Existence of continuous derivatives of partial functions and total differentiability We know that for a function $\mathbb{R}^m\to \mathbb{R}^n$ the existence and continuity of partial derivatives implies the differentiability of that function.
Will this hold true for functions on product spaces of arbitrary Banach spaces? I.e., given $f\colon E_1\times E_2\to F$ ($E_1,E_2,F$ being Banach spaces) and $(x,y)\in E_1\times E_2$, does the existence and continuity of $(x,y)\mapsto Df(\cdot,y)(x)$ and $(x,y)\mapsto Df(x,\cdot)(y)$ locally at $(x,y)$ imply $f$ to be differentiable at $(x,y)$?
I guess it won't because the proof in the euclidean case goes from $f(p)$ to $f(p+h)$ in finitely many steps along the coordinate axes... So, maybe it will hold in product spaces of Hilbert spaces? What would a counter-example be?
 A: This is still true, even in the Banach case. Because you have a finite number of factors, and the whole differentiable continuity on each of them.
To show this, apply the mean value equality two times to join $f (x,y)$ to $f (0,0)$ to see differentiability at 0 : 
$$f (x,y) = f (0,0) + \int_0^1 {D_E f (tx, 0)(x)dt } + \int_0^1 {D_F f (x, ty)(y)dt } $$
$$= f (0,0) + \int_0^1 {D_E f (tx, 0)(x)dt } + \int_0^1 {D_F f (0, ty)(y)dt } + \int_0^1 {D_F f (x, ty)(y) -  D_F f (0, ty)(y) dt} $$
Where the last term is $o (|(x,y)|)$. EDIT : this is true because you should notice that continuity implies local uniform continuity. 
You might want to know if a function $E \rightarrow F$ is continuously diffentiable on each finite subspace $G $, then it is differentiable ? 
EDIT : I am sorry, my counter example was not valid. Still, I think it should be false.
EDIT 2 : I have found one. Take $E = F = L^\infty$. Take the functions $(f_n)$ where $(f_n)$ is the sequence of functions from $\mathbf{R}$ to $[0,1]$ defined by $f_n(x) = cos(x)$ (I've just took a random smooth bounded function which is not $0$). Let $f$ the following function : it sends an element $(x_n)$ of $L^\infty$ to $(f_n(x_n))$ if $(x_n)$ is almost everywhere vanishing (let us note $F'$ the subspace of such sequences), $0$ on a supplementary $F''$ of $F'$, and extend $f$ by the relation $f(a + b) = f(a) + f(b)$ for $a \in F'$, $b \in F''$. Then $f$ not differentiable (not even continuous) ; furthermore, it is continuously differentiable on each finitely generated space of $L^\infty$. (I have still difficulties to find an example where $f$ is continuous, not differentiable, and continuously differentiable on each subspace of finite dimension ; it would be more interesting)
Nevertheless the result can still be generalized. Let $(E_i)$ subspaces of your Banach space $E$ which generates it in a Banach sense (i.e the closure of the space geneated by the $E_i$ is E). Let $D_i $ denote the partial differentials.
If you have existence of partial differentials of $f $ and :


*

*For all $x$, the majorations $|D_i f(x) | \leq M_{i,x}$ where $\sum_i M_{i,x}$ converges .

*uniform majorations $|D_if (x) - D_if (y)| \leq M_{i,U}$ over numerous open sets $U $ of $E$ whose union is $E$, where the sum of the $M_{i,U}$ over $i $ is convergent (kind of equicontinuity hypothesis)
You will have differentiability of f ! 
It is essentially the same proof of the case of two factors, in which case these hypothesis are trivially true.
