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Chapter 8: Q29E (page 437)

Question: 29. Prove that the open ball \(B\left( {{\rm{p}},\delta } \right) = \left\{ {{\rm{x:}}\left\| {{\rm{x - p}}} \right\| < \delta } \right\}\)is a convex set. (Hint: Use the Triangle Inequality).

Short Answer

Expert verified

It is shown that the open ball\(B\left( {{\rm{p}},\delta } \right) = \left\{ {{\rm{x}}:\left\| {{\rm{x}} - {\rm{p}}} \right\| < \delta } \right\}\)is a convex set.

Step by step solution

01

Assume some vectors in an open ball set

Assume \({\rm{x,y}} \in B\left( {{\rm{p}},\delta } \right)\) and for \(0 \le t \le 1\), the vector \(z\) satisfies \(z = \left( {1 - t} \right)x + ty\).

02

Evaluate \(\left\| {z - p} \right\|\)

\(\begin{array}{c}\left\| {z - {\rm{p}}} \right\| = \left\| {\left( {\left( {1 - t} \right){\rm{x}} + t{\rm{y}}} \right) - {\rm{p}}} \right\|\\ = \left\| {\left( {\left( {1 - t} \right)\left( {{\rm{x}} - {\rm{p}}} \right) + t\left( {y - {\rm{p}}} \right)} \right)} \right\|\end{array}\)

03

Use Triangle inequality and \(x,y \in B\left( {p,\delta } \right)\)

According to triangle inequality \(\left( {1 - t} \right)\left\| {\left( {{\rm{x}} - {\rm{p}}} \right)} \right\| + t\left\| {\left( {{\rm{y}} - {\rm{p}}} \right)} \right\| < \left( {1 - t} \right)\delta + t\delta \) and as \({\rm{x,y}} \in B\left( {{\rm{p}},\delta } \right)\).

So, \(\left\| {{\rm{z}} - {\rm{p}}} \right\| < \left( {1 - t} \right)\delta + t\delta \).

04

Draw a conclusion

From \(\left\| {z - p} \right\| < \left( {1 - t} \right)\delta + t\delta \), it can be concluded that \(z \in B\left( {{\rm{p}},\delta } \right)\) and \(B\left( {{\rm{p}},\delta } \right)\) is a convex set of the ball.

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Most popular questions from this chapter

Suppose that\(\left\{ {{p_1},{p_2},{p_3}} \right\}\)is an affinely independent set in\({\mathbb{R}^{\bf{n}}}\)and q is an arbitrary point in\({\mathbb{R}^{\bf{n}}}\). Show that the translated set\(\left\{ {{p_1} + q,{p_2} + q,{p_3} + {\bf{q}}} \right\}\)is also affinely independent.

Question: 18. Choose a set \(S\) of four points in \({\mathbb{R}^3}\) such that aff \(S\) is the plane \(2{x_1} + {x_2} - 3{x_3} = 12\). Justify your work.

Question: Suppose that the solutions of an equation \(A{\bf{x}} = {\bf{b}}\) are all of the form \({\bf{x}} = {x_{\bf{3}}}{\bf{u}} + {\bf{p}}\), where \({\bf{u}} = \left( {\begin{array}{*{20}{c}}{\bf{5}}\\{\bf{1}}\\{ - {\bf{2}}}\end{array}} \right)\) and \({\bf{p}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{3}}}\\{\bf{4}}\end{array}} \right)\). Find points \({{\bf{v}}_{\bf{1}}}\) and \({{\bf{v}}_{\bf{2}}}\) such that the solution set of \(A{\bf{x}} = {\bf{b}}\) is \({\bf{aff}}\left\{ {{{\bf{v}}_{\bf{1}}},\,{{\bf{v}}_{\bf{2}}}} \right\}\).

Let\({v_1} = \left[ {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{1}}\end{array}} \right]\),\({v_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}{\bf{1}}\\{\bf{5}}\end{array}} \right]\),\({v_{\bf{3}}} = \left[ {\begin{array}{*{20}{c}}{\bf{4}}\\{\bf{3}}\end{array}} \right]\),\({p_1} = \left[ {\begin{array}{*{20}{c}}{\bf{3}}\\{\bf{5}}\end{array}} \right]\),\({p_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}{\bf{5}}\\{\bf{1}}\end{array}} \right]\),\({p_{\bf{3}}} = \left[ {\begin{array}{*{20}{c}}{\bf{2}}\\{\bf{3}}\end{array}} \right]\),\({p_{\bf{4}}} = \left[ {\begin{array}{*{20}{c}}{ - {\bf{1}}}\\{\bf{0}}\end{array}} \right]\),\({p_{\bf{5}}} = \left[ {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{4}}\end{array}} \right]\),\({p_{\bf{6}}} = \left[ {\begin{array}{*{20}{c}}{\bf{1}}\\{\bf{2}}\end{array}} \right]\),\({p_{\bf{7}}} = \left[ {\begin{array}{*{20}{c}}{\bf{6}}\\{\bf{4}}\end{array}} \right]\)and let\(S = \left\{ {{v_1},{v_2},{v_3}} \right\}\).

  1. Show that the set is affinely independent.
  2. Find the barycentric coordinates of\({p_1}\),\({p_{\bf{2}}}\), and\({p_{\bf{3}}}\)with respect to S.
  3. On graph paper, sketch the triangle\(T\)with vertices\({v_1}\),\({v_{\bf{2}}}\), and\({v_{\bf{3}}}\), extend the sides as in Figure 8, and plot the points\({p_{\bf{4}}}\),\({p_{\bf{5}}}\),\({p_{\bf{6}}}\), and\({p_{\bf{7}}}\). Without calculating the actual values, determine the signs of the barycentric coordinates of points\({p_{\bf{4}}}\),\({p_{\bf{5}}}\),\({p_{\bf{6}}}\), and\({p_{\bf{7}}}\).

In Exercises 13-15 concern the subdivision of a Bezier curve shown in Figure 7. Let \({\mathop{\rm x}\nolimits} \left( t \right)\) be the Bezier curve, with control points \({{\mathop{\rm p}\nolimits} _0},...,{{\mathop{\rm p}\nolimits} _3}\), and let \({\mathop{\rm y}\nolimits} \left( t \right)\) and \({\mathop{\rm z}\nolimits} \left( t \right)\) be the subdividing Bezier curves as in the text, with control points \({{\mathop{\rm q}\nolimits} _0},...,{{\mathop{\rm q}\nolimits} _3}\) and \({{\mathop{\rm r}\nolimits} _0},...,{{\mathop{\rm r}\nolimits} _3}\), respectively.

15. Sometimes only one-half of a Bezier curve needs further subdividing. For example, subdivision of the “left” side is accomplished with parts (a) and (c) of Exercise 13 and equation (8). When both halves of the curve \({\mathop{\rm x}\nolimits} \left( t \right)\) are divided, it is possible to organize calculations efficiently to calculate both left and right control points concurrently, without using equation (8) directly.

a. Show that the tangent vector \(y'\left( 1 \right)\) and \(z'\left( 0 \right)\) are equal.

b. Use part (a) to show that \({{\mathop{\rm q}\nolimits} _3}\) (which equals \({{\mathop{\rm r}\nolimits} _0}\)) is the midpoint of the segment from \({{\mathop{\rm q}\nolimits} _2}\) to \({{\mathop{\rm r}\nolimits} _1}\).

c. Using part (b) and the results of Exercises 13 and 14, write an algorithm that computes the control points for both \({\mathop{\rm y}\nolimits} \left( t \right)\) and \({\mathop{\rm z}\nolimits} \left( t \right)\) in an efficient manner. The only operations needed are sums and division by 2.
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