Proof of Basic Properties of Weak Convergence in the Lebesgue Space of Square-Integrable Vector-Valued Functions

Throughout we use claims 4 and 5 of the inner-product lemma: the pairing $\langle\cdot,\cdot\rangle_{L^{2}}$ is symmetric and linear in each argument, $\langle u,u\rangle_{L^{2}}=\lVert u\rVert_{L^{2}}^{2}$, the norm vanishes only at the zero element, and the Cauchy-Schwarz inequality $|\langle u,v\rangle_{L^{2}}|\le\lVert u\rVert_{L^{2}}\lVert v\rVert_{L^{2}}$ holds. We also record an elementary fact used below: if a real sequence $(a_{n})$ has limit $a$ and $a_{n}\le M$ for every $n$, then $a\le M$; for if $a>M$, choosing $n$ with $|a_{n}-a|<a-M$ would give $a_{n}>M$.

Claim 1. Let $v\in H$. The real sequence $\bigl(\langle u_{n},v\rangle_{L^{2}}\bigr)_{n}$ has limit $\langle u,v\rangle_{L^{2}}$ and also limit $\langle u',v\rangle_{L^{2}}$, so these are equal by uniqueness of limits of real sequences. By linearity, $\langle u-u',v\rangle_{L^{2}}=0$ for every $v\in H$. Taking $v=u-u'$ gives $\lVert u-u'\rVert_{L^{2}}^{2}=0$, hence $u-u'=0$, that is $u=u'$.

Claim 2. Let $v\in H$ and let $\varepsilon>0$ be real. By linearity and Cauchy-Schwarz,

If $\lVert v\rVert_{L^{2}}=0$ the right-hand side is $0$ for every $n$. Otherwise choose $N$ with $\lVert u_{n}-u\rVert_{L^{2}}<\varepsilon/\lVert v\rVert_{L^{2}}$ for $n\ge N$, so that the left-hand side is less than $\varepsilon$ for $n\ge N$. In either case the sequence $\bigl(\langle u_{n},v\rangle_{L^{2}}\bigr)_{n}$ has limit $\langle u,v\rangle_{L^{2}}$; as $v$ was arbitrary, $u_{n}\rightharpoonup u$.

Claim 3. Note first that $C\ge\lVert u_{1}\rVert_{L^{2}}\ge0$. Applying the definition of weak convergence with the test element $u$, the real sequence $\bigl(\langle u_{n},u\rangle_{L^{2}}\bigr)_{n}$ has limit $\langle u,u\rangle_{L^{2}}=\lVert u\rVert_{L^{2}}^{2}$. For every $n$, Cauchy-Schwarz gives

By the elementary fact recorded above, $\lVert u\rVert_{L^{2}}^{2}\le C\lVert u\rVert_{L^{2}}$. If $\lVert u\rVert_{L^{2}}=0$ then $\lVert u\rVert_{L^{2}}=0\le C$; otherwise dividing by the positive number $\lVert u\rVert_{L^{2}}$ gives $\lVert u\rVert_{L^{2}}\le C$.

Claim 4. As in claim 3, $C\ge0$. By linearity of the pairing in its first argument,

using linearity in the second argument for the first term. By Cauchy-Schwarz, $\bigl|\langle u_{n},v_{n}-v\rangle_{L^{2}}\bigr|\le C\,\lVert v_{n}-v\rVert_{L^{2}}$.

Let $\varepsilon>0$ be real. If $C=0$ the first term vanishes for every $n$; if $C>0$, choose $N_{1}$ with $\lVert v_{n}-v\rVert_{L^{2}}<\varepsilon/(2C)$ for $n\ge N_{1}$, so that the first term is less than $\varepsilon/2$ in absolute value. Since $u_{n}\rightharpoonup u$, choose $N_{2}$ with $\bigl|\langle u_{n},v\rangle_{L^{2}}-\langle u,v\rangle_{L^{2}}\bigr|<\varepsilon/2$ for $n\ge N_{2}$. For $n\ge\max(N_{1},N_{2})$ the displayed difference is less than $\varepsilon$ in absolute value. Hence $\bigl(\langle u_{n},v_{n}\rangle_{L^{2}}\bigr)_{n}$ has limit $\langle u,v\rangle_{L^{2}}$.

Claim 5. As in claim 3, $C\ge0$. Let $v\in H$ and let $\varepsilon>0$ be real. Put $M=C+\lVert u\rVert_{L^{2}}+1$, a real number with $M>0$. The set $\{\zeta\in H:d_{L^{2}}(\zeta,v)<\varepsilon/(2M)\}$ is a nonempty open subset of $(H,d_{L^{2}})$, so by density of $E$ it contains an element $w\in E$; thus $\lVert v-w\rVert_{L^{2}}<\varepsilon/(2M)$.

For every $n$, writing $\langle u_{n},v\rangle_{L^{2}}-\langle u,v\rangle_{L^{2}}$ as

and applying Cauchy-Schwarz to the first and third terms gives

and $\lVert u_{n}\rVert_{L^{2}}+\lVert u\rVert_{L^{2}}\le C+\lVert u\rVert_{L^{2}}<M$. Since $w\in E$, choose $N$ with $\bigl|\langle u_{n},w\rangle_{L^{2}}-\langle u,w\rangle_{L^{2}}\bigr|<\varepsilon/2$ for $n\ge N$. Then for $n\ge N$,

So $\bigl(\langle u_{n},v\rangle_{L^{2}}\bigr)_{n}$ has limit $\langle u,v\rangle_{L^{2}}$ for every $v\in H$, that is $u_{n}\rightharpoonup u$.