Proof of Bounded Sequences in the Lebesgue Space of Square-Integrable Vector-Valued Functions Have Weakly Convergent Subsequences

Throughout, claim numbers for $H$ refer to The Lebesgue Space of Square-Integrable Vector-Valued Functions is a Real Inner Product Space: by claims 4 and 5 the pairing is symmetric and linear in each argument and satisfies the Cauchy-Schwarz inequality $|\langle v,w\rangle_{L^{2}}|\le\lVert v\rVert_{L^{2}}\lVert w\rVert_{L^{2}}$. Note $C\ge0$, since $0\le\lVert u_{1}\rVert_{L^{2}}\le C$.

Step 1 (a countable dense set, enumerated). By Separability of the Lebesgue Space of Square-Integrable Vector-Valued Functions there is a subset $E$ of $H$ that is dense in the metric space $(H,d_{L^{2}})$, whose metric open sets form a topology by Metric Open Sets Form a Topology, and that is countable. Since $0\in H$ and a dense subset meets every open ball by Characterization of the Closure in a Metric Space by Open Balls, $E$ is nonempty, so there is a sequence $(w_{r})_{r\in\mathbb{N}}$ whose set of terms is $E$.

Step 2 (nested selectors). For every $r$ and every $n$, Cauchy-Schwarz gives $|\langle u_{n},w_{r}\rangle_{L^{2}}|\le C\lVert w_{r}\rVert_{L^{2}}$, so each real sequence $\bigl(\langle u_{n},w_{r}\rangle_{L^{2}}\bigr)_{n\in\mathbb{N}}$ is bounded.

We construct strictly increasing maps $\sigma_{r}:\mathbb{N}\to\mathbb{N}$, each selecting a subsequence of the one selected by its predecessor. Since one selector is chosen at every stage, the construction is an application of Axiom of Dependent Choice: let $P$ be the set of pairs $(r,\sigma)$ with $r\in\mathbb{N}$ and $\sigma:\mathbb{N}\to\mathbb{N}$ strictly increasing such that $\bigl(\langle u_{\sigma(n)},w_{s}\rangle_{L^{2}}\bigr)_{n}$ converges for every $s\le r$, and let a pair $(r,\sigma)$ be related to $(r+1,\sigma\circ\tau)$ whenever $\tau:\mathbb{N}\to\mathbb{N}$ is strictly increasing and $\bigl(\langle u_{\sigma(\tau(n))},w_{r+1}\rangle_{L^{2}}\bigr)_{n}$ converges. The paragraph that follows produces a starting element of $P$ and shows that every element of $P$ has a related successor in $P$, so that axiom yields a sequence of pairs whose second entries are the maps $\sigma_{r}$ below. By Bolzano-Weierstrass Theorem for Real Sequences the bounded sequence $\bigl(\langle u_{n},w_{1}\rangle_{L^{2}}\bigr)_{n}$ has a convergent subsequence; let $\sigma_{1}$ be a strictly increasing map with $\bigl(\langle u_{\sigma_{1}(n)},w_{1}\rangle_{L^{2}}\bigr)_{n}$ convergent. If $\sigma_{r}$ has been constructed, apply the same theorem to the bounded sequence $\bigl(\langle u_{\sigma_{r}(n)},w_{r+1}\rangle_{L^{2}}\bigr)_{n}$ to obtain a strictly increasing $\tau$ with $\bigl(\langle u_{\sigma_{r}(\tau(n))},w_{r+1}\rangle_{L^{2}}\bigr)_{n}$ convergent, and set $\sigma_{r+1}=\sigma_{r}\circ\tau$, again strictly increasing.

For all $s\le r$ there is a strictly increasing $\theta_{s,r}:\mathbb{N}\to\mathbb{N}$ with $\sigma_{r}=\sigma_{s}\circ\theta_{s,r}$: by induction on $r$ for fixed $s$, taking $\theta_{s,s}$ the identity and, if $\sigma_{r}=\sigma_{s}\circ\theta_{s,r}$, noting $\sigma_{r+1}=\sigma_{r}\circ\tau=\sigma_{s}\circ(\theta_{s,r}\circ\tau)$ with $\theta_{s,r}\circ\tau$ strictly increasing. Consequently, for $s\le r$, the sequence $\bigl(\langle u_{\sigma_{r}(n)},w_{s}\rangle_{L^{2}}\bigr)_{n}$ is a subsequence of a subsequence, hence a subsequence, of the convergent sequence $\bigl(\langle u_{\sigma_{s}(n)},w_{s}\rangle_{L^{2}}\bigr)_{n}$, so it converges by the fact that a subsequence of a convergent sequence has the same limit.

Step 3 (the diagonal subsequence). Put $n_{j}=\sigma_{j}(j)$. Writing $\sigma_{j+1}=\sigma_{j}\circ\rho$ with $\rho$ strictly increasing, we have $\rho(j+1)\ge j+1$ by the growth bound for strictly increasing sequences of natural numbers, so $n_{j+1}=\sigma_{j}(\rho(j+1))\ge\sigma_{j}(j+1)>\sigma_{j}(j)=n_{j}$; thus $n_{1}<n_{2}<\dots$ and $(u_{n_{j}})_{j}$ is a subsequence of $(u_{n})_{n}$.

Fix $r$. For $j\ge r$ we have $n_{j}=\sigma_{r}\bigl(\theta_{r,j}(j)\bigr)$, and the index map $j\mapsto\theta_{r,j}(j)$ is strictly increasing on $\{j\in\mathbb{N}:j\ge r\}$: if $j'>j\ge r$ then $\sigma_{r}(\theta_{r,j'}(j'))=n_{j'}>n_{j}=\sigma_{r}(\theta_{r,j}(j))$, and a strictly increasing map is order-reflecting: if $\theta_{r,j'}(j')\le\theta_{r,j}(j)$ then $\sigma_{r}(\theta_{r,j'}(j'))\le\sigma_{r}(\theta_{r,j}(j))$, contradicting the strict inequality just displayed. Writing $j=r+i-1$ with $i\in\mathbb{N}$, the sequence $\bigl(\langle u_{n_{r+i-1}},w_{r}\rangle_{L^{2}}\bigr)_{i\in\mathbb{N}}$ is obtained from $\bigl(\langle u_{\sigma_{r}(n)},w_{r}\rangle_{L^{2}}\bigr)_{n}$ through the strictly increasing index map $i\mapsto\theta_{r,r+i-1}(r+i-1)$, so it is a subsequence of a convergent sequence and converges, by A Subsequence of a Convergent Sequence Has the Same Limit. The sequence $\bigl(\langle u_{n_{j}},w_{r}\rangle_{L^{2}}\bigr)_{j\in\mathbb{N}}$ agrees with it from index $r$ onwards, and changing finitely many terms affects neither convergence nor the limit, since the condition of the definition of the limit concerns only sufficiently large indices. Hence $\bigl(\langle u_{n_{j}},w_{r}\rangle_{L^{2}}\bigr)_{j\in\mathbb{N}}$ converges for every $r\in\mathbb{N}$.

Step 4 (a bounded linear limit functional). Let $v\in H$. We claim that $\bigl(\langle u_{n_{j}},v\rangle_{L^{2}}\bigr)_{j}$ is a Cauchy sequence. Let $\varepsilon>0$ be real and put $\delta=\varepsilon\bigl(3(C+1)\bigr)^{-1}$, a positive real number. Since $E$ is dense, Characterization of the Closure in a Metric Space by Open Balls provides $r\in\mathbb{N}$ with $\lVert v-w_{r}\rVert_{L^{2}}<\delta$. For all $i,j$, linearity of the pairing, the triangle inequality for the absolute value and Cauchy-Schwarz give

and the first and third terms are at most $C\delta<\varepsilon/3$ each. The middle term is less than $\varepsilon/3$ for all large $i,j$: by step 3 the sequence $\bigl(\langle u_{n_{j}},w_{r}\rangle_{L^{2}}\bigr)_{j}$ converges, say to $a$, so there is $J$ with $|\langle u_{n_{j}},w_{r}\rangle_{L^{2}}-a|<\varepsilon/6$ for $j\ge J$, and for $i,j\ge J$ the middle term is less than $\varepsilon/3$. Hence the sequence is Cauchy, and by Every Cauchy Sequence of Real Numbers Converges it converges. Define $\Lambda(v)$ to be its limit; this defines a map $\Lambda:H\to\mathbb{R}$.

$\Lambda$ is linear: for $v,v'\in H$ and reals $s,s'$ one has $\langle u_{n_{j}},sv+s'v'\rangle_{L^{2}}=s\langle u_{n_{j}},v\rangle_{L^{2}}+s'\langle u_{n_{j}},v'\rangle_{L^{2}}$ by claim 4, and Arithmetic of Limits of Real Sequences gives $\Lambda(sv+s'v')=s\Lambda(v)+s'\Lambda(v')$. Moreover $-C\lVert v\rVert_{L^{2}}\le\langle u_{n_{j}},v\rangle_{L^{2}}\le C\lVert v\rVert_{L^{2}}$ for every $j$ by Cauchy-Schwarz, so comparing with the two constant sequences and using claim 1 (comparison) of Order Properties of Limits of Real Sequences gives $|\Lambda(v)|\le C\lVert v\rVert_{L^{2}}$.

Step 5 (conclusion). By Riesz-Frechet Representation of Bounded Linear Functionals on the Lebesgue Space of Square-Integrable Vector-Valued Functions there is $u\in H$ with $\Lambda(v)=\langle u,v\rangle_{L^{2}}$ for every $v\in H$ and $\lVert u\rVert_{L^{2}}\le C$. By the definition of $\Lambda$, for every $v\in H$ the real sequence $\bigl(\langle u_{n_{j}},v\rangle_{L^{2}}\bigr)_{j}$ has limit $\langle u,v\rangle_{L^{2}}$, which is precisely weak convergence of $(u_{n_{j}})_{j}$ to $u$.