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Let $H$ be a Hilbert space and $C \subset H $ be a convex, closed and nonempty subset of $H$.

Prove that there exists a unique $x_0\in C$ with minimum norm among the elements of $C$.

I don't know how to prove this. I tried to consider the set $\{y\in \mathbb{R}:y=||x||, x\in C\}$, and since this set is bounded below, there exists an infimum of the set, say $s$. Then we can find a sequence $x_n\in C$ such that $s\le ||x_n|| \le s+1/n$. By the continuity of the norm function and closedness of $C$, we can find the limit of $x_n$, say, $x$ to be in $C$.

But I don't know how to show that this must be unique. How can I prove this? I would greatly appreciate any help.


I've come up with a solution.

By convexity $\frac{1}{2}(x_1+x_2)\in C$, and so $s\le ||\frac{1}{2}(x_1+x_2)||\le 1/2||x_1||+1/2||x_2||=s$.

Hence we have $||x_1+x_2||=2s$. By Parallelogram law,

$||x_1+x_2||^2+||x_1-x_2||^2=2||x_1||^2+2||x_2||^2$. And so $||x_1-x_2||^2=4s^2-4s^2=0.$ Thus $x_1=x_2$.

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  • $\begingroup$ Google search with keyword POCS (Projection On Convex Sets). $\endgroup$ – Jean Marie Apr 24 '16 at 13:27
  • $\begingroup$ Can you check if my solution is correct? $\endgroup$ – nomadicmathematician Apr 24 '16 at 13:34
  • $\begingroup$ I extended my answer. Your uniqueness proof is fine as well, regarding existince I added a comment to my answer. $\endgroup$ – Thomas Apr 24 '16 at 14:12
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There are two cases: $0\in C$ and $0\notin C$. In the first case the minimum norm is $=0$, and $x_0=0$ is the unique element with this property.

If $x_0$ with minimum norm has already been found (and your question indicates that this is already known to you) assume $x_1\neq x_0$ has the same norm and it is an element of $C$. It is easy to see that $x_1$ and $x_0$ cannot be collinear, since then they either would not have the same norm or $0\in C$. By convexity $tx_1+ (1-t)x_0 \in C $ for each $t\in [0,1]$. Now, for $t=1/2$ $$|| \frac{1}{2} (x_1 + x_0) ||^2 = \frac{1}{4} (||x_1||^2 + ||x_0||^2) +\frac{1}{2} \langle x_0, x_1\rangle $$ By Cauchy-Schwartz the right hand side is strictly $<||x_0||^2$, since $x_0,x_1$ are not collinear, contradicting the minimum property of $x_0$.

Edit: Additional remarks in response to comments: Actually I have to admit I'm assuming a real hilbert space here.

If $x\neq y$ are collinear, $x= \lambda y$. If $\lambda = \pm 1$, then either $x=y$ which has been excluded by assumption, or $x=-y$, which means the line joining $x$ and $y $ passes through $0$.

Your reasoning for $||x_1|| = \frac{1}{2} ||x_1+x_0||$ uses the fact that $||x_1||$ has minimal norm in $C$. My estimate shows that this cannot be correct, which is exactly the contradiction I'm aiming at (there are several ways to get to this conclusion)

(To understand what I'm doing geometrically consider $C\subset \mathbb{R}^2$. If two points have the same distance to the origin they are both lying on the boundary of a sphere, and the convex set lies outside, which contradits the intuition of convexity. The line joining the points will, of course, intersect the ball the boundary of which is the sphere).

Regarding existence, just a sketch (sorry, I'm short of time): The existence parts needs a bit additional reasoning. The usual reasoning is that a minimizing sequence is norm bounded, hence has a weakly convergent subsequence. Then first the norm is lower semicontinuos wrt to weak convergence (so the limit element has lower norm) and convex sets are closed wrt to weak convergence (so the limit is in $C$)

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    $\begingroup$ Sorry I don't follow some parts of your argument. First, why do $x_1$ and $x_2$ not have the same norm when they are collinear? Actually, we should have $||1/2(x_1+x_2)||=||x_1||$, since $||x_1||\le ||1/2(x_1+x_2)||\le 1/2||x_1||+1/2||x_2||=||x_1||$. And we're assuming complex Hilbert space here so the last should be $1/2<x_0,x_1>+1/2<x_1,x_0>$. $\endgroup$ – nomadicmathematician Apr 24 '16 at 13:28
  • $\begingroup$ Also, the question also asked to show that there is a minimum norm. Is my argument using sequences correct? $\endgroup$ – nomadicmathematician Apr 24 '16 at 13:29
  • $\begingroup$ And I've just added my answer to the uniqueness part. Can you see verify if it is correct? $\endgroup$ – nomadicmathematician Apr 24 '16 at 13:35
  • $\begingroup$ @takecare the uniqueness part looks fine to as mentioned in my comment and even covers the complex case (which mine does not do out of the box, need to check whether this can be fixed). $\endgroup$ – Thomas Apr 24 '16 at 14:24
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It seems that the uniqueness part has already been settled. In your argument regarding the existence part, you have proved that if the sequence $(x_n)_{n=1}^\infty$ you have constructed converges, then its limit belongs to $C$. But so far you have only that $\|x_n\| \rightarrow s$. It remains to be shown that $(x_n)_{n=1}^\infty$ does converge. To this end, I suggest you proceed as follows: let $\epsilon > 0$ be given. Choose $r$ to be the positive root of the equation $$r^2 + 2rs - \frac{\epsilon^2}{4} = 0.$$ Since $\|x_n\| \rightarrow s$, as $n \rightarrow \infty$, there is an integer $N$ for which $s \leqq \|x_n\| < s + r$ for all $n \geqq N$. If $m \geqq N$ and $n \geqq N$, then $$\Big\|\frac{x_m - x_n}{2}\Big\|^2 = 2\Big\|\frac{x_m}{2}\Big\|^2 + 2\Big\|\frac{x_n}{2}\Big\|^2 - \Big\|\frac{x_m+x_n}{2}\Big\|^2 < \frac{(s+r)^2}{2} + \frac{(s+r)^2}{2} - s^2 = 2sr + r^2 =\frac{\epsilon^2}{4}. $$ This shows that $(x_n)_{n=1}^\infty$ is a Cauchy sequence. Since $H$ is a Hilbert space, this sequence must converge.

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