In the setting of an introduction to functional analysis course, I have read the following statement:
Let $H$ be a Hilbert space and let $A\subseteq H$ be a closed convex set.
Then there exist a projection $$ P_{A}:\, H\to A $$
s.t for every $x\in H$, $P_{A}(x)$ is the only point in $A$ that satisfy $$ d(x,A)=d(P_{A}(x),x) $$
I have three problems with the proof given, I don't understand one technical transition and I don't understand where did the convexity of $A$ came into play, and the uniqueness.
The proof given was:
Take a sequence $\{y_{n}\}\subseteq A$ s.t $d=d(x,A)=\lim_{n\to\infty}d(x,y_{n})$. We will prove that $\{y_{n}\}$is a Cauchy sequence and get both existence and uniqueness. Define $P_{A}(x)=y$, apply the parallelogram equality for $y_{n}-x,y_{m}-x$: $$ 4d^{2}\leftarrow_{\text{as {m,n\to\infty}}}2||y_{n}-x||^{2}+2||y_{m}-x||^{2}=||y_{n}-y_{m}||^{2}+||y_{n}+y_{m}-2x||^{2} $$ $$ =||y_{n}-y_{m}||^{2}+4||\frac{y_{n}+y_{m}}{2}-x||^{2}\geq||y_{n}-y_{m}||^{2}+4d^{2} $$
hence $||y_{n}-y_{m}||$ is as small as we want.
I have $3$ questions that I would appreciate any help with:
1) How was the transition of the first equality made ? I don't see how this transition have anything to to with he parallelogram equality
2) Where did we use that $A$ is convex ? (I can see how we used that $A$ is closed when we took the limit and and it defined an element of $A$, but I don't see where did we need the other condition on $A$ that is the convexity)
3) Why is the point $y$ unique ? at first I thought it was because a normed space is in particular a metric space and thus a Hausdorff space and so the limit is unique, but that just say that the $\{y_{n}\}$ have a unique limit.
But maybe there is a different sequence $\{a_{n}\}$ s.t $$d=d(x,A)=\lim_{n\to\infty}d(x,a_{n})$$ and $a_{n}$ converges to a different element of $A$ rather than $y$.