21-356 Lecture 4

# Erentiability of f at x0 lets prove that these

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## This document was uploaded on 03/31/2014.

Ask a homework question - tutors are online