Fréchet derivative of $\|Au-f\|^2$

Question

$A$ is a bounded linear operator on an infinite dimensional Hilbert space. I have been told the answer is supposed to be $2A^{*}(Au-f)$. Here is my progress:

$\|A(u+h)-f\|^2-\|Au-f\|^2=\langle A(u+h)-f, A(u+h)-f \rangle - \langle Au-f, Au-f \rangle = \langle Au-f,Ah \rangle + \langle Ah,Au-f \rangle + \|Ah\|^2$

Now $\|Ah\|^2 \leq \|A\|^2 \|h\|^2$ so this term is $o(h)$. I am not quite sure how to get to the last step.

Answer 1

you have : $$\|A(u+h)-f\|^2-\|Au-f\|^2= \langle Au-f,Ah \rangle + \langle Ah,Au-f \rangle + \|Ah\|^2\\ =\langle Ah,2(Au-f)\rangle \text{ by symmetry }\\ = \langle h,2A^*(Au-f)\rangle \text{ by the proprety of the adjoint}$$

Then, you have the result : $$\|A(u+h)-f\|^2-\|Au-f\|^2=\langle df,h\rangle+o(||h||) $$

Hence you get your derivative : $$ 2A^*(Au-f)$$