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Question: Find an example to show that the convexity of S necessary in Exercise 19.

Short Answer

Expert verified

For \(c = 1\), \(d = 1\) and \(S = \left\{ {0,1} \right\} \subset \mathbb{R}\), the identity \(cS + dS = \left( {c + d} \right)S\) is not true as S is not convex.

Step by step solution

01

Consider the values of constants

Let \(c = 1\), \(d = 1\) and \(S = \left\{ {0,1} \right\} \subset \mathbb{R}\) (not a convex set).

Check for the expression \(cS + dS\) as shown below:

\(\begin{array}{c}cS + dS = S + S\\ = \left\{ {0 + 0,\,0 + 1,\,1 + 1} \right\}\\ = \left\{ {0,\,1,\,\,2} \right\}\end{array}\)

02

Find the value for \(\left( {c + d} \right)S\)

Obtain the value of \(\left( {c + d} \right)S\) as shown below:

\(\begin{array}{c}\left( {c + d} \right)S = \left( {1 + 1} \right)S\\ = 2S\\ = \left\{ {0,2} \right\}\end{array}\)

As it can be observed, that \(cS + dS \ne \left( {c + d} \right)S\).

So, S should be convex to hold the relation \(cS + dS \ne \left( {c + d} \right)S\).

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Most popular questions from this chapter

Repeat Exercise 25 with\({v_1} = \left[ {\begin{array}{*{20}{c}}1\\{\bf{2}}\\{ - {\bf{4}}}\end{array}} \right]\),\({v_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}{\bf{8}}\\{\bf{2}}\\{ - {\bf{5}}}\end{array}} \right]\), \({v_{\bf{3}}} = \left[ {\begin{array}{*{20}{c}}{\bf{3}}\\{{\bf{10}}}\\{ - {\bf{2}}}\end{array}} \right]\), \({\bf{a}} = \left[ {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{0}}\\{\bf{8}}\end{array}} \right]\), and \({\bf{b}} = \left[ {\begin{array}{*{20}{c}}{.{\bf{9}}}\\{{\bf{2}}.{\bf{0}}}\\{ - {\bf{3}}.{\bf{7}}}\end{array}} \right]\).

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b. If\(f\)is a linear functional defined on\({\mathbb{R}^n}\), then there exists a real number\(k\)such that\(f\left( x \right) = kx\)for all\(x\)in\({\mathbb{R}^n}\).

c. If a hyper plane strictly separates sets\(A\)and\(B\), then\(A \cap B = \emptyset \)

d. If\(A\)and\(B\)are closed convex sets and\(A \cap B = \emptyset \), then there exists a hyper plane that strictly separate\(A\)and\(B\).

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