Does weak convergence in $L^2$ plus uniform convergence imply strong convergence in $L^2$?

Question

If we have $\{u_n\}_{n \ge 1} \subseteq C^{1}[0,1]$ weakly converging to $u$ in $L^{2}([0,1])$, and $\{u_n\}_{n \ge 1}$ is uniformly convergent to some function $v$ on $[0,1]$, can we conclude that $\{u_n\}_{n \ge 1}$ strongly converges to $u$ in $L^{2}([0,1])?$

Answer 1

Note that $$\int\limits_0^1 |u_n(x)-v(x)|^2\, dx\leq \left(\sup_{x\in [0,1]} |u_n(x)-v(x)|\right)^2\int\limits_0^1 \, dx=\left(\sup_{x\in [0,1]} |u_n(x)-v(x)|\right)^2\rightarrow 0$$ as $n\rightarrow\infty.$ So, $u_n\rightarrow v$ strongly in $L^2.$ This implies that $u_n\rightarrow v$ weakly, as well, and since weak limits are unique, it follows that $u=v.$