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.
 Spring '14
 Derivative

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