Where $F$ is a complete subspace of an inner product space $X$ and $x\in X$, is there a unique closest element of $F$ to $x$?

Question

That is: where $F$ is a complete subspace of an inner product space $X$ and $x\in X$, can one prove that $\exists! z\in F$ such that $\lVert x-z\rVert = \inf_{y\in F}\lVert x-y \rVert$? This is a question on an old preliminary exam in Analysis at my institution; I'm prepping for the prelim.

Clearly there is a sequence of elements $\{y_n\}\subseteq F$ such that $$\lVert x-y_n\rVert < \inf_{y\in F}\lVert x-y \rVert + \frac1n$$ But since we're not given that $F$ is compact, I don't know how to show that the sequence converges.

I take it from what I have read about similar problems that it would be best to try to show that $\{y_n\}$ is bounded. However, I don't see why it must be bounded.

Answer 1

You should use the parallelogram identity and convexity argument (the statement is true for every convex and closet set, which is a slight generalization with respect to subspaces). Try!