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s7color - ECE 580 Math 587 SPRING 2011 Correspondence 19...

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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 (0 , 1) , which is also the unique maximum of | x ( t ) | , and let h be an arbitrary element of C [0 , 1] . Then, lim α 0 1 α {∥ x + αh ∥ − | x ( t 0 ) + αh ( t 0 ) |} = 0 . Toward this end assume, without any loss of generality, that x ( t 0 ) > 0, and let ϵ > 0 be a sufficiently small scalar (in particular, ϵ < t 0 , t 0 + ϵ < 1). Then, for sufficiently small α , x + αh = max 0 t 1 | x ( t ) + αh ( t ) | = max t 0 ϵ t t 0 + ϵ | x ( t ) + αh ( t ) | which follows because of the continuity of x and h . Furthermore, since x ( t 0 ) > 0, for suffi- ciently small ϵ and α , x ( t ) + αh ( t ) > 0 t [ t 0 ϵ, t 0 + ϵ ] . Case 1. α > 0 Then, max t 0 ϵ t t 0 + ϵ | x ( t ) + αh ( t ) | = max t ∈N ϵ ( t 0 ) [ x ( t ) + αh ( t )] max t ∈N ϵ ( t 0 ) x ( t ) + α max t ∈N ϵ ( t 0 ) h ( t ) = x ( t 0 ) + α max t ∈N ϵ ( t 0 ) h ( t ) . Similarly, replacing the maximum of h with minimum at the 2nd line: max t ∈N ϵ ( t 0 ) | x ( t ) + αh ( t ) | x ( t 0 ) + α min t ∈N ϵ ( t 0 ) h ( t ) . Hence, x ( t 0 ) + α min t ∈N ϵ ( t 0 ) h ( t ) max t ∈N ϵ ( t 0 ) [ x ( t ) + αh ( t ) | ≤ x ( t 0 ) + max t ∈N ϵ ( t 0 ) h ( t ) · α. This now leads to lim α 0 α> 0 x + αh ∥ − | x ( t 0 ) + αh ( t 0 ) | α = lim α 0 α> 0 x + αh ∥ − x ( t 0 ) αh ( t 0 ) α ̸ α ̸ α [ max t ∈N ϵ ( t 0 ) h ( t ) h ( t 0 ) ] = max t ∈N ϵ ( t 0 ) h ( t ) h ( t 0 ) .
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Likewise, the limit above can be bounded below by min t ∈N ϵ ( t 0 ) h ( t ) h ( t 0 ) . But since ϵ > 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 α 0 on the positive side. Case 2. α < 0. Let β = α , ¯ h = h in [0 , 1]. Then lim α 0 α> 0 x + αh ∥ − | x ( t 0 ) + αh ( t 0 ) | α = lim β 0 x + β ¯ h ∥ − | x ( t 0 ) + β ¯ h ( t 0 ) | β 0 by the argument of Case 1. Remark. If t 0 is a boundary point, say t 0 = 0, then replace N ϵ ( t 0 ) by { t : 0 t ϵ } ; 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 0 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 0 be the point where | x ( t ) | achieves its unique maximum on [0 , 1]. We now claim that, δf ( x ; h ) = sgn[ x ( t 0 )] h ( t 0 ) To show this, simply note that lim α 0 1 α [ x + αh ∥ − | x ( t 0 ) | − αδf ( x ; h )] = lim α 0 1 α [ x + αh ∥ − | x ( t 0 ) + αh ( t 0 ) | ] = 0 where the first equality follows because for sufficiently small α (since clearly x ( t 0 ) ̸ = 0), the sign of x ( t 0 ) + αh ( t 0 ) is determined by sgn[ x ( t 0 )], and the equality to zero follows from the Lemma just proved.
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