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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?
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 $$\|p-f\| = \min_{v\in V} \|v-f\| \leq \lim_{n\to\infty}\|v_n-f\| = 0 \\ \implies f = p \in V$$
