If $x_n \in M$ for all $n$ where $M$ is closed subspace of $H$ and $x_n \rightarrow x$ weakly then $x \in M$.

Question

Let $H$ be infinite dimensional Hilbert space.

If $x_n \in M$ for all $n$ where $M$ is closed subspace of $H$ and $x_n \rightarrow x$ weakly then $x \in M$.

By the definition of a closed subspace we have that $x_n \in M$ converges to some $x \in H$ such that $x \in M$. Why do we need the weak convergence condition here? Isnt the statement obvious from the definition of the closed subspace?

Answer 1

Suppose that $x \not\in M$ then we have that $dist(x,M) > 0$. By the Hanh-Banach theorem there exists a bounded linear operator $f$ such that $f(M) = \{0\}$, $\|f\| = 1$ and $<x,f> = dist(x,M)$, but this is a contradiction since $<x,f> = \lim_{n \rightarrow \infty}<x_n,f> = 0$, hence $x \in M$.