Smallest Norm in a Hilbert Space

[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 $c \in C$

 

[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.

 

[thepatientmental]

.




The concept of the smallest norm in a Hilbert Space can be quite challenging to understand, but with some practice and guidance, it can be easily grasped. Let's break down the question into smaller parts to better understand it.

Firstly, we are given a nonempty, closed, and convex set C in a Hilbert Space H. This means that C is a subset of H, and it contains all its limit points (closed) and any line segment connecting two points in C is also within C (convex).

(i) To prove the existence of a unique point c_0 with the smallest norm, we need to use the properties of the Hilbert Space. By definition, a Hilbert Space is a complete inner product space, meaning that it satisfies the Cauchy-Schwarz inequality and every Cauchy sequence in H converges to a point in H. Using this, we can show that the set C is bounded, since the norms of all its elements are bounded by the norms of its limit points. Then, by using the Bolzano-Weierstrass theorem, we can find a sequence of points in C that converges to a point c_0 with the smallest norm. This point is unique because if there were two points with the smallest norm, their average would also have the smallest norm, contradicting the uniqueness.

Furthermore, we can show that \Re \langle c_0 \, | \, c-c_0\rangle \geq 0 for all c \in C, by using the Cauchy-Schwarz inequality. This inequality states that for any two vectors x and y in a Hilbert Space, we have \Re \langle x \, | \, y\rangle \leq \|x\| \|y\|. Therefore, for any c \in C, we have \Re \langle c_0 \, | \, c-c_0\rangle \leq \|c_0\| \|c-c_0\| = \|c_0\|^2, which shows that \Re \langle c_0 \, | \, c-c_0\rangle \geq 0.

(ii) Now, for any point x_0 \in H, we need to show that there is a unique closest point c_0 of C to x_0, and that it satisfies the variational inequality \Re \langle