21-356 Lecture 4

Erentiability of f at x0 lets prove that these

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Unformatted text preview: the di!erentiability of f at x0 . Let’s prove that these conditions are not, however, su"cient. Example 14 Let f (x, y ) := % x if y = x2 , x %= 0, 0 otherwise. Given a direction v = (v1 , v2 ), the line L through 0 in the direction v intersects the parabola y = x2 only in 0 and in at most one point. Hence, if t is very small, f (0 + tv1 , 0 + tv2 ) = 0. 9 It follows that !f f (0 + tv1 , 0 + tv2 ) # f (0, 0) 0#0 (0, 0) = lim = lim = 0. t!0 t!0 !v t t Thus formula (1) holds. Moreover, f is continuous in 0, since |f (x, y )| & |x| ' 0 as (x, y ) ' (0, 0). We claim that f is not di!erentiable at (0, 0). We need to consider to prove that the quotient f (x, y ) # f (0, 0) # "f (0, 0) · (x, y ) & 2 2 (x # 0) + (y # 0) does not tend to zero as (x, y ) ' (0, 0). Take y = x2 . Then ' ( ' ( ' ( f x, x2 # f (0, 0) # "f (0, 0) · x, x2 x # 0 # 0 · x, x2 & =& 2 2 2 2 (x # 0) + (x2 # 0) (x # 0) + (x2 # 0) x x ( =( = ! 0. x2 + x4 |x| 1 + x2 Exercise 15 Let f : E ' R be Lipschitz and let x0 ! E " . !f (i) Assume that all the dire...
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This document was uploaded on 03/31/2014.

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