Understanding a proof that a bounded sequence in a separable Hilbert space contains a weakly convergent subsequence

Question

I am trying to understand a proof that a bounded sequence in a separable Hilbert space $\mathcal H$ contains a weakly convergent subsequence, as in Theorem 5.12 of Gilbarg & Trudinger's "Elliptic Partial Differential Equations of Second Order". There are three points in the proof which I cannot justify.

I put the condition of separability because then the authors prove the general case from the separable one, and I understand how to make that transition, so my problems are with the separable case.

For ease of following here, I will give the whole proof and I will highlight with bold and a question number where my confusions lie.

Let $\{ y_m \}$ be a countable dense set in $\mathcal H$ and $x_n$ a bounded sequence. If I understand right, the authors claim that, for each $m$, by the Cantor diagonal process we can find a subsequence $x_{n_k^m}$ such that $(x_{n_k^m}, y_m) \xrightarrow{k\rightarrow \infty} \alpha_m$ for some $\alpha_m \in \mathbb R$.

Question 1. How is Cantor's diagonalisation process applied to obtained such sequences?

Then they define a mapping $f: \{ y_m \} \rightarrow \mathbb R$ by $f(y_m):= \alpha_m$. They claim that this $f$ can be extended to a bounded linear map on $\mathcal H$. Naturally, I would suspect that we extend it as follows: For $x \in \mathcal H$, we find $y_{m_i} \rightarrow x$ and we put $f(x):= \lim_{i \rightarrow \infty} \alpha_{m_i}$.

Question 2. Why is this definition good? I.e. why does $\alpha_{m_i}$ converge, why does it not depend on the choice of the sequence $y_{m_i}$ converging to $x$, and why is does it yield a continuous linear map?

Finally, they apply the Riesz representation theorem to this extended $f$, obtaining an $x$ such that $f(y) = (x, y)$ for any $y \in \mathcal H$. Then they claim that a subsequence $x_{n_k}$ of the original sequence $\{ x_n \}$ satisfies $(x_{n_k}, y) \xrightarrow{k \rightarrow \infty} f(y)$. I fail to see why. We know that: $(x_{n_k^i}, y_{m_i}) \rightarrow f(y_{m_i})$. We pick $y_{m_i} \rightarrow y$ and then we have $$ \lim_i (\lim_k (x_{n_k^i}, y_{m_i})) = \lim_i (f(y_{m_i})) = f(y) $$ But what we need is different, namely we need $\lim_k \lim_i$, which is equal to $\lim_k (\lim_i (x_{n_k^i}), y)$ since $\lim_i y_{m_i} = y$. But for this argument to work, there remains:

Question 3. Why is it that $\lim_i (x_{n_k^i})$ exists in $\mathcal H$, is an element of the original sequence $x_n$, and why is it that we can permute $\lim_i$ with $\lim_k$ i.e. that $$ \lim_i (\lim_k (x_{n_k^i}, y_{m_i})) = \lim_k ( \lim_i (x_{n_k^i}, y_{m_i})) $$?

Now that I have redacted this question, I have a feeling it would be largely solved by understanding how the Cantor argument is used.

Answer 1

Your question is quite long and requires a long answer. I will try to be as precise as possible. Let $(x_k)_k \subset \mathcal H$ bounded, meaning that $$\|x_k\|_\mathcal H \le C$$ uniformly in $k$, and $\{y_m\} \subset \mathcal H$ a countable dense subset of $\mathcal H$.

Question 1. Here I will use the following notation: The sequence $(x_{n_k})_k$ is a subsequence of $(x_n)_n$ if and only if there exists a strictly increasing function $\phi: \mathbb N \to \mathbb N$ such that $n_k = \phi(k)$. The subsequence is therefore denoted by $(x_{\phi(k)})_k$.

For each given $m\in \mathbb N$, there holds $$(x_k, y_m) =: (\alpha_{m, k})_k \le \|x_k\|_\mathcal H\|y_m\|_\mathcal H \le C\|y_m\|_\mathcal H.$$ Hence, for each $m $ one can apply Bolzano-Weierstrass Theorem, so that there exists a function $\phi_m: \mathbb N \to \mathbb N$, satisfying $$\alpha_{m, \phi_m(k)} = (x_{\phi_m(k)}, y_m) \to \alpha_m$$ as $k \to \infty$. The problem here is that the subsequence itself depends on $m$. This means that for two different elements of this sequence, say $y_{m_1}, y_{m_2}$, the subsequence given by Bolzano-Weierstrass might have no element in common, i.e. $$\phi_{m_1}(\mathbb N) \cap \phi_{m_2}(\mathbb N) = \varnothing.$$

A way to avoid this problem is to use Cantor's diagonal process: Let $\phi_0:\mathbb N \to \mathbb N$ such that $$\alpha_{0, \phi_0(k)} := (x_{\phi_0(k)}, y_0) \to \alpha_0.$$ Now since the sequence $((x_{\phi_0(k)}, y_1))_{k}$ is still bounded in $\mathbb R$, there exists $\phi_1:\mathbb N \to \mathbb N$ such that $$\alpha_{1, (\phi_0\circ\phi_1)(k)} := (x_{(\phi_0\circ\phi_1)(k)}, y_1) \to \alpha_1~\text{and}~\alpha_{0, (\phi_0\circ\phi_1)(k)} \to \alpha_0, \quad k \to +\infty.$$ By induction, one can prove that $$(x_{(\phi_0\circ\ldots \circ \phi_m)(k)}, y_j) \to \alpha_j, \quad \text{for any } j \le m.$$ Let us set $\varphi(n) = (\phi_0\circ\ldots \circ \phi_n)(n)$. Exercice: Prove that $\varphi:\mathbb N \to \mathbb N$ is strictly increasing and that $$\lim_{k \to +\infty}(x_{\varphi(k)}, y_j) = \alpha_j, \quad \forall j \in \mathbb N.$$

Question 2. Let $f: V := \langle y_m \rangle \to \mathbb R$ be defined on the vector space $V$ generated by $\{y_m\}$ such that $$f(y_m) = \lim_{k \to \infty} (x_{\varphi(k)}, y_m),$$ and extended linearly on $V$. Remark that $f$ is continuous on $V$ since $$f(y_m) \le C\|y_m\|_\mathcal H$$ and as $V$ is dense in $\mathcal H$, $f$ can be uniquely extended to $\mathcal H$. See this question for a proof of this result (or try to prove it yourself, it's a good exercise).

Question 3. If you have a doubt about limits, always come back to espilon-delta. Let $y \in \mathcal H$ and $(y_j)_j \subset \{y_m\}$ such that $y_j$ converges to $y$ (note that $(y_j)_j$ is a subsequence of $\{y_m\}$, but as I don't need Cantor's argument here I don't use the $\phi$ notation). One gets \begin{align} |f(y) - (x_{\varphi(n)}, y)| &\le |f(y) - f(y_j)| + |f(y_j) - (x_{\varphi(n)}, y_j)| + |(x_{\varphi(n)}, y_j) - (x_{\varphi(n)}, y)| \end{align} and I leave as an exercise the proof that the right hand side can be bounded by any $\varepsilon > 0$ for $j$ and $n$ large enough.