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).

boundedlinear operators, so it is equipped with the standard supremum norm of linear operators. $\endgroup$ – TheGeekGreek Jan 30 at 8:15