Question

Let \(A\) be a subset of a Banach space \(X .\) Denote by sconv \((A)\) the set of all \(x \in X\) that can be written as \(x=\sum_{i=1}^{\infty} \lambda_{i} x_{i}\), where \(x_{i} \in A, \lambda_{i} \geq 0\), and \(\sum \lambda_{i}=1\). Show that if \(A\) is bounded, then sconv \((A) \subset \overline{\operatorname{conv}}(A)\) Let \(A\) be the set of all standard unit vectors \(e_{i}\) in \(\ell_{2} .\) Show that \(0 \in\) \(\overline{\operatorname{conv}}(A)\) and \(0 \notin \operatorname{sconv}(A)\) Hint: If \(\lambda_{i} \geq 0\) and \(\sum_{i=1}^{\infty} \lambda_{i}=1\), approximate by a finite linear combination given by \(\tilde{\lambda}_{1}=\lambda_{1}, \ldots, \tilde{\lambda}_{n}=\lambda_{n}, \tilde{\lambda}_{n+1}=1-\sum_{i=1}^{n} \lambda_{i}\) If \(\sum \lambda_{i} e_{i}=0\), then \(\lambda_{i}=0\) for every \(i\)

Short answer

If A is bounded, then \(\operatorname{sconv}(A) \subset \overline{\operatorname{conv}}(A)\). For the unit vectors in \(\ell_2\), 0 is in the closure but not in the infinite convex hull.

Step-by-step solution

- Definition of sconv(A)

Recall the definition: the set sconv (A) consists of all points in the Banach space X that can be expressed as an infinite convex combination of points in A, specifically, \(x = \sum_{i=1}^{\infty} \lambda_i x_i\) where \(x_i \in A\), \(\lambda_i \geq 0\), and \(\sum_{i=1}^{\infty} \lambda_i = 1\).

- Definition of conv(A)

The convex hull of a set A, denoted as conv(A), is the set of all finite convex combinations of points in A. The closed convex hull, \(\overline{\operatorname{conv}}(A)\), is the closure of conv(A). It consists of all limits of sequences of points in conv(A).

- Boundedness of A

Next, observe that since A is bounded, there exists a constant M such that \(||x_i|| \leq M\) for all \(x_i \in A\). This implies that any convex combination of points in \(A\) will also have a norm bounded by M.

- Inclusion in \(\overline{\operatorname{conv}}(A)\)

To show that \(\operatorname{sconv}(A) \subset \overline{\operatorname{conv}}(A)\), consider any \(x \in \operatorname{sconv}(A)\). Since \(x = \sum_{i=1}^{\infty} \lambda_i x_i\) and A is bounded, approximate \(x\) by finite sums \(\sum_{i=1}^{n} \lambda_i x_i + \tilde{\lambda}_{n+1} x_{n+1}\) where \(\tilde{\lambda}_{n+1} = 1 - \sum_{i=1}^{n} \lambda_i\). This finite sum lies in \(\operatorname{conv}(A)\), and by taking limits, \(x\) belongs to \(\overline{\operatorname{conv}}(A)\).

- Unit Vectors in \(\ell_2\)

Now let A be the set of all standard unit vectors \(e_i\) in \(\ell_2\). Each \(e_i\) has norm 1 and they form an orthonormal basis for \(\ell_2\).

- Convex Combinations of \(e_i\)

Any finite convex combination of \(e_i\)’s will lie in the convex hull of A. Since the sum of any finite number of \(e_i\)’s (with the coefficients summing to 1) is bounded in \(\ell_2\), the limit of such finite combinations can approach 0. Hence, \(0 \in \overline{\operatorname{conv}}(A)\).

- Check if 0 belongs to \(\operatorname{sconv}(A)\)

Assume \(0 \in \operatorname{sconv}(A)\). Then there exists an infinite sum \(\sum_{i=1}^{\infty} \lambda_i e_i = 0\), where \(\lambda_i \geq 0\) and \(\sum_{i=1}^{\infty} \lambda_i = 1\). However, the only way for this infinite series to sum to 0 is if all \(\lambda_i\)'s are 0. Since this contradicts \(\sum_{i=1}^{\infty} \lambda_i = 1\), it follows that \(0 otin \operatorname{sconv}(A)\).

Key concepts

Banach space

A Banach space is a special type of vector space equipped with a norm. The norm allows us to measure the size or length of vectors in the space. Additionally, Banach spaces are complete, meaning that any Cauchy sequence (a series of elements that get arbitrarily close to each other) in the space will always have a limit that also lies within the space. This completeness property is crucial for many functional analysis applications.
Common examples of Banach spaces include spaces of continuous functions and \( \ell_2\) space.

convex hull

The convex hull of a set A, denoted as conv(A), is the smallest convex set that contains all the points of A. A set is called convex if, for any two points in the set, the line segment connecting them also lies entirely within the set.
To understand convex hulls:

• Imagine stretching a rubber band around the outermost points of A; the shape formed is the convex hull.
• Any point within this hull can be written as a convex combination of points in A, i.e., as \(x = \sum_{i=1}^n \lambda_i x_i\) where \( \lambda_i \geq 0\) and \( \sum_{i=1}^n \lambda_i = 1\).

The closed convex hull, denoted \( \overline{\operatorname{conv}}(A)\), includes all the points that can be approached by sequences within conv(A).

sconv(A)

The set \( \operatorname{sconv}(A) \) (or the strong convex hull of A) is a bit more complicated than the regular convex hull. It includes all points in the Banach space X that can be expressed as an infinite convex combination of points in A.
This means that:

• Any point \( \operatorname{sconv}(A)\) can be written as \(x = \sum_{i=1}^\infty \lambda_i x_i\) where \(x_i \in A\).
• The coefficients \( \lambda_i \geq 0\) and they must sum to 1, \( \sum_{i=1}^\infty \lambda_i = 1\).

An example of such a combination is when each \(x_i\) belongs to a bounded set A.

bounded set

A set A in a Banach space is called bounded if there exists a constant M such that the norm of every element in A is less than or equal to M, i.e., \(||x_i|| \leq M\) for all \(x_i \in A\).
Bounded sets ensure that:

• All points within the set A do not spread out infinitely.
• Any convex combination of points in A will also be bounded by M.

This property is useful when working with summations and limits in functional analysis, allowing for better control over the resulting vectors.

ell_2 space

The \( \ell_2\) space, also known as the space of square-summable sequences, is a classic example of a Banach space. It consists of all sequences of real or complex numbers \( \{ x_i \}_{i=1}^\infty \) such that the sum \( \sum_{i=1}^\infty |x_i|^2 \) is finite.
Properties of \( \ell_2\) space:

• Vectors in this space are infinite sequences where the sum of the squares of their elements is finite.
• The norm in \( \ell_2\) space is given by \(orm{x} = \sqrt{\sum_{i=1}^\infty |x_i|^2}\).
• Standard unit vectors \( \{ e_i \}_{i=1}^\infty \) form an orthonormal basis for the \( \ell_2\) space, meaning each vector has only one element '1' at the i-th position and '0' elsewhere.

These properties make \( \ell_2\) space fundamental in many areas of analysis and applied mathematics.