Let $M$ be a closed subspace of the Banach space $X$ and let $x_{n}\in M$ converge weakly to $x$. Show that $x \in M$.
We use the following definition $x_n\rightharpoonup x$ in $X$, $x_n$ converges weakly to $x$ iff
$$
\lim_{n\to\infty}(\xi,x_n)=(\xi,x),\qquad\forall\xi\in X^*\;.
$$
I have seen some similar questions posted here, but not in this form, and I cannot seem to put the pieces together. My first thought was that since $M$ is closed it is a Banach space w.r.t. the norm inherited from $X$, therefore every Cauchy sequence converges in $M$. Maybe $\{x_n\}$ is Cauchy? But, I can't come anywhere from here.
Another thought I had was whether it isn't best to prove this by contradiction. Assume $x\notin M$, then $\|x_n-x\|>\varepsilon$, for all $n$, but I don't know where the contradiction would be.
We have some other tools as Hahn-Banach, but we don't have the uniform boundedness theorem (yet). All hints are welcome.
Best Answer
Note that $x$ lies in the weak closure of $M$. By Mazur's theorem (see corollary of theorem 3.12 in Rudin's Functional analysis) weak closure coincides with strong closure. The rest is clear.
A little more down to Earth solution. Assume $x$ is not in the strong closure of $M$, then $d(x,M)>0$. Then we can find $\xi\in X^*$ such that $(\xi,x)\neq 0$ while $\xi|_M=0$. From weak convergence we infer $(\xi, x)=\lim_n(\xi,x_n)=0$. Contradiction, so $x\in M$ because $M$ is closed.