Math Forum Discussions

Kiuhnm

Posts: 88
Registered: 8/17/07

Frechét Derivative
Posted: Feb 9, 2013 12:17 PM

Plain Text

I have trouble with a theorem about the Fréchet derivative. I think the
best thing to do is to write the entire proof -- since it isn't too long
-- and then ask my questions.

~~~~~
Definition. Let f:D->Y be a mapping from an open set D in a normed
linear space X into a normed linear space Y. Let x in D. If there is a
bounded linear map A:X->Y such that
lim_{h->0} ||f(x+h)-f(x)-Ah||/||h|| = 0 (1)
then f is said to be Fréchet differentiable at x, or simply
differentiable at x. Furthermore, A is called the (Fréchet) derivative
of f at x.

Example. Let X=Y=C[0,1] and let phi:R->R be continuously differentiable.
Define f:X->Y by the equation f(x) = phi o x, where x is any element of
C[0,1]. What is f'(x)? To answer this, we undertake a calculation of
f(x+h)-f(x), using the classical mean value theorem:

[f(x+h)-f(x)](t) =
phi(x(t)+h(t)) - phi(x(t)) =
phi'( x(t)+theta(t)h(t) ) h(t)
where 0 < theta(t) < 1.

This suggests that we define A by
Ah = (phi' o x)h
With this definition, we shall have at every point t,
[f(x+h)-f(x)-Ah](t) = phi'( x(t)+theta(t)h(t) ) h(t) - phi'(x(t))h(t)
Hence, upon taking the supremum norm, we have
||f(x+h)-f(x)-Ah|| <= ||phi' o (x + theta h) - phi' o x|| ||h||
By comparing this to Equation (1) and invoking the continuity of phi',
we see that A is indeed the derivative of f at x. Hence f'(x) = phi' o x.
~~~~~

There are a few things I don't understand. First of all, if I'm not
mistaken, A is a bounded linear operator and Ah is really A(h), i.e. A
evaluated at h.
But then Ah is defined as (phi' o x)h. What does it mean? Is that (phi'
o x) times h or (phi' o x) evaluated at h?
I'll ask the other questions later.

Thanks in advance.

Kiuhnm