# Weak convergence in closed subspace of a Hilbert Space

I get stuck in following problems

Let $$(x_n)_n \subset Y$$ be such that $$x_n \xrightarrow{w} x$$ (converges weakly), and Y a closed subspace of a Hilbert Space. Show that $$x \in Y$$

My try I didn't make a great progress.

I just know that For every $$y \in H$$ we have that $$ \rightarrow $$ iff $$| - | \rightarrow 0$$.

I think that I have to show a subsequence of $$(x_n)$$ which strongly converges to $$x$$. (I get stuck here)

Also I've to show

Let $$T \in B(H)$$ and $$x_n \xrightarrow{w} x$$ (converges weakly) then $$Tx_n \xrightarrow{w} Tx$$ (converges weakly).

I think that I've to solve the previous problem before this.

Thank you.

If $$Y$$ is closed, $$Y=\{\cap_{f\in H^*}Kerf, Y\subset Ker f\}$$. To see this, remark that $$Y\subset \{\cap_{f\in H^*}Kerf, Y\subset Ker f\}$$. Let $$y\in \{\cap_{f\in H^*}Kerf, Y\subset Ker f\}$$, suppose that $$y$$ is not in $$Y$$, consider $$f$$ defined on $$Z=Vect(Y,y)$$ such that $$f(y)=1, f(Y)=0$$, $$f$$ is bounded on $$Z$$. By Hahn Banach you can extend it to $$H$$. Contradiction.
Suppose that $$f\in H^*$$ and $$Y\subset Ker f$$, $$f(x_n)=0$$ implies that $$f(x)=0$$, implies that $$x\in \{\cap_{f\in H^*}Kerf, Y\subset Ker f\}=Y$$.