A proof on Frechet derivative!

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Let $X,Y$ be Banach spaces, define by $F:X\times Y \rightarrow \mathbb{R}$ be a functional, $F_u,F_v$ be Frechet derivative of $F$ with respect to $u$ and $v$ variables. We show that $F$ is Frechet differentiable.

We have

\begin{align} P&=\Vert F\left(u+h,v+l\right)-F\left(u,v\right)-F_u\left(u,v\right)h-F_v\left(u,v\right)l \Vert \\ &= \Vert F\left(u+h,v+l\right)-F\left(u,v+l\right)-F_u\left(u,v\right)h+F\left(u,v+l\right)-F\left(u,v\right)-F_v\left(u,v\right)l \Vert \\ &\leq \Vert F\left(u+h,v+l\right)-F\left(u,v+l\right)-F_u\left(u,v\right)h \Vert +\Vert F\left(u,v+l\right)-F\left(u,v\right)-F_v\left(u,v\right)l \Vert \\ &\leq \Vert F\left(u+h,v+l\right)-F\left(u,v+l\right)-F_u\left(u,v\right)h \Vert +\epsilon\Vert l\Vert \end{align} We need to show that $$\Vert F\left(u+h,v+l\right)-F\left(u,v+l\right)-F_u\left(u,v\right)h \Vert \leq \epsilon' \Vert h\Vert \quad ???$$

It follows that $F$ admits $F'\left(u,v\right)=F_u\left(u,v\right)h+F_v\left(u,v\right)l$ is the Frechet derivative. Besides, we have that

> $$\Vert F\left(u+h,v+l\right)-F\left(u,v+l\right)-F_u\left(u,v+l\right)h \Vert \leq \epsilon'' \Vert h\Vert \quad $$ Therefore, \begin{align} Q&=\Vert F\left(u+h,v+l\right)-F\left(u,v+l\right)-F_u\left(u,v\right)h \Vert \\ &=\Vert F\left(u+h,v+l\right)-F\left(u,v+l\right)-F_u\left(u,v+l\right)h +F_u\left(u,v+l\right)h -F_u\left(u,v\right)h \Vert \\ &\leq \Vert F\left(u+h,v+l\right)-F\left(u,v+l\right)-F_u\left(u,v+l\right)h \Vert+\Vert F_u\left(u,v+l\right)h -F_u\left(u,v\right)h \Vert \\ &\leq \epsilon''\Vert h\Vert +\Vert F_u\left(u,v+l\right)-F_u\left(u,v\right)\Vert \Vert h \Vert \end{align} On the other hand, since $F_u$ is continuous with respect to $v$ variable it follows that for every $\epsilon''' >0$ there exists a $\delta >0$ such that $\Vert F_u\left(u,v+l\right)-F_u\left(u,v\right)\Vert <\epsilon'''$ for all $\Vert \left(h,l \right) \Vert < \delta.$

Therefore, $$Q \leq \left(\epsilon'' +\epsilon'''\right) \Vert h\Vert.$$

By Cauchy-Schwarz's inequality It leads to that \begin{align} P&\leq \epsilon'\Vert h\Vert+\epsilon \Vert l\Vert , \quad \epsilon'=\left(\epsilon'' +\epsilon'''\right) \\ &\leq \sqrt{{\epsilon'}^2+{\epsilon}^2}\sqrt{\Vert h\Vert^2 + \Vert l \Vert^2}\\ &= \sqrt{2}\epsilon\Vert \left(h,l\right)\Vert, \quad \epsilon=\epsilon' \end{align}

Consequently, $F$ is Frechet differentiable!

Can you tell me whether or not there is any mistake in this proof?

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There are two mistakes in this proof, that can be resolved by imposing the assumption of continuity on both $F_u$ and $F_v$.

First in $$\Vert F\left(u+h,v+l\right)-F\left(u,v+l\right)-F_u\left(u,v+l\right)h \Vert \leq \epsilon'' \Vert h\Vert \quad$$

Here $\delta$ depends on $l$. we know $u,v$ are fix but not $l$ ! as $l$ gets small $\delta $ may get small too.

Second, Absolutely $F_u$ is not linear (necessarily) nor continuous!