Proof of Closed Convex Subsets of the Lebesgue Space of Square-Integrable Vector-Valued Functions are Weakly Sequentially Closed

The set $K$ is nonempty, since $u_{1}\in K$. By claim 1 of Nearest-Point Projection onto a Nonempty Closed Convex Subset of the Lebesgue Space of Square-Integrable Vector-Valued Functions the nearest-point projection $p=\pi_{K}(u)$ of $u$ onto $K$ exists, and by claim 2 of that theorem

$$\langle u-p,\;v-p\rangle_{L^{2}}\le0\qquad\text{for every }v\in K .$$

Taking $v=u_{n}$ and expanding by linearity of the pairing in the second argument (claim 4 of The Lebesgue Space of Square-Integrable Vector-Valued Functions is a Real Inner Product Space) gives

$$\langle u-p,\;u_{n}\rangle_{L^{2}}\le\langle u-p,\;p\rangle_{L^{2}}\qquad(n\in\mathbb{N}).$$

By symmetry of the pairing (claim 4 of that lemma), $\langle u-p,u_{n}\rangle_{L^{2}}=\langle u_{n},u-p\rangle_{L^{2}}$, and weak convergence of $(u_{n})_{n\in\mathbb{N}}$ to $u$, applied to the test element $u-p$, says that the real sequence $\bigl(\langle u_{n},u-p\rangle_{L^{2}}\bigr)_{n\in\mathbb{N}}$ has limit $\langle u,u-p\rangle_{L^{2}}$.

The constant sequence with value $\langle u-p,p\rangle_{L^{2}}$ has that value as its limit, and weak inequalities between convergent sequences are preserved in the limit by claim 1 (comparison) of Order Properties of Limits of Real Sequences. Hence

$$\langle u-p,\;u\rangle_{L^{2}}\le\langle u-p,\;p\rangle_{L^{2}},$$

that is, again by linearity in the second argument, $\langle u-p,\;u-p\rangle_{L^{2}}\le0$. By claim 4 of the inner-product lemma the left-hand side equals $\lVert u-p\rVert^{2}_{L^{2}}$, which is nonnegative; therefore $\lVert u-p\rVert^{2}_{L^{2}}=0$, so $\lVert u-p\rVert_{L^{2}}=0$ and $u=p$ by claim 4 of that lemma. Since $p\in K$, this gives $u\in K$.