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Chapter 1: Problem 3

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 \(X\) be the normed space obtained by taking \(c_{0}\) with the norm \(\|x\|_{0}=\sum 2^{-i}\left|x_{i}\right| .\) Show that \(X\) is not a Banach space. Note that this shows that \(\|\cdot\|_{0}\) is not an equivalent norm on \(c_{0}\). Hint: The sequence \(\\{(1,1, \ldots, 1,0, \ldots)\\}_{n=1}^{\infty}\) is Cauchy and not convergent since the only candidate for the limit would be \((1,1, \ldots) \notin c_{0}\).

Let \(M\) be a dense subset (not necessarily countable) of a Banach space \(X\). Show that for every \(x \in X \backslash\\{0\\}\) there are \(x_{k} \in M\) such that \(x=\sum x_{k}\) and \(\left\|x_{k}\right\| \leq \frac{3\|x\|}{2^{k}} .\) Hint: Find \(x_{1} \in M\) such that \(\left\|x-x_{1}\right\| \leq \frac{\|x\|}{2} ;\) then by induction find \(x_{k} \in M\) such that \(\left\|x-\left(x_{1}+\ldots+x_{k-1}\right)-x_{k}\right\| \leq \frac{\|x\|}{2^{k}} .\) Then \(x=\sum x_{k}\) and \(\left\|x_{k}\right\|=\left\|\left(x-\sum_{n=1}^{k-1} x_{n}\right)-\left(x-\sum_{n=1}^{k} x_{n}\right)\right\| \leq \frac{\|x\|}{2^{k-1}}+\frac{\|x\|}{2^{k}}=\frac{3\|x\|}{2^{k}}\)

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 \(\mathbf{R}^{n_{0}}\).

Let \(C^{n}[0,1]\) be the space of all real-valued functions on \([0,1]\) that have \(n\) continuous derivatives on \([0,1]\), with the norm $$ \|f\|=\max _{0 \leq k \leq n}\left(\max \left\\{\left|f^{k}(t)\right| ; t \in[0,1]\right\\}\right) $$

Let \(C\) be a compact set in a finite-dimensional Banach space \(X\). Show that \(\operatorname{conv}(C)\) is compact. Hint: If a point \(x\) lies in the \(\operatorname{conv}(E) \subset \mathbf{R}^{n}\), then \(x\) lies in the convex hull of some subset of \(E\) that has at most \(n+1\) points. Indeed, assume that \(r>n\) and \(x=\sum t_{i} x_{i}\) is a convex combination of some \(r+1\) vectors \(x_{i} \in E\). We will show that then \(x\) is actually a convex combination of some \(r\) of these vectors. Assume that \(t_{i}>0\) for \(1 \leq i \leq r+1\). The \(r\) vectors \(x_{i}-x_{r+1}\) for \(1 \leq i \leq r\) are linearly dependent since \(r>n\). Thus, there are real numbers \(a_{i}\) not all zero such that \(\sum_{i=1}^{r+1} a_{i} x_{i}=0\) and \(\sum_{i=1}^{r+1} a_{i}=0 .\) Choose \(m\) so that \(\left|a_{i} / t_{i}\right| \leq\left|a_{m} / t_{m}\right|\) for \(1 \leq i \leq r+1\) and define \(c_{i}=t_{i}-\frac{a_{i} t_{m}}{a_{m}}\) for \(1 \leq i \leq r+1\). Then \(c_{i} \geq 0, \sum c_{i}=\sum t_{i}=1, x=\sum c_{i} x_{i}\), and \(c_{m}=0\) Having this, we can use the fact that in \(\mathbf{R}^{n}, \operatorname{conv}(C)\) is an image of the compact set \(S \times C^{n+1}\) in \(\mathbf{R}^{n+1} \times C^{n+1}\), where \(S\) is formed by points \(\left\\{\lambda_{i}\right\\}_{1}^{n+1}\) such that \(\lambda_{i} \geq 0\) and \(\sum \lambda_{i}=1\), under the \(\operatorname{map}\left(\\{\lambda\\},\left\\{x_{i}\right\\}\right) \mapsto \sum \lambda_{i} x_{i}\).
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