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](https://doi.org/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>).



Fréchet vs. Carathéodory differentiability on Banach spaces
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
$$\|\varphi(x)-\varphi(x\_0)\|\_{\mathcal{L}(X,Y)} \leq \frac{\|f(x) - f(x\_0) - f'(x\_0)(x-x\_0)\|\_Y}{\|x-x\_0\|\_X} \quad \xrightarrow{~x\to x\_0~} \quad 0.$$


