The Principle of Uniform Boundedness

Here is another proof of the PUB using the Baire Category Theorem. With the notation of (14.1) -- that is, $\mathcal{X}$ is a Banach space, $\mathcal{Y}$ is a normed space, and $\mathcal{A} \subseteq \mathcal{B}(\mathcal{X}, \mathcal{Y})$ is such that for each $x$ in $\mathcal{X}$, $\sup \left\{ \left\| Ax\right\| : A \in \mathcal{A}\right\} < \infty$ -- let $B_n \equiv \left\{ x \in \mathcal{X} : \left\| Ax\right\| \leq n \text{ for all } A \text{ in } \mathcal{A}\right\}$. By hypothesis, $\bigcup_{n=1}^{\infty } B_n = \mathcal{X}$. Now apply the Baire Category Theorem.

If $1 < p < \infty$ and $\left\{ x_n\right\} \subseteq l^p$, then $\sum_{j=1}^{\infty } x_n(j) y(j) \to 0$ for every $y$ in $l^q$, $1/p + 1/q = 1$, if and only if $\sup_n \left\| x_n\right\|_{p} < \infty$ and $x_n(j) \to 0$ for every $j \geq 1$.

If $\left\{ x_n\right\} \subseteq l^1$, then $\sum_{j=1}^{\infty } x_n(j) y(j) \to 0$ for every $y$ in $c_0$ if and only if $\sup_n \left\| x_n\right\|_{1} < \infty$ and $x_n(j) \to 0$ for every $j \geq 1$.

If $(X, \Omega , \mu )$ is a measure space, $1 < p < \infty$, and $\left\{ f_n\right\} \subseteq L^p(X, \Omega , \mu )$, then $\int f_n g \, d\mu \to 0$ for every $g$ in $L^q(\mu )$, $1/p + 1/q = 1$, if and only if $\sup \left\{ \left\| f_n\right\|_{p} : n \geq 1\right\} < \infty$ and for every set $E$ in $\Omega$ with $\mu (E) < \infty$, $\int_E f_n \, d\mu \to 0$ as $n \to \infty$.

If $(X, \Omega , \mu )$ is a $\sigma$-finite measure space and $\left\{ f_n\right\}$ is a sequence in $L^1(X, \Omega , \mu )$, then $\int f_n g \, d\mu \to 0$ for every $g$ in $L^\infty (\mu )$ if and only if $\sup \left\{ \left\| f_n\right\|_{1} : n \geq 1\right\} < \infty$ and $\int_E f_n \, d\mu \to 0$ for every $E$ in $\Omega$.

Let $\mathcal{H}$ be a Hilbert space and let $\mathcal{E}$ be an orthonormal basis for $\mathcal{H}$. Show that a sequence $\left\{ h_n\right\}$ in $\mathcal{H}$ satisfies $\left\langle h_n, h \right\rangle \to 0$ for every $h$ in $\mathcal{H}$ if and only if $\sup \left\{ \left\| h_n\right\| : n \geq 1\right\} < \infty$ and $\left\langle h_n, e \right\rangle \to 0$ for every $e$ in $\mathcal{E}$.

If $X$ is locally compact and $\left\{ \mu_n\right\}$ is a sequence in $M(X)$, then $L(\mu_n) \to 0$ for every $L$ in $M(X)^*$ if and only if $\sup \left\{ \left\| \mu_n\right\| : n \geq 1\right\} < \infty$ and $\mu_n(E) \to 0$ for every Borel set $E$.

In (14.6) -- the Banach--Steinhaus Theorem, which states: if $\mathcal{X}$ and $\mathcal{Y}$ are Banach spaces and $\left\{ A_n\right\}$ is a sequence in $\mathcal{B}(\mathcal{X}, \mathcal{Y})$ with the property that for every $x$ in $\mathcal{X}$ there is a $y$ in $\mathcal{Y}$ such that $\left\| A_n x - y\right\| \to 0$, then there is an $A$ in $\mathcal{B}(\mathcal{X}, \mathcal{Y})$ such that $\left\| A_n x - Ax\right\| \to 0$ for every $x$ in $\mathcal{X}$ and $\sup \left\| A_n\right\| < \infty$ -- show that $\left\| A\right\| \leq \liminf \left\| A_n\right\|$.

If $(S, d)$ is a metric space and $\mathcal{X}$ is a normed space, say that a function $f : S \to \mathcal{X}$ is a Lipschitz function if there is a constant $M > 0$ such that $\left\| f(s) - f(t)\right\| \leq Md(s,t)$ for all $s, t$ in $S$. Show that if $f : S \to \mathcal{X}$ is a function such that for all $L$ in $\mathcal{X}^*$, $L \circ f : S \to \mathbb {F}$ is Lipschitz, then $f : S \to \mathcal{X}$ is a Lipschitz function.

Let $\mathcal{X}$ be a Banach space and suppose $\left\{ x_n\right\}$ is a sequence in $\mathcal{X}$ such that for each $x$ in $\mathcal{X}$ there are unique scalars $\left\{ \alpha_n\right\}$ such that $\lim_{n \to \infty } \left\| x - \sum_{k=1}^{n} \alpha_k x_k\right\| = 0$. Such a sequence is called a Schauder basis.