📂Banach Space

Uniform Convexity

Definitions¹

Let’s call $(X, \left\| \cdot \right\|)$ a normed space. We say that the norm $\left\| \cdot \right\|$ of $X$ is uniformly convex if it satisfies the following condition:

• For every $\epsilon$ where $0 \lt \epsilon \le 2$, there exists a positive number $\delta (\epsilon) \gt 0$ such that if $x,y \in X$ and $\| x \| = \|y\| = 1$, $\| x-y\| \ge \epsilon$ then it satisfies $\|( x+y)/2 \| \le 1-\delta (\epsilon)$.

In this case, the normed space $X$ itself is also considered to be uniformly convex. If a normable space has a uniformly convex norm, it is also said to be uniformly convex.

Explanation

It is important to note that just because some norm defined on $X$ is uniformly convex, does not mean that another equivalent norm is also uniformly convex.

Theorem

Hilbert space is uniformly convex.

Proof

Let us assume that a positive number $\epsilon$ is given where $0< \epsilon \le 2$. Assume that $H$ is a Hilbert space, and that $x,y \in H$, $\| x\|=\| y\|=1$, and $\| x-y\| \ge \epsilon$.

Parallelogram Law(../1842)

$$\| x+ y \|^2 + \| x - y \|^2 = 2 \left( \| x \|^2 + \| y \|^{2} \right)$$

Substituting $x, y$ into the parallelogram law gives

$$\| x+ y \|^2 + \| x - y \|^2 = 2 \left( \| x \|^2 + \| y \|^{2} \right)$$

$$\implies \|x+y\|^2 =4-\|x-y\|^2\le 4-\epsilon^2$$

$$\implies \|x+y\|^2\le 4-\epsilon^2$$

Summarizing,

$$\| x+y\| \le \sqrt{4-\epsilon^2} \quad \implies \left\| \frac{x+y}{2} \right\| = \dfrac{1}{2}\|x+y \| \le \frac{1}{2}\sqrt{4-\epsilon^2}$$

Since $0 \lt \epsilon \le 2$, the range on the right-hand side is $0 \le \dfrac{1}{2}\sqrt{4-\epsilon^2} \lt 1$. Therefore, as the Hilbert space is a complete space, it can be represented with some positive number $\delta (\epsilon)$ depending on $\epsilon$ as follows.

$$\left\| \frac{x+y}{2} \right\| \le 1-\delta (\epsilon)$$

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