# Smallest Norm in a Hilbert Space

1. Apr 25, 2005

### Oxymoron

I have this problem which I want to do before I go back to uni. The context was not covered in class before the break, but I want to get my head around the problem before we resume classes. So any help on this is greatly appreciated.

Question

Suppose $C$ is a nonempty closed convex set in a Hilbert Space $H$.

(i) Prove that there exists a unique point $c_0 \in C$ of smallest norm, and that we then have $\Re \langle c_0 \, | \, c-c_0\rangle \geq 0$ for all $c \in C$.

(ii) For any point $x_0 \in H$ show that there is a unique closest point $c_0$ of $C$ to $x_0$, and that it satisfies the
variational inequality $\Re \langle x_0 - c_0 \,|\, c -c_0\rangle \leq 0$ for all $c \in C$

Last edited: Apr 25, 2005
2. Apr 25, 2005

### Oxymoron

Solution (i)

Since we are in a Hilbert space I figured that using the distance metric and perhaps the parallelogram law will be used.

Let $d$ be the smallest distance from $C$ to $c$. This can also be written as

$$d = \inf d(x,y)$$

The first thing I thought was, how can I make such a bold statement? Does $d$ even exist? But I assured myself that there is such a $d$ because we made the condition that $C$ was a closed convex set, therefore we can make $d$ as small as we like. If $C$ wasn't closed convex then there would be trouble at the boundaries.

Anyway, the next step involves setting up a sequence of elements of $C$, like the sequence of $y_n \in C$. This enables us to introduce the parallelogram law.

$$\|c - y_n\| = d(x,y_n) \rightarrow d$$.

So the distance from $c$ to every element $y_n$ in the sequence approaches the smallest distance $d$ the further along the sequence we go. But this can only happen if $(y_n)$ is a Cauchy sequence. We know that Cauchy sequences behave like this in Hilbert Spaces, and the easiest way I know of proving $(y_n)$ is a Cauchy sequence is to use the Parallelogram Law and prove that it approaches 0 as $n \rightarrow 0$. So we have

$$\|y_n - y_m\|^2 &=& 2\|y_n - c\|^2 + 2\|y_m - c\|^2 - \|(y_n+y_m)-2c\|^2$$

$$\leq 2\|y_n - c\|^2 + 2\|y_m - c\|^2 - 4d^2$$

$$\rightarrow 0$$

Remember that $d$ is our minimum distance. So we know $(y_n)$ is a Cauchy sequence. And we also know that $C$ is closed. Hence $C$ is a complete space and so $y_n \rightarrow c_0 \in C$. That is, the Cauchy sequence approaches the unique point $c_0$ as $n \rightarrow \infty$.

It follows, by the continuity of the norm, that

$$\|c-c_0\| = d$$

Which says that there exists a distance $d$ from any arbitrary point $c \in C$ to the point $c_0 \in C$ that is minimum, that is there exists a smallest norm.

Last edited: Apr 25, 2005