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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 $y$.

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Ad 1), the parallelogram identity is

$$\lVert a+b\rVert^2 + \lVert a-b\rVert^2 = 2\lVert a\rVert^2 + 2\lVert b\rVert^2$$

for all $a,b$. The first line applies that with $a = y_n - x$ and $b = y_m - x$.

2) Where did we use that $A$ is convex?

In the step where it was deduced that

$$\left\lVert \frac{y_n + y_m}{2} - x\right\rVert \geqslant d,$$

because of convexity, the midpoint of $y_m$ and $y_n$ lies in $A$.

3) Why is the point $y$ unique?

Let $p_1, p_2 \in A$ with $\lVert p_i - x\rVert = d$. Then

$$4d^2 = 2\lVert p_1 - x\rVert^2 + 2\lVert p_2 - x\rVert^2 = \lVert p_1 - p_2\rVert^2 + 4 \left\lVert \frac{p_1+p_2}{2} - x\right\rVert^2 \geqslant 4d^2,$$

hence $\lVert p_1 - p_2\rVert^2 = 0$.

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  • $\begingroup$ Those were very clear explanations, thank you. $\endgroup$ – Belgi Sep 12 '13 at 10:39

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