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 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 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 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 αf 0 k = sup x [ a,b ] | αf 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 ] ² ² ( f + g ) 0 ( x ) ² ² = 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 = ³ f ∈ C 2 [ a,b ] : f ( a ) = 0 ´ , then in fact ρ is a norm on X . 7.2(D)
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PS5-Solutions - MAT 337 Problem Set 5 Sample Solutions Ian...

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