frechet derivative in space with inner product

Question

If $V$ is a space with inner product ($\cdot,\cdot$). If $f:V\rightarrow \mathbb{R}$, $$f(u)=(u,u)$$ find Frechet derivative $f'(u)$ Can anybody help me? Thanks

Answer 1

$$f(u+h)=(u+h,u+h)=(u,u)+(u,h)+(h,u)+(h,h)$$ and $$f(u)=(u,u)$$ so $$f(u+h)-f(u)=(u,h)+(h,u)+(h,h)$$ so $$\lim\limits_{||h||\rightarrow 0} \frac{||f(u+h)-f(u)-f'(u)h||}{||h||}=\lim\limits_{||h||\rightarrow 0} \frac{||(u,h)+(h,u)+(h,h)-f'(u)h||}{||h||}$$ i am wrong?

Answer 2

if we set $f'(u)=(2u,1)$ then $\frac{|| 2(u,h) +(h,h) - (2u,1)h||}{\sqrt{|(h,h)|}} =\frac{|| 2(u,h) +(h,h) - (2u,h)||}{\sqrt{|(h,h)|}} = \frac{||(h,h)||}{\sqrt{|(h,h)|}} = \frac{|(h,h)|}{\sqrt{|(h,h)|}}=\sqrt{|(h,h)|}$ wich goes to $0$ as $h \rightarrow 0$

Answer 3

If it exists, since your $f$ is a quadratic form derived from a bilinear function, the only possible candidate is $$df(u)v = (u,v)+(v,u)$$