

Note: \(f\) is a function \({\mathbb{R}}\to {\mathbb{R}}\) in the original problem, but here I’ve assumed \(f:{\mathbb{R}}^n\to {\mathbb{R}}\).

• Since \(f\geq 0\), set \begin{align*} E\mathrel{\vcenter{:}}=\left\{{(x, t) \in {\mathbb{R}}^{n} \times{\mathbb{R}}{~\mathrel{\Big|}~}f(x) > t}\right\} = \left\{{(x, t) \in {\mathbb{R}}^n \times{\mathbb{R}}{~\mathrel{\Big|}~}0 \leq t < f(x)}\right\} .\end{align*}

• Claim: since \(f\) is measurable, \(E\) is measurable and thus \(m(E)\) makes sense.

  • Since \(f\) is measurable, \(F(x, t) \mathrel{\vcenter{:}}= t - f(x)\) is measurable on \({\mathbb{R}}^n \times{\mathbb{R}}\).
  • Then write \(E = \left\{{F < 0}\right\} \cap\left\{{t\geq 0}\right\}\) as an intersection of measurable sets.

• We have slices \begin{align*} E^t &\mathrel{\vcenter{:}}=\left\{{x\in {\mathbb{R}}^n {~\mathrel{\Big|}~}(x, t) \in E}\right\} = \left\{{x\in {\mathbb{R}}^n {~\mathrel{\Big|}~}0 \leq t < f(x)}\right\} \\ E^x &\mathrel{\vcenter{:}}=\left\{{t\in {\mathbb{R}}{~\mathrel{\Big|}~}(x, t) \in E}\right\} = \left\{{t\in {\mathbb{R}}{~\mathrel{\Big|}~}0 \leq t \leq f(x)}\right\} = [0, f(x)] .\end{align*}

  • \(E_t\) is precisely the set that appears in the original RHS integrand.
  • \(m(E^x) = f(x)\).

• Claim: \(\chi_E\) satisfies the conditions of Tonelli, and thus \(m(E) = \int \chi_E\) is equal to any iterated integral.

  • Non-negative: clear since \(0\leq \chi_E \leq 1\)
  • Measurable: characteristic functions of measurable sets are measurable.

• Conclude:

  1. For almost every \(x\), \(E^x\) is a measurable set, \(x\mapsto m(E^x)\) is a measurable function, and \(m(E) = \int_{{\mathbb{R}}^n} m(E^x) \, dx\)
  2. For almost every \(t\), \(E^t\) is a measurable set, \(t\mapsto m(E^t)\) is a measurable function, and \(m(E) = \int_{{\mathbb{R}}} m(E^t) \, dt\)

• On one hand, \begin{align*} m(E) &= \int_{{\mathbb{R}}^{n+1}} \chi_E(x, t) \\ &= \int_{{\mathbb{R}}} \int_{{\mathbb{R}}^n} \chi_E(x, t) \,dt \,dx \quad\text{by Tonelli}\\ &= \int_{{\mathbb{R}}^n} m(E^x) \,dx \quad\text{first conclusion}\\ &= \int_{{\mathbb{R}}^n} f(x) \,dx .\end{align*}

• On the other hand, \begin{align*} m(E) &= \int_{{\mathbb{R}}^{n+1}} \chi_E(x, t) \\ &= \int_{\mathbb{R}}\int_{{\mathbb{R}}^n} \chi_E(x, t) \, dx \,dt \quad\text{by Tonelli} \\ &= \int_{\mathbb{R}}m(E^t) \,dt \quad\text{second conclusion} .\end{align*}

• Thus \begin{align*} \int_{{\mathbb{R}}^n} f \,dx = m(E) = \int_{\mathbb{R}}m(E^t) \,dt = \int_{\mathbb{R}}m\qty{\left\{{x{~\mathrel{\Big|}~}f(x) > t}\right\}} .\end{align*}

10.1.3 a

Claim: \(\ell^1({\mathbb{Z}}) \subseteq \ell^2({\mathbb{Z}})\).

• Set \(\mathbf{c} = \left\{{c_k {~\mathrel{\Big|}~}k\in {\mathbb{Z}}}\right\} \in \ell^1({\mathbb{Z}})\).
• It suffices to show that if \(\sum_{k\in {\mathbb{Z}}} {\left\lvert {c_k} \right\rvert} < \infty\) then \(\sum_{k\in {\mathbb{Z}}} {\left\lvert {c_k} \right\rvert}^2 < \infty\).
• Let \(S = \left\{{c_k {~\mathrel{\Big|}~}{\left\lvert {c_k} \right\rvert} \leq 1}\right\}\), then \(c_k \in S \implies {\left\lvert {c_k} \right\rvert}^2 \leq {\left\lvert {c_k} \right\rvert}\)
• Claim: \(S^c\) can only contain finitely many elements, all of which are finite.
  • If not, either \(S^c \mathrel{\vcenter{:}}=\left\{{c_j}\right\}_{j=1}^\infty\) is infinite with every \({\left\lvert {c_j} \right\rvert} > 1\), which forces \begin{align*}\sum_{c_k\in S^c} {\left\lvert {c_k} \right\rvert} = \sum_{j=1}^\infty {\left\lvert {c_j} \right\rvert} > \sum_{j=1}^\infty 1 = \infty.\end{align*}
  • If any \(c_j = \infty\), then \(\sum_{k\in {\mathbb{Z}}} {\left\lvert {c_k} \right\rvert} \geq c_j = \infty\).
• So \(S^c\) is a finite set of finite integers, let \(N = \max \left\{{{\left\lvert {c_j} \right\rvert}^2 {~\mathrel{\Big|}~}c_j \in S^c}\right\} < \infty\).
• Rewrite the sum \begin{align*} \sum_{k\in {\mathbb{Z}}} {\left\lvert {c_k} \right\rvert}^2 &= \sum_{c_k\in S} {\left\lvert {c_k} \right\rvert}^2 + \sum_{c_k \in S^c} {\left\lvert {c_k} \right\rvert}^2 \\ &\leq \sum_{c_k\in S} {\left\lvert {c_k} \right\rvert} + \sum_{c_k \in S^c} {\left\lvert {c_k} \right\rvert}^2 \\ &\leq \sum_{k\in {\mathbb{Z}}} {\left\lvert {c_k} \right\rvert} + \sum_{c_k \in S^c} {\left\lvert {c_k} \right\rvert}^2 \quad\text{since the ${\left\lvert {c_k} \right\rvert}$ are all positive} \\ &= {\left\lVert {\mathbf{c}} \right\rVert}_{\ell^1} + \sum_{c_k \in S^c} {\left\lvert {c_k} \right\rvert}^2 \\ &\leq {\left\lVert {\mathbf{c}} \right\rVert}_{\ell^1} + {\left\lvert {S^c} \right\rvert}\cdot N \\ &< \infty .\end{align*}

Claim: \(L^2([0, 1]) \subseteq L^1([0, 1])\).

• It suffices to show that \(\int {\left\lvert {f} \right\rvert}^2 < \infty \implies \int {\left\lvert {f} \right\rvert} < \infty\).

• Define \(S = \left\{{x\in [0, 1] {~\mathrel{\Big|}~}{\left\lvert {f(x)} \right\rvert} \leq 1}\right\}\), then \(x\in S^c \implies {\left\lvert {f(x)} \right\rvert}^2 \geq {\left\lvert {f(x)} \right\rvert}\).

• Break up the integral: \begin{align*} \int_{\mathbb{R}}{\left\lvert {f} \right\rvert} &= \int_S {\left\lvert {f} \right\rvert} + \int_{S^c} {\left\lvert {f} \right\rvert} \\ &\leq \int_S {\left\lvert {f} \right\rvert} + \int_{S^c} {\left\lvert {f} \right\rvert}^2 \\ &\leq \int_S {\left\lvert {f} \right\rvert} + {\left\lVert {f} \right\rVert}_2 \\ &\leq \sup_{x\in S}\left\{{{\left\lvert {f(x)} \right\rvert}}\right\} \cdot \mu(S) + {\left\lVert {f} \right\rVert}_2 \\ &= 1 \cdot \mu(S) + {\left\lVert {f} \right\rVert}_2 \quad\text{by definition of } S\\ &\leq 1 \cdot \mu([0, 1]) + {\left\lVert {f} \right\rVert}_2 \quad\text{since } S\subseteq [0, 1] \\ &= 1 + {\left\lVert {f} \right\rVert}_2 \\ &< \infty .\end{align*}

Note: this proof shows \(L^2(X) \subseteq L^1(X)\) whenever \(\mu(X) < \infty\).

10.2.1 a

Lemma: If \(\phi \in C_c^1\), then \((f \ast \phi)' = f \ast \phi'\) almost everywhere.

Silly Proof:

\begin{align*} \mathcal{F}( (f \ast \phi)' ) &= 2\pi i \xi ~\mathcal{F}(f\ast \phi) \\ &= 2\pi i \xi ~ \mathcal{F}(f) ~ \mathcal{F}(\phi) \\ &= \mathcal{F}(f) \cdot \left( 2\pi i \xi ~\mathcal{F}(\phi)\right) \\ &= \mathcal{F}(f) \cdot \mathcal{F}(\phi') \\ &= \mathcal{F}(f\ast \phi') .\end{align*}

Actual proof:

\begin{align*} (f\ast \phi)'(x) &= (\phi\ast f)'(x) \\ &= \lim_{h\to 0} \frac{(\phi\ast f)'(x+h) - (\phi\ast f)'(x)}{h} \\ &= \lim_{h\to 0} \int \frac{\phi(x + h - y) - \phi(x - y)}{h} f(y) \\ &\overset{DCT}= \int \lim_{h\to 0} \frac{\phi(x + h - y) - \phi(x - y)}{h} f(y) \\ &= \int \phi'(x-y) f(y) \\ &= (\phi' \ast f)(x) \\ &= (f \ast \phi')(x) .\end{align*}

To see that the DCT is justified, we can apply the MVT on the interval \([0, h]\) to \(f\) to obtain

\begin{align*} \frac{\phi(x + h - y) - \phi(x - y)}{h} &= \phi'(c) \quad c\in [0, h] ,\end{align*}

and since \(\phi'\) is continuous and compactly supported, \(\phi'\) is bounded by some \(M < \infty\) by the extreme value theorem and thus \begin{align*} \int {\left\lvert {\frac{\phi(x + h - y) - \phi(x - y)}{h} f(y)} \right\rvert} &= \int {\left\lvert {\phi'(c) f(y)} \right\rvert} \\ &\leq \int {\left\lvert {M} \right\rvert}{\left\lvert {f} \right\rvert} \\ &= {\left\lvert {M} \right\rvert} \int {\left\lvert {f} \right\rvert} < \infty ,\end{align*}

since \(f\in L^1\) by assumption, so we can take \(g\mathrel{\vcenter{:}}={\left\lvert {M} \right\rvert} {\left\lvert {f} \right\rvert}\) as the dominating function.

Applying this theorem infinitely many times shows that \(f\ast \phi\) is smooth.

To see that \(f\ast \phi\) is compactly supported, approximate \(f\) by a continuous compactly supported function \(h\), so \({\left\lVert {h - f} \right\rVert}_1 \overset{L^1}\to 0\).

Now let \(g_x(y) = \phi(x-y)\), and note that \(\mathrm{supp}(g) = x - \mathrm{supp}(\phi)\) which is still compact.

But since \(\mathrm{supp}(h)\) is bounded, there is some \(N\) such that \begin{align*} {\left\lvert {x} \right\rvert} > N \implies A_x\mathrel{\vcenter{:}}=\mathrm{supp}(h) \cap\mathrm{supp}(g_x) = \emptyset \end{align*}

and thus \begin{align*} (h\ast \phi)(x) &= \int_{\mathbb{R}}\phi(x-y) h(y)~dy \\ &= \int_{A_x} g_x(y) h(y) \\ &= 0 ,\end{align*}

so \(\left\{{x {~\mathrel{\Big|}~}f\ast g(x) = 0}\right\}\) is open, and its complement is closed and bounded and thus compact.

10.2.2 b

\begin{align*} {\left\lVert {f\ast K_j - f} \right\rVert}_1 &= \int {\left\lvert {\int f(x-y) K_j(y) ~dy - f(x)} \right\rvert}~dx \\ &= \int {\left\lvert {\int f(x-y) K_j(y) ~dy - \int f(x) K_j(y) ~ dy} \right\rvert}~dx \\ &= \int {\left\lvert {\int ( f(x-y) - f(x) ) K_j(y) ~dy } \right\rvert} ~dx \\ &\leq \int \int {\left\lvert {(f(x-y) - f(x))} \right\rvert} \cdot {\left\lvert {K_j(y)} \right\rvert} ~ dy~dx \\ &\overset{FT}= \int \int {\left\lvert {(f(x-y) - f(x))} \right\rvert} \cdot {\left\lvert {K_j(y)} \right\rvert} \mathbf{~ dx~dy}\\ &= \int {\left\lvert {K_j(y)} \right\rvert} \left( \int {\left\lvert {(f(x-y) - f(x))} \right\rvert} ~ dx\right) ~dy \\ &= \int {\left\lvert {K_j(y)} \right\rvert} \cdot {\left\lVert {f - \tau_y f} \right\rVert}_1 ~dy .\end{align*}

We now split the integral up into pieces.

1. Chose \(\delta\) small enough such that \({\left\lvert {y} \right\rvert} < \delta \implies {\left\lVert {f - \tau_y f} \right\rVert}_1 < \varepsilon\) by continuity of translation in \(L^1\), and

2. Since \(\phi\) is compactly supported, choose \(J\) large enough such that \begin{align*} j > J \implies \int_{{\left\lvert {y} \right\rvert} \geq \delta} {\left\lvert {K_j(y)} \right\rvert} ~dy = \int_{{\left\lvert {y} \right\rvert} \geq \delta} {\left\lvert {j\phi(jy)} \right\rvert} = 0 \end{align*}

Then \begin{align*} {\left\lVert {f\ast K_j - f} \right\rVert}_1 &\leq \int {\left\lvert {K_j(y)} \right\rvert} \cdot {\left\lVert {f - \tau_y f} \right\rVert}_1 ~dy \\ &= \int_{{\left\lvert {y} \right\rvert} < \delta} {\left\lvert {K_j(y)} \right\rvert} \cdot {\left\lVert {f - \tau_y f} \right\rVert}_1 ~dy + \int_{{\left\lvert {y} \right\rvert} \geq \delta} {\left\lvert {K_j(y)} \right\rvert} \cdot {\left\lVert {f - \tau_y f} \right\rVert}_1 ~dy \\ &= \varepsilon \int_{{\left\lvert {y} \right\rvert} \geq \delta} {\left\lvert {K_j(y)} \right\rvert} + 0 \\ &\leq \varepsilon(1) \to 0 .\end{align*}

11.1.3 a

Claim: \begin{align*} 0 \leq \left\|x-\sum_{n=1}^{N}\left\langle x, u_{n}\right\rangle u_{n}\right\|^{2} &= \|x\|^{2}-\sum_{n=1}^{N}\left|\left\langle x, u_{n}\right\rangle\right|^{2} \\ &\implies \sum_{n=1}^{\infty}\left|\left\langle x, u_{n}\right\rangle\right|^{2} \leq\|x\|^{2} .\end{align*}

Proof: Let \(S_N = \sum_{n=1}^N {\left\langle {x},~{u_n} \right\rangle} u_n\). Then \begin{align*} 0 &\leq {\left\lVert {x - S_N} \right\rVert}^2 \\ &= {\left\langle {x - S_n},~{x - S_N} \right\rangle} \\ &= {\left\lVert {x} \right\rVert}^2 - \sum_{n=1}^N {\left\lvert {{\left\langle {x},~{u_n} \right\rangle}} \right\rvert}^2 \\ &\xrightarrow{N\to\infty} {\left\lVert {x} \right\rVert}^2 - \sum_{n=1}^N {\left\lvert {{\left\langle {x},~{u_n} \right\rangle}} \right\rvert}^2 .\end{align*}

11.1.4 b

1. Fix \(\left\{{a_n}\right\} \in \ell^2\), then note that \(\sum {\left\lvert {a_n} \right\rvert}^2 < \infty \implies\) the tails vanish.

2. Define \begin{align*} x \mathrel{\vcenter{:}}=\displaystyle\lim_{N\to\infty} S_N = \lim_{N\to \infty} \sum_{k=1}^N a_k u_k \end{align*}

3. \(\left\{{S_N}\right\}\) Cauchy (by 1) and \(H\) complete \(\implies x\in H\).

4. \begin{align*} {\left\langle {x},~{u_n} \right\rangle} = {\left\langle {\sum_k a_k u_k},~{u_n} \right\rangle} = \sum_k a_k {\left\langle {u_k},~{u_n} \right\rangle} = a_n \quad \forall n\in {\mathbb{N}} \end{align*} since the \(u_k\) are all orthogonal.

5. \begin{align*} {\left\lVert {x} \right\rVert}^2 = {\left\lVert {\sum_k a_k u_k} \right\rVert}^2 = \sum_k {\left\lVert {a_k u_k} \right\rVert}^2 = \sum_k {\left\lvert {a_k} \right\rvert}^2 \end{align*} by Pythagoras since the \(u_k\) are orthogonal, where we’ve used normality in the last equality.

Bonus: We didn’t use completeness here, so the Fourier series may not actually converge to \(x\). If \(\left\{{u_n}\right\}\) is complete (so \(x = 0 \iff {\left\langle {x},~{u_n} \right\rangle} = 0 ~\forall n\)) then the Fourier series does converge to \(x\) and \(\sum_{n=1}^{\infty}\left|\left\langle x, u_{n}\right\rangle\right|^{2}=\|x\|^{2}\) for all \(x \in H\).

11.2.3 a

• Note \(X = [0, 1] \implies m(X) = 1\).

• By Holder’s inequality with \(p=q=2\), \begin{align*} {\left\lVert {f} \right\rVert}_1 = {\left\lVert {f\cdot 1} \right\rVert}_1 \leq {\left\lVert {f} \right\rVert}_2 \cdot {\left\lVert {1} \right\rVert}_2 = {\left\lVert {f} \right\rVert}_2 \cdot m(X)^{\frac 1 2} = {\left\lVert {f} \right\rVert}_2, \end{align*}

• Thus \(L^2(X) \subseteq L^1(X)\)

• Since they share a common dense subset (simple functions), \(L^2\) is dense in \(L^1\)

11.2.4 b

Let \(\Lambda \in L^1(X)^\vee\) be arbitrary.

11.2.4.1 1: Existence of \(g\) Representing \(\Lambda\).

Let \(f\in L^2\subseteq L^1\) be arbitrary.

Claim: \(\Lambda\in L^1(X)^\vee\implies \Lambda \in L^2(X)^\vee\).

• Suffices to show that \({\left\lVert {\Gamma} \right\rVert}_{L^2(X)^\vee} \mathrel{\vcenter{:}}=\sup_{{\left\lVert {f} \right\rVert}_2 = 1} {\left\lvert {\Gamma(f)} \right\rvert} < \infty\), since bounded implies continuous.

• By the lemma, \({\left\lVert {f} \right\rVert}_1 \leq C{\left\lVert {f} \right\rVert}_2\) for some constant \(C \approx m(X)\).

• Note \begin{align*}{\left\lVert {\Lambda} \right\rVert}_{L^1(X)^\vee} \mathrel{\vcenter{:}}=\displaystyle\sup_{{\left\lVert {f} \right\rVert}_1 = 1} {\left\lvert {\Lambda(f)} \right\rvert}\end{align*}

• Define \(\widehat{f} = {f\over {\left\lVert {f} \right\rVert}_1}\) so \({\left\lVert {\widehat{f}} \right\rVert}_1 = 1\)

• Since \({\left\lVert {\Lambda} \right\rVert}_{1^\vee}\) is a supremum over all \(f \in L^1(X)\) with \({\left\lVert {f} \right\rVert}_1 =1\), \begin{align*} {\left\lvert {\Lambda(\widehat{f})} \right\rvert} \leq {\left\lVert {\Lambda} \right\rVert}_{(L^1(X))^\vee} ,\end{align*}

• Then \begin{align*} \frac{{\left\lvert {\Lambda(f)} \right\rvert}}{{\left\lVert {f} \right\rVert}_1} &= {\left\lvert {\Lambda(\widehat{f})} \right\rvert} \leq {\left\lVert {\Lambda} \right\rVert}_{L^1(X)^\vee} \\ \implies {\left\lvert {\Lambda(f)} \right\rvert} &\leq {\left\lVert {\Lambda} \right\rVert}_{1^\vee} \cdot {\left\lVert {f} \right\rVert}_1 \\ &\leq {\left\lVert {\Lambda} \right\rVert}_{1^\vee} \cdot C {\left\lVert {f} \right\rVert}_2 < \infty \quad\text{by assumption} ,\end{align*}

• So \(\Lambda \in (L^2)^\vee\).

Now apply Riesz Representation for \(L^2\): there is a \(g \in L^2\) such that \begin{align*}f\in L^2 \implies \Lambda(f) = {\left\langle {f},~{g} \right\rangle} \mathrel{\vcenter{:}}=\int_0^1 f(x) \mkern 1.5mu\overline{\mkern-1.5mug(x)\mkern-1.5mu}\mkern 1.5mu\, dx.\end{align*}

11.2.4.2 2: \(g\) is in \(L^\infty\)

• It suffices to show \({\left\lVert {g} \right\rVert}_{L^\infty(X)} < \infty\).

• Since we’re assuming \({\left\lVert {\Gamma} \right\rVert}_{L^1(X)^\vee} < \infty\), it suffices to show the stated equality.

• Claim: \({\left\lVert {\Lambda} \right\rVert}_{L^1(X)^\vee} ={\left\lVert {g} \right\rVert}_{L^\infty(X)}\)

  • The result will follow since \(\Lambda\) was assumed to be in \(L^1(X)^\vee\), so \({\left\lVert {\Lambda} \right\rVert}_{L^1(X)^\vee} < \infty\).

  • \(\leq\): \begin{align*} {\left\lVert {\Lambda} \right\rVert}_{L^1(X)^\vee} &= \sup_{{\left\lVert {f} \right\rVert}_1 = 1} {\left\lvert {\Lambda(f)} \right\rvert} \\ &= \sup_{{\left\lVert {f} \right\rVert}_1 = 1} {\left\lvert {\int_X f \mkern 1.5mu\overline{\mkern-1.5mug\mkern-1.5mu}\mkern 1.5mu} \right\rvert} \quad\text{by (i)}\\ &= \sup_{{\left\lVert {f} \right\rVert}_1 = 1} \int_X {\left\lvert {f \mkern 1.5mu\overline{\mkern-1.5mug\mkern-1.5mu}\mkern 1.5mu} \right\rvert} \\ &\mathrel{\vcenter{:}}=\sup_{{\left\lVert {f} \right\rVert}_1 = 1} {\left\lVert {fg} \right\rVert}_1 \\ &\leq \sup_{{\left\lVert {f} \right\rVert}_1 = 1} {\left\lVert {f} \right\rVert}_1 {\left\lVert {g} \right\rVert}_\infty \quad\text{by Holder with } p=1,q=\infty\\ &= {\left\lVert {g} \right\rVert}_\infty ,\end{align*}

  • \(\geq\):

    • Suppose toward a contradiction that \({\left\lVert {g} \right\rVert}_\infty > {\left\lVert {\Lambda} \right\rVert}_{1^\vee}\).

    • Then there exists some \(E\subseteq X\) with \(m(E) > 0\) such that \begin{align*}x\in E \implies {\left\lvert {g(x)} \right\rvert} > {\left\lVert {\Lambda} \right\rVert}_{L^1(X)^\vee}.\end{align*}

    • Define \begin{align*} h = \frac{1}{m(E)} \frac{\overline{g}}{{\left\lvert {g} \right\rvert}} \chi_E .\end{align*}

    • Note \({\left\lVert {h} \right\rVert}_{L^1(X)} = 1\).

    • Then \begin{align*} \Lambda(h) &= \int_X hg \\ &\mathrel{\vcenter{:}}=\int_X \frac{1}{m(E)} \frac{g \overline g}{{\left\lvert {g} \right\rvert}} \chi_E \\ &= \frac{1}{m(E)} \int_E {\left\lvert {g} \right\rvert} \\ &\geq \frac{1}{m(E)} {\left\lVert {g} \right\rVert}_\infty m(E) \\ &= {\left\lVert {g} \right\rVert}_\infty \\ &> {\left\lVert {\Lambda} \right\rVert}_{L^1(X)^\vee} ,\end{align*} a contradiction since \({\left\lVert {\Lambda} \right\rVert}_{L^1(X)^\vee}\) is the supremum over all \(h_\alpha\) with \({\left\lVert {h_\alpha} \right\rVert}_{L^1(X)} = 1\).

12.1.1 a

• Let \(\left\{{f_n}\right\}\) be a Cauchy sequence in \(C(I, {\left\lVert {{\,\cdot\,}} \right\rVert}_\infty)\), so \(\lim_n\lim_m {\left\lVert {f_m - f_n} \right\rVert}_\infty = 0\), we will show it converges to some \(f\) in this space.

• For each fixed \(x_0 \in [0, 1]\), the sequence of real numbers \(\left\{{f_n(x_0)}\right\}\) is Cauchy in \({\mathbb{R}}\) since \begin{align*} x_0\in I \implies {\left\lvert {f_m(x_0) - f_n(x_0)} \right\rvert} \leq \sup_{x\in I} {\left\lvert {f_m(x) - f_n(x)} \right\rvert} \mathrel{\vcenter{:}}={\left\lVert {f_m - f_n} \right\rVert}_\infty \overset{m>n\to\infty}\to 0, \end{align*}

• Since \({\mathbb{R}}\) is complete, this sequence converges and we can define \(f(x) \mathrel{\vcenter{:}}=\lim_{k\to \infty} f_n(x)\).

• Thus \(f_n\to f\) pointwise by construction

• Claim: \({\left\lVert {f - f_n} \right\rVert} \overset{n\to\infty}\to 0\), so \(f_n\) converges to \(f\) in \(C([0, 1], {\left\lVert {{\,\cdot\,}} \right\rVert}_\infty)\).

  • Proof:
    • Fix \(\varepsilon> 0\); we will show there exists an \(N\) such that \(n\geq N \implies {\left\lVert {f_n - f} \right\rVert} < \varepsilon\)
    • Fix an \(x_0 \in I\). Since \(f_n \to f\) pointwise, choose \(N_1\) large enough so that \begin{align*}n\geq N_1 \implies {\left\lvert {f_n(x_0) - f(x_0)} \right\rvert} < \varepsilon/2.\end{align*}
    • Since \({\left\lVert {f_n - f_m} \right\rVert}_\infty \to 0\), choose and \(N_2\) large enough so that \begin{align*}n, m \geq N_2 \implies {\left\lVert {f_n - f_m} \right\rVert}_\infty < \varepsilon/2.\end{align*}
    • Then for \(n, m \geq \max(N_1, N_2)\), we have \begin{align*} {\left\lvert {f_n(x_0) - f(x_0)} \right\rvert} &= {\left\lvert {f_n(x_0) - f(x_0) + f_m(x_0) - f_m(x_0)} \right\rvert} \\ &= {\left\lvert {f_n(x_0) - f_m(x_0) + f_m(x_0) - f(x_0)} \right\rvert} \\ &\leq {\left\lvert {f_n(x_0) - f_m(x_0)} \right\rvert} + {\left\lvert {f_m(x_0) - f(x_0)} \right\rvert} \\ &< {\left\lvert {f_n(x_0) - f_m(x_0)} \right\rvert} + {\varepsilon\over 2} \\ &\leq \sup_{x\in I} {\left\lvert {f_n(x) - f_m(x)} \right\rvert} + {\varepsilon\over 2} \\ &< {\left\lVert {f_n - f_m} \right\rVert}_\infty + {\varepsilon\over 2} \\ &\leq {\varepsilon\over 2} + {\varepsilon\over 2} \\ \implies {\left\lvert {f_n(x_0) - f(x_0)} \right\rvert} &< \varepsilon\\ \implies \sup_{x\in I} {\left\lvert {f_n(x_0) - f(x_0)} \right\rvert} &\leq \sup_{x\in I} \varepsilon\quad\text{by order limit laws} \\ \implies {\left\lVert {f_n - f} \right\rVert} &\leq \varepsilon\\ .\end{align*}

• \(f\) is the uniform limit of continuous functions and thus continuous, so \(f\in C([0, 1])\).

12.1.2 b

• It suffices to produce a Cauchy sequence that does not converge to a continuous function.

• Take the following sequence of functions:

  • \(f_1\) increases linearly from 0 to 1 on \([0, 1/2]\) and is 1 on \([1/2, 1]\)
  • \(f_2\) is 0 on \([0, 1/4]\) increases linearly from 0 to 1 on \([1/4, 1/2]\) and is 1 on \([1/2, 1]\)
  • \(f_3\) is 0 on \([0, 3/8]\) increases linearly from 0 to 1 on \([3/8, 1/2]\) and is 1 on \([1/2, 1]\)
  • \(f_3\) is 0 on \([0, (1/2 - 3/8)/2]\) increases linearly from 0 to 1 on \([(1/2 - 3/8)/2, 1/2]\) and is 1 on \([1/2, 1]\)

  Idea: take sequence starting points for the triangles: \(0, 0 + {1\over 4}, 0 + {1 \over 4} + {1\over 8}, \cdots\) which converges to \(1/2\) since \(\sum_{k=1}^\infty{1\over 2^k} = -{1\over 2} + \sum_{k=0}^\infty {1\over 2^k}\).

• Then each \(f_n\) is clearly integrable, since its graph is contained in the unit square.

• \(\left\{{f_n}\right\}\) is Cauchy: geometrically subtracting areas yields a single triangle whose area tends to 0.

• But \(f_n\) converges to \(\chi_{[{1\over 2}, 1]}\) which is discontinuous.

• Denote this norm \({\left\lVert {{\,\cdot\,}} \right\rVert}_u\)

• Let \(f_n\) be a Cauchy sequence in this space, so \({\left\lVert {f_n} \right\rVert}_u < \infty\) for every \(n\) and \({\left\lVert {f_j - f_k} \right\rVert}_u \overset{j, k\to\infty}\to 0\).

and define a candidate limit: for each \(x\in I\), set \begin{align*}f(x) \mathrel{\vcenter{:}}=\lim_{n\to\infty} f_n(x).\end{align*}

• Note that \begin{align*} {\left\lVert {f_n} \right\rVert}_\infty &\leq {\left\lVert {f_n} \right\rVert}_u < \infty \\ {\left\lVert {f_n'} \right\rVert}_\infty &\leq {\left\lVert {f_n} \right\rVert}_u < \infty .\end{align*}

  • Thus both \(f_n, f_n'\) are Cauchy sequences in \(C^0([a, b], {\left\lVert {{\,\cdot\,}} \right\rVert}_\infty)\), which is a Banach space, so they converge.

• So

  • \(f_n \to f\) uniformly (by uniqueness of limits),
  • \(f_n' \to g\) uniformly for some \(g\), and
  • \(f, g\in C^0([a, b])\).

• Claim: \(g = f'\)

  • For any fixed \(a\in I\), we have \begin{align*} f_n(x) - f_n(a) \quad &\overset{u}\to f(x) - f(a) \\ \int_a^x f'_n \quad &\overset{u}\to \int_a^x g .\end{align*}
  • By the FTC, the left-hand sides are equal.
  • By uniqueness of limits so are the right-hand sides, so \(f' = g\).

• Claim: the limit \(f\) is an element in this space.

  • Since \(f, f'\in C^0([a, b])\), they are bounded, and so \({\left\lVert {f} \right\rVert}_u < \infty\).

• Claim: \({\left\lVert {f_n - f} \right\rVert}_u \overset{n\to\infty}\to 0\)

• Thus the Cauchy sequence \(\left\{{f_n}\right\}\) converges to a function \(f\) in the \(u{\hbox{-}}\)norm where \(f\) is an element of this space, making it complete.

Let \(\left\{{f_k}\right\}\) be a Cauchy sequence, so \({\left\lVert {f_k} \right\rVert} < \infty\) for all \(k\). Then for a fixed \(x\), the sequence \(f_k(x)\) is Cauchy in \({\mathbb{R}}\) and thus converges to some \(f(x)\), so define \(f\) by \(f(x) \mathrel{\vcenter{:}}=\lim_{k\to\infty} f_k(x)\).

Then \({\left\lVert {f_k - f} \right\rVert} = \max_{x\in X}{\left\lvert {f_k(x) - f(x)} \right\rvert} \overset{k\to\infty}\to 0\), and thus \(f_k \to f\) uniformly and thus \(f\) is continuous. It just remains to show that \(f\) has bounded norm.

Choose \(N\) large enough so that \({\left\lVert {f - f_N} \right\rVert} < \varepsilon\), and write \({\left\lVert {f_N} \right\rVert} \mathrel{\vcenter{:}}= M < \infty\)

\begin{align*} {\left\lVert {f} \right\rVert} \leq {\left\lVert {f - f_N} \right\rVert} + {\left\lVert {f_N} \right\rVert} < \varepsilon + M < \infty .\end{align*}