In the following $E,F$ are Banach spaces. A standard result from analysis is
Let $f: \mathbb{R}^n \to F$ with $n \in \mathbb{N^{\times}}$. Then $f$ is continuously Fréchet-differentiable if and only if all the partial derivatives $\mathbb{R}^n\ni h \mapsto \frac{\partial f}{\partial x_j}(h) \in F$ for $j \in \{1,…,n\}$ exist and are continuous.
Does there exist a generalisation of this for maps with a general Banach space as a domain? A direct generalisation without appealing to basis vectors as in the finite-dimensional case would be
Let $f: E \to F$. Then $f$ is continuously Fréchet-differentiable if and only if for every $e\in E$ the Gâteaux-derivative $E\ni h \mapsto Df(h)e\in F$ exists and is continuous.
However, I suspect that this is false, since the proof of the finite-dimensional statement pretty much leans on the fact that $\mathbb{R}^n$ has a finite basis. What would be a counter-example to this?
Can the finite-dimensional case still be generalised in some way? What about if $E$ is a (separable?) Hilbert space and we replace the basis vectors by a maximal orthonormal system? Finally, what about the following statement:
Let $f: E \to F$. Then $f$ is continuously Fréchet-differentiable if and only if the map $E\oplus E \ni (h,e) \mapsto Df(h)e \in F$ exists and is continuous, where $Df(\cdot)$ denotes the Gâteaux-derivative.
Best Answer
Your first statement is true. If you have a continuous Gâteaux derivative, then your function is Fréchet differentiable. Here, it is important that you equip $\mathcal L(E,F)$ with the usual operator norm. The proof is not hard, it uses the weak mean value inequality.
There is also a concept of partial derivatives: if $E = E_1 \times E_2$ is a product of Banach spaces, Fréchet differentiability follows from partial differentiability with continuous partial derivatives, see Theorem 3.7.1 in the book ``Differential calculus'' by Cartan.
Your last statement is false. For a counterexample consider $E = L^2(0,1)$ (usual Lebesgue measure) and \begin{equation*} f(u) := \int_0^1 \sin(u(t)) \, \mathrm dt. \end{equation*} Then, $f \colon L^2(0,1) \to \mathbb R$ is Gâteaux differentiable with \begin{equation*} f'(u)\,h = \int_0^1 \cos(u(t)) \, h(t) \, \mathrm dt, \end{equation*} but this is not a Fréchet derivative. Moreover, $(u,h) \mapsto f'(u)\,h$ is continuous.