subspace of Hilbert space is closed if and only if it is weakly closed Any hints for this question, thank you!
Prove that a subspace of Hilbert space is closed if and only if it is
weakly closed.
 A: Note that this result is true for any convex set, not just subspaces.
(Very loosely, you can think of the norm closure and weak closure as a primal and dual sort of thing.)
The norm closure of $C$ is the intersection of all norm closed sets containing $C$.
The weak closure of $C$ is the intersection of all weakly closed sets containing $C$.
Let $C$ be a convex set.
In general, since a weakly closed set is norm closed, we have
$C \subset \overline{C} \subset \overline{C}^w$ (the latter being the weak closure).
If $C = \overline{C}^w $, it follows immediately that $C$ is closed.
Now suppose $C = \overline{C} $ (the norm closure). Now suppose $y \notin C$ (note that $\{y\}$ is compact). The Hahn Banach theorem shows that there exists a continuous linear functional $\phi$ and some constant $\alpha$ such that $\phi(y) < \alpha < \phi(c) $ for all $c \in C$. Hence $W =\phi^{-1}([\alpha, \infty))$ is weakly closed, $y \notin W$ and $C \subset W$. Hence
$y \notin \overline{C}^w $. Hence $\overline{C}^w  \subset C$, hence they are equal.
Addendum: If you are dealing with linear subspaces of a Hilbert space, then there is a simpler approach (albeit the same idea, it is just that we can give an 'explicit' functional with relying directly on the Hahn Banach theorem):
It follows from the above that weak closed implies norm closed.
Suppose $V$ is a norm closed subspace. Then it has an orthogonal complement $V^\bot$. Again, from the above, it follows that
$V \subset \overline{V}^w$, the weak closure. Suppose $y \notin V$, then $y = v_0+w_0$, where $v_0 \in V, w_0 \in V^\bot$. Then $H=\{ x | \langle x, w_0 \rangle \le {1 \over 2} \|w_0\|^2 \}$ is a weakly closed set, and $V \subset H$ and $y \notin H$. Hence $y \notin \overline{V}^w$, and so $V = \overline{V}^w$, and the set is weakly closed.
