the image of the closed unit ball by a continuous homomorphism of hilbert spaces is closed Let E and F be two hilbert spaces. and $u$ a continuous homomorphism from E to F.that is u continuous, $u(x+y)=u(x)+u(y)$ and $u(\alpha x)=\alpha u(x)$ .
Let $B$ be the unit closed ball in $E$.
$$B=\{x\in E\ ,\ ||x||\leq 1\}$$
The question is to show that $u(B)$ the image of the closed unit ball is closed in F.
Im stuck with this. Im wondering if even this is true after seeing this
Any hint would be appreciated.
 A: Before seeing the solution note that $B $ is closed convex and bounded, and every Hilbert space is reflexive. Thus $B$ is weakly compact, based on Eberlein-Smulyan theorem. Therefore   every sequence   in $B$, like say $\{z_n\}$ ,  has a subsequence, say $ \{z_{n_{j}}\}  $, weakly convergent to a point $x \in B$ that means for every continuous linear operator $\phi:E \longrightarrow \mathbb{R}$ we have that $ \phi(z_{n_{j}}) \longrightarrow \phi(x)$. \
Now let's solve the problem: Suppose that $(y_n )$ is an arbitrary sequence in $u(B)$ which converges to some $y \in F$. We want to show $y \in u(B)$. For every $n$, we infer from $y_n \in u(B)$ that there is $x_n \in B$ such that $y_n = u(x_n)$. Because $B$ is weakly compact $\{x_n\}$ has a   subsequence say $x_{n_{j}}$ which weakly converges to a point $x \in B$.
Now, let $\psi : F  \longrightarrow \mathbb{R}$  be an arbitrary continuous functional. So, $ \psi o u : E \longrightarrow \mathbb{R}$ is a continuous functionl. Thus
$$\psi(y_n ) = (\psi o u)(x_n) \longrightarrow (\psi o u)(x) = \psi( u(x)).  $$
Because $\psi : F  \longrightarrow \mathbb{R}$ is arbitrary,  $\{ y_n \}$ converges to $u(x)$ weakly. We also know   $ y_n \longrightarrow y$ strongly, therefore $y_n$ goes to $y$ weakly as well. Because the limit of weak-convergence is unique we infer that $y= u(x) \in u(B)$. THus $u(B)$ is closed
