I’d like to extend my gratitude to Peter Woolfitt for supplying many solutions and checking many proofs of the rest in problem sessions. Many other solutions contain input and ideas from other graduate students and faculty members at UGA, along with questions and answers posted on Math Stack Exchange or Math Overflow.
Let \(f(x) = \frac 1 x\). Show that \(f\) is uniformly continuous on \((1, \infty)\) but not on \((0,\infty)\).
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 \(\delta < \varepsilon c^2\).
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 \(c=0\).
Let \begin{align*} f(x) = \sum _{n=0}^{\infty} \frac{x^{n}}{n !}. \end{align*}
Describe the intervals on which \(f\) does and does not converge uniformly.
Set \(f_N(x) = \sum_{n=1}^N {x^n \over n!}\).
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 \({\mathbb{R}}\).
Let \(\left\{{f_n}\right\}\) be a sequence of continuous functions such that \(\sum f_n\) converges uniformly.
Prove that \(\sum f_n\) is also continuous.
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*}
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.
Let \(f(x, y)\) on \([-1, 1]^2\) be defined by \begin{align*} f(x, y) = \begin{cases} \frac{x y}{\left(x^{2}+y^{2}\right)^{2}} & (x, y) \neq (0, 0) \\ 0 & (x, y) = (0, 0) \end{cases} \end{align*} Determine if \(f\) is integrable.
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.
Let \((X, d)\) and \((Y, \rho)\) be metric spaces, \(f: X\to Y\), and \(x_0 \in X\).
Prove that the following statements are equivalent:
\(1\implies 2\):
\(2\implies 1\):
Note that we need a \(\delta\) for every sequence, so picking a sequence for the forward implication is not a good idea here.
Let \(I\) be an index set and \(\alpha: I \to (0, \infty)\).
Show that \begin{align*} \sum_{i \in I} a(i):=\sup _{\substack{ J \subset I \\ J \text { finite }}} \sum_{i \in J} a(i)<\infty \implies I \text{ is countable.} \end{align*}
Suppose \(I = {\mathbb{Q}}\) and \(\sum_{q \in \mathbb{Q}} a(q)<\infty\). Define \begin{align*} f(x):=\sum_{\substack{q \in \mathbb{Q}\\ q \leq x}} a(q). \end{align*} Show that \(f\) is continuous at \(x \iff x\not\in {\mathbb{Q}}\).
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.
\todo \begin{align*}inline\end{align*} {Not sure.}
Let \(\left\{{a_n}\right\}\) be a sequence of real numbers such that \begin{align*} \left\{{b_n}\right\} \in \ell^2({\mathbb{N}}) \implies \sum a_n b_n < \infty. \end{align*} Show that \(\sum a_n^2 < \infty\).
Note: Assume \(a_n, b_n\) are all non-negative.
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 \(\sum a_n^2 < \infty\).
Prove that if \(f: [0, 1] \to {\mathbb{R}}\) is continuous then \begin{align*} \lim_{k\to\infty} \int_0^1 kx^{k-1} f(x) \,dx = f(1) .\end{align*}
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*}
So the result holds when \(f\) is a polynomial.
Now use the Weierstrass approximation theorem:
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, \(p_\varepsilon(1) \overset{\varepsilon\to 0}\to f(1)\).
Let \(\{a_n\}_{n=1}^\infty\) be a sequence of real numbers.
Prove that if \(\displaystyle\lim_{n\to \infty } a_n = 0\), then \begin{align*} \lim _{n \rightarrow \infty} \frac{a_{1}+\cdots+a_{n}}{n}=0 \end{align*}
Prove that if \(\displaystyle\sum_{n=1}^{\infty} \frac{a_{n}}{n}\) converges, then \begin{align*} \lim _{n \rightarrow \infty} \frac{a_{1}+\cdots+a_{n}}{n}=0 \end{align*}
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*}
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*}
Let \(f\in L^1([0, 1])\). Prove that \begin{align*} \lim_{n \to \infty} \int_{0}^{1} f(x) {\left\lvert {\sin n x} \right\rvert} ~d x= \frac{2}{\pi} \int_{0}^{1} f(x) ~d x \end{align*}
Hint: Begin with the case that \(f\) is the characteristic function of an interval.
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
We can compute the integral over one full period (which is independent of which period is chosen)
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 \({n-1 \over n} \to 1\).
\(R_n \to 0\):
General case:
By linearity of the integral, the result holds for simple functions:
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.
Let \begin{align*} f_{n}(x) = n x(1-x)^{n}, \quad n \in {\mathbb{N}}. \end{align*}
Hint: Consider the maximum of \(f_n\).
\(f_n\to 0\) 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*}
Thus \({\left\lVert {f_n - 0} \right\rVert}_\infty = {\left\lVert {f_n} \right\rVert}_\infty > e^{-1} > 0\).
Let \begin{align*} f_{n}(x) = a e^{-n a x} - b e^{-n b x} \quad \text{ where } 0 < a < b. \end{align*}
Show that
\(\sum_{n=1}^{\infty} \left|f_{n}\right|\) is not in \(L^{1}([0, \infty), m)\)
Hint: \(f_n(x)\) has a root \(x_n\).
\begin{align*} \sum_{n=1}^{\infty} f_{n} \text { is in } L^{1}([0, \infty), m) {\quad \operatorname{and} \quad} \int _{0}^{\infty} \sum _{n=1}^{\infty} f_{n}(x) \,dm = \ln \frac{b}{a} \end{align*}
\(f_n\) 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*}
?
Define \begin{align*} f(x) = \sum_{n=1}^{\infty} \frac{1}{n^{x}}. \end{align*}
Show that \(f\) converges to a differentiable function on \((1, \infty)\) and that \begin{align*} f'(x) =\sum_{n=1}^{\infty}\left(\frac{1}{n^{x}}\right)^{\prime}. \end{align*}
Hint: \begin{align*} \left(\frac{1}{n^{x}}\right)' = -\frac{1}{n^{x}} \ln n \end{align*}
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 \(\phi\in L^\infty({\mathbb{R}})\). Show that the following limit exists and satisfies the equality \begin{align*} \lim _{n \to \infty} \left(\int _{\mathbb{R}} \frac{|\phi(x)|^{n}}{1+x^{2}} \, dx \right) ^ {\frac{1}{n}} = {\left\lVert {\phi} \right\rVert}_\infty. \end{align*}
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: \(R\leq L\).
Let \(f, g \in L^2({\mathbb{R}})\). Show that \begin{align*} \lim _{n \to \infty} \int _{{\mathbb{R}}} f(x) g(x+n) \,dx = 0 \end{align*}
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*}
For \(n\in {\mathbb{N}}\), define \begin{align*} e_{n} = \left (1+ {1\over n} \right)^{n} {\quad \operatorname{and} \quad} E_{n} = \left( 1+ {1\over n} \right)^{n+1} \end{align*}
Show that \(e_n < E_n\), and prove Bernoulli’s inequality:
\begin{align*}
(1+x)^{n} \geq 1+n x \text { for }-1
Use this to show the following:
Define \begin{align*} f(x)=c_{0}+c_{1} x^{1}+c_{2} x^{2}+\ldots+c_{n} x^{n} \text { with } n \text { even and } c_{n}>0. \end{align*}
Show that there is a number \(x_m\) such that \(f(x_m) \leq f(x)\) for all \(x\in {\mathbb{R}}\).
Let \(m_*\) denote the Lebesgue outer measure on \({\mathbb{R}}\).
Prove that for every \(E\subseteq {\mathbb{R}}\) there exists a Borel set \(B\) containing \(E\) such that \begin{align*} m_*(B) = m_*(E) .\end{align*}
Prove that if \(E\subseteq {\mathbb{R}}\) has the property that \begin{align*} m_*(A) = m_*(A\cap E) + m_*(A\cap E^c) \end{align*} for every set \(A\subseteq {\mathbb{R}}\), then there exists a Borel set \(B\subseteq {\mathbb{R}}\) such that \(E = B\setminus N\) with \(m_*(N) = 0\).
Be sure to address the case when \(m_*(E) = \infty\).
\(m_*(Q) \leq {\left\lvert {Q} \right\rvert}\):
Since \(Q\subseteq Q\), \(Q\rightrightarrows Q\) and \(m_*(Q) \leq {\left\lvert {Q} \right\rvert}\) since \(m_*\) is an infimum over such coverings.
\({\left\lvert {Q} \right\rvert} \leq m_*(Q)\):
Fix \(\varepsilon> 0\).
Let \(\left\{{Q_i}\right\}_{i=1}^\infty \rightrightarrows Q\) be arbitrary, it suffices to show that \begin{align*}{\left\lvert {Q} \right\rvert} \leq \qty{\sum_{i=1}^\infty {\left\lvert {Q_i} \right\rvert}} + \varepsilon.\end{align*}
Pick open cubes \(S_i\) such that \(Q_i\subseteq S_i\) and \({\left\lvert {Q_i} \right\rvert} \leq {\left\lvert {S_i} \right\rvert} \leq (1+\varepsilon){\left\lvert {Q_i} \right\rvert}\).
Then \(\left\{{S_i}\right\} \rightrightarrows Q\), so by compactness of \(Q\) pick a finite subcover with \(N\) elements.
Note \begin{align*} Q \subseteq \cup_{i=1}^N S_i \implies {\left\lvert {Q} \right\rvert} \leq \sum_{i=1}^N {\left\lvert {S_i} \right\rvert} \leq \sum_{i=1}^N (1+\varepsilon) {\left\lvert {Q_j} \right\rvert} \leq (1+\varepsilon)\sum_{i=1}^\infty {\left\lvert {Q_i } \right\rvert} .\end{align*}
Taking an infimum over coverings on the RHS preserves the inequality, so \begin{align*}{\left\lvert {Q} \right\rvert} \leq (1+\varepsilon) m_*(Q)\end{align*}
Take \(\varepsilon\to 0\) to obtain final inequality.
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 \(B \mathrel{\vcenter{:}}=\cap_{n=1}^\infty B_n\).
This forces \(m_*(E) = m_*(B)\).
Suppose \(m_*(E) < \infty\).
If \(m_*(E) = \infty\):
Let \((X, \mathcal B, \mu)\) be a measure space with \(\mu(X) = 1\) and \(\{B_n\}_{n=1}^\infty\) be a sequence of \(\mathcal B\)-measurable subsets of \(X\), and \begin{align*} B \mathrel{\vcenter{:}}=\left\{{x\in X {~\mathrel{\Big|}~}x\in B_n \text{ for infinitely many } n}\right\}. \end{align*}
Argue that \(B\) is also a \(\mathcal{B} {\hbox{-}}\)measurable subset of \(X\).
Prove that if \(\sum_{n=1}^\infty \mu(B_n) < \infty\) then \(\mu(B)= 0\).
Prove that if \(\sum_{n=1}^\infty \mu(B_n) = \infty\) and the sequence of set complements \(\left\{{B_n^c}\right\}_{n=1}^\infty\) satisfies \begin{align*} \mu\left(\bigcap_{n=k}^{K} B_{n}^{c}\right)=\prod_{n=k}^{K}\left(1-\mu\left(B_{n}\right)\right) \end{align*} for all positive integers \(k\) and \(K\) with \(k < K\), then \(\mu(B) = 1\).
Hint: Use the fact that \(1 - x ≤ e^{-x}\) for all \(x\).
Borel-Cantelli: for a sequence of sets \(X_n\), \begin{align*} \limsup_n X_n &= \left\{{x {~\mathrel{\Big|}~}x\in X_n \text{ for infinitely many $n$} }\right\} &= \cap_{m\in {\mathbb{N}}} \cup_{n\geq m} X_n \\ \liminf_n X_n &= \left\{{x {~\mathrel{\Big|}~}x\in X_n \text{ for all but finitely many $n$} }\right\} &= \cup_{m\in {\mathbb{N}}} \cap_{n\geq m} X_n .\end{align*}
Properties of logs and exponentials: \begin{align*} \prod_n e^{x_n} = e^{\Sigma_n x_n} \quad\text{and} \quad \sum_n \log(x_n) = \log\left(\prod_n x_n\right) .\end{align*}
Tails of convergent sums vanish.
Continuity of measure: \(B_n \searrow B\) and \(\mu(B_0)<\infty\) implies \(\lim_n \mu(B_n) = \mu(B)\), and \(B_n\nearrow B \implies \lim_n \mu(B_n) = \mu(B)\).
\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*}
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 \(B_K^c \xrightarrow{K\to\infty} B^c\).
Let \(\mathcal B\) denote the set of all Borel subsets of \({\mathbb{R}}\) and \(\mu : \mathcal B \to [0, \infty)\) denote a finite Borel measure on \({\mathbb{R}}\).
Prove that if \(\{F_k\}\) is a sequence of Borel sets for which \(F_k \supseteq F_{k+1}\) for all \(k\), then \begin{align*} \lim _{k \rightarrow \infty} \mu\left(F_{k}\right)=\mu\left(\bigcap_{k=1}^{\infty} F_{k}\right) \end{align*}
Suppose \(\mu\) has the property that \(\mu (E) = 0\) for every \(E \in \mathcal B\) with Lebesgue measure \(m(E) = 0\).
Prove that for every \(\epsilon > 0\) there exists \(\delta > 0\) so that if \(E \in \mathcal B\) with \(m(E) < δ\), then \(\mu(E) < ε\).
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*}
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*}
Let \(E\subset {\mathbb{R}}\) be a Lebesgue measurable set. Show that there is a Borel set \(B \subset E\) such that \(m(E\setminus B) = 0\).
Define \begin{align*} E:=\left\{x \in \mathbb{R}:\left|x-\frac{p}{q}\right|
Prove that \(m(E) = 0\).
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 \(m(E \cap[0, 1]) = 0\).
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*}
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 \(m(E) = 0\).
Let \(f(x) = x^2\) and \(E \subset [0, \infty) \mathrel{\vcenter{:}}={\mathbb{R}}^+\).
Show that \begin{align*} m^*(E) = 0 \iff m^*(f(E)) = 0. \end{align*}
Deduce that the map
\begin{align*} \phi: \mathcal{L}({\mathbb{R}}^+) &\to \mathcal{L}({\mathbb{R}}^+) \\ E &\mapsto f(E) \end{align*} is a bijection from the class of Lebesgue measurable sets of \([0, \infty)\) to itself.
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. \(\hfill\blacksquare\)
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 \(\phi\).
Let \(\mu\) be a measure on a measurable space \((X, \mathcal M)\) and \(f\) a positive measurable function.
Define a measure \(\lambda\) by \begin{align*} \lambda(E):=\int_{E} f ~d \mu, \quad E \in \mathcal{M} \end{align*}
Show that for \(g\) any positive measurable function, \begin{align*} \int_{X} g ~d \lambda=\int_{X} f g ~d \mu \end{align*}
Let \(E \subset {\mathbb{R}}\) be a measurable set such that \begin{align*} \int_{E} x^{2} ~d m=0. \end{align*} Show that \(m(E) = 0\).
Strategy: use approximation by simple functions to show absolute continuity and apply Radon-Nikodym
Claim: \(\lambda \ll \mu\), i.e. \(\mu(E) = 0 \implies \lambda(E) = 0\).
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 \(\lambda(E) = 0\).
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 \(m(E) = 0\).
Let \((X, \mathcal M, \mu)\) be a measure space and suppose \(\left\{{E_n}\right\} \subset \mathcal M\) satisfies \begin{align*} \lim _{n \rightarrow \infty} \mu\left(X \backslash E_{n}\right)=0. \end{align*}
Define \begin{align*} G \mathrel{\vcenter{:}}=\left\{{x\in X {~\mathrel{\Big|}~}x\in E_n \text{ for only finitely many } n}\right\}. \end{align*}
Show that \(G \in \mathcal M\) and \(\mu(G) = 0\).
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*}
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: \(\mu(G) = 0\).
Let \(f\) be Lebesgue measurable on \({\mathbb{R}}\) and \(E \subset {\mathbb{R}}\) be measurable such that \begin{align*} 0
Show that for every \(0 < t < 1\), there exists a measurable set \(E_t \subset E\) such that \begin{align*} \int_{E_{t}} f(x) d x=t A. \end{align*}
Let \((X, \mathcal M, \mu)\) be a measure space. For \(f\in L^1(\mu)\) and \(\lambda > 0\), define \begin{align*} \phi(\lambda)=\mu(\{x \in X | f(x)>\lambda\}) \quad \text { and } \quad \psi(\lambda)=\mu(\{x \in X | f(x)<-\lambda\}) \end{align*}
Show that \(\phi, \psi\) are Borel measurable and \begin{align*} \int_{X}|f| ~d \mu=\int_{0}^{\infty}[\phi(\lambda)+\psi(\lambda)] ~d \lambda \end{align*}
Let \(f: {\mathbb{R}}\to {\mathbb{R}}\) be Lebesgue measurable.
Let \(\mu\) be a finite Borel measure on \({\mathbb{R}}\) and \(E \subset {\mathbb{R}}\) Borel. Prove that the following statements are equivalent:
Let \(f: {\mathbb{R}}\to {\mathbb{R}}\) and suppose \begin{align*} \forall x\in {\mathbb{R}},\quad f(x) \geq \limsup _{y \rightarrow x} f(y) \end{align*} Prove that \(f\) is Borel measurable.
Let \((X, \mathcal M, \mu)\) be a measure space and suppose \(f\) is a measurable function on \(X\). Show that \begin{align*} \lim _{n \rightarrow \infty} \int_{X} f^{n} ~d \mu = \begin{cases} \infty & \text{or} \\ \mu(f^{-1}(1)), \end{cases} \end{align*} and characterize the collection of functions of each type.
Let \(K\) be the set of numbers in \([0, 1]\) whose decimal expansions do not use the digit \(4\).
We use the convention that when a decimal number ends with 4 but all other digits are different from 4, we replace the digit \(4\) with \(399\cdots\). For example, \(0.8754 = 0.8753999\cdots\).
Show that \(K\) is a compact, nowhere dense set without isolated points, and find the Lebesgue measure \(m(K)\).
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 \({\mathbb{R}}\)).
Identify \(K^c\) as the set of real numbers in \([0, 1]\) whose decimal expansion does contain a 4.
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:
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
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 \(B_r(x) \subsetneq K^c\).
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.
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 \(K\).
Let \(0 < \lambda < 1\) and construct a Cantor set \(C_\lambda\) by successively removing middle intervals of length \(\lambda\).
Prove that \(m(C_\lambda) = 0\).
Let \(f, g: [a, b] \to {\mathbb{R}}\) be measurable with \begin{align*} \int_{a}^{b} f(x) ~d x=\int_{a}^{b} g(x) ~d x. \end{align*}
Show that either
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
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*}
Let \(E \subset {\mathbb{R}}\) be measurable with \(m(E) < \infty\). Define \begin{align*} f(x)=m(E \cap(E+x)). \end{align*}
Show that
Hint: \begin{align*} \chi_{E \cap(E+x)}(y)=\chi_{E}(y) \chi_{E}(y-x) \end{align*}
Prove that \begin{align*} \left| \frac{d^{n}}{d x^{n}} \frac{\sin x}{x}\right| \leq \frac{1}{n} \end{align*}
for all \(x \neq 0\) and positive integers \(n\).
Hint: Consider \(\displaystyle\int_0^1 \cos(tx) dt\)
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*}
Compute the following limit and justify your calculations: \begin{align*} \lim_{n\to\infty} \int_0^n \qty{1 + {x^2 \over n}}^{-(n+1)} \,dx .\end{align*}
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:
Let \(\{f_k\}\) be any sequence of functions in \(L^2([0, 1])\) satisfying \({\left\lVert {f_k} \right\rVert}_2 ≤ M\) for all \(k ∈ {\mathbb{N}}\).
Prove that if \(f_k \to f\) almost everywhere, then \(f ∈ L^2([0, 1])\) with \({\left\lVert {f} \right\rVert}_2 ≤ M\) and \begin{align*} \lim _{k \rightarrow \infty} \int_{0}^{1} f_{k}(x) dx = \int_{0}^{1} f(x) d x \end{align*}
Hint: Try using Fatou’s Lemma to show that \({\left\lVert {f} \right\rVert}_2 ≤ M\) and then try applying Egorov’s Theorem.
\(L^2\) bound:
\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*}
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.
Compute the following limit and justify your calculations: \begin{align*} \lim_{n \rightarrow \infty} \int_{1}^{n} \frac{d x}{\left(1+\frac{x}{n}\right)^{n} \sqrt[n]{x}} \end{align*}
Suppose \(f(x)\) and \(xf(x)\) are integrable on \({\mathbb{R}}\). Define \(F\) by \begin{align*} F(t)\mathrel{\vcenter{:}}=\int _{-\infty}^{\infty} f(x) \cos (x t) dx \end{align*} Show that \begin{align*} F'(t)=-\int _{-\infty}^{\infty} x f(x) \sin (x t) 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.
Suppose that
Show that \(\int f_{n} \rightarrow \int f\).
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*}
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
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 \(\lim_{n\to\infty} \int f_n = \int f\).
Let \begin{align*} f_{n}(x):=\frac{x}{1+x^{n}}, \quad x \geq 0. \end{align*}
Show that this sequence converges pointwise and find its limit. Is the convergence uniform on \([0, \infty)\)?
Compute \begin{align*} \lim _{n \rightarrow \infty} \int_{0}^{\infty} f_{n}(x) d x \end{align*}
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 \((0, \infty)\).
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.
Let \(f\in L^1({\mathbb{R}})\). Show that \begin{align*} \lim _{x \to 0} \int _{{\mathbb{R}}} {\left\lvert {f(y-x)-f(y)} \right\rvert} \, dy = 0 \end{align*}
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*}
Compute the following limit: \begin{align*} \lim _{n \rightarrow \infty} \int_{1}^{n} \frac{n e^{-x}}{1+n x^{2}} \, \sin \left(\frac x n\right) \, dx \end{align*}
Let \(f: [1, \infty) \to {\mathbb{R}}\) such that \(f(1) = 1\) and \begin{align*} f^{\prime}(x)= \frac{1} {x^{2}+f(x)^{2}} \end{align*}
Show that the following limit exists and satisfies the equality \begin{align*} \lim _{x \rightarrow \infty} f(x) \leq 1 + \frac \pi 4 \end{align*}
Let \(f\) be a non-negative measurable function on \([0, 1]\).
Show that \begin{align*} \lim _{p \rightarrow \infty}\left(\int_{[0,1]} f(x)^{p} d x\right)^{\frac{1}{p}}=\|f\|_{\infty}. \end{align*}
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*}
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*}
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 \({\left\lVert {f} \right\rVert}_p \geq {\left\lVert {f} \right\rVert}_\infty\).
Let \(f\in L^2([0, 1])\) and suppose \begin{align*} \int _{[0,1]} f(x) x^{n} d x=0 \text { for all integers } n \geq 0. \end{align*} Show that \(f = 0\) almost everywhere.
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.
\(\hfill\blacksquare\)
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*}
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*}
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.
Let \(f: {\mathbb{R}}\to {\mathbb{C}}\) be continuous with period 1. Prove that \begin{align*} \lim _{N \rightarrow \infty} \frac{1}{N} \sum_{n=1}^{N} f(n \alpha)=\int_{0}^{1} f(t) d t \quad \forall \alpha \in {\mathbb{R}}\setminus{\mathbb{Q}}. \end{align*}
Hint: show this first for the functions \(f(t) = e^{2\pi i k t}\) for \(k\in {\mathbb{Z}}\).
Let \(g\in L^\infty([0, 1])\) Prove that \begin{align*} \int _{[0,1]} f(x) g(x)\, dx = 0 \quad\text{for all continuous } f:[0, 1] \to {\mathbb{R}} \implies g(x) = 0 \text{ almost everywhere. } \end{align*}
Prove that if \(g\in L^1({\mathbb{R}})\) then \begin{align*} \lim_{N\to \infty} \int _{{\left\lvert {x} \right\rvert} \geq N} {\left\lvert {f(x)} \right\rvert} \, dx = 0 ,\end{align*} and demonstrate that it is not necessarily the case that \(f(x) \to 0\) as \({\left\lvert {x} \right\rvert}\to \infty\).
Prove that if \(f\in L^1([1, \infty])\) and is decreasing, then \(\lim_{x\to\infty}f(x) =0\) and in fact \(\lim_{x\to \infty} xf(x) = 0\).
If \(f: [1, \infty) \to [0, \infty)\) is decreasing with \(\lim_{x\to \infty} xf(x) = 0\), does this ensure that \(f\in L^1([1, \infty))\)?
Stated integral equality:
To see that this doesn’t force \(f(x)\to 0\) as \({\left\lvert {x} \right\rvert} \to \infty\):
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.
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.
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 \(L>0\):
If \(L<0\):
Showing \(xf(x) \overset{x\to \infty}\to 0\).
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
Show that if \(f\) is continuous with compact support on \({\mathbb{R}}\), then \begin{align*} \lim _{y \rightarrow 0} \int_{\mathbb{R}}|f(x-y)-f(x)| d x=0 \end{align*}
Let \(f\in L^1({\mathbb{R}})\) and for each \(h > 0\) let \begin{align*} \mathcal{A}_{h} f(x):=\frac{1}{2 h} \int_{|y| \leq h} f(x-y) d y \end{align*}
Prove that \(\left\|\mathcal{A}_{h} f\right\|_{1} \leq\|f\|_{1}\) for all \(h > 0\).
Prove that \(\mathcal{A}_h f \to f\) in \(L^1({\mathbb{R}})\) as \(h \to 0^+\).
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*}
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*}
Let \begin{align*} S = {\operatorname{span}}_{\mathbb{C}}\left\{{\chi_{(a, b)} {~\mathrel{\Big|}~}a, b \in {\mathbb{R}}}\right\}, \end{align*} the complex linear span of characteristic functions of intervals of the form \((a, b)\).
Show that for every \(f\in L^1({\mathbb{R}})\), there exists a sequence of functions \(\left\{{f_n}\right\} \subset S\) such that \begin{align*} \lim _{n \rightarrow \infty}\left\|f_{n}-f\right\|_{1}=0 \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 \(L^1\).
Define \begin{align*} f(x, y):=\left\{\begin{array}{ll}{\frac{x^{1 / 3}}{(1+x y)^{3 / 2}}} & {\text { if } 0 \leq x \leq y} \\ {0} & {\text { otherwise }}\end{array}\right. \end{align*}
Carefully show that \(f \in L^1({\mathbb{R}}^2)\).
Let \(f\in L^1({\mathbb{R}})\). Show that \begin{align*} \forall\varepsilon > 0 \exists \delta > 0 \text{ such that } \qquad m(E) < \delta \implies \int _{E} |f(x)| \, dx < \varepsilon \end{align*}
Give an example of a continuous \(f\in L^1({\mathbb{R}})\) such that \(f(x) \not\to 0\) as\({\left\lvert {x} \right\rvert} \to \infty\).
Show that if \(f\) is uniformly continuous, then \begin{align*} \lim_{{\left\lvert {x} \right\rvert} \to \infty} f(x) = 0. \end{align*}
Let \(f, g\in L^1({\mathbb{R}})\). Argue that \(H(x, y) \mathrel{\vcenter{:}}= f(y) g(x-y)\) defines a function in \(L^1({\mathbb{R}}^2)\) and deduce from this fact that \begin{align*} (f\ast g)(x) \mathrel{\vcenter{:}}=\int_{\mathbb{R}}f(y) g(x-y) \,dy \end{align*} defines a function in \(L^1({\mathbb{R}})\) that satisfies \begin{align*} {\left\lVert {f\ast g} \right\rVert}_1 \leq {\left\lVert {f} \right\rVert}_1 {\left\lVert {g} \right\rVert}_1 .\end{align*}
\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*}
Note: Fubini is not needed, since we’re not calculating the actual integral, just showing \(H\) is integrable.
Let \(f\) be a non-negative function on \({\mathbb{R}}^n\) and \(\mathcal A = \{(x, t) ∈ {\mathbb{R}}^n \times {\mathbb{R}}: 0 ≤ t ≤ f (x)\}\).
Prove the validity of the following two statements:
\(f\) is a Lebesgue measurable function on \({\mathbb{R}}^n \iff \mathcal A\) is a Lebesgue measurable subset of \({\mathbb{R}}^{n+1}\)
If \(f\) is a Lebesgue measurable function on \({\mathbb{R}}^n\), then \begin{align*} m(\mathcal{A})=\int _{{\mathbb{R}}^{n}} f(x) d x=\int_{0}^{\infty} m\left(\left\{x \in {\mathbb{R}}^{n}: f(x) \geq t\right\}\right) dt \end{align*}
\(\implies\):
\(\impliedby\):
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.
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.
Let \(f \geq 0\) be a measurable function on \({\mathbb{R}}\). Show that \begin{align*} \int _{{\mathbb{R}}} f = \int _{0}^{\infty} m(\{x: f(x)>t\}) dt \end{align*}
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.
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*}
Claim: \(\chi_E\) satisfies the conditions of Tonelli, and thus \(m(E) = \int \chi_E\) is equal to any iterated integral.
Conclude:
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*}
Let \(f, g \in L^1({\mathbb{R}})\) be Borel measurable.
Let \(f, g \in L^1([0, 1])\) and for all \(x\in [0, 1]\) define \begin{align*} F(x) \mathrel{\vcenter{:}}=\int _{0}^{x} f(y) \, dy {\quad \operatorname{and} \quad} G(x)\mathrel{\vcenter{:}}=\int _{0}^{x} g(y) \, dy. \end{align*}
Prove that \begin{align*} \int _{0}^{1} F(x) g(x) \, dx = F(1) G(1) - \int _{0}^{1} f(x) G(x) \, dx \end{align*}
Show that \begin{align*} L^2([0, 1]) \subseteq L^1([0, 1]) {\quad \operatorname{and} \quad} \ell^1({\mathbb{Z}}) \subseteq \ell^2({\mathbb{Z}}) .\end{align*}
For \(f\in L^1([0, 1])\) define \begin{align*} \widehat{f}(n) \mathrel{\vcenter{:}}=\int _0^1 f(x) e^{-2\pi i n x} \, dx .\end{align*}
Prove that if \(f\in L^1([0, 1])\) and \(\left\{{\widehat{f}(n)}\right\} \in \ell^1({\mathbb{Z}})\) then \begin{align*} S_N f(x) \mathrel{\vcenter{:}}=\sum_{{\left\lvert {n} \right\rvert} \leq N} \widehat{f} (n) e^{2 \pi i n x} .\end{align*} converges uniformly on \([0, 1]\) to a continuous function \(g\) such that \(g = f\) almost everywhere.
Hint: One approach is to argue that if \(f\in L^1([0, 1])\) with \(\left\{{\widehat{f} (n)}\right\} \in \ell^1({\mathbb{Z}})\) then \(f\in L^2([0, 1])\).
Claim: \(\ell^1({\mathbb{Z}}) \subseteq \ell^2({\mathbb{Z}})\).
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 \(\mu(X) < \infty\).
Let \(\phi\) be a compactly supported smooth function that vanishes outside of an interval \([-N, N]\) such that \(\int _{{\mathbb{R}}} \phi(x) \, dx = 1\).
For \(f\in L^1({\mathbb{R}})\), define \begin{align*} K_{j}(x) \mathrel{\vcenter{:}}= j \phi(j x), \qquad f \ast K_{j}(x) \mathrel{\vcenter{:}}=\int_{{\mathbb{R}}} f(x-y) K_{j}(y) \, dy \end{align*} and prove the following:
Each \(f\ast K_j\) is smooth and compactly supported.
\begin{align*} \lim _{j \to \infty} {\left\lVert {f * K_{j}-f} \right\rVert}_{1} = 0 \end{align*}
Hint: \begin{align*} \lim _{y \to 0} \int _{{\mathbb{R}}} |f(x-y)-f(x)| dy = 0 \end{align*}
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.
\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.
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
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*}
Let \(f, g \in L^2({\mathbb{R}})\). Prove that the formula \begin{align*} h(x) \mathrel{\vcenter{:}}=\int _{-\infty}^{\infty} f(t) g(x-t) \, dt \end{align*} defines a uniformly continuous function \(h\) on \({\mathbb{R}}\).
Let \(f \in L^1({\mathbb{R}})\) and \(g\) be a bounded measurable function on \({\mathbb{R}}\).
Let \(f \in C_c^0({\mathbb{R}}^n)\), and show \begin{align*} \lim _{t \to 0} \int_{{\mathbb{R}}^n} |f(x+t) - f(x)| \, dx = 0 .\end{align*}
Extend the above result to \(f\in L^1({\mathbb{R}}^n)\) and show that \begin{align*} f\in L^1({\mathbb{R}}^n), \quad g\in L^\infty({\mathbb{R}}^n) \quad \implies f \ast g \text{ is bounded and uniformly continuous. } \end{align*}
Let \(\{u_n\}_{n=1}^∞\) be an orthonormal sequence in a Hilbert space \(\mathcal{H}\).
Prove that for every \(x ∈ \mathcal H\) one has \begin{align*} \displaystyle\sum_{n=1}^{\infty}\left|\left\langle x, u_{n}\right\rangle\right|^{2} \leq\|x\|^{2} \end{align*}
Prove that for any sequence \(\{a_n\}_{n=1}^\infty \in \ell^2({\mathbb{N}})\) there exists an element \(x\in\mathcal H\) such that \begin{align*} a_n = {\left\langle {x},~{u_n} \right\rangle} \text{ for all } n\in {\mathbb{N}} \end{align*} and \begin{align*} {\left\lVert {x} \right\rVert}^2 = \sum_{n=1}^{\infty}\left|\left\langle x, u_{n}\right\rangle\right|^{2} \end{align*}
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*}
Fix \(\left\{{a_n}\right\} \in \ell^2\), then note that \(\sum {\left\lvert {a_n} \right\rvert}^2 < \infty \implies\) the tails vanish.
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*}
\(\left\{{S_N}\right\}\) Cauchy (by 1) and \(H\) complete \(\implies x\in H\).
\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.
\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 \(x \in H\).
Show that \(L^2([0, 1]) ⊆ L^1([0, 1])\) and argue that \(L^2([0, 1])\) in fact forms a dense subset of \(L^1([0, 1])\).
Let \(Λ\) be a continuous linear functional on \(L^1([0, 1])\).
Prove the Riesz Representation Theorem for \(L^1([0, 1])\) by following the steps below:
Hint: You may use, without proof, the Riesz Representation Theorem for \(L^2([0, 1])\).
Holders’ inequality: \({\left\lVert {fg} \right\rVert}_1 \leq {\left\lVert {f} \right\rVert}_p {\left\lVert {f} \right\rVert}_q\)
Riesz Representation for \(L^2\): If \(\Lambda \in (L^2)^\vee\) then there exists a unique \(g\in L^2\) such that \(\Lambda(f) = \int fg\).
\({\left\lVert {f} \right\rVert}_{L^\infty(X)} \mathrel{\vcenter{:}}=\inf \left\{{t\geq 0 {~\mathrel{\Big|}~}{\left\lvert {f(x)} \right\rvert} \leq t \text{ almost everywhere} }\right\}\).
Lemma: \(m(X) < \infty \implies L^p(X) \subset L^2(X)\).
Write Holder’s inequality as \({\left\lVert {fg} \right\rVert}_1 \leq {\left\lVert {f} \right\rVert}_a {\left\lVert {g} \right\rVert}_b\) where \(\frac 1 a + \frac 1 b = 1\), then \begin{align*} {\left\lVert {f} \right\rVert}_p^p = {\left\lVert {{\left\lvert {f} \right\rvert}^p} \right\rVert}_1 \leq {\left\lVert {{\left\lvert {f} \right\rvert}^p} \right\rVert}_a ~{\left\lVert {1} \right\rVert}_b .\end{align*}
Now take \(a = \frac 2 p\) and this reduces to \begin{align*} {\left\lVert {f} \right\rVert}_p^p &\leq {\left\lVert {f} \right\rVert}_2^p ~m(X)^{\frac 1 b} \\ \implies {\left\lVert {f} \right\rVert}_p &\leq {\left\lVert {f} \right\rVert}_2 \cdot O(m(X)) < \infty .\end{align*}
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 \(L^1\)
Let \(\Lambda \in L^1(X)^\vee\) be arbitrary.
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*}
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 \({\left\lVert {h_\alpha} \right\rVert}_{L^1(X)} = 1\).
Without using the Riesz Representation Theorem, compute \begin{align*} \sup \left\{\left|\int_{0}^{1} f(x) e^{x} d x\right| {~\mathrel{\Big|}~}f \in L^{2}([0,1], m),~~ \|f\|_{2} \leq 1\right\} \end{align*}
Let \(\mathcal H\) be a Hilbert space.
Let \(f: [0, 1] \to {\mathbb{R}}\) be continuous. Show that \begin{align*} \sup \left\{\|f g\|_{1} {~\mathrel{\Big|}~}g \in L^{1}[0,1],~~ \|g\|_{1} \leq 1\right\}=\|f\|_{\infty} \end{align*}
Let \(1 \leq p,q \leq \infty\) be conjugate exponents, and show that \begin{align*} f \in L^p({\mathbb{R}}^n) \implies \|f\|_{p} = \sup _{\|g\|_{q}=1}\left|\int f(x) g(x) d x\right| \end{align*}
Let \(C([0, 1])\) denote the space of all continuous real-valued functions on \([0, 1]\).
Prove that \(C([0, 1])\) is complete under the uniform norm \({\left\lVert {f} \right\rVert}_u := \displaystyle\sup_{x\in [0,1]} |f (x)|\).
Prove that \(C([0, 1])\) is not complete under the \(L^1{\hbox{-}}\)norm \({\left\lVert {f} \right\rVert}_1 = \displaystyle\int_0^1 |f (x)| ~dx\).
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 \(C([0, 1], {\left\lVert {{\,\cdot\,}} \right\rVert}_\infty)\).
\(f\) is the uniform limit of continuous functions and thus continuous, so \(f\in C([0, 1])\).
It suffices to produce a Cauchy sequence that does not converge to a continuous function.
Take the following sequence of functions:
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.
Show that the space \(C^1([a, b])\) is a Banach space when equipped with the norm \begin{align*} \|f\|:=\sup _{x \in[a, b]}|f(x)|+\sup _{x \in[a, b]}\left|f^{\prime}(x)\right|. \end{align*}
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*}
So
Claim: \(g = f'\)
Claim: the limit \(f\) is an element in this space.
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 \(X\) be a complete metric space and define a norm \begin{align*} \|f\|:=\max \{|f(x)|: x \in X\}. \end{align*}
Show that \((C^0({\mathbb{R}}), {\left\lVert {{\,\cdot\,}} \right\rVert} )\) (the space of continuous functions \(f: X\to {\mathbb{R}}\)) is 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*}
Show that if \(x_n\) is a decreasing sequence of positive real numbers such that \(\sum_{n=1}^\infty x_n\) converges, then \begin{align*} \lim_{n\to\infty} n x_n = 0. \end{align*}
Let \(f: {\mathbb{R}}\to {\mathbb{R}}\). Prove that \begin{align*} f(x) \leq \liminf_{y\to x} f(y)~ \text{for each}~ x\in {{\mathbb{R}}} \iff \{ x\in {{\mathbb{R}}} \mathrel{\Big|}f(x) > a \}~\text{is open for all}~ a\in {{\mathbb{R}}} \end{align*}
Recall that a function \(f: {{\mathbb{R}}} \to {{\mathbb{R}}}\) is called lower semi-continuous iff it satisfies either condition in part (a) above.
Prove that if \(\mathcal{F}\) is an y family of lower semi-continuous functions, then \begin{align*} g(x) = \sup\{ f(x) \mathrel{\Big|}f\in \mathcal{F}\} \end{align*} is Borel measurable.
Note that \(\mathcal{F}\) need not be a countable family.
Let \(f\) be a non-negative Lebesgue measurable function on \([1, \infty)\).
Prove that \begin{align*} 1 \leq \qty{ {1 \over b-a} \int_a^b f(x) \,dx }\qty{ {1\over b-a} \int_a^b {1 \over f(x)}\, dx } \end{align*} for any \(1\leq a < b <\infty\).
Prove that if \(f\) satisfies \begin{align*} \int_1^t f(x) \, dx \leq t^2 \log(t) \end{align*} for all \(t\in [1, \infty)\), then \begin{align*} \int_1^\infty {1\over f(x) \,dx} = \infty .\end{align*}
Hint: write \begin{align*} \int_1^\infty {1\over f(x) \, dx} = \sum_{k=0}^\infty \int_{2^k}^{2^{k+1}} {1 \over f(x)}\,dx .\end{align*}
Prove that if \(xf(x) \in L^1({\mathbb{R}})\), then \begin{align*} F(y) \mathrel{\vcenter{:}}=\int f(x) \cos(yx)\, dx \end{align*} defines a \(C^1\) function.
Suppose \(\phi\in L^1({\mathbb{R}})\) with \begin{align*} \int \phi(x) \, dx = \alpha .\end{align*} For each \(\delta > 0\) and \(f\in L^1({\mathbb{R}})\), define \begin{align*} A_\delta f(x) \mathrel{\vcenter{:}}=\int f(x-y) \delta^{-1} \phi\qty{\delta^{-1} y}\, dy .\end{align*}
Prove that for all \(\delta > 0\), \begin{align*} {\left\lVert {A_\delta f} \right\rVert}_1 \leq {\left\lVert {\phi} \right\rVert}_1 {\left\lVert {f} \right\rVert}_1 .\end{align*}
Prove that \begin{align*} A_\delta f \to \alpha f \text{ in } L^1({\mathbb{R}}) {\quad \operatorname{as} \quad} \delta\to 0^+ .\end{align*}
Hint: you may use without proof the fact that for all \(f\in L^1({\mathbb{R}})\), \begin{align*} \lim_{y\to 0} \int_{\mathbb{R}}{\left\lvert {f(x-y) - f(x)} \right\rvert}\, dx = 0 .\end{align*}