Question

Let \(N\) be a maximal \(\varepsilon\) -separated set in the unit sphere of a Banach space \(X\). Show that \((1-\varepsilon) B_{X} \subset \overline{\operatorname{conv}}(N)\) Hint: Otherwise, by the separation theorem, we find \(x \in X\) and \(f \in S_{X^{*}}\) with \(\|x\| \leq 1-\varepsilon\) and \(f(x)>\sup _{\operatorname{conv}(N)}(f)>\sup _{N}(f) .\) For \(\delta>0\), choose \(y \in S_{X}\) such that \(f(y)>1-\delta .\) By the maximality of \(N\), there exists \(z \in N\) with \(\varepsilon>\|y-z\| \geq f(y)-f(z)\). Thus \(\sup _{N}(f) \geq f(z)>f(y)-\varepsilon>1-\delta-\varepsilon\) This holds for any \(\delta>0\), so we have \(1-\varepsilon \leq \sup _{N}(f)

Short answer

(1-\varepsilon) B_{X} \subset \overline{\operatorname{conv}}(N).

Step-by-step solution

Understand the problem

We're required to show that \text{N\( (1-\varepsilon) B_{X} \subset \overline{\operatorname{conv}}(N) \) } in the context of Banach spaces and \text{ \(\varepsilon \)-separated sets. Let's break it down.

Assume the contrary

Assume that \text{ \( (1-\varepsilon) B_{X} ot\subset \overline{\operatorname{conv}}(N) \). } According to this assumption, there exists some \text{ \( x \in (1-\varepsilon) B_{X} \) } such that \text{ \( x otin \overline{\operatorname{conv}}(N) \). }

Separation theorem

By the separation theorem, there exists \text{ \( f \in S_{X^{*}} \) } such that \text{ \( f(x) > \sup _{\operatorname{conv}(N)}(f) > \sup _{N}(f) \). } Given \text{ \( \|x\| \leq 1-\varepsilon \). }

Choosing \text{ y }

For any \text{ \( \delta > 0 \), choose \( y \in S_{X} \) such that } \text{ \( f(y) > 1 - \delta \). }

Maximality of \text{ N}

Due to the maximality of the \text{ \( \varepsilon \)-separated set \( N \), there exists \( z \in N \) such that }\text{\( \varepsilon > \|y - z\| \geq f(y) - f(z). \)}

Inequality relations

Thus, it happens that \text{ \( \sup _{N}(f) \geq f(z) > f(y) - \varepsilon > 1 - \delta - \varepsilon \). This applies for any \( \delta > 0 \) meaning }\text{ \(1-\varepsilon \leq \sup _{N}(f) < f(x) \leq \|x\| \leq 1-\varepsilon.\)}

Conclusion

The inequality \text{ \( 1-\varepsilon < 1-\varepsilon \) } leads to a contradiction, hence implying \text{ \( (1-\varepsilon) B_{X} \subset \overline{\operatorname{conv}}(N) \). }

Key concepts

separation theorem

The Separation Theorem is a fundamental concept in functional analysis. This theorem states that given two disjoint convex sets, there exists a hyperplane that separates them. In simpler terms, we can always find a linear functional that takes different values on these sets.

In the context of the problem, the separation theorem was used to demonstrate that if \((1-\varepsilon) B_{X}\) was not a subset of the closure of the convex hull of \(N\), then there would be a functional \(f\) in the dual space \(S_{X^{*}}\) that could differentiate between an element of \((1-\varepsilon) B_{X}\) and the elements of \(\text{conv}(N)\).

This leads to a logical contradiction, confirming that \((1-\varepsilon) B_{X} \subset \overline{\text{conv}}(N)\).

unit sphere in Banach space

A Banach space is a complete normed vector space. The unit sphere in a Banach space \(X\) is the set of all vectors with a norm of 1. Formally, it is defined as \S_X = \{ x \in X : \|x\| = 1 \}.

In this problem, we consider a set of points on the unit sphere of a Banach space that is maximal and \(\text{ \varepsilon\}-separated\). This means that within the unit sphere, no two points in this set are closer than a distance \(\text{ \varepsilon\}\) apart, and you cannot add another point to this set without violating this condition.

Such sets help in approximating and studying the geometric structure of Banach spaces. In our proof, they provide a framework to use functional analysis concepts like the separation theorem.

convex hull

The convex hull of a set of points \N\ in a vector space is the smallest convex set that contains \N\. Essentially, it is formed by taking all convex combinations of points in \N\. If \N\ consists of points \{x_1, x_2, \dots, x_k \}\, then the convex hull can be denoted as
\[ \text{conv}(N) = \left\{ \sum_{i=1}^k \lambda_i x_i \mid \lambda_i \geq 0, \sum_{i=1}^k \lambda_i = 1 \right\}. \]
In the problem, we denote the closure of the convex hull as \overline{\text{conv}}(N)\. This means including all limit points of sequences of points in \text{conv}(N)\.

Understanding the convex hull is crucial here because proving that \( (1-\text{ \varepsilon\}) B_X \subset \overline{\text{conv}}(N)\) essentially shows that even if we take the set of points maximally spaced by \(\text{ \varepsilon\}\) on the unit sphere, their convex combinations are dense enough to approximate elements scaled by \(1-\text{ \varepsilon\} \) within the unit ball of the Banach space.