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Unformatted text preview: s di!erentiable at (0, 0) but the partial derivatives are not continuous at (0, 0).
Deﬁnition 18 Given an open set U " RN , we denote by C 1 (U ) the space of
!f
all continuous functions f : U # R whose partial derivatives ! xi , i = 1, . . . , N ,
exist at all x $ U and are continuous.
Remark 19 The hypothesis that f is continuous is redundant. It is enough to
deﬁne C 1 (U ) the space of all functions f : U # R whose partial derivatives
!f
! xi , i = 1, . . . , N , exist at all x $ U and are continuous. Indeed, for any such
function f , by Theorem 16, we have that f must be di!erentiable at all x $ U ,
and, in turn, by Theorem 9, f must be continuous.
Next we extend the mean value theorem to functions of several variables.
Theorem 20 (Mean Value Theorem) Let x, y $ RN , with x != y, let S be
the segment of endpoints x and y, that is,
S = {tx + (1 % t) y : t $ [0, 1]} ,
and let f : S # R be such that f is continuous in S and there exists the
!f
directional derivative ! v (z) for all z $ S except at most x and y, where v :=
x!y
"x!y" . Then there exists ! $ (0, 1) such that
f (x) % f (y) = "f
(!x + (1 % !) y) &x % y& .
"v (5) Lemma 21 Under the hypotheses of the previous theorem, the function g (t) :=
f (tx + (1 % t) y), t $ [0, 1], is di!erentiable for all t $ (0, 1), with
g # (t) = "f
(tx + (1 % t) y) &x % y& .
"v Proof. Fix t0 $ (0, 1) and consider
g (t) % g (t0 )
f (tx + (1 % t) y) % f (t0 x + (1 % t0 ) y)
=
t % t0
t % t0
$
%
f t0 x + (1 % t0 ) y+ (t % t0 ) &x % y& "x!y" % f (t0 x + (1 % t0 ) y)
x!y
=
&x % y&...
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This document was uploaded on 03/31/2014.
 Spring '14

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