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).

  • $\begingroup$ Which topology are you putting on $\mathcal{L}(X,Y)$? $\endgroup$ Jan 30, 2021 at 1:03
  • $\begingroup$ @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. $\endgroup$ Jan 30, 2021 at 8:15

1 Answer 1


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 $$\|\varphi(x)-\varphi(x_0)\|_{\mathcal{L}(X,Y)} \leq \frac{\|f(x) - f(x_0) - f'(x_0)(x-x_0)\|_Y}{\|x-x_0\|_X} \quad \xrightarrow{~x\to x_0~} \quad 0.$$

  • $\begingroup$ This is brilliant! Thanks. And much nicer than the proof in the paper. $\endgroup$ Jan 30, 2021 at 14:19
  • $\begingroup$ Brilliant answer indeed. It makes me curious, does the equivalence between both derivatives still hold if we move to a locally convex topological vector space? $\endgroup$ Jan 14, 2022 at 18:51

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