Background
I'm studying Gateaux derivatives and I find some difficulties to underestand the general case where the order of the derivative is $n>1$. First, I'm considering the following definition for the first-order derivative at $h(x):\mathbb{R}^d\mapsto \mathbb{R}$ and in direction $g(x):\mathbb{R}^d\mapsto\mathbb{R}$ of the generic functional $F[h]$, \begin{equation*} F'[h;g]\triangleq \left[\frac{\text{d}}{\text{d}\varepsilon}F[h+\varepsilon g]\right]_{\varepsilon=0} \end{equation*} we can simplfy this expression by removing the variable $\varepsilon$ (since in the end it will be fixed to the value $\varepsilon=0$). Since the internal derivative is a "conventional" derivative in $\varepsilon$, we have \begin{equation*} F'[h;g]=\left[\lim_{\tau\to 0} \frac{F[h+(\varepsilon+\tau)g]-F[h+\varepsilon g]}{\tau}\right]_{\varepsilon=0} \end{equation*} then setting $\varepsilon=0$ gives \begin{equation*} F'[h;g]=\lim_{\tau\to 0} \frac{F[h+\tau g]-F[h]}{\tau} \end{equation*}
Questions
The $n$-th order derivative is defined as \begin{equation*} F^{(n)}[h;g]\triangleq \left[\frac{\text{d}^n}{\text{d}\varepsilon^n}F[h+\varepsilon g]\right]_{\varepsilon=0} \end{equation*}
Currently I have two questions about this expression.
- I'm wondering if we can do, like in the previous case, a simplification where we eliminate $\varepsilon$
- I'm wondering if, as it happens with conventional derivatives, we can write a recursion of the type \begin{equation*} F^{(k+1)}[h;g]=F^{(k)}[h;g] \end{equation*}
If the answer of the previous questions is yes, how do you prove it? It will be enough the simple case with $n=2$.