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Unformatted text preview: ECE 580 / Math 587 SPRING 2011 Correspondence # 19 May 3, 2011 Solution Set 7 48. Let us first prove the following Lemma, which was given as a hint: Lemma. Let x C [0 , 1] have a unique maximum at t (0 , 1) , which is also the unique maximum of  x ( t )  , and let h be an arbitrary element of C [0 , 1] . Then, lim 1 { x + h  x ( t ) + h ( t ) } = 0 . Toward this end assume, without any loss of generality, that x ( t ) > 0, and let > 0 be a suciently small scalar (in particular, < t , t + < 1). Then, for suciently small , x + h = max t 1  x ( t ) + h ( t )  = max t t t +  x ( t ) + h ( t )  which follows because of the continuity of x and h . Furthermore, since x ( t ) > 0, for su ciently small and , x ( t ) + h ( t ) > t [ t ,t + ] . Case 1. > Then, max t t t +  x ( t ) + h ( t )  = max t N ( t ) [ x ( t ) + h ( t )] max t N ( t ) x ( t ) + max t N ( t ) h ( t ) = x ( t ) + max t N ( t ) h ( t ) . Similarly, replacing the maximum of h with minimum at the 2nd line: max t N ( t )  x ( t ) + h ( t )  x ( t ) + min t N ( t ) h ( t ) . Hence, x ( t ) + min t N ( t ) h ( t ) max t N ( t ) [ x ( t ) + h ( t )  x ( t ) + max t N ( t ) h ( t ) . This now leads to lim > x + h  x ( t ) + h ( t )  = lim > x + h x ( t ) h ( t ) [ max t N ( t ) h ( t ) h ( t ) ] = max t N ( t ) h ( t ) h ( t ) . Likewise, the limit above can be bounded below by min t N ( t ) h ( t ) h ( t ) . But since > 0 was arbitrary, we can let it go to zero, which makes both bounds go to zero, since h is continuous. This then verifies the desired result as 0 on the positive side. Case 2. < 0. Let = , h = h in [0 , 1]. Then lim > x + h  x ( t ) + h ( t )  = lim x + h  x ( t ) + h ( t )  0 by the argument of Case 1. Remark. If t is a boundary point, say t = 0, then replace N ( t ) by { t : 0 t } ; and observe that similar arguments as above apply to prove an extension of the Lemma to this more general case. Now, coming back to the original problem, where we are given the function f ( x ) = max t 1  x ( t )  = x , x C [0 , 1] , let us take D C [0 , 1] to be the class of continuous functions x with the property that  x ( t )  has a unique maximum on [0 , 1]. Given x D , let t be the point where  x ( t )  achieves its unique maximum on [0 , 1]. We now claim that, f ( x ; h ) = sgn[ x ( t )] h ( t ) To show this, simply note that lim 1 [ x + h  x ( t )  f ( x ; h )] = lim 1 [ x + h  x ( t ) + h ( t )  ] = 0 where the first equality follows because for suciently small...
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This note was uploaded on 10/11/2011 for the course ECE 580 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.
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