A Nice Diagonalization Argument

Kreyszig Functional Analysis

Theorem. Let $X$ be a separable Banach space and $M\subset X^{\prime }$ be a bounded subset of its dual space. Show that every sequence of elements in $M$ contains a subsequence which is weak* convergent to an element of $X^{\prime }$.

Proof. Note it suffices to show there is a subsequence which is strong convergent when treated as a sequence of operators. As $X$ is Banach, the limit operator (which is a functional) is guaranteed to be both linear and bounded, making it an element of $X^{\prime }$. Furthermore, as the scalar field ($\mathbb{R}$ or $\mathbb{C}$) is complete, it suffices to show a subsequence $m_i$ exists such that for any $x$, $m_{i}x$ is Cauchy.

Let $S=\{s_j\}$ be an at most countable dense subset of $X$. We first show there exists a subsequence $m_i$ such that $m_i{s}_j$ is Cauchy for all $s_j\in S$, which we will see is sufficient to complete the proof. Let $a_i$ denote the original bounded sequence in $M$; then, for $s_1$, $a_i{s}_1$ is bounded, and thus by Bolzano-Weierstrass, there is a subsequence $a_{p(1,i)}$ such that $a_{p(1,i)}{s}_1$ converges in the scalar field. Repeating this logic for $s_2$ gives a subsequence $a_{p(2,i)}$ of $a_{p(1,i)}$ such that $a_{p(2,i)}{m}_2$ converges. We can take the “diagonal” subsequence $m_i=a_{p(i,i)}$. Then, for any $s_j\in S$, the sequence from $m_j$ onwards is a subsequence of $a_{p(j,i)}$, and thus $m_i{s}_j$ is convergent.

It suffices to show this guarantees $m_{i}x$ is Cauchy for any $x\in X$. Fix $ϵ>0$. As $m_i$ is bounded, there is some fixed $c>0$ such that $\Vert m_i\Vert <c$. Furthermore, as $S$ is dense, we can take some $s\in S$ such that $\Vert x-s\Vert <ϵ/(3c)$. There exists some $N$ such that $\alpha ,\beta >N$ implies $|m_{\alpha }s-m_{\beta }s|<ϵ/3$. The triangle inequality gives

$$|m_{\alpha }x-m_{\beta }x|<|m_{\alpha }x-m_{\alpha }s|+|m_{\alpha }s-m_{\beta }s|+|m_{\beta }s-m_{\beta }x|$$ $$<\Vert m_{\alpha }\Vert \Vert x-s\Vert +\Vert m_{\beta }\Vert \Vert x-s\Vert +\Vert m_{\alpha }s-m_{\beta }s\Vert <ϵ$$

Thus, $m_{i}x$ is Cauchy for any $x\in X$, and we are done. Note this can be seen as taking the canonical mapping $\Phi :X\to X^{\prime \prime }$ and configuring a subsequence $m_i$ in $M$ such that $\Phi (s)m_i$ converges for all $s\in S$. We then show this is sufficient to show $m_i$ weakly converges as a sequence of operators (which is equivalent to strong convergence for functionals), which completes the proof as the limit operator must be bounded and in the dual space.