Question

Let \(X\) be a Banach space and \(C\) be a compact set in \(X .\) Is it true that \(\operatorname{conv}(C)\) is compact? Hint: Not in general. Consider \(C=\left\\{\frac{1}{i} e_{i}\right\\} \cup\\{0\\}\) in \(\ell_{2}\), where \(e_{i}\) are the standard unit vectors. Clearly, \(C\) is compact. The vector \(\left(2^{-i \frac{1}{i}}\right)\) is in \(\overline{\operatorname{conv}(C)}\) and it is not in \(\operatorname{conv}(C)\), since any point in \(\operatorname{conv}(C)\) is finitely supported.

Short answer

No, \(\backslash\text{conv}(C)\) is not necessarily compact.

Step-by-step solution

Understand the Problem Statement

The goal is to determine whether the convex hull of a compact set in a Banach space is also compact. We are given a specific hint to consider a set in \(\ell_2\) space to analyze this.

Define the Given Variables

Let \(X\) be a Banach space and \(C\) be a compact set in \(X\). We need to check whether \(\backslash\text{conv}(C)\) is compact.

Consider the Specific Hint

Consider \(C=\{ \frac{1}{i} e_i \}\backslashcup\{0\}\) in \ell_2\, where \(e_i\) are standard unit vectors. The set \(C\) is compact because it is finite (and the finite union of compact sets is compact).

Identify a Point in \(\backslash\text{conv}(C)\)

The convex hull \(\backslash\text{conv}(C)\) consists of all finite convex combinations of elements of \(C\). However, consider a particular sequence \(\left(2^{-i \frac{1}{i}}\right)\).

Analyze the Sequence \((2^{-i \frac{1}{i}})\)

Even though this sequence is in the closure \(\backslash\text{conv}(C)\), it is not in \(\backslash\text{conv}(C)\) itself because elements in \(\backslash\text{conv}(C)\) are finitely supported (i.e., have only finitely many non-zero components). Thus, \((2^{-i \frac{1}{i}})\) cannot be represented as a finite convex combination of elements of \(C\).

Conclude the Solution

Since there exists a point in the closure of the convex hull that is not in the convex hull itself, \(\backslash\text{conv}(C)\) is not compact. Thus, the convex hull of a compact set in a Banach space is not necessarily compact.

Key concepts

compact sets

In mathematics, a set is called 'compact' if every open cover of the set has a finite subcover. This is a bit technical, so let’s break it down. Think of an open cover as a way to 'cover' the set using open intervals or regions. If you can find a finite number of these open intervals that still cover the entire set, then the set is compact.
Compact sets have several nice properties. For one, they are always closed and bounded in \(\mathbb{R}^n\). They are also important because continuous functions, when applied to compact sets, yield very manageable and predictable results.
In this exercise, we considered a specific compact set \(C\) in a Banach space, defined by \(C=\{ \frac{1}{i} e_i \} \cup \{ 0 \}\), which is compact because it consists of a finite or countable union of points and includes the zero vector, making it closed and bounded.

Banach spaces

A Banach space is a complete normed vector space. Let's unpack this:
1. **Normed Vector Space**: This is a vector space equipped with a function called a 'norm' that measures the 'size' or 'length' of the vectors. An example of a norm could be the Euclidean norm, \(||x||_2 = \sqrt{x_1^2 + x_2^2 + ... + x_n^2} \).
2. **Complete**: Completeness means that every Cauchy sequence in the space converges to a limit that is also within the space. A Cauchy sequence is one where the elements get arbitrarily close to each other as the sequence progresses.
In our context, \( \ell_2 \) (the space of square-summable sequences) is an example of a Banach space. Elements in \( \ell_2 \) are sequences whose infinite series of squared elements sum to a finite value.

convex combinations

A convex combination involves creating a new point by taking a weighted average of other points, where the weights (or coefficients) are non-negative and sum up to 1. Formally, a convex combination of points \(x_1, x_2, ..., x_n\) is \( \lambda_1 x_1 + \lambda_2 x_2 + ... + \lambda_n x_n\) where \( \lambda_i \geq 0\) and \( \sum_{i=1}^n \lambda_i = 1\).
In our problem, the convex hull \( \operatorname{conv}(C) \) consists of all possible convex combinations of the elements of set \C\. This means any point within \( \operatorname{conv}(C) \) can be written as a sum of the elements of \C\, weighted by some non-negative coefficients that sum to 1.
However, an important subtlety is that while \( \operatorname{conv}(C) \) contains all finite convex combinations, it doesn't necessarily include all possible sequences of such combinations, which ties into the idea of compactness in this specific exercise.

sequence analysis

Sequence analysis is the study of sequences and their properties. In this context, we’re particularly interested in how sequences behave inside a Banach space. Let’s consider the sequence \( \left(2^{-i \frac{1}{i}}\right) \):
1. **Finiteness and Support**: A key term here is 'finitely supported', meaning a sequence has only a finite number of non-zero elements. The sequences in \( \operatorname{conv}(C) \) must be finitely supported.
2. **Limit Points**: The point \( \left(2^{-i \frac{1}{i}}\right) \), while in the closure of \( \operatorname{conv}(C) \), is not finitely supported. This sequence can’t be represented as a finite convex combination of points in \C\.
3. **Closure**: The closure of \( \operatorname{conv}(C) \) includes all limit points of sequences in \( \operatorname{conv}(C) \). However, since \( \left(2^{-i \frac{1}{i}}\right) \) is not in \( \operatorname{conv}(C) \) itself, we conclude that \( \operatorname{conv}(C) \) is not compact.