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Show that \(\overline{\operatorname{span}}(L)=\overline{\operatorname{span}(L)}\) and \(\overline{\operatorname{conv}}(M)=\overline{\operatorname{conv}(M)}\). Hint: By the definition, \(\overline{\operatorname{span}}(L)\) is the intersection of all closed subspaces containing \(L\), and \(\overline{\operatorname{span}(L)}\) is one such subspace. From this, one inclusion follows. The other inclusion follows similarly because \(\operatorname{span}(L)\) is the intersection of all subspaces containing \(L .\)

Short Answer

Expert verified
\(\overline{\operatorname{span}}(L) = \overline{\operatorname{span}(L)}\) and \(\overline{\operatorname{conv}}(M) = \overline{\operatorname{conv}(M)}\).

Step by step solution

01

Understanding \(\text{span}(L)\)

The span of a set \(L\), denoted as \(\text{span}(L)\), is the set of all finite linear combinations of elements in \(L\). It represents the smallest subspace containing \(L\).
02

Closure of \(\text{span}(L)\)

The closure of the span of \(L\), denoted as \(\overline{\text{span}(L)}\), is the smallest closed subspace containing \(L\).
03

Definition of \(\overline{\text{span}}(L)\)

By definition, \(\overline{\text{span}}(L)\) is the intersection of all closed subspaces containing \(L\).
04

Showing \(\text{span}(L) \subseteq \overline{\text{span}(L)}\)

Since \(\overline{\text{span}}(L)\) is the smallest closed subspace containing \(L\), and \(\overline{\text{span}(L)}\) is one such closed subspace, it follows that \(\overline{\text{span}}(L) \subseteq \overline{\text{span}(L)}\).
05

Completeness of Closed Span

Knowing that \(\overline{\text{span}}(L)\) is closed and should contain \(\text{span}(L)\), we have \(\text{span}(L) \subseteq \overline{\text{span}}(L)\). Because \(\overline{\text{span}}(L)\) is the smallest closed subspace that contains \(L\), it implies \(\text{span}(L) \subseteq \overline{\text{span}}(L)\).
06

Definition of \(\text{conv}(M)\)

The convex hull of a set \(M\), denoted as \(\text{conv}(M)\), is the set of all convex combinations of elements in \(M\). It represents the smallest convex set containing \(M\).
07

Closure of \(\text{conv}(M)\)

The closure of the convex hull of \(M\), denoted as \(\overline{\text{conv}(M)}\), is the smallest closed convex set containing \(M\).
08

Definition of \(\overline{\text{conv}}(M)\)

By definition, \(\overline{\text{conv}}(M)\) is the intersection of all closed convex sets containing \(M\).
09

Showing \(\text{conv}(M) \subseteq \overline{\text{conv}(M)}\)

Since \(\overline{\text{conv}}(M)\) is the smallest closed convex set containing \(M\), and \(\overline{\text{conv}(M)}\) is one such closed convex set, it follows that \(\overline{\text{conv}}(M) \subseteq \overline{\text{conv}(M)}\).
10

Completeness of Closed Convex Hull

Knowing that \(\overline{\text{conv}}(M)\) is closed and should contain \(\text{conv}(M)\), we have \(\text{conv}(M) \subseteq \overline{\text{conv}}(M)\). Because \(\overline{\text{conv}}(M)\) is the smallest closed convex set that contains \(M\), it implies \(\text{conv}(M) \subseteq \overline{\text{conv}}(M)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Span
In linear algebra, the concept of 'span' is fundamental. The span of a set of vectors, denoted as \text{span}(L)\, refers to all possible linear combinations of these vectors. Essentially, if you can represent any vector in the span of a set via multiplication and addition of the original set members, you have found the smallest subspace covering these vectors. This is crucial for understanding the structure and boundaries of vector spaces. Further, it helps in determining the dimensions and dependencies among vectors.

For instance, consider a set of vectors in \mathbb{R}^2\:\[ L = \{ \mathbf{v_1}, \mathbf{v_2} \} \]. The span of \( L \) is defined as: \[ \text{span}(L) = \{ a\mathbf{v_1} + b\mathbf{v_2} \ | \ a, b \in \mathbb{R} \} \]

This means any vector in the plane can be expressed as a combination of \mathbf{v_1}\ and \mathbf{v_2}\ given appropriate coefficients \text{a}\ and \text{b}\.
Convex Hull
The 'convex hull' of a set is another important concept, especially in functional analysis and optimization. It involves taking all convex combinations of a set of points. To visualize, imagine stretching a rubber band to enclose a set of points; the shape formed is the convex hull. The set of all convex combinations of points \[M = \{ \mathbf{v_1}, \mathbf{v_2}, \ldots, \mathbf{v_n} \} \]\, is denoted by \text{conv}(M)\.

Formally, the convex hull \[ \text{conv}(M) \] is defined as: \[\text{conv}(M) = \{ \lambda_1 \mathbf{v_1} + \lambda_2 \mathbf{v_2} + \ldots + \lambda_n \mathbf{v_n} \ | \ \lambda_i \geq 0, \sum_{i=1}^{n} \lambda_i = 1 \} \]

Every point in the convex hull can be expressed as a weighted average (convex combination) of the original points in \(M\), with weights summing to 1. This property makes the convex hull extremely useful in regions like optimization and game theory.
Closure
The 'closure' of a set adds a layer of completeness to the concept of span and convex hull. It involves including all limit points of the set, leading to the smallest closed set containing the original set. For any set \(A\), the closure is denoted as \overline{A}\. For example, the closure of \text{span}(L)\, written as \overline{\text{span}(L)}\, is the smallest closed subspace encompassing \text{span}(L)\.

This ensures that even sequences converging within the span are included. In simpler terms, you can't escape from a closed set via convergence. The same goes for the closure of the convex hull, \[ \overline{\text{conv}(M)} \], which is the smallest closed convex set encompassing all the convex combinations of \(M\). This approach is essential in analysis, ensuring that we have a thorough and rigorous determination of complete subspaces and convex sets.

Conclusively, understanding and applying span, convex hull, and closure are foundational in functional analysis, helping build a solid grasp of vector spaces, linear combinations, and convergence properties.

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Most popular questions from this chapter

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 \(i\)

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.

The Hilbert cube \(C\) is defined as \(\left\\{x=\left(x_{i}\right) \in \ell_{2} ; \forall i:\left|x_{i}\right| \leq 2^{-i}\right\\}\). Show that the Hilbert cube is a compact set in \(\ell_{2}\). Hint: Given \(\varepsilon>0\), there is \(n_{0}\) such that \(\sum_{i=n_{0}+1}^{\infty}\left|x_{i}\right| \leq \varepsilon\) for every \(x \in C .\) Then use finite \(\varepsilon\) -nets in \(\mathbf{R}^{n_{0}}\).

Show that the set \(A=\left\\{x \in \ell_{2} ; \sum\left(1+\frac{1}{i}\right) x_{i}^{2} \leq 1\right\\}\) does not contain an element with norm equal to \(\sup \\{\|x\| ; x \in A\\}\). Hint: If \(x \in A\), then \(\|x\|^{2}=\sum x_{i}^{2}<\sum\left(1+\frac{1}{i}\right) x_{i}^{2} \leq 1=\sup _{A}\|x\|\).

Let \(X\) be a normed linear space. Prove that for any \(x, y \in X\) we have \(|\|x\|-\|y\|| \leq\|x-y\|\) Hint: Triangle inequality, \(\|x\|=\|(x-y)+y\|\).

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