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Topic:
Frechét Derivative
Replies:
3
Last Post:
Feb 11, 2013 5:16 AM



Kiuhnm
Posts:
88
Registered:
8/17/07


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


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



