Question: If \(c \in \mathbb{R}\) and S is a set, define \(cS = \left\{ {c{\bf{x}}:{\bf{x}} \in S} \right\}\). Let S be a convex set and suppose \(c > {\bf{0}}\) and \(d > {\bf{0}}\). Prove that \(cS + dS = \left( {c + d} \right)S\).

Short Answer

Expert verified

It is proved that \(cS + dS = \left( {c + d} \right)S\).

Step by step solution

01

Simplify the expression \(cS + dS\)

As S is a convex set, therefore it can be represented as shown below:

\(\begin{array}{c}cS + dS = \left\{ {cx + dy:\,\,x,y \in S} \right\}\\\left( {c + d} \right)S = \left\{ {\left( {c + d} \right)z:\,z \in S} \right\}\\\left( {c + d} \right)S \subset cS + dS\end{array}\)

If the above equation is true, then \(x = y = z\).

Let, \(z = \frac{c}{{c + d}}x + \frac{d}{{c + d}}y\).

Since c and d both are positive. Therefore z can be represented as aconvex combination of x and y.

02

Prove the given identity

As it was established that S is convex, therefore it can be observed that:

\(\begin{array}{c}\left( {c + d} \right)z \in \left( {c + d} \right)S\\\left( {c + d} \right)z = cz + dz\\\left( {c + d} \right)z \in cS + dS\end{array}\)

Hence, it is proved that \(cS + dS = \left( {c + d} \right)S\).

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Most popular questions from this chapter

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: 11. In Exercises 11 and 12, mark each statement True or False. Justify each answer.

11. a. The set of all affine combinations of points in a set \(S\) is called the affine hull of \(S\).

b. If \(\left\{ {{{\rm{b}}_{\rm{1}}}{\rm{,}}.......{{\rm{b}}_{\rm{2}}}} \right\}\) is a linearly independent subset of \({\mathbb{R}^n}\) and if \({\bf{p}}\) is a linear combination of \({{\rm{b}}_{\rm{1}}}.......{{\rm{b}}_{\rm{k}}}\), then \({\rm{p}}\) is an affine combination of \({{\rm{b}}_{\rm{1}}}.......{{\rm{b}}_{\rm{k}}}\).

c. The affine hull of two distinct points is called a line.

d. A flat is a subspace.

e. A plane in \({\mathbb{R}^3}\) is a hyper plane.

In Exercises 21–24, a, b, and c are non-collinear points in\({\mathbb{R}^{\bf{2}}}\)and p is any other point in\({\mathbb{R}^{\bf{2}}}\). Let\(\Delta {\bf{abc}}\)denote the closed triangular region determined by a, b, and c, and let\(\Delta {\bf{pbc}}\)be the region determined by p, b, and c. For convenience, assume that a, b, and c are arranged so that\(\left[ {\begin{array}{*{20}{c}}{\overrightarrow {\bf{a}} }&{\overrightarrow {\bf{b}} }&{\overrightarrow {\bf{c}} }\end{array}} \right]\)is positive, where\(\overrightarrow {\bf{a}} \),\(\overrightarrow {\bf{b}} \) and\(\overrightarrow {\bf{c}} \)are the standard homogeneous forms for the points.

22. Let p be a point on the line through a and b. Show that\(det\left[ {\begin{array}{*{20}{c}}{\overrightarrow {\bf{a}} }&{\overrightarrow {\bf{b}} }&{\overrightarrow {\bf{p}} }\end{array}} \right] = 0\).

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.

14.a. Justify each equal sign:

\(3\left( {{{\mathop{\rm r}\nolimits} _3} - {{\mathop{\rm r}\nolimits} _2}} \right) = z'\left( 1 \right) = .5x'\left( 1 \right) = \frac{3}{2}\left( {{{\mathop{\rm p}\nolimits} _3} - {{\mathop{\rm p}\nolimits} _2}} \right)\)

b. Show that \({{\mathop{\rm r}\nolimits} _2}\) is the midpoint of the segment from \({{\mathop{\rm p}\nolimits} _2}\) to \({{\mathop{\rm p}\nolimits} _3}\).

c. Justify each equal sign: \(3\left( {{{\mathop{\rm r}\nolimits} _1} - {{\mathop{\rm r}\nolimits} _0}} \right) = z'\left( 0 \right) = .5x'\left( {.5} \right)\).

d. Use part (c) to show that \(8{{\mathop{\rm r}\nolimits} _1} = - {{\mathop{\rm p}\nolimits} _0} - {{\mathop{\rm p}\nolimits} _1} + {{\mathop{\rm p}\nolimits} _2} + {{\mathop{\rm p}\nolimits} _3} + 8{{\mathop{\rm r}\nolimits} _0}\).

e. Use part (d) equation (8), and part (a) to show that \({{\mathop{\rm r}\nolimits} _1}\) is the midpoint of the segment from \({{\mathop{\rm r}\nolimits} _2}\) to the midpoint of the segment from \({{\mathop{\rm p}\nolimits} _1}\) to \({{\mathop{\rm p}\nolimits} _2}\). That is, \({{\mathop{\rm r}\nolimits} _1} = \frac{1}{2}\left( {{{\mathop{\rm r}\nolimits} _2} + \frac{1}{2}\left( {{{\mathop{\rm p}\nolimits} _1} + {{\mathop{\rm p}\nolimits} _2}} \right)} \right)\).

Let \({\bf{x}}\left( t \right)\) be a B-spline in Exercise 2, with control points \({{\bf{p}}_o}\), \({{\bf{p}}_1}\) , \({{\bf{p}}_2}\) , and \({{\bf{p}}_3}\).

a. Compute the tangent vector \({\bf{x}}'\left( t \right)\) and determine how the derivatives \({\bf{x}}'\left( 0 \right)\) and \({\bf{x}}'\left( 1 \right)\) are related to the control points. Give geometric descriptions of the directions of these tangent vectors. Explore what happens when both \({\bf{x}}'\left( 0 \right)\)and \({\bf{x}}'\left( 1 \right)\)equal 0. Justify your assertions.

b. Compute the second derivative and determine how and are related to the control points. Draw a figure based on Figure 10, and construct a line segment that points in the direction of . [Hint: Use \({{\bf{p}}_2}\) as the origin of the coordinate system.]

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