/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 5 Suppose that the two compact-val... [FREE SOLUTION] | 91影视

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Chapter 14: Problem 5

Suppose that the two compact-valued correspondences \(F, G: X \subseteq \mathbb{R}^{n} \rightarrow \mathbb{R}^{m}\) are upper hemicontinuous at a point \(\mathrm{x}^{0}\) in \(X\). Consider the summation correspondence \(H: X \rightarrow \mathbb{R}^{m}\) defined by \(H(\mathbf{x})=F(\mathrm{x})+G(\mathrm{x})\) for all \(\mathrm{x}\) in \(X\). Prove that \(H\) is upper hemicontinuous at \(\mathbf{x}^{0}\). What may go wrong if \(F\) and \(G\) are not compact-valued?

Short Answer

Expert verified
The summation correspondence \(H\) is upper hemicontinuous at \(\mathbf{x}^0\) if the correspondences \(F\) and \(G\) are compact-valued and upper hemicontinuous at that point. Non-compact values may disrupt this property.

Step by step solution

01

Define Upper Hemicontinuity

A correspondence \(F: X \subseteq \mathbb{R}^n \rightarrow \mathbb{R}^m\) is upper hemicontinuous at \(\mathbf{x}^0\) if for every open set \(V\) containing \(F(\mathbf{x}^0)\), there exists an open neighborhood \(U\) of \(\mathbf{x}^0\) such that \(F(\mathbf{x}) \subseteq V\) for all \(\mathbf{x} \in U\).
02

Understand the Given Conditions

Given \(F\) and \(G\) are compact-valued and upper hemicontinuous at \(\mathbf{x}^0\). We also have the summation correspondence \(H(\mathbf{x}) = F(\mathbf{x}) + G(\mathbf{x})\). The objective is to prove \(H\) is upper hemicontinuous at \(\mathbf{x}^0\).
03

Define a Neighborhood for the Summation

Consider any open set \(W\) that contains \(H(\mathbf{x}^0) = F(\mathbf{x}^0) + G(\mathbf{x}^0)\). To prove upper hemicontinuity, it needs to be shown that there exists a neighborhood \(U\) of \(\mathbf{x}^0\) such that \(H(\mathbf{x}) \subseteq W\) for all \(\mathbf{x} \in U\).
04

Use Compactness and Upper Hemicontinuity of \(F\) and \(G\)

Since \(F\) and \(G\) are compact-valued and upper hemicontinuous at \(\mathbf{x}^0\), for each element \(y \in F(\mathbf{x}^0)\) and \(z \in G(\mathbf{x}^0)\), there exist open neighborhoods \(U_y\) and \(U_z\) of \(\mathbf{x}^0\) such that \(F(\mathbf{x}) \subseteq V_y\) and \(G(\mathbf{x}) \subseteq V_z\), where \(V_y\) and \(V_z\) are open sets containing \(y\) and \(z\).
05

Construct the Desired Neighborhood for \(H\)

Take \(U = U_y \cap U_z\). This intersection is also a neighborhood of \(\mathbf{x}^0\). For any \(\mathbf{x} \in U\), \(F(\mathbf{x}) \subseteq V_y\) and \(G(\mathbf{x}) \subseteq V_z\). Consequently, \(H(\mathbf{x}) = F(\mathbf{x}) + G(\mathbf{x}) \subseteq V_y + V_z\).
06

Conclude Upper Hemicontinuity of \(H\)

Given \(V_y\) and \(V_z\) are open sets, the Minkowski sum \(V_y + V_z\) is also open. Hence, for each element \(y + z \in F(\mathbf{x}^0) + G(\mathbf{x}^0)\), there exists a neighborhood \(U\) such that \(F(\mathbf{x}) + G(\mathbf{x}) \subseteq W\), proving \(H(\mathbf{x})\) is upper hemicontinuous at \(\mathbf{x}^0\).
07

Discuss the Role of Compact-Valuedness

If either \(F\) or \(G\) were not compact-valued, the resulting set from addition may fail to be bounded or closed, which can disrupt the upper hemicontinuity property. Non-compact sets can lead to sums that do not adhere to the openness requirements necessary for upper hemicontinuity.

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Key Concepts

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

Compact-Valued Correspondences
When working with correspondences, it鈥檚 essential to understand the concept of compact-valued correspondences. A correspondence is a function that assigns a set of values from one space to another. In the context of this exercise, a compact-valued correspondence means that for every point in the domain, the set of values it maps to is a compact set in the codomain. Compact sets are sets that are both closed (contain all their limit points) and bounded (can be contained within some large sphere). The compactness ensures nice properties like closure and boundedness, which are crucial in proving upper hemicontinuity.
Neighborhood
A neighborhood in mathematical terms refers to a set containing a point where one can find an 'open set' around that point. Open sets are collections of points that do not include their boundary, meaning you can move a little bit in any direction from any point in the set and still remain inside the set. In the context of upper hemicontinuity, if a correspondence is upper hemicontinuous at a point, there exists a neighborhood around this point where the correspondence's values remain within an open set in the codomain. This ensures that small changes in the input only cause small, contained changes in the output.
Minkowski Sum
The Minkowski sum of two sets is a fundamental operation in set theory. If you have two sets, A and B, their Minkowski sum, denoted as A+B, is the set of all points you get by adding each point in A to each point in B. Mathematically, it is expressed as: ewline ewline ewline ewline ewline ewline A+B = ewline ewline ewline ewline ewline ewline { a + b ewline ewline ewline | a ewline A, b ewline B}.ewline ewline ewline In this exercise, the Minkowski sum is used to combine the sets from two upper hemicontinuous correspondences to prove that their sum is also upper hemicontinuous. Understanding Minkowski sums helps in grasping how two combined sets behave in terms of open and closed sets, leading to the results needed for the proof.
Boundedness
Boundedness refers to a set that can be contained within some large sphere or finite measure in space. For a set to be bounded, there has to be a limit to how far the points in the set can be from each other. This property is critical in the context of compact-valued correspondences because compact sets are always bounded. If a set isn鈥檛 bounded, it can鈥檛 form a compact or closed set, making it difficult to analyze continuity properties like upper hemicontinuity. Thus, compact-valued correspondences ensure boundedness essential for the properties used in the proof for upper hemicontinuity.
Closed Set
A closed set contains all of its limit points, meaning that if a sequence of points in the set converges, its limit point is also in the set. This property is significant when dealing with upper hemicontinuity because it prevents the escape of limit points out of the set. When a correspondence or its image is closed, it ensures that the behavior of sequences and limits within the domain is well-behaved. This is why compactness, which includes closedness, is so important. If the sets weren鈥檛 closed, ensuring the containment conditions needed for upper hemicontinuity would be much trickier, thus showcasing why both compactness and closedness are vital for the proof.

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Most popular questions from this chapter

Let \(f(x, y)=-y^{4}+x\left(y^{2}-1\right)\) for all \(x\) and \(-1 \leq y \leq 1\), and consider the maximization problem \(\max _{-1 \leq y \leq 1} f(x, y)\). Determine the value function for this problem, and describe the correspondence \(Y^{*}(x)=\\{y \in[-1,1]: y\) maximizes \(f(x, y)\) over \([-1,1]\\}\). Show that \(Y^{*}\) has the closed graph property.

Let \(F: X \subseteq \mathbb{R}^{n} \rightarrow \mathbb{R}^{m}\) be a compact- valued correspondence and let \(G: X \rightarrow \mathbb{R}^{m}\) be the correspondence whose value at each \(\mathrm{x}\) in \(X\) is the convex hull \(\operatorname{co}(F(\mathbf{x}))\). Prove that if \(F\) is upper hemicontinuous at \(\mathbf{x}^{0}\), then \(G\) is also upper hemicontinuous at \(\mathbf{x}^{0}\). (Hint: Apply Carath茅odory's theorem.)

Let \(F(x)=\\{1 / x\\}\) for \(x \neq 0\), with \(F(0)=\\{0\\} .\) Prove that \(f\) has the closed graph property, but is not upper hemicontinuous.

Assume that \(f:[0,1] \rightarrow[0,1]\) satisfies \(\sqrt{\lim _{s \rightarrow x^{-}}} f(s) \leq f(x) \leq \varlimsup_{s \rightarrow x^{+}} f(s)\) for all \(x\) in \([0,1]\). (Only the right-hand (left-hand) inequality is required to hold for \(x=0(x=1)\).) Prove that \(f\) has a fixed point. (Hint: Consider \(x^{*}=\sup A\), where \(A=\\{x \in[0,1]: f(x) \geq x\\} .)\)

Consider the function \(f\) defined for all \(x\) in \((0,1)\) by \(f(x)=\frac{1}{2}(x+1)\). Prove that \(f\) maps \((0,1)\) into itself, but \(f\) has no fixed point. Why does Brouwer's theorem not apply?
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