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I'm trying to do the following exercise:

Let $H$ a Hilbert space (Vector space complete with inner product), $F \subseteq H$ a closed convex set and $a \in H$. Show that $\exists! b \in F \subseteq H$ such that $d(a,b) = d(a,F)$

My teacher gave a hint that I can't see where it'll lead. He said this:

If $x_n \in F$ is a sequence such that $||a-x_n|| \to d(a,F)$, then use the parallelogram identity to show $(x_n)$ is Cauchy. I know that in vector spaces with inner product the parallelogram identity is true: $||x+y||^{2} + ||x-y||^{2} = 2||x||^{2} + 2||y||^{2}$. But I don't see how should I use it to prove $(x_n)$ is Cauchy. Since the space is complete it will converge, but why would this help me to prove existence or uniqueness of $b$?

Any help would be appreciated. Thanks.

2 Answers2

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Hint: For the Cauchyness, put $x= a - x_n$, $y= a-x_m$ (as in your notation) for some positive integers $m>n$ and then use convexity! Limit of this sequence is a candidate for your $b$.

ters
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Hint: The trick is that $(x_n+x_m)/2\in F$ as well (since $F$ is convex). Use this on $$ 2\|a-x_n\|^2 + 2\|a-x_m\|^2 = \|2a -(x_n+x_m)\|^2 + \|x_n-x_m\|^2$$

H. H. Rugh
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  • $ 2|a-x_n|^2 + 2|a-x_m|^2 = |2a -(x_n+x_m)|^2 + |x_n-x_m|^2 \Rightarrow \frac{|a-x_n|^2}{2} + \frac{|a-x_m|^2}{2} = |a -\frac{(x_n+x_m)}{2}|^2 + \frac{|x_n-x_m|^2}{4} $. Taking the limit (Since square is continuous): $ \frac{d(a,F)^{2}}{2} + \frac{d(a,F)^{2}}{2} = d(a,F)^{2} + \lim \frac{ | x_n - x_m | ^{2}}{4} \Rightarrow 0 = \lim |x_n - x_m |$, then it is Cauchy, Then $(x_n)$ converges to some element of $F$. But I still can't see why this will imply existence and uniqueness of $b$. –  Oct 20 '16 at 11:37
  • Is this right? $(x_n)$ is Cauchy, then $x_n \to b \in F$, then $| a - x_n | \to | a - b | = d(a,b)$. But $ | a - x_n | \to d(a,F)$, then by uniqueness of limit (Metric space is Hausdorff), $d(a,b) = d(a,F)$ –  Oct 20 '16 at 11:48
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    If any minimizing sequence is Cauchy then the limit must be unique. Suppose $b$ and $b'$ are two possible limits. Then construct a sequence $x_n$ so that $x_{2n} \rightarrow b$ and $x_{2n+1}\rightarrow b'$. Now use Cauchy to conclude that $b=b'$. – H. H. Rugh Oct 20 '16 at 12:38
  • What I did on my second comment was wrong? –  Oct 20 '16 at 12:41
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    Sorry, forgot to answer: That was ok and shows that you attain the minimum (but didn't show uniqueness of $b$). – H. H. Rugh Oct 20 '16 at 13:11