Suppoce for every $u$, $f(u)$ is a square integrable function, namely, $f(u)\in L^2([0,1])$. We can compute the Frechet derivative of this functional $f$, or in other words, the condition for $f$ to be Frechet differentiable is $$ \lim_{h\to 0}\frac{1}{h}\left( \left\|f(u+h)-f(u)-hf'(u) \right\| \right)=0. $$
However, we can also compute the derivative for $f$ with the index $s$ fixed. Say, for a fixed $s\in[0,1]$, $f(u;s)$ will only be a mapping from $[0,1]$ to $R$. We can also compute the partial derivative of $u$ and the differentiable condition will be $$ \lim_{h\to 0}\frac{1}{h}\left(f(u+h;s)-f(u;s)-hf'(u;s) \right)=0. $$ We can therefore require the above equation (with $s$ fixed) exists for every $s\in[0,1]$ and also obtain the derivative $f'(u;s)$. Namely, we compute this derivative pointwisely.
My question is: what is the relationship between these two differentiable conditions???
Based on my intuition, I believe the later is sufficient to prove the former, but the contray may not exist. Moreover, the second equation should exist for almost everywhere $s\in[0,1]$ instead of the strictly every $s\in[0,1]$. (It is just my guess).
To prove my guess, note that the first condition needs $$ \lim_{h\to 0}\int_0^1\frac{1}{h^2}\left(f(u+h;s)-f(u;s)-hf'(u;s) \right)^2 ds=0. $$ The second simply interchanges the $\lim$ and the integral.(With $\lim$ first and integral second). We can use Lebesgue dominated convergence theorem to prove this problem. However, since our limit is simply $0$, I believe the dominated convergence theorem will strengthen the condtion. In fact, we should not need such a strict condition.(I guess it...)
I wonder also is there any reference talking about this question comprehensively?
Thanks for every kind of help!!!It torments me for a long time(cry...)
