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PS5-Solutions

# PS5-Solutions - MAT 337 Problem Set 5 Sample Solutions Ian...

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MAT 337 Problem Set 5 Sample Solutions Ian Zwiers - April 14, 2009 This document is for demonstration purposes; it is not a marking key. 7.1(A) Show that k ( x, y, z ) k = | x | + 2 p y 2 + z 2 is a norm on R 3 . Sketch the unit ball. positivite definite k ( x, y, z ) k = 0 | x | = 0 and 2 p y 2 + z 2 = 0 | x | = 0 , y 2 = 0 and z 2 = 0 . That is, k ( x, y, z ) k = 0 x = y = z = 0 ( x, y, z ) = 0. homogeneous Let α R . Then, k α ( x, y, z ) k = | αx | + q ( αy ) 2 + ( αz ) 2 = | α | | x | + p y 2 + z 2 = | α | k ( x, y, z ) k . triangle inequality Note that, ( y + b ) 2 + ( z + c ) 2 = ( y 2 + 2 y b + b 2 ) + ( z 2 + 2 z c + c 2 ) = ( y 2 + z 2 ) + 2[ y z ] · [ b c ] + ( b 2 + c 2 ) ( y 2 + z 2 ) + 2 p y 2 + z 2 p b 2 + c 2 + ( b 2 + c 2 ) = p y 2 + z 2 + p b 2 + c 2 2 , where the inequality is due to the Cauchy-Schwarz inequality, popularly known as the regular R 2 dot product formula. Then, it is straightforward, k ( x, y, z ) + ( a, b, c ) k = | x + a | + 2 p ( y + b ) 2 + ( z + c ) 2 ( | x | + | a | ) + 2 p y 2 + z 2 + 2 p b 2 + c 2 = k ( x, y, z ) k + k ( a, b, c ) k . The unit ball is two open cones, each with height one and with base radius 1 2 , where the base of one cone has been glued to the base of the other. 7.1(C) For f ∈ C 1 [ a, b ] , define ρ ( f ) = k f 0 k . Show that ρ is nonnegative, homogeneous, and satisfies the triangle inequality. Why is it not a norm? nonnegative ρ is nonnegative by definition of k·k . For f ∈ C 2 [ a, b ], derivative f 0 is continuous on a compact set, and must achieve its extrema. This proves ρ ( f ) < + . 1

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homogeneous Let α R , f ∈ C 1 [ a, b ]. Then, ρ ( αf ) = ( αf ) 0 = k α f 0 k = sup x [ a,b ] | α f 0 ( x ) | = | α | sup x [ a,b ] | f 0 ( x ) | = | α | k f 0 k = | α | ρ ( f ) . triangle inequality Let f, g ∈ C 1 [ a, b ]. Then, ρ ( f + g ) = sup x [ a,b ] ( f + g ) 0 ( x ) = sup x [ a,b ] | f 0 ( x ) + g 0 ( x ) | sup x [ a,b ] | f 0 ( x ) | + | g 0 ( x ) | sup x [ a,b ] | f 0 ( x ) | + sup z [ a,b ] | g 0 ( z ) | = ρ ( f ) + ρ ( g ) Note that the constant functions y = 1 and y = 2 both have ρ = 0. Hence, ρ is not a norm because it is not positive definite. Please Note: This counter example is essentially the only one. If you “anchor” the functions, you can make ρ a norm. For example: let X = f ∈ C 2 [ a, b ] : f ( a ) = 0 , then in fact ρ is a norm on X . 7.2(D) Show that if A is an arbitrary subset of a normed space V and U is an open subset, then A + U = { a + u : a A, u U } is open. Let x A + U , then a A and u U such that x = a + u . Since U is open, δ > 0 such that B δ ( u ) U . Consider arbitrary y B δ ( x ); note then that, k ( y - a ) - u k = k y - ( a + u ) k = k y - x k < δ, hence y - a B δ ( u ). That is, y - a U . Since a A , this proves y A + U . Since y was arbitrary in B δ ( x ), this proves and hence B δ ( x ) A + U . Then, since x was arbitrary in A + U
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PS5-Solutions - MAT 337 Problem Set 5 Sample Solutions Ian...

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