Relation of $n$-th variation, Gâteaux and Fréchet derivatives

Question

Let $E$ be a Banach space, $U \subset E$ be an open subset of $E$ and $f: U \to \mathbb{R}$ a given function. For fixed $x_{0}, h \in E$, there exists $t_{0} \ge 0$ such that $x_{0} + (-t_{0},t_{0})h \subset U$. Thus, we can introduce a new function $F_{h}: (-t_{0},t_{0}) \to \mathbb{R}$ by setting $F_{h}(t) = f(x_{0}+th)$. Suppose $F_{h}$ is $N$ times differentiable on $(-t_{0},t_{0})$, so we can consider its Taylor expansion: $$F_{h}(t) = F_{h}(0) + \sum_{n=1}^{N}\frac{t^{n}}{n!}F_{h}^{(n)}(0) + R_{N} $$ for all $t \in (-t_{0},t_{0})$. Here, $R_{N}$ is just the remainder coming from Taylor's expansion. The term $F_{h}^{(k)}(0)$ is called the $n$-th variation of $f$ at $x_{0}$ in the direction of $h$.

My question is: what is the relation between the $n$-th variation of $f$ at $x_{0}$ in the direction of $h$ and the Gâteaux and Fréchet derivative of $f$? I am guessing that, in order to establish such a relation, I have to assume that $f$ is $N$ times Fréchet or Gâteaux differentiable or that the $n$-th variation exists for all $h$?

Answer 1

It is easy to establish the equivalence of the following two statements:

• $f$ is Gâteaux differentiable at $x_0$.
• the first variation exists for all $h \in E$ and $h \mapsto F_h^{(1)}(0)$ is linear and continuous.

For higher order derivatives, we merely have that

• $f$ is Gâteaux differentiable of order $n$ at $x_0$

implies that

• the $n$-th variation of $f$ at $x_0$ exists for all $h \in E$ and $F_h^{(n)} = q(h,h,\ldots,h)$, where $q \colon E^n \to \mathbb R$ is multilinear, symmetric and continuous.

The converse does not hold, because we cannot infer differentiability of $f$ at points in a neighborhood of $x_0$. Indeed, consider $E = \mathbb R$ and something like $$ x \mapsto x^3 g(x), $$ where $g \colon \mathbb R \to [-1,1]$ is arbitrary. Then, $g$ possesses a second-order expansion at $x_0 = 0$, but may fail to be differentiable at $x \ne 0$.