1 Preface

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.

2 Undergraduate Analysis: Uniform Convergence

2.1 Fall 2018 # 1 \(\done\)

Let \(f(x) = \frac 1 x\). Show that \(f\) is uniformly continuous on \((1, \infty)\) but not on \((0,\infty)\).

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  • Uniform continuity: \begin{align*} \forall \varepsilon>0, \exists \delta(\varepsilon)>0 {\quad \operatorname{such that} \quad} {\left\lvert {x-y} \right\rvert}<\delta \implies {\left\lvert {f(x) - f(y)} \right\rvert} < \varepsilon .\end{align*}
  • Negating uniform continuity: \(\exists \varepsilon> 0\) such that \(\forall \delta(\varepsilon)\) there exist \(x, y\) such that \({\left\lvert {x-y} \right\rvert} < \delta\) and \({\left\lvert {f(x) - f(y)} \right\rvert} > \varepsilon\).

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

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

2.2 Fall 2017 # 1 \(\done\)

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.

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  • \(f_N\to f\) uniformly \(\iff\) \({\left\lVert {f_N - f} \right\rVert}_\infty \to 0\).
  • \(\sum_{n=0}^\infty c_n x^n \mathrel{\vcenter{:}}=\lim_{N\to \infty} \sum_{n=0}^N c_n x^n\)
    • I.e. an infinite sum is defined as the pointwise limit of its partial sums.
  • If \(\sum_{n=0}^\infty g_n(x)\) converges uniformly on a set \(A\), then \(\sup_{x\in A} {\left\lvert {f_n(x)} \right\rvert} \to 0\).

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

2.3 Fall 2014 # 1 \(\done\)

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.

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

2.4 Spring 2017 # 4 \(\done\)

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.

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  • Just Calculus.

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.

2.5 Spring 2015 # 1 \(\done\)

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. For every \(\varepsilon > 0 \quad \exists \delta > 0\) such that \(\rho( f(x), f(x_0) ) < \varepsilon\) whenever \(d(x, x_0) < \delta\).
  2. The sequence \(\left\{{f(x_n)}\right\}_{n=1}^\infty \to f(x_0)\) for every sequence \(\left\{{x_n}\right\} \to x_0\) in \(X\).
(Click to expand)
  • What it means for a sequence to converge.

\(1\implies 2\):

\(2\implies 1\):

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

2.6 Fall 2014 # 2 \(\work\)

Let \(I\) be an index set and \(\alpha: I \to (0, \infty)\).

2.6.1 1

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*}

2.6.2 2

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 \(x \iff x\not\in {\mathbb{Q}}\).

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2.6.3 1

2.6.4 2

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

2.7 Spring 2014 # 2 \(\done\)

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.

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3 General Analysis

3.1 Spring 2020 # 1 \(\done\)

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*}

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  • DCT
  • Weierstrass Approximation Theorem

3.2 Fall 2019 # 1 \(\done\)

Let \(\{a_n\}_{n=1}^\infty\) be a sequence of real numbers.

3.2.1 a

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*}

3.2.2 b

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*}

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  • Cesaro mean/summation.
  • Break series apart into pieces that can be handled separately.
  • Idea: once \(N\) is large enough, \(a_k \approx S\), and all smaller terms will die off as \(N\to \infty\).

3.2.3 a

3.2.4 b

3.3 Fall 2018 # 4 \(\done\)

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.

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  • Converting floor/ceiling functions to inequalities: \(x-1 \leq {\left\lfloor x \right\rfloor} \leq x\).

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

General case:

3.4 Fall 2017 # 4 \(\done\)

Let \begin{align*} f_{n}(x) = n x(1-x)^{n}, \quad n \in {\mathbb{N}}. \end{align*}

  1. Show that \(f_n \to 0\) pointwise but not uniformly on \([0, 1]\).

Hint: Consider the maximum of \(f_n\).

  1. \begin{align*} \lim _{n \to \infty} \int _{0}^{1} n(1-x)^{n} \sin x \, dx = 0 \end{align*}
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  • \(\sum f_n < \infty \iff \sup f_n \to 0\).
  • Negating uniform convergence: \(f_n\not\to f\) uniformly iff \(\exists \varepsilon\) such that \(\forall N(\varepsilon)\) there exists an \(x_N\) such that \({\left\lvert {f(x_N) - f(x)} \right\rvert} > \varepsilon\).
  • Exponential inequality: \(1+y \leq e^y\) for all \(y\in {\mathbb{R}}\).

3.4.1 a

\(f_n\to 0\) pointwise:

The convergence is not uniform:

3.4.2 b

3.5 Spring 2017 # 3 \(\work\)

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

3.5.1 a

\(\sum_{n=1}^{\infty} \left|f_{n}\right|\) is not in \(L^{1}([0, \infty), m)\)

Hint: \(f_n(x)\) has a root \(x_n\).

3.5.2 b

\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*}

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3.5.3 a

3.5.4 b


3.6 Fall 2016 # 1 \(\done\)

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*}

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  • ?

3.7 Fall 2016 # 5 \(\done\)

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*}

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  • ?

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: \(R\leq L\).

3.8 Fall 2016 # 6 \(\done\)

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*}

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  • Cauchy Schwarz: \({\left\lVert {fg} \right\rVert}_1 \leq {\left\lVert {f} \right\rVert}_1 {\left\lVert {g} \right\rVert}_1\).
  • Small tails.

3.9 Spring 2016 # 1 \(\work\)

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:

  1. The sequence \(e_n\) is increasing.
  2. The sequence \(E_n\) is decreasing.
  3. \(2 < e_n < E_n < 4\).
  4. \(\lim _{n \to \infty} e_{n} = \lim _{n \to \infty} E_{n}\).

3.10 Fall 2015 # 1 \(\work\)

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 \(x\in {\mathbb{R}}\).

4 Measure Theory: Sets

4.1 Spring 2020 # 2 \(\done\)

Let \(m_*\) denote the Lebesgue outer measure on \({\mathbb{R}}\).

4.1.1 a.

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*}

4.1.2 b.

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 \(m_*(E) = \infty\).

(Click to expand)
  • Definition of outer measure: \begin{align*} m_*(E) = \inf_{\left\{{Q_j}\right\} \rightrightarrows E} \sum {\left\lvert {Q_j} \right\rvert} \end{align*} where \(\left\{{Q_j}\right\}\) is a countable collection of closed cubes.
  • Break \({\mathbb{R}}\) into \({\coprod}_{n\in {\mathbb{Z}}} [n, n+1)\), each with finite measure.
  • Theorem: \(m_*(Q) = {\left\lvert {Q} \right\rvert}\) for \(Q\) a closed cube (i.e. the outer measure equals the volume).
  • \(m_*(Q) \leq {\left\lvert {Q} \right\rvert}\):

  • 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.

4.1.3 a

4.1.4 b

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

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

4.2 Fall 2019 # 3. \(\done\)

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*}

  1. Argue that \(B\) is also a \(\mathcal{B} {\hbox{-}}\)measurable subset of \(X\).

  2. Prove that if \(\sum_{n=1}^\infty \mu(B_n) < \infty\) then \(\mu(B)= 0\).

  3. 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 \(x\).

(Click to expand)
  • 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 \(B_n\nearrow B \implies \lim_n \mu(B_n) = \mu(B)\).

4.2.1 a

4.2.2 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*}

4.2.3 c

4.3 Spring 2019 # 2 \(\done\)

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 \({\mathbb{R}}\).

4.3.1 a

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*}

4.3.2 b

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 \(\mu(E) < ε\).

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  • ?

4.3.3 a

See Folland p.26

4.3.4 b

4.4 Fall 2018 # 2 \(\done\)

Let \(E\subset {\mathbb{R}}\) be a Lebesgue measurable set. Show that there is a Borel set \(B \subset E\) such that \(m(E\setminus B) = 0\).

(Click to expand)
  • Definition of measurability: there exists an open \(O\supset E\) such that \(m_*(O\setminus E) < \varepsilon\) for all \(\varepsilon> 0\).
  • Theorem: \(E\) is Lebesgue measurable iff there exists a closed set \(F\subseteq E\) such that \(m_*(E\setminus F) < \varepsilon\) for all \(\varepsilon>0\).
  • Every \(F_\sigma, G_\delta\) is Borel.
  • Claim: \(E\) is measurable \(\iff\) for every \(\varepsilon\) there exist \(F_\varepsilon \subset E \subset G_\varepsilon\) with \(F_\varepsilon\) closed and \(G_\varepsilon\) open and \(m(G_\varepsilon \setminus E)< \varepsilon\) and \(m(E\setminus F_\varepsilon) < \varepsilon\).
    • Proof: existence of \(G_\varepsilon\) is the definition of measurability.
    • Existence of \(F_\varepsilon\): ?
  • Claim: \(E\) is measurable \(\implies\) there exists an open \(O\supseteq E\) such that \(m(O\setminus E) = 0\).
    • Since \(E\) is measurable, for each \(n\in {\mathbb{N}}\) choose \(G_n \supseteq E\) such that \(m_*(G_n\setminus E) < {1\over n}\).
    • Set \(O_N \mathrel{\vcenter{:}}=\cap_{n=1}^N G_n\) and \(O\mathrel{\vcenter{:}}=\cap_{n=1}^\infty G_n\).
    • Suppose \(E\) is bounded.
      • Note \(O_N \searrow O\) and \(m_*(O_1) < \infty\) if \(E\) is bounded, since in this case \begin{align*} m_*(G_n\setminus E) = m_*(G_1) - m_*(E) < 1 \iff m_*(G_1) < m_*(E) + {1\over n} < \infty .\end{align*}
      • Note \(O_N \setminus E \searrow O \setminus E\) since \(O_N\setminus E \mathrel{\vcenter{:}}= O_N \cap E^c \supseteq O_{N+1} \cap E^c\) for all \(N\), and again \(m_*(O_1 \setminus E) < \infty\).
      • So it’s valid to apply continuity of measure from above: \begin{align*} m_*(O\setminus E) &= \lim_{N\to\infty} m_*(O_N\setminus E) \\ &\leq \lim_{N\to \infty} m_*(G_N\setminus E) \\ &= \lim_{N\to\infty} {1\over N} = 0 ,\end{align*} where the inequality uses subadditivity on \(\cap_{n=1}^N G_n \subseteq G_N\)
    • Suppose \(E\) is unbounded.
      • Write \(E^k = E \cap[k, k+1]^d \subset {\mathbb{R}}^d\) as the intersection of \(E\) with an annulus, and note that \(E = {\coprod}_{k\in {\mathbb{N}}} E_k\).
      • Each \(E_k\) is bounded, so apply the previous case to obtain \(O_k \supseteq E_k\) with \(m(O_k\setminus E_k) = 0\).
      • So write \(O_k = E_k {\coprod}N_k\) where \(N_k \mathrel{\vcenter{:}}= O_k \setminus E_k\) is a null set.
      • Define \(O = \cup_{k\in {\mathbb{N}}} O_k\), note that \(E\subseteq O\).
      • Now note \begin{align*} O\setminus E &= \qty{{\coprod}_k O_k}\setminus \qty{{\coprod}_K E_k} \\ &\subseteq {\coprod}_k \qty{O_k \setminus E_k} \\ \implies m_*(O\setminus E) &\leq m_*\qty{{\coprod}\qty{O_k \setminus E_k} } = 0 ,\end{align*} since any countable union of null sets is again null.
    • So \(O\supseteq E\) with \(m(O\setminus E) = 0\).
  • Theorem: since \(E\) is measurable, \(E^c\) is measurable
    • Proof: It suffices to write \(E^c\) as the union of two measurable sets, \(E^c = S \cup(E^c - S)\), where \(S\) is to be determined.
    • We’ll produce an \(S\) such that \(m_*(E^c - S) = 0\) and use the fact that any subset of a null set is measurable.
    • Since \(E\) is measurable, for every \(\varepsilon> 0\) there exists an open \({\mathcal{O}}_\varepsilon\supseteq E\) such that \(m_*({\mathcal{O}}_\varepsilon\setminus E) < \varepsilon\).
    • Take the sequence \(\left\{{\varepsilon_n \mathrel{\vcenter{:}}={1\over n}}\right\}\) to produce a sequence of sets \({\mathcal{O}}_n\).
    • Note that each \({\mathcal{O}}_n^c\) is closed and \begin{align*} {\mathcal{O}}_n \supseteq E \iff {\mathcal{O}}_n^c \subseteq E^c .\end{align*}
    • Set \(S \mathrel{\vcenter{:}}=\cup_n {\mathcal{O}}_n^c\), which is a union of closed sets, thus an \(F_\sigma\) set, thus Borel, thus measurable.
    • Note that \(S\subseteq E^c\) since each \({\mathcal{O}}_n \subseteq E^c\).
    • Note that \begin{align*} E^c\setminus S &\mathrel{\vcenter{:}}= E^c \setminus \qty{\cup_{n=1}^\infty {\mathcal{O}}_n^c} \\ &\mathrel{\vcenter{:}}= E^c \cap\qty{\cup_{n=1}^\infty {\mathcal{O}}_n^c}^c \quad\text{definition of set minus} \\ &= E^c \cap\qty{\cap_{n=1}^\infty {\mathcal{O}}_n}^c \quad \text{De Morgan's law}\\ &= E^c \cup\qty{\cap_{n=1}^\infty {\mathcal{O}}_n} \\ &\mathrel{\vcenter{:}}=\qty{ \cap_{n=1}^\infty {\mathcal{O}}_n} \setminus E \\ & \subseteq {\mathcal{O}}_N \setminus E \quad \text{for every } N\in {\mathbb{N}} .\end{align*}
    • Then by subadditivity, \begin{align*} m_*(E^c\setminus S) \leq m_*({\mathcal{O}}_N \setminus E) \leq {1\over N} \quad \forall N \implies m_*(E^c\setminus S) = 0 .\end{align*}
    • Thus \(E^c\setminus S\) is measurable.

4.4.1 Indirect Proof

4.4.2 Direct Proof (Todo)

4.5 Spring 2018 # 1 \(\done\)

Define \begin{align*} E:=\left\{x \in \mathbb{R}:\left|x-\frac{p}{q}\right|

Prove that \(m(E) = 0\).

(Click to expand)
  • Borel-Cantelli: If \(\left\{{E_k}\right\}_{k\in{\mathbb{Z}}}\subset 2^{\mathbb{R}}\) is a countable collection of Lebesgue measurable sets with \(\sum_{k\in {\mathbb{Z}}} m(E_k) < \infty\), then almost every \(x\in {\mathbb{R}}\) is in at most finitely many \(E_k\).
    • Equivalently (?), \(m(\limsup_{k\to\infty} E_k) = 0\), where \(\limsup_{k\to\infty} E_k = \cap_{k=1}^\infty \cup_{j\geq k} E_j\), the elements which are in \(E_k\) for infinitely many \(k\).

4.6 Fall 2017 # 2 \(\done\)

Let \(f(x) = x^2\) and \(E \subset [0, \infty) \mathrel{\vcenter{:}}={\mathbb{R}}^+\).

  1. Show that \begin{align*} m^*(E) = 0 \iff m^*(f(E)) = 0. \end{align*}

  2. 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.

(Click to expand)
  • ?

4.6.1 a

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

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

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

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

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

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

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

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

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

4.6.2 b

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

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

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

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

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

4.7 Spring 2017 # 2 \(\done\)

4.7.1 a

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*}

4.7.2 b

Let \(E \subset {\mathbb{R}}\) be a measurable set such that \begin{align*} \int_{E} x^{2} ~d m=0. \end{align*} Show that \(m(E) = 0\).

(Click to expand)
  • Absolute continuity of measures: \(\lambda \ll \mu \iff E\in\mathcal{M}, \mu(E) = 0 \implies \lambda(E) = 0\).
  • Radon-Nikodym: if \(\lambda \ll \mu\), then there exists a measurable function \({\frac{\partial \lambda}{\partial \mu}\,} \mathrel{\vcenter{:}}= f\) where \(\lambda(E) = \int_E f \,d\mu\).
  • Chebyshev’s inequality: \begin{align*} A_c \mathrel{\vcenter{:}}=\left\{{ x\in X {~\mathrel{\Big|}~}{\left\lvert {f(x)} \right\rvert} \geq c }\right\} \implies \mu(A_c) \leq c^{-p} \int_{A_c} {\left\lvert {f} \right\rvert}^p \,d\mu \quad \forall 0 < p < \infty .\end{align*}

4.7.3 a

4.7.4 b

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

4.8 Fall 2016 # 4 \(\done\)

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 \(\mu(G) = 0\).

(Click to expand)
  • ?

4.9 Spring 2016 # 3 \(\work\)

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*}

4.10 Spring 2016 # 5 \(\work\)

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*}

4.11 Fall 2015 # 2 \(\work\)

Let \(f: {\mathbb{R}}\to {\mathbb{R}}\) be Lebesgue measurable.

  1. Show that there is a sequence of simple functions \(s_n(x)\) such that \(s_n(x) \to f(x)\) for all \(x\in {\mathbb{R}}\).
  2. Show that there is a Borel measurable function \(g\) such that \(g = f\) almost everywhere.

4.12 Spring 2015 # 3 \(\work\)

Let \(\mu\) be a finite Borel measure on \({\mathbb{R}}\) and \(E \subset {\mathbb{R}}\) Borel. Prove that the following statements are equivalent:

  1. \(\forall \varepsilon > 0\) there exists \(G\) open and \(F\) closed such that \begin{align*} F \subseteq E \subseteq G \quad \text{and} \quad \mu(G\setminus F) < \varepsilon. \end{align*}
  2. There exists a \(V \in G_\delta\) and \(H \in F_\sigma\) such that \begin{align*} H \subseteq E \subseteq V \quad \text{and}\quad \mu(V\setminus H) = 0 \end{align*}

4.13 Spring 2014 # 3 \(\work\)

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.

4.14 Spring 2014 # 4 \(\work\)

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.

4.15 Spring 2017 # 1 \(\done\)

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 \(m(K)\).

(Click to expand)
  • Definition: \(A\) is nowhere dense \(\iff\) every interval \(I\) contains a subinterval \(S \subseteq A^c\).
    • Equivalently, the interior of the closure is empty, \(\qty{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}^\circ = \emptyset\).

Claim: \(K\) is compact.

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

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

4.16 Spring 2016 # 2 \(\work\)

Let \(0 < \lambda < 1\) and construct a Cantor set \(C_\lambda\) by successively removing middle intervals of length \(\lambda\).

Prove that \(m(C_\lambda) = 0\).

5 Measure Theory: Functions

5.1 Fall 2016 # 2 \(\done\)

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

  1. \(f(x) = g(x)\) almost everywhere, or
  2. There exists a measurable set \(E \subset [a, b]\) such that \begin{align*}] \int _{E} f(x) \, dx > \int _{E} g(x) \, dx \end{align*}
(Click to expand)
  • ?

5.2 Spring 2016 # 4 \(\work\)

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

  1. \(f\in L^1({\mathbb{R}})\).
  2. \(f\) is uniformly continuous.
  3. \(\lim _{|x| \to \infty} f(x) = 0\).

Hint: \begin{align*} \chi_{E \cap(E+x)}(y)=\chi_{E}(y) \chi_{E}(y-x) \end{align*}

6 Integrals: Convergence

6.1 Fall 2019 # 2 \(\done\)

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 \(\displaystyle\int_0^1 \cos(tx) dt\)

(Click to expand)
  • DCT
  • Bounding in the right place. Don’t evaluate the actual integral!

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*}

6.2 Spring 2020 # 5 \(\done\)

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*}

(Click to expand)
  • DCT
  • Passing limits through products and quotients

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:

6.3 Spring 2019 # 3 \(\done\)

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.

(Click to expand)
  • Definition of \(L^+\): space of measurable function \(X\to [0, \infty]\).
  • Fatou: For any sequence of \(L^+\) functions, \(\int \liminf f_n \leq \liminf \int f_n\).
  • Egorov’s Theorem: If \(E\subseteq {\mathbb{R}}^n\) is measurable, \(m(E) > 0\), \(f_k:E\to {\mathbb{R}}\) a sequence of measurable functions where \(\lim_{n\to\infty} f_n(x)\) exists and is finite a.e., then \(f_n\to f\) almost uniformly: for every \(\varepsilon>0\) there exists a closed subset \(F_\varepsilon\subseteq E\) with \(m(E\setminus F) < \varepsilon\) and \(f_n\to f\) uniformly on \(F\).

\(L^2\) 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:

6.4 Fall 2018 # 6 \(\done\)

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*}

(Click to expand)
  • ?

6.5 Fall 2018 # 3 \(\done\)

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*}

(Click to expand)
  • Mean Value Theorem
  • DCT

\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.

6.6 Spring 2018 # 5 \(\done\)

Suppose that

Show that \(\int f_{n} \rightarrow \int f\).

(Click to expand)
  • \(\int {\left\lvert {f_n - f} \right\rvert} \to \iff \int f_n = \int f\).
  • Fatou: \begin{align*} \int \liminf f_n \leq \liminf \int f_n \\ \int \limsup f_n \geq \limsup \int f_n .\end{align*}

6.7 Spring 2018 # 2 \(\done\)

Let \begin{align*} f_{n}(x):=\frac{x}{1+x^{n}}, \quad x \geq 0. \end{align*}

  1. Show that this sequence converges pointwise and find its limit. Is the convergence uniform on \([0, \infty)\)?

  2. Compute \begin{align*} \lim _{n \rightarrow \infty} \int_{0}^{\infty} f_{n}(x) d x \end{align*}

(Click to expand)
  • ?

6.7.1 a

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

6.7.2 b

6.8 Fall 2016 # 3 \(\done\)

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*}

(Click to expand)
  • \(C_c^\infty \hookrightarrow L^p\) is dense.
  • If \(f\)…?

6.9 Fall 2015 # 3 \(\work\)

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*}

6.10 Fall 2015 # 4 \(\work\)

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*}

7 Integrals: Approximation

7.1 Spring 2018 # 3 \(\done\)

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*}

(Click to expand)
  • \({\left\lVert {f} \right\rVert}_\infty \mathrel{\vcenter{:}}=\inf_t {\left\{{ t{~\mathrel{\Big|}~}m\qty{\left\{{x\in {\mathbb{R}}^n {~\mathrel{\Big|}~}f(x) > t}\right\}} = 0 }\right\} }\), i.e. this is the lowest upper bound that holds almost everywhere.

7.2 Spring 2018 # 4 \(\done\)

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.

(Click to expand)

7.2.1 Proof 1: Using Fourier Transforms

  • Weierstrass Approximation: A continuous function on a compact set can be uniformly approximated by polynomials.


7.2.2 Alternative Proof

  • \(C^1([0, 1])\) is dense in \(L^2([0, 1])\)
  • Polynomials are dense in \(L^p(X, \mathcal{M}, \mu)\) for any \(X\subseteq {\mathbb{R}}^n\) compact and \(\mu\) a finite measure, for all \(1\leq p < \infty\).
    • Use Weierstrass Approximation, then uniform convergence implies \(L^p(\mu)\) convergence by DCT.

7.3 Spring 2015 # 2 \(\work\)

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 \(k\in {\mathbb{Z}}\).

7.4 Fall 2014 # 4 \(\work\)

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*}

8 \(L^1\)

8.1 Spring 2020 # 3 \(\done\)

  1. 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 \({\left\lvert {x} \right\rvert}\to \infty\).

  2. Prove that if \(f\in L^1([1, \infty])\) and is decreasing, then \(\lim_{x\to\infty}f(x) =0\) and in fact \(\lim_{x\to \infty} xf(x) = 0\).

  3. If \(f: [1, \infty) \to [0, \infty)\) is decreasing with \(\lim_{x\to \infty} xf(x) = 0\), does this ensure that \(f\in L^1([1, \infty))\)?

(Click to expand)
  • Limits
  • Cauchy Criterion for Integrals: \(\int_a^\infty f(x) \,dx\) converges iff for every \(\varepsilon>0\) there exists an \(M_0\) such that \(A,B\geq M_0\) implies \({\left\lvert {\int_A^B f} \right\rvert} < \varepsilon\), i.e. \({\left\lvert {\int_A^B f} \right\rvert} \overset{A\to\infty}\to 0\).
  • Integrals of \(L^1\) functions have vanishing tails: \(\int_{N}^\infty {\left\lvert {f} \right\rvert} \overset{N\to\infty}\to 0\).
  • Mean Value Theorem for Integrals: \(\int_a^b f(t)\, dt = (b-a) f(c)\) for some \(c\in [a, b]\).

8.1.1 a

Stated integral equality:

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

8.1.2 b Solution 1 (“Trick”) Solution 2 (Variation on the Trick) Solution 3 (Contradiction)

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

Showing \(xf(x) \overset{x\to \infty}\to 0\). Solution 4 (Akos’s Suggestion)

For \(x\geq 1\), \begin{align*} {\left\lvert {xf(x)} \right\rvert} = {\left\lvert { \int_x^{2x} f(x) \, dt } \right\rvert} \leq \int_x^{2x} {\left\lvert {f(x)} \right\rvert} \, dt \leq \int_x^{2x} {\left\lvert {f(t)} \right\rvert}\, dt \leq \int_x^{\infty} {\left\lvert {f(t)} \right\rvert} \,dt \overset{x\to\infty}\to 0 \end{align*} where we’ve used Solution 5 (Peter’s)

8.1.3 c

8.2 Fall 2019 # 5. \(\done\)

8.2.1 a

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*}

8.2.2 b

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*}

  1. Prove that \(\left\|\mathcal{A}_{h} f\right\|_{1} \leq\|f\|_{1}\) for all \(h > 0\).

  2. Prove that \(\mathcal{A}_h f \to f\) in \(L^1({\mathbb{R}})\) as \(h \to 0^+\).

(Click to expand)
  • Continuity in \(L^1\) (recall that DCT won’t work! Notes 19.4, prove it for a dense subset first).
  • Lebesgue differentiation in 1-dimensional case. See HW 5.6.

8.2.3 a

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

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

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

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

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

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

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

8.2.4 b

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

and (rough sketch)

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

8.3 Fall 2017 # 3 \(\done\)

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*}

(Click to expand)
  • From homework: \(E\) is Lebesgue measurable iff there exists a finite union of closed cubes \(A\) such that \(m(E\Delta A) < \varepsilon\).

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 \(L^1\).

8.4 Spring 2015 # 4 \(\work\)

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 \(f \in L^1({\mathbb{R}}^2)\).

8.5 Fall 2014 # 3 \(\work\)

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*}

8.6 Spring 2014 # 1 \(\work\)

  1. Give an example of a continuous \(f\in L^1({\mathbb{R}})\) such that \(f(x) \not\to 0\) as\({\left\lvert {x} \right\rvert} \to \infty\).

  2. Show that if \(f\) is uniformly continuous, then \begin{align*} \lim_{{\left\lvert {x} \right\rvert} \to \infty} f(x) = 0. \end{align*}

9 Fubini-Tonelli

9.1 Spring 2020 # 4 \(\done\)

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*}

(Click to expand)
  • Tonelli: non-negative and measurable yields measurability of slices and equality of iterated integrals
  • Fubini: \(f(x, y) \in L^1\) yields integrable slices and equality of iterated integrals
  • F/T: apply Tonelli to \({\left\lvert {f} \right\rvert}\); if finite, \(f\in L^1\) and apply Fubini to \(f\)

\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.

9.2 Spring 2019 # 4 \(\done\)

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:

  1. \(f\) is a Lebesgue measurable function on \({\mathbb{R}}^n \iff \mathcal A\) is a Lebesgue measurable subset of \({\mathbb{R}}^{n+1}\)

  2. 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*}

(Click to expand)
  • See S&S p.82.

9.2.1 a



9.2.2 b

9.3 Fall 2018 # 5 \(\done\)

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*}

(Click to expand)
  • Claim: If \(E\subseteq {\mathbb{R}}^a \times{\mathbb{R}}^b\) is a measurable set, then for almost every \(y\in {\mathbb{R}}^b\), the slice \(E^y\) is measurable and \begin{align*} m(E) = \int_{{\mathbb{R}}^b} m(E^y) \,dy .\end{align*}
    • Set \(g = \chi_E\), which is non-negative and measurable, so apply Tonelli.
    • Conclude that \(g^y = \chi_{E^y}\) is measurable, the function \(y\mapsto \int g^y(x)\, dx\) is measurable, and \(\int \int g^y(x)\,dx \,dy = \int g\).
    • But \(\int g = m(E)\) and \(\int\int g^y(x) \,dx\,dy = \int m(E^y)\,dy\).

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

9.4 Fall 2015 # 5 \(\work\)

Let \(f, g \in L^1({\mathbb{R}})\) be Borel measurable.

  1. Show that
  1. Show that \(f\ast g \in L^1({\mathbb{R}})\) and \begin{align*} \|f * g\|_{1} \leq \|f\|_{1} \|g\|_{1} \end{align*}

9.5 Spring 2014 # 5 \(\work\)

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*}

10 \(L^2\) and Fourier Analysis

10.1 Spring 2020 # 6 \(\done\)

10.1.1 a

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*}

10.1.2 b

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 \(f\in L^2([0, 1])\).

(Click to expand)
  • For \(e_n(x) \mathrel{\vcenter{:}}= e^{2\pi i n x}\), the set \(\left\{{e_n}\right\}\) is an orthonormal basis for \(L^2([0, 1])\).
  • For any orthonormal sequence in a Hilbert space, we have Bessel’s inequality: \begin{align*} \sum_{k=1}^{\infty}\left|\left\langle x, e_{k}\right\rangle\right|^{2} \leq\|x\|^{2} .\end{align*}
  • When \(\left\{{e_n}\right\}\) is a basis, the above is an equality (Parseval)
  • Arguing uniform convergence: since \(\left\{{\widehat{f}(n)}\right\} \in \ell^1({\mathbb{Z}})\), we should be able to apply the \(M\) test.

10.1.3 a

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

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

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

10.2 Fall 2017 # 5 \(\done\)

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:

  1. Each \(f\ast K_j\) is smooth and compactly supported.

  2. \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*}

(Click to expand)
  • ?

10.2.1 a

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

Silly Proof:

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

Actual proof:

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

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

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

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

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

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

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

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

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

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

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

10.2.2 b

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

We now split the integral up into pieces.

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

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

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

10.3 Spring 2017 # 5 \(\work\)

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 \({\mathbb{R}}\).

10.4 Spring 2015 # 6 \(\work\)

Let \(f \in L^1({\mathbb{R}})\) and \(g\) be a bounded measurable function on \({\mathbb{R}}\).

  1. Show that the convolution \(f\ast g\) is well-defined, bounded, and uniformly continuous on \({\mathbb{R}}\).
  2. Prove that one further assumes that \(g \in C^1({\mathbb{R}})\) with bounded derivative, then \(f\ast g \in C^1({\mathbb{R}})\) and \begin{align*} \frac{d}{d x}(f * g)=f *\left(\frac{d}{d x} g\right) \end{align*}

10.5 Fall 2014 # 5 \(\work\)

  1. 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*}

  2. 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*}

11 Functional Analysis: General

11.1 Fall 2019 # 4 \(\done\)

Let \(\{u_n\}_{n=1}^∞\) be an orthonormal sequence in a Hilbert space \(\mathcal{H}\).

11.1.1 a

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*}

11.1.2 b

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*}

(Click to expand)
  • Bessel’s Inequality
  • Pythagoras
  • Surjectivity of the Riesz map
  • Parseval’s Identity
  • Trick – remember to write out finite sum \(S_N\), and consider \({\left\lVert {x - S_N} \right\rVert}\).

11.1.3 a

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

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

11.1.4 b

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

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

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

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

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

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

11.2 Spring 2019 # 5 \(\done\)

11.2.1 a

Show that \(L^2([0, 1]) ⊆ L^1([0, 1])\) and argue that \(L^2([0, 1])\) in fact forms a dense subset of \(L^1([0, 1])\).

11.2.2 b

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:

  1. Establish the existence of a function \(g ∈ L^2([0, 1])\) which represents \(Λ\) in the sense that \begin{align*} Λ(f ) = f (x)g(x) dx \text{ for all } f ∈ L^2([0, 1]). \end{align*}

Hint: You may use, without proof, the Riesz Representation Theorem for \(L^2([0, 1])\).

  1. Argue that the \(g\) obtained above must in fact belong to \(L^∞([0, 1])\) and represent \(Λ\) in the sense that \begin{align*} \Lambda(f)=\int_{0}^{1} f(x) \overline{g(x)} d x \quad \text { for all } f \in L^{1}([0,1]) \end{align*} with \begin{align*} \|g\|_{L^{\infty}([0,1])} = \|\Lambda\|_{L^{1}([0,1])^\vee} \end{align*}
(Click to expand)
  • 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*}

11.2.3 a

11.2.4 b

Let \(\Lambda \in L^1(X)^\vee\) be arbitrary. 1: Existence of \(g\) Representing \(\Lambda\).

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

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

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*} 2: \(g\) is in \(L^\infty\)

11.3 Spring 2016 # 6 \(\work\)

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*}

11.4 Spring 2015 # 5 \(\work\)

Let \(\mathcal H\) be a Hilbert space.

  1. Let \(x\in \mathcal H\) and \(\left\{{u_n}\right\}_{n=1}^N\) be an orthonormal set. Prove that the best approximation to \(x\) in \(\mathcal H\) by an element in \({\operatorname{span}}_{\mathbb{C}}\left\{{u_n}\right\}\) is given by \begin{align*} \widehat{x} \mathrel{\vcenter{:}}=\sum_{n=1}^N {\left\langle {x},~{u_n} \right\rangle}u_n. \end{align*}
  2. Conclude that finite dimensional subspaces of \(\mathcal H\) are always closed.

11.5 Fall 2015 # 6 \(\work\)

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*}

11.6 Fall 2014 # 6 \(\work\)

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*}

12 Functional Analysis: Banach Spaces

12.1 Spring 2019 # 1 \(\done\)

Let \(C([0, 1])\) denote the space of all continuous real-valued functions on \([0, 1]\).

  1. Prove that \(C([0, 1])\) is complete under the uniform norm \({\left\lVert {f} \right\rVert}_u := \displaystyle\sup_{x\in [0,1]} |f (x)|\).

  2. Prove that \(C([0, 1])\) is not complete under the \(L^1{\hbox{-}}\)norm \({\left\lVert {f} \right\rVert}_1 = \displaystyle\int_0^1 |f (x)| ~dx\).

(Click to expand)
  • ?

12.1.1 a

12.1.2 b

12.2 Spring 2017 # 6 \(\done\)

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*}

(Click to expand)

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*}

12.3 Fall 2017 # 6 \(\done\)

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.

(Click to expand)
  • ?

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*}

13 Fall 2020

13.1 1

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*}

13.2 2

13.2.1 a

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*}

13.2.2 b

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.

13.3 3

Let \(f\) be a non-negative Lebesgue measurable function on \([1, \infty)\).

13.3.1 a

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 \(1\leq a < b <\infty\).

13.3.2 b

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*}

13.4 4

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.

13.5 5

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*}

13.6 a

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*}

13.6.1 b

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*}

14 Bibliography