Weak convergence in $L^2$

Question

I'm studying weak convergence in $L^p$ space and I got trouble in understanding the following identity:

If $u_n \rightharpoonup u$ in $L^2(\mathbb{R}^N)$, then $$ \|u \|^2 + \limsup_{n \to \infty} \| u_n - u \|^2 = \limsup_{n \to \infty} \|u_n\|^2 .$$

Can anyone show me how we get that one?

Thanks for your helps.

Answer 1

First we have $$ \| u_n-u \|^2 = \langle u_n-u, u_n-u \rangle = \|u_n\|^2 + \|u\|^2 - 2 \Re \langle u_n, u \rangle, $$ where $\Re$ denotes the real part.

Now, since $u_n \rightharpoonup u,$ for any $\epsilon>0$ there is some $N$ such that $|\langle u_n, u \rangle - \langle u, u \rangle| < \epsilon$ whenever $n>N.$ Therefore $\langle u_n, u \rangle = \|u\|^2 + \delta,$ where $|\delta|<\epsilon.$ Thus, $$ \| u_n-u \|^2 = \|u_n\|^2 + \|u\|^2 - 2 \Re (\|u\|^2 + \delta) = \|u_n\|^2 - \|u\|^2 - 2 \Re\delta $$ i.e. $$ \|u\|^2 + \| u_n-u \|^2 = \|u_n\|^2 - 2 \Re\delta $$

Taking limits as $n \to \infty$ we end up with $$ \|u\|^2 + \lim_{n\to\infty} \| u_n-u \|^2 = \lim_{n\to\infty} \|u_n\|^2 $$ and therefore also $$ \|u\|^2 + \limsup_{n\to\infty} \| u_n-u \|^2 = \limsup_{n\to\infty} \|u_n\|^2 $$

I'm not sure if I'm missing something. As you can see I get a stronger result with ordinary limits, not just limit superior.

Answer 2

If $\left(a_n\right)_{n\geqslant 1}$, $\left(b_n\right)_{n\geqslant 1}$ and $\left(\delta_n\right)_{n\geqslant 1}$ are sequences of real numbers such that $a_n=b_n+\delta_n$ and $\delta_n\to 0$, then $\limsup_{n\to +\infty} a_n=\limsup_{n\to +\infty} b_n$.

Apply this to $a_n:=\left\lVert u_n-u\right\rVert_2$, $b_n:=\left\lVert u_n-u\right\rVert_2+\left\lVert u\right\rVert_2$ and $\delta_n:=-2\left\langle u_n,u\right\rangle$.