Some selected problems from Gerald Folland’s Real Analysis: Modern Techniques and Their Applications that I’ve done.

Chapter 1

Problem 14

Problem: If \(\mu\) is a semifinite measure and \(\mu(E) = \infty\), prove that for any \(C > 0\) there exists \(F\subseteq E\) such that \(C < \mu(F) < \infty\).

Solution: Let \(\alpha := \sup\{\mu(F):F\subseteq E,\, \mu(F)<\infty\}\). It suffices to show that \(\alpha = \infty\). Suppose \(\alpha < \infty\). Choose \(G_n\) such that

\begin{equation*} \alpha -\frac{1}{n} < \mu(G_n) < \infty. \end{equation*}

Note that \(\mu(G_1\cup\dots\cup G_n) < \infty\), so

\begin{equation*} \alpha - \frac{1}{n} < \mu(G_1\cup\dots\cup G_n) \leq \alpha. \end{equation*}

Let \(G:=\bigcup_{n=1}^\infty G_n\) so that

\begin{equation*} \mu(G) = \lim_{n\to\infty}\mu(G_1\cup\dots\cup G_n) = \lim_{n\to\infty}(\alpha-\tfrac{1}{n}) = \alpha. \end{equation*}

Thus, \(\mu(G) = \alpha\).

Let \(E’ = E \,\backslash\, G\) so \(\mu(E’) = \infty\). Because \(\mu^*\) is semifinite, there exists \(F\subseteq E’\) with \(0<\mu(F) <\infty\). Thus,

\begin{equation*} \alpha = \mu(G) < \mu(G) + \mu(F) = \mu(G\cup F) < \infty, \end{equation*}

but this contradictions the definition of \(\alpha\). Thus, \(\alpha = \infty\).

Problem 17

Problem: If \(\mu^*\) is an outer measure on \(X\), and \(\{A_j\}_{j=1}^\infty\) is a collection of disjoint \(\mu^* \)-measurable sets, prove that for any \(E\subseteq X\), \(\mu^*\left(E\cap\bigcup_{j=1}^\infty A_j\right) = \sum_{j=1}^\infty \mu^*(E\cap A_j)\).

Solution: Firstly, note that if \(A_1\cap A_2=\varnothing\) and both are \(\mu^*\)-measurable,

\begin{align*} \mu^*(E\cap (A_1\cup A_2)) &= \mu^*((E\cap (A_1\cup A_2))\cap A_1) +\mu^*((E\cap (A_1\cup A_2))\cap A_1^\complement) \\ &= \mu^*(E\cap A_1) + \mu^*(E\cap A_2). \end{align*}

By induction, we have the result for finite collections of disjoint \(\mu^*\)-measurable sets.

Then, note that for all \(k\in\mathbb{N}\),

\begin{align*} \sum_{j=1}^k\mu^*(E\cap A_j) = \mu^*\left(E\cap\bigcup_{j=1}^k A_j\right) \leq \mu^*\left(E\cap\bigcup_{j=1}^\infty A_j\right) \leq \sum_{j=1}^\infty\mu^*(E\cap A_j). \end{align*}

Letting \(k\to\infty\) we obtain the desired result.

Problem 18

Problem: Let $\mathcal{A}\subset P(X)$ be an algebra, $\mathcal{A}_\sigma$ the collection of countable unions of sets in $\mathcal{A}$, and $\mathcal{A}_{\sigma\delta}$ the collection of countable intersections of sets in $\mathcal{A}_\sigma$. Let $\mu_{0}$ be a premeasure on $\mathcal{A}$ and $\mu^*$ the induced outer measure.

1. For any $E\subset X$ and $\varepsilon > 0$ there exists $A\in \mathcal{A}_\sigma$ with $E\subset A$ and $\mu^*(A) \leq \mu^*(E) + \varepsilon$.
2. If $\mu^{*}(E) < \infty$, then $E$ is $\mu^{*}$-measurable if and only if there exists $B\in A_{\sigma\delta}$ with $E\subset B$ and $\mu^{*}(B\setminus E) = 0$.
3. If $\mu_0$ is $\sigma$-finite, the restriction $\mu^{*}(E) < \infty$ in (b) is superfluous.

Solution:

1. From the definition, \begin{align*} \mu^*(E) = \inf \left\{\sum_{n=1}^\infty \mu_0(A_n) : A_n\in\mathcal{A}, \, E \subseteq \bigcup_{n=1}^\infty A_n\right\}. \end{align*} Simply applying the definition of infimum yields a collection $\{A_n\}_{n=1}^\infty\subseteq\mathcal{A}$ with $E \subseteq \bigcup_{n=1}^\infty A_n \in\mathcal{A}_\sigma$ and $\sum_{n=1}^\infty \mu_0(A_n) \leq \mu^*(E) + \varepsilon$. Then, by Proposition 1.13, we have \begin{align*} \mu^*\left(\bigcup_{n=1}^\infty A_n\right) \leq \sum_{n=1}^\infty \mu^*(A_n) =\sum_{n=1}^\infty \mu_0(A_n) \leq \mu^*(E) + \varepsilon. \end{align*}

2. Suppose $E$ is $\mu^*$-measurable.

   From part (a), choose $B_n\in A_{\sigma}$ such that $E\subseteq B_n$ and $\mu^*(B_n)\leq\mu^*(E)+\frac{1}{n}$. Then, \begin{align*} \mu^*(E)+\frac{1}{n}\geq \mu^*(B_n) = \mu^*(B_n\cap E) + \mu^*(B_n\cap E^\complement) = \mu^*(E) + \mu^*(B_n \setminus E). \end{align*} Because $\mu^*(E)<\infty$, we have $\mu^*(B_n \setminus E)\leq \frac{1}{n}$. Let $B = \bigcap_{n=1}^\infty B_n$, so that $\mu^*(B\setminus E) \leq \mu^*(B_n\setminus E) \leq \frac{1}{n}$ for all $n$. Thus, $\mu^*(B\setminus E) =0$, and since $E\subseteq B\in A_{\sigma\delta}$, we have completed the proof of the forwards direction.

   Suppose there exists $B\in A_{\sigma\delta}$ such that $E\subseteq B$ and $\mu^*(B\setminus E) = 0$. Let $A\subseteq X$ be any set. Note that $\mu^*(A\cap E) \leq \mu^*(A\cap B)$. Also, \begin{align*} \mu^*(A\cap B) &= \mu^*(A\cap E \cup A\cap (B\setminus E)) \leq \mu^*(A\cap E) + \mu^*(A\cap (B\setminus E)) = \mu^*(A\cap E). \end{align*} Thus, $\mu^*(A\cap E) = \mu^*(A\cap B)$.

   Similarly, \begin{align*} \mu^*(A\cap E^\complement) &= \mu^*(A\cap B^\complement \cup A\cap (E^\complement\setminus B^\complement)) \leq \mu^*(A\cap B^\complement) + \mu^*(A\cap (B\setminus E)) = \mu^*(A\cap B^\complement). \end{align*} Since $\mu^*(A\cap B^\complement)\leq \mu^*(A\cap E^\complement)$, we have $\mu^*(A\cap B^\complement) = \mu^*(A\cap E^\complement)$. We also used the fact that $E^\complement\setminus B^\complement = B\setminus E$.

   Lastly, by Proposition 1.13 every set in $\mathcal{A}$ is $\mu^*$-measurable. By Theorem 1.11 (Carathéodory's Theorem), the set of $\mu^*$-measurable sets forms a $\sigma$-algebra. Thus, $B$ is $\mu^*$-measurable. Combining all of the results, \begin{align*} \mu^*(A) = \mu^*(A\cap B) + \mu^*(A\cap B^\complement) = \mu^*(A\cap E) + \mu^*(A\cap E^\complement). \end{align*} Since $A$ was arbitrary, we have that $E$ is $\mu^*$-measurable. This completes the proof of the reverse direction.

3. Let $\{B_n\}_{n=1}^\infty$ be a collection of sets in $\mathcal{A}$ such that $X = \bigcup_{n=1}^\infty B_n$ and $\mu_0(B_n)<\infty$ for all $n$, where $X$ is the ambient space.

   Now, suppose $E$ is $\mu^*$-measurable. Define $E_n:=E\cap \bigcup_{i=1}^n B_i$ so that \begin{align*} \mu^*(E_n) = \mu^*\left(E\cap \bigcup_{i=1}^n B_i\right) \leq \mu^*\left( \bigcup_{i=1}^n B_i\right) \leq \sum_{i=1}^n\mu^*(B_i) = \sum_{i=1}^n\mu_0(B_i) < \infty.\end{align*} Note that we use Proposition 1.13 in the last equality.

   By part (a), choose $A_n^{(k)}\in\mathcal{A}_\sigma$ so that $E_n\subseteq A_n^{(k)}$ and \begin{align*} \mu^*(A_n^{(k)}) \leq \mu^*(E_k)+ \frac{1}{2^{k+n}}.\end{align*} By Theorem 1.11 (Carathéodory's Theorem) and Proposition 1.13, $E_n$ is $\mu^*$-measurable, Repeating a computation from part (b), we find that $\mu^*(A_n^{(k)} \setminus E_k) \leq \frac{1}{2^{k+n}}$. Then, $E \subseteq \bigcup_{k=1}^\infty A_n^{(k)}$, and \begin{align*} \mu^*\left(\bigcup_{k=1}^\infty A_n^{(k)}\setminus E\right) &= \mu^*\left(\bigcup_{k=1}^\infty (A_n^{(k)}\setminus E)\right) \\ &\leq \sum_{k=1}^\infty \mu^*\left(A_n^{(k)}\setminus E\right) \\ &\leq \sum_{k=1}^\infty \mu^*\left(A_n^{(k)}\setminus E_k\right) \\ &\leq \frac{1}{2^n}\sum_{k=1}^\infty \frac{1}{2^{k}} \\ &= \frac{1}{2^n}. \end{align*} Define $A = \bigcap_{n=1}^\infty \bigcup_{k=1}^\infty A_n^{(k)}$, so that \begin{align*} \mu^*(A\setminus E) \leq \mu^*\left(\bigcup_{k=1}^\infty A_n^{(k)}\setminus E\right) \leq \frac{1}{2^n}\end{align*} for all $n$. Thus, $\mu^*(A\setminus E) = 0$. This proves the forwards direction.

   Suppose there exists $G\in \mathcal{A}_{\sigma\delta}$ such that $E\subseteq G$ and $\mu^*(G\setminus E) = 0$. Let $A\subseteq X$ be any set. Define $E_n = E\cap\bigcup_{i=1}^n B_i$, and $G_n = G\cap\bigcup_{i=1}^n B_i$. Note that $G_n\in\mathcal{A}_{\sigma\delta}$ and that for any sets $A,B$, and $C$, \begin{align*} (A\cap C)\setminus (B\cap C) = (A\cap C)\cap (B\cap C)^\complement = (A\cap B^\complement\cap C)\cup (A\cap C^\complement\cap C)=(A\setminus B)\cap C.\end{align*} Applying this fact, we find \begin{align*} \mu^*(G_n\setminus E_n) &= \mu^*\left((G\setminus E)\cap\bigcup_{i=1}^n B_i \right) \leq \mu^*(G\setminus E) = 0. \end{align*} By part (b), $E_n$ is $\mu^*$-measurable. Since $E = \bigcup_{n=1}^\infty E_n$ and from Theorem 1.11 (Carathéodory's Theorem), we have that $E$ is $\mu^*$-measurable. This completes the proof of the reverse direction.

Chapter 5

AKA: I got bored with measure theory AKA functional ``analysis’’ (algebra).

Problem 57

Problem: Suppose $H$ is a Hilbert space and $T\in L(H,H)$.

1. There exists a unique $T^*\in L(H,H)$ such that $\langle Tx, y\rangle = \langle x, T^*y \rangle$ for all $x,y\in H$. $T^*$ is called the adjoint of $T$
2. $\| T^*\| = \|T\|$, $T^** = T$, $\|T^*T\| = \|T\|^2$ $T\mapsto T^*$ is conjugate-linear, and $(ST)^* = T^*S^*$.
3. $(\im T )^\perp = \ker T^*$ and $(\ker T)^\perp = \mathrm{cl}(\im T^*)$
4. $T$ is unitary if and only if $T$ is invertible and $T^{-1} = T^*$.

Solution:

1. Define $T^\top:H^*\to H^*$ by $T^\topf(x) = f(Tx)$. Let $V:H\to H^*$ be the conjugate linear isomorphism. Then $T^* = V^{-1}T^\top V$ and \begin{align*} \langle x, T^*y\rangle = \langle x, V^{-1}T^\dag V y\rangle = \langle x, V^{-1}(Vy\circ T)\rangle = (Vy\circ T)(x) = \langle Tx, y\rangle . \end{align*}

   Suppose $T^*_1, T^*_2$ are such that for all $x,y\in H$, \begin{align*} \langle Tx, y\rangle = \langle x, T_1^*y\rangle = \langle x, T_2^*y\rangle. \end{align*} Then, \begin{align*} 0 = \langle Tx, y\rangle -\langle Tx, y\rangle = \langle x, T_1^*y\rangle - \langle x, T_2^*y\rangle = \langle x, (T_1^*-T_2^*)y\rangle. \end{align*} Since $x$ was arbitrary, $(T_1^*-T_2^*)y = 0$. But since $y$ was also arbitrary, this implies that $T_1^*-T_2^* = 0$.

2. Note that \begin{align*} \langle T^*x, y\rangle = \overline{\langle y,T^*x\rangle} = \overline{\langle Ty,x\rangle} = \langle x,Ty\rangle. \end{align*} Thus, $(T^*)^* = T$.

   Then, \begin{align*} \|Tx\|^2 = \langle Tx, Tx\rangle = \langle x, T^*Tx\rangle \leq \|x\|\|T^*Tx\|\leq \|T^*\|\|x\|\cdot \|Tx\| \end{align*} Thus, $\|Tx\|\leq \|T^*\|\|x\|$, so $\|T\|\leq \|T^*\|$. Using the fact that $(T^*)^* = T$, we see that $\|T\| = \|T^*$.

   Also, \begin{align*} \|T^*Tx\|^2 = \langle T^*Tx, T^*Tx\rangle = \langle TT^*Tx, Tx\rangle \leq \|T\|^2\|T^*Tx\|\|x\|. \end{align*} Thus, $\|T^*T\|\leq\|T\|^2$. Moreover, \begin{align*} \|Tx\|^2 = \langle Tx, Tx\rangle = \langle x, T^*Tx\rangle \leq \|T^*T\|\|x\|^2. \end{align*} Thus, $\|T\|^2\leq \|T^*T\|$ so $\|T\|^2 = \|T^*T\|$.

   Also, \begin{align*} \langle x, (\overline{a}S^*+\overline{b}T^*)y\rangle = \langle x,\overline{a}S^*y\rangle + \langle x, \overline{b}T^*y\rangle = \langle \overline{a}x,S^*y\rangle + \langle \overline{b}x, T^*y\rangle &= \langle \overline{a}Sx, y\rangle + \langle \overline{b}Tx, y\rangle \\ &= \langle (aS+bT)x, y\rangle. \end{align*} Thus, $({a}S+{b}T)^* = \overline{a}S^*+\overline{b}T^*$.

   Lastly, \begin{align*} \langle x, T^*S^*y\rangle = \langle Tx, S^*y\rangle = \langle STx, y\rangle, \end{align*} so $(ST)^* = T^*S^*$.

3. Note that \begin{align*} R(T)^\perp = \left\{ y\in H : \langle Tx, y\rangle = 0 \ \forall \ x\in H \right\} &= \left\{ y\in H : \langle x, T^*y\rangle = 0 \ \forall \ x\in H \right\} \\ &= \left\{ y\in H : T^*y = 0 \right\} \\ &= N(T^*). \end{align*} We used the fact that $\langle x, y\rangle =0$ for all $x$ if and only if $y = 0$.

   Now, applying the previous result and the fact that $(T^*)^* = T$, we see that $R(T^*)^\perp = N(T)$. Thus, $(R(T^*)^\perp)^\perp = N(T)^\perp$. By problem 56, $(R(T^*)^\perp)^\perp$ is the smallest closed subspace containing $R(T^*)$. Since $R(T^*)$ is a subspace, and the closure of a subspace is still a subspace, we have that $(R(T^*)^\perp)^\perp = \overline{R(T^*)}$. Thus, $N(T)^\perp = \overline{R(T^*)}$.

4. Suppose $T$ is unitary. Then, for all $x,y$, \begin{align*} \langle x, y\rangle = \langle Tx, Ty\rangle = \langle x, T^*Ty\rangle. \end{align*} This implies that $0 = \langle x, T^*Ty -y\rangle$ for all $x$. Choosing $x = T^*Ty -y$ shows that $T^*Ty -y = 0$. Thus, $T^*Ty = y$ for all $y$, so $T^*T = I$. But then, since $T$ is unitary and thus invertible, \begin{align*} T^{-1} = T^*T T^{-1} = T^*. \end{align*}

   Now suppose $T^* = T^{-1}$. Then, \begin{align*} \langle Tx, Ty\rangle = \langle x, T^*Ty\rangle = \langle x,y\rangle. \end{align*} This shows that $T$ is unitary.

Problem 67 (Mean Ergodic Theorem)

Problem: Let $U$ be a unitary operator on a Hilbert space $H$. Let $M = {x \in H: Ux = x}$ (these are the fixed points of $U$). Show that if $S_n = \frac{1}{n}\sum_{k=0}^{n-1} U^k$, then $S_n$ converges to $\Pi_M$ (the orthogonal projection onto $M$) in the strong operator topology.

Solution: We need to first show that $\Pi_M$ exists.