Generalisation of characterisation of continuous Fréchet-differentiability to infinite dimensions

Question

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.

Answer 1

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.

Answer 2

In fact to define a Fréchet derivative for infinite dimensional spaces, you're facing two problem.

First for a fixed point $x_0 \in E$, $Df(x_0)$ needs to be continuous. In finite dimensional spaces, all norms are equivalent. So defining the continuity doesn't depend on the norm. This is no more true for infinite dimensional spaces. So you have to define $Df(x_0)$ and its continuity for a given norm... Knowing that $Df(x_0)$ might be discontinuous for another norm.

Second, and more complicated... $Df(x)$ is a function from $E$ to $\mathcal L(E,F)$ so to say that $Df(x)$ is continuous, you have to define a topology on $\mathcal L(E,F)$. Not an easy question.