The Hilbert space projetion theorem is the following theorem:

Let $H$ be a Hilbert space and $C$ any closed convex subset. Then for $h \in H$ there exists a unique $c_0 \in C$ such that $\|h-c_0\| = \inf_{c\in C}\|h-c\|$.

Please could someone show me how to finish my idea for a proof?


Note that $C$ is a closed set therefore $\inf_{c\in C}\|h-c\| = \min_{c\in C}\|h-c\|$, or in other words: there exists $c_0 \in C$ with $\|h-c_0\|=\min_{c\in C}\|h-c\|$. This shows existence.

We have to show uniqueness next. Let $c_0' \in C$ be another element with $\|h-c_0'\| = \min_{c\in C}\|h-c\|$.

We have not yet used that $C$ is convex and that $H$ is a Hilbert space. Convexity could mean we have to use that ${1\over 2}(c_0 + c_0')\in C$ but I can't proceed from there. That $H$ is a Hilbert space could mean we have to use the parallelogram identity. I tried to apply it to $\|c_0 - c_0'\|^2 = \|(c_0 - h) - (c_0'-h)\|^2$ to show it is zero but couldn't finish the idea either.

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    $\begingroup$ Unfortunately your existence proof is incorrect; it is not true in general that the infimum over $c\in C$ where $C$ is closed of $\|h-c\|$ is attained. The proof of the projection theorem relies on the fact that if $s,t\in C$ then $\frac12(s+t)\in C$ (this is the convexity bit) and uses Cauchy sequences and a completeness argument. $\endgroup$ – Jason Jan 19 '15 at 3:40
  • $\begingroup$ @Jason, do you know of a counterexample illustrating this point? I am having a difficult time imagining a scenario in Euclidean space where the minimum distance to a closed set is not attained. Possibly a counterexample requires a more exotic Hilbert space? $\endgroup$ – Him Nov 26 '19 at 18:04
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    $\begingroup$ @Scott You always attain a minimum distance in Euclidean space since bounded closed sets are compact. For a counterexample in a Hilbert space, let $H=\ell^2$, $h=0$, and $C=\{(1+\frac1n)e_n:n\in\mathbb N\}$ where $\{e_n\}$ is the standard basis. Then $C$ is closed, and $\inf_{c\in C}\|h-c\|=1$, but this distance is never attained. $\endgroup$ – Jason Nov 27 '19 at 18:03

I alluded to a method in my comment, but I guess I'll leave it here as an answer. Let $\delta=\inf_{c\in C}\|c-h\|$. Suppose $x,y\in C$ are such that $\|x-h\|=\|y-h\|=\delta$. Using the parallelogram identity, we have $$\|x-y\|^2=2\|x-h\|^2+2\|y-h\|^2-\|(x-h)+(y-h)\|^2=4\delta^2-4\|\frac{x+y}2-h\|^2.$$ Using the fact that $C$ is convex, $\frac{x+y}2\in C$ so $\|\frac{x+y}2-h\|\geq\delta$. From here we can deduce $\|x-y\|^2=0$.

Let $(x_n)$ be a sequence in $C$ with $\|x_n-h\|\to\delta$. Fix $\epsilon>0$. There exists $N>1$ such that if $n\geq N$, $\|x_n-h\|^2<\delta^2+\frac{\epsilon^2}2$. From here, you should be able to use similar steps to the above to show that $(x_n)$ is Cauchy. If $x=\lim_{n\to\infty}x_n$, it is quite simple to show $\|x-h\|=\delta$. I'll leave the last few details to you.

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  • $\begingroup$ Thank you... maybe I see my mistake: to argue for existence like I did we need $C$ to be compact, not closed, right? $\endgroup$ – user167889 Jan 19 '15 at 4:13
  • $\begingroup$ Yes, that's right. If $C$ is compact, then by the continuity of $f(x)=\|x-h\|$ we have that $f(C)\subset\mathbb{R}$ is compact, and thus $\inf f(C)\in f(C)$. $\endgroup$ – Jason Jan 19 '15 at 6:45
  • $\begingroup$ Why does the sequence $(x_n)$ exist? $\endgroup$ – plebmatician Dec 10 '17 at 12:57
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    $\begingroup$ @plebmatician By definition of infimum, there exists $x_n\in C$ with $\|x_n-h\|\le\delta+\frac1n$. $\endgroup$ – Jason Dec 10 '17 at 18:10

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