Question

Let \(X\) be a uniformly convex space with modulus of convexity \(\delta(\varepsilon)\). Take \(f \in S_{X^{*}}\) and consider the affine hyperplane \(H=\left(f^{-1}(0)+z\right)\) for some \(z \in X\). Show that if \(\operatorname{dist}(0, H) \geq 1-\delta(\varepsilon) / 2\), then \(\operatorname{diam}\left(B_{X} \cap H\right) \leq \varepsilon\).

Short answer

Since \(\operatorname{dist}(0,H) \geq 1-\frac{\delta(\varepsilon)}{2}\), the uniform convexity property ensures that \(\operatorname{diam}(B_X \cap H) \leq \varepsilon\).

Step-by-step solution

Understand the Problem

We are given a uniformly convex space with modulus of convexity \(\delta(\varepsilon)\) and we have to show that if \(\operatorname{dist}(0, H) \geq 1-\frac{\delta(\varepsilon)}{2}\), then \(\operatorname{diam}(B_X \cap H) \leq \varepsilon\).

Recall Key Definitions and Properties

Uniformly convex spaces have the property that for every \(\varepsilon > 0\) there exists a \(\delta(\varepsilon)\) such that if \(\Vert x \Vert \leq 1\), \(\Vert y \Vert \leq 1\), and \(\Vert \frac{x + y}{2} \Vert \geq 1 - \delta(\varepsilon)\), then \(\Vert x - y \Vert \leq \varepsilon\).

Calculate the Distance

The distance from the origin to the hyperplane \(H\) is given by \(\operatorname{dist}(0, H) = \operatorname|f(z)\|\). Given \(\operatorname|f(z)\| \geq 1 - \frac{\delta(\varepsilon)}{2}\), we use this to understand how far the affine hyperplane is.

Introduce the Balls Intersection

Consider \(x, y \in B_X \cap H\) with \(\Vert x \Vert \leq 1\) and \(\Vert y \Vert \leq 1\), we seek to show \(\operatorname{diam}(B_X \cap H) = \sup \{ \Vert x - y \Vert : x, y \in B_X \cap H \} \leq \varepsilon \).

Apply the Modulus of Convexity

Use the property of the uniformly convex space: if \(\Vert \frac{x + y}{2} \Vert \leq 1\) and \(\Vert \frac{x + y}{2} \Vert \geq 1 - \delta(\varepsilon)\), then \(\Vert x - y \Vert \leq \varepsilon\). As \(x + y \in H\) and since the distance condition is satisfied, it enforces \(\Vert x - y \Vert \leq \varepsilon\).

Conclude

Given the conditions and the application of the modulus of convexity, we derive that \(\operatorname{diam}(B_X \cap H) \leq \varepsilon\).

Key concepts

Modulus of Convexity

Let’s dive into the modulus of convexity first, which is a key property defining uniformly convex spaces. A space is uniformly convex if, for every \(\varepsilon \gt 0\), there exists a \(\delta(\varepsilon)\) such that whenever two points \(x\) and \(y\) within the unit ball (i.e., their norms are less than or equal to 1) are averaged and their midpoint stays sufficiently within the unit ball, the distance between the two points is restricted to stay within \(\varepsilon\).
In simpler terms, if means that if we take any two points in the space, we can't have them be too far from each other unless their average gets very close to the boundary of the ball. This concept ensures the