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 $<x_n, y> \rightarrow <x, y>$ iff $|<x_n, y> - <x, y>| \rightarrow 0$.
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 $x\in \{\cap_{f\in H^*}Kerf, Y\subset Ker f\}=Y$.