# Smallest Norm in a Hilbert Space

• Oxymoron
In summary, we can prove the existence and uniqueness of a point c_0 \in C that has the smallest norm d, given that C is a nonempty closed convex set in a Hilbert Space H. Additionally, we can also prove that for any c \in C, we have \Re \langle c_0 \, | \, c-c_0\rangle \geq 0, satisfying the variational inequality.
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 $c \in C$

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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.

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.

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

## 1. What is the definition of the smallest norm in a Hilbert space?

The smallest norm in a Hilbert space is the smallest possible value for the norm of a vector in that space. It represents the shortest distance between the origin and the vector, and is calculated by taking the square root of the sum of the squares of the vector's components.

## 2. How is the smallest norm in a Hilbert space used in mathematical analysis?

The smallest norm in a Hilbert space is used to measure the size or magnitude of a vector in that space. It is often used in mathematical analysis to determine the convergence or divergence of a sequence of vectors, as well as in optimization problems to find the minimum value of a function.

## 3. Can the smallest norm in a Hilbert space be negative?

No, the smallest norm in a Hilbert space is always a positive value. This is because the norm is defined as the length or magnitude of a vector, and length cannot be negative. The smallest norm represents the shortest distance between the origin and the vector, so it cannot be negative.

## 4. What is the relationship between the smallest norm and the inner product in a Hilbert space?

The smallest norm and the inner product in a Hilbert space are closely related. The inner product is used to calculate the norm of a vector, and the smallest norm is the result of taking the square root of the inner product. This means that the smallest norm is a function of the inner product, and the two are inseparable in Hilbert spaces.

## 5. How is the smallest norm in a Hilbert space affected by orthogonal projections?

Orthogonal projections have a significant impact on the smallest norm in a Hilbert space. When a vector is projected onto a subspace, the smallest norm is reduced because the projected vector is closer to the origin. In fact, the smallest norm is minimized when the vector is projected onto an orthogonal basis for the subspace.

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