Prove $\liminf_{n\to \infty } \|u_n\|_{L^2}\geq \|u\|_{L^2}$ if $u_n\to u$ weakly.

Question

Prove $$\liminf_{n\to \infty } \|u_n\|_{L^2}\geq \|u\|_{L^2}$$ if $u_n\to u$ weakly in $L^2$.

Attempts

Since $u_n\to u$ weakly in $L^2$ in particularly, $$\lim_{n\to \infty }\int (u_n-u)u=0.$$ I tried to play with this and use the fact that $$0=\int u_n^2=\int u_n(u_n-u)+\int u_nu,$$ but it's not conclusif.

I also tried :

$$\int(u_n^2-u^2)=\int(u_n-u)u_n+\int u(u_n-u)$$ but I can't prove that the limit is positive.

Answer 1

We have \begin{align*} \|u_{n}\|_{L^{2}}&=\sup\left\{\left|\int u_{n}v\right|: v\in L^{2},~\|v\|_{L^{2}}\leq 1\right\}\\ &\geq\left|\int u_{n}v\right| \end{align*} for all such $v$, then taking $\liminf$, we have \begin{align*} \liminf_{n}\|u_{n}\|_{L^{2}}\geq\left|\lim_{n}\int u_{n}v\right|=\left|\int uv\right|, \end{align*} once again the formula \begin{align*} \|u\|_{L^{2}}=\sup\left\{\left|\int uv\right|: v\in L^{2},~\|v\|_{L^{2}}\leq 1\right\}, \end{align*} the result follows.

Answer 2

$$\|u_n\|^2_{L^2}=\int u_n^2=\int(u_n-u+u)^2=\int(u_n-u)^2+2\int u(u_n-u)+\int u^2$$ $$\geq 2\int (u_n-u)u+\int u^2$$ By taking the $\liminf$, the claim follow.