# Fréchet vs. Carathéodory differentiability on Banach spaces

It is well-known that in the finite-dimensional case one can use the notion of Fréchet differentiability and Carathéodory differentiability interchangeably. See for example the 194 AMM article Frechet vs. Carathéodory by Acosta and Delgado (doi:10.2307/2975625). I am trying to obtain similar results in the case of general Banach spaces. So I suggest the following definition.

Definition. Let $$X$$ and $$Y$$ be Banach spaces. A function $$f \colon X \to Y$$ is called Carathéodory differentiable at some $$x_0 \in X$$, iff there exists $$\varphi \colon X \to \mathcal{L}(X,Y)$$ which is continuous at $$x_0$$ such that $$f(x) - f(x_0) = \varphi(x)(x - x_0) \qquad \forall x \in X.$$

Now it is easy to show that Carathéodory differentiability of $$f$$ at $$x_0$$ implies Fréchet differentiablity of $$f$$ at $$x_0$$. However, I do not quite see if the converse is true. The proof in the aforementioned paper uses the tensor product and the standard Euclidean inner product, as well as coordinates on $$\mathbb{R}^n$$. Is it possible to show the implication Fréchet differentiable implies Carathéodory differentiable or is there a counterexample in this general setting?

Edit. Apart from the very good answer, I've found a reference in the article An alternative Approach to Fréchet Derivatives appearing in the Journal of the Australian Mathematical Society on the 24th of May 2020 (doi:https://doi.org/10.1017/S1446788720000166).

• Which topology are you putting on $\mathcal{L}(X,Y)$? Jan 30, 2021 at 1:03
• @JamesHanson I denote by $\mathcal{L}(X, Y)$ the space of bounded linear operators, so it is equipped with the standard supremum norm of linear operators. Jan 30, 2021 at 8:15

We replace coordinates in $$\mathbb{R}^n$$ with the Hahn-Banach theorem in infinite-dimensional spaces. Suppose that $$f$$ is Fréchet-differentiable at $$x_0 \in X$$ with derivative $$f'(x_0) \in \mathcal{L}(X,Y)$$, so $$\lim_{x\to x_0} \frac{\|f(x)-f(x_0) - f'(x_0)(x-x_0)\|_Y}{\|x-x_0\|_X} = 0.$$
Set first $$\varphi(x_0) = f'(x_0)$$.
For $$x \neq x_0$$, we use the Hahn-Banach theorem. It is a classical consequence of this theorem that for every $$x \neq 0$$ there exists a functional $$\xi_x \in X'$$ such that $$\xi_x(x) = \|x\|_X$$ and $$\|\xi_x\|_{X'} = 1$$. For every $$x\in X\setminus\{x_0\}$$ and $$h \in X$$, define $$\varphi(x)h := \left[\frac{f(x) - f(x_0) - f'(x_0)(x-x_0)}{\|x-x_0\|_X}\right]\xi_{x-x_0}(h) + f'(x_0)h.$$ Then $$\varphi(x) \in \mathcal{L}(X,Y)$$ for every $$x\in X$$, $$\varphi(x)(x-x_0) = f(x)-f(x_0),$$ and $$\|\varphi(x)-\varphi(x_0)\|_{\mathcal{L}(X,Y)} \leq \frac{\|f(x) - f(x_0) - f'(x_0)(x-x_0)\|_Y}{\|x-x_0\|_X} \quad \xrightarrow{~x\to x_0~} \quad 0.$$