TITLE: Is projection on a convex closed weakly-sequentially continuous?
QUESTION [2 upvotes]: I think to have proved the following: 
Given $K$ a convex closed(maybe also limited is needed)subset(also curve not just subspaces) of an Hilbert space $H$, is well defined the projection operator $p_K:H \to K$. This operator is continuos for the norm topology. If i didn't miss something, my guess (is not a very formal proof) shows that is sequentially continuous for the weak topology. Is that true?
My guess goes more or less like this: i write my convex set as intersection of half spaces, and i reduce in some way to those.
Before to try to work it out more clearly i'd like to know if anybody knows if the result is actually true.
Thanks

REPLY [2 votes]: Here is a counter-example regarding continuity of the projection in the weak topology. Is is taken from Zarantonello's article I mentioned in the comments.
Let $H=l^2$, $K$ be the closed unit ball. Let $e_k=(0,\dots, 0,1,0,\dots)$ be the $k$-the unit element.
Set
$$
x_n := e_1 + e_n.
$$
Then $x_n\rightharpoonup e_1$. The projection of $x_n$ satisfies
$$
p_K(x_n) = \frac1{\sqrt 2}(e_1 + e_n) \rightharpoonup \frac1{\sqrt 2}e_1.
$$
Thus the projection of the weak limit of $x_n$ is $e_1$, and it is different from the weak limit of the projections, which is $\frac1{\sqrt 2}e_1$. And $p_K$ is not continuous with respect to the weak topology.
{"set_name": "stack_exchange", "score": 2, "question_id": 829845}