Suppose that C is a closed convex subset of $\mathbb R^n$ and $x \in \mathbb R^n$. The projection of $\mathbf x$ onto C is the closest point $\mathbf y \in C : \mathbf z = \mathbf y$ minimizes ||$\mathbf z - \mathbf x$||$_2$ over all $\mathbf z \in C$.
(a) Show that the minimum exists.
(b) Show that there is only one minimizer: that is, show that if $\mathbf y_1$ and $\mathbf y_2$ both minimize ||$\mathbf x - \mathbf z$|| then $\mathbf y_1 = \mathbf y_2$
My ideas:
For (a), to show that the minimum exists I will need to show that it is closed and bounded. We already know it's closed because C is a closed convex subset so I just need to show it's bounded. Am I wanting to minimize ||$\mathbf x - \mathbf z$||$_2$? This is just the Euclidean norm which would be $\sqrt{(\mathbf z - \mathbf x)^2}$=$\sqrt{\mathbf z^2 -\mathbf z \mathbf x -\mathbf x \mathbf z + \mathbf x^2}$. I'm not sure where to go from here...
For (b) I want to use proof by contradiction- so I will assume that $\mathbf y_1$ and $\mathbf y_2$ both minimize the function and then I'll somehow see that they are in fact equal. How do I set up this argument?
For (a), the set $C$ may not be bounded so you need to do a little more work. I make the additional assumption that $C$ is nonempty. Let $z_0\in C$. If $\|x-z\|_2<\|x-z_0\|$ then $z\in \overline{B_{\|x-z_0\|}(z_0)}\cap C$, which is closed and bounded. Hence $\|z-x\|$ has a minimum in $\overline{B_{\|x-z_0\|}(z_0)}\cap C$, which must be the minimum in $C$.
For (b), suppose $\|x-z_1\|$ and $\|x-z_2\|$ are minima, and consider the line $f(t)=tz_2+(1-t)z_1$ which is contained in $C$ by convexity. But $\|x-f(t)\|^2$ is a nonnegative quadratic, so it has a unique minimum which must be between $0$ and $1$ since $\|x-f(0)\|^2=\|x-f(1)\|^2$. But this contradicts the minimality of $\|x-z_1\|$ and $\|x-z_2\|$.