2
$\begingroup$

To prove: If for every $f\in H$ ($H$ is a Hilbert space) there is a $p\in V$ such that $\|p−f\|=\min_{v\in V}\|v−f\|$ then $V$ is closed.

I was able to prove the converse of this statement, but not this. I am not able to write down the proof coherently, and A subspace $X$ is closed iff $X =( X^\perp)^\perp$ is confusing me even more! Can someone write down steps of hints to complete this proof?

$\endgroup$
4
$\begingroup$

Well, since a Hilbert space is in particular a metric space, $V$ being closed can be characterized by convergent sequences.

That is, suppose $v_n \in V$ converges to some $f \in H.$ Then we have that there is some $p \in V$ such that $$\|p-f\| = \min_{v\in V} \|v-f\| \leq \lim_{n\to\infty}\|v_n-f\| = 0 \\ \implies f = p \in V$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.