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Let \(A\) be an open set in a normed space \(X\). Show that \(\operatorname{conv}(A)\) is open. Hint: Given \(\left\\{x_{i}\right\\}_{i=1}^{n} \in A\) and \(\left\\{\lambda_{i}\right\\}_{i=1}^{n}\) such that \(\lambda_{i} \geq 0\) for every \(i\) and \(\sum \lambda_{i}=1\), note that if \(\lambda_{1}>0\) and \(O_{1}\) is an open set containing \(x_{1}\) and contained in \(A\), then \(\operatorname{conv}(A)\) contains the open set \(\lambda_{1} O_{1}+\lambda_{2} x_{2}+\ldots+\lambda_{n} x_{n}\)

Short Answer

Expert verified
\(\operatorname{conv}(A)\) is open because every point in \(\operatorname{conv}(A)\) has an open neighborhood contained within \(\operatorname{conv}(A)\).

Step by step solution

01

Understand the hint

The hint suggests considering a finite subset \{x_{i}\}_{i=1}^{n} of points in the open set A and coefficients \{\lambda_{i}\}_{i=1}^{n} such that \(\lambda_{i} \geq 0\) for all i and \(\sum \lambda_{i} = 1\).
02

Consider a point in \(\operatorname{conv}(A)\)

Consider a point \(y \in \operatorname{conv}(A)\). By the definition of convex hull, \(y\) can be written as \(y = \sum \lambda_{i} x_{i}\) where \(x_{i} \in A\) and \(\lambda_{i} \geq 0\) with \(\sum \lambda_{i} = 1\).
03

Use openness of \(A\)

Since \(A\) is open, for each point \(x_{1} \in A\) and \(\lambda_{1} > 0\), there exists an open set \(O_{1}\) containing \(x_{1}\) such that \(O_{1} \subseteq A\).
04

Construct an open set in \(\operatorname{conv}(A)\)

Consider the set \(\lambda_{1} O_{1} + \lambda_{2} x_{2} + \ldots + \lambda_{n} x_{n}\). This set is open because it is a translation and scaling of the open set \(O_{1}\).
05

Containment in \(\operatorname{conv}(A)\)

Since \(O_{1} \subseteq A\), for any \(z \in \lambda_{1} O_{1} + \lambda_{2} x_{2} + \ldots + \lambda_{n} x_{n}\), \(z\) can be written as a convex combination of points in \(A\). Hence, \(z \in \operatorname{conv}(A)\).
06

Conclude that \(\operatorname{conv}(A)\) is open

Every point \(y \in \operatorname{conv}(A)\) has an open set around it contained in \(\operatorname{conv}(A)\). Thus, \(\operatorname{conv}(A)\) is open.

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Key Concepts

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

Convex Hull
To understand why the convex hull (\(\operatorname{conv}(A)\)) is open, we need to first understand what a convex hull is. The convex hull of a set \(A\) in a normed space \(X\) is the smallest convex set that contains \(A\). Essentially, it is formed by taking all convex combinations of points in \(A\). A convex combination of points \(\{x_i\}_{i=1}^n\) is a point \(y\) that can be written as: \[ y = \sum \lambda_{i} x_{i}, \] where \(x_{i} \in A\), \(\lambda_{i} \geq 0\), and \(\sum \lambda_{i} = 1\). This definition ensures that \(y\) lies 'inside' the set formed by \(x_{i}\) points. Understanding this helps in grasping why open sets remain open after taking convex combinations.
Openness in Normed Spaces
In the context of normed spaces, an open set is a set where for every point \(x\) in the set, there is an epsilon neighborhood around \(x\) that is entirely contained within the set. Formally, a set \(A\) is open if for every \(x \in A\), there exists \(\epsilon > 0\) such that the ball \(B(x,\epsilon)\) is contained in \(A\). In simpler terms, you can move a little bit in any direction from any point in an open set and still remain inside the set. This property of openness is crucial when we deal with convex combinations and convex hulls. If \(A\) is an open set, we can use small open neighborhoods around points to show that these neighborhoods translate to larger open sets in their convex combinations.
Finite Convex Combinations
Finite convex combinations are the building blocks of defining the convex hull of a set. They are formed by taking a finite number of points \(\{x_{i}\}_{i=1}^{n}\) and coefficients \(\{\lambda_{i}\}_{i=1}^{n}\) such that \(\lambda_{i} \geq 0\) and \(\sum \lambda_{i} = 1\). The point \(y\) formed by this combination is: \[ y = \sum \lambda_{i} x_{i} \] Each \(\lambda_{i}\) acts as a weight, determining how much each point \(x_{i}\) contributes to \(y\). If one of \(\lambda_{i}\) is greater than zero, say \(\lambda_{1} > 0\), we can find an open set around \(x_{1}\) deep enough to maintain the openness when forming the convex combination. By taking appropriate open neighborhoods and using these 'weights', we ensure the point \(y\) remains within the convex hull, proving \(\operatorname{conv}(A)\) is indeed open.

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Most popular questions from this chapter

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.

Let \(X\) be a Banach space and \(G\) be a subspace of \(X\) that is a \(G_{\delta}\) set in \(X\). Show that \(G\) is closed in \(X\). Hint: Let \(S=\bar{G} \backslash G\), where \(\bar{G}\) denotes the closure of \(G .\) Since \(G\) is a \(G_{\delta}\) set in \(X, G\) is \(G_{\delta}\) in \(\bar{G}\). Hence \(G=\bigcap G_{n}\), where \(G_{n}\) are open subsets in \(\bar{G}\). Therefore, \(S=\bigcup\left(\bar{G} \backslash G_{n}\right)\) and each \(G_{n}\) is dense in \(\bar{G}\) because it contains \(G .\) Thus, each \(\bar{G} \backslash G_{n}\) is nowhere dense in \(\bar{G}\), so \(S\) is of first category in \(\bar{G}\). We will show that \(S\) is an empty set. If \(x_{0} \in S\), consider the set \(G^{*}=\left\\{x_{0}+x ; x \in G\right\\}\). Note that \(G^{*} \subset S\). Indeed, any point \(x_{0}+x, x \in G\), is in \(\bar{G}\) since \(\bar{G}\) is a linear set. If for some \(x \in G\) we have \(x_{0}+x \in G\), then by linearity of \(G\) we have \(x_{0} \in G_{1}\) a contradiction. Therefore, \(G^{*} \subset S\) and \(G^{*}\) is of first category in \(\bar{G} .\) Thus, the shift \(G\) of \(G^{*}\) is also of first category in \(\bar{G}\). Hence, the whole \(\bar{G}\) as a union of \(G\) and \(S\), which are both of first category in \(\bar{G}\), is of first category in itself. This is a contradiction, since \(\bar{G}\) is a complete metric space.

We say that \(\sum x_{i}\) is unconditionally Cauchy if, given \(\varepsilon>0\), there is a finite set \(F\) in \(\mathbf{N}\) such that \(\left\|\sum_{F^{\prime}} x_{i}\right\|<\varepsilon\) whenever \(F^{\prime}\) is a finite set in \(\mathbf{N}\) satisfying \(F^{\prime} \cap F=\emptyset\) Show that a series \(\sum x_{i}\) in a Banach space \(X\) is unconditionally Cauchy if and only if it is unconditionally convergent. Hint: If \(\sum x_{i}\) is unconditionally Cauchy, then it is Cauchy and thus it converges to some \(x \in X\). Given \(\varepsilon>0\), find a finite \(F_{1}\) such that \(\left\|\sum_{\vec{F}^{\prime}} x_{i}\right\|<\varepsilon\) for every finite \(F^{\prime}\) satisfying \(F^{\prime} \cap F_{1}=\emptyset\). Then find \(n_{0}>\max \left(F_{1}\right)\) such that \(\left\|\sum_{i=1}^{n_{0}} x_{i}-x\right\|<\varepsilon\) and set \(F=\left\\{1, \ldots, n_{0}\right\\}\). If \(F^{\prime} \supset F\), then \(\left\\{i \in F^{\prime} ; i\right\rangle\) \(\left.n_{0}\right\\} \cap F_{1}=\emptyset\), so \(\left\|\sum_{F^{\prime}} x_{i}-x\right\| \leq\left\|_{i \in F^{\prime}, i>n_{0}} x_{i}-x\right\|+\left\|\sum_{i=1}^{n_{0}} x_{i}\right\|<2 \varepsilon\)

Let \(X\) be a Banach space whose norm \(\|\cdot\|\) satisfies the parallelogram equality. Define \((x, y)\) by the polarization identity, and prove that \((x, y)\) is an inner product. Hint: Clearly, \(\left(\cdot_{1} \cdot\right)\) is continuous in both coordinates, \((x, y)=\overline{(y, x)}\) and \((-x, y)=-(x, y)\). Using the parallelogram equality, show that \((x+y, z)=\) \((x, z)+(y, z)\). Then by induction, \((n x, y)=n(x, y)\) for all \(n \in \mathbf{N}\), and hence also for all integers \(n\). Given \(\frac{n}{m}\), we have \(\left(\frac{n}{m} x, y\right)=n\left(\frac{1}{m} x, y\right)=\frac{n}{m} m\left(\frac{1}{m} x, y\right)=\frac{n}{m}\left(\frac{m}{m} x, y\right)=\) \(\frac{n}{m}(x, y)\). By continuity, we get \((\alpha x, y)=\alpha(x, y)\) for all \(\alpha \in \mathbf{R}\).

Let \(A\) and \(B\) be two convex sets in a normed space \(X\). Show that \(\operatorname{conv}(A \cup B)=\\{\lambda x+(1-\lambda) y ; x \in A, y \in B, \lambda \in[0,1]\\}\) Hint: Show first that the set on the right-hand side is convex.

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