Given $E, F$ two normed space, $f : E \to F$. I got quite confused about the notation $d_xf(\cdot)$ and $df(\cdot)(x)$. It seems to me that the differential of the map $x\mapsto d_xf (v)$ at some point $x_0\in E$ is the map $d_{x_0}^2f(v)$, and I can prove that by decomposing the original map into an evaluation map and $x \mapsto d_xf$.
But I struggled to generalize this to higher derivatives. Say we have a map
$$x \mapsto d_x^nf(v_1, v_2,\dots, v_n)$$
for $v_i\in E$. Is it true that the derivative of this map at some point $x_0$ is $d_{x_0}^nf(v_1, v_2,\dots, v_n)$? And how would one prove it explicitly?
Thanks in advance.
Best Answer
Fix a tuple $(v_1,\dots, v_n)\in E^n$, and let $\mathcal{L}^n(E;F)$ be the Banach space of bounded $n$-multilinear maps $\underbrace{E\times\cdots \times E}_{\text{$n$ times}}\to F$, and now consider the evaluation mapping $\epsilon:\mathcal{L}^n(E;F)\to F$, \begin{align} \epsilon(T):= T(v_1,\dots, v_n) \end{align} Then, $\epsilon$ is a bounded linear transformation. The mapping $x\mapsto d^n_xf(v_1,\dots, v_n)$ is the composition of $d^nf:E\to \mathcal{L}^n(E;F)$ with $\epsilon:\mathcal{L}^n(E;F)\to F$ to yield $\epsilon \circ d^nf:E\to F$. By the chain rule, it's derivative at a point $x_0$ is (because $\epsilon$ is linear) $\epsilon\circ d(d^nf)_{x_0}\in \mathcal{L}(E,F)$. So, unwinding the definitions, its value on a vector $\xi$ is \begin{align} d(d^nf)_{x_0}(\xi)[v_1,\dots, v_n], \end{align} or after identifying $d(d^nf)_{x_0}$ with a $(n+1)$ multilinear map, denoted $d^{n+1}f_{x_0}$, in the canonical way, simply \begin{align} d^{n+1}f_{x_0}(\xi,v_1,\dots, v_n) \end{align} (the ordering doesn't matter since higher Frechet derivatives are symmetric multilinear maps).