I am trying to solve this problem.
Let $E$,$F$ be Banach spaces and $f:E\to F$ be an n-times differentialble function on $a \in E$ $D^nf(a)$ with $n\geq 2$ is a multilinear, bounded and symmetrical map.
So in general I know that the space of multilinear bounded maps can be identified with the derivatives of order n, for example the second derivative is a bilinear map. And according to Schwarz's theorem it is a symmetric application, since it proceeds by induction,
For the case $n=2$ the base case is valid.
I don't know how to relate my inductive hypotheses to the proof of the case $n+1$
Big thx for the Help.
Here is a sketch of one argument with some of the trickier parts explained in more detail. Suppose the statement is true for $n$ and suppose $f$ is $(n+1)$-times differentiable at the point $a$. The statement follows from proving the following: If $h_{1}, \ldots , h_{n+1}\in E$, then
$$[D^{n+1}f(a)](h_{2}, h_{1}, h_{3}, \ldots , h_{n+1}) = [D^{n+1}f(a)](h_{1}, h_{2}, h_{3}, \ldots , h_{n+1})$$
and if, in addition, $\sigma$ is a permutation of $\{2, \ldots , n+1\}$,
$$[D^{n+1}f(a)](h_{1}, h_{\sigma (2)}, \ldots , h_{\sigma (n+1)}) = [D^{n+1}f(a)](h_{1}, h_{2}, \ldots , h_{n+1}).$$
For if those statements have been proved, the general case follows from noting that any permutation of $\{1, \ldots , n+1\}$ is a composition of permutations of $\{1,2\}$ and $\{2, \ldots , n+1\}$.
Throughout the rest of the discussion, consider $h_{1}, \ldots , h_{n+1}$ all as fixed elements of $E$.
For the first statement, the function $g$ defined on an open neighbourhood of $a$ by $g(x) := [D^{n-1}f(x)](h_{3}, \ldots , h_{n+1})$ is twice differentiable at $a$ with $[D^{2}g(a)](t_{1}, t_{2}) = [D^{n+1}f(a)](t_{1}, t_{2}, h_{3}, \ldots , h_{n+1})$ for all $t_{1}, t_{2}\in E$. You can apply the base case to $g$ to conclude that $[Dg(a)](h_{1}, h_{2}) = [Dg(a)](h_{2}, h_{1})$, and the first statement follows.
For the second statement, note that by the induction hypothesis, if $\sigma$ is a permutation of $\{2, \ldots , n+1\}$, then for all $x$ in some open neighbourhood of $a$,
$$[D^{n}f(x)](h_{\sigma (2)}, \ldots , h_{\sigma (n+1)}) = [D^{n}f(x)](h_{2}, \ldots , h_{n+1}).$$
Define $\phi$ on an open neighbourhood of $a$ by $\phi (x) := [D^{n}f(x)](h_{2}, \ldots , h_{n+1})$ and $\psi$ on an open neighbourhood of $a$ by $\psi (x) := [D^{n}f(x)](h_{\sigma (2)}, \ldots , h_{\sigma (n+1)})$. It was shown above that $\phi (x) = \psi (x)$ for all $x$ in some open neighbourhood of $a$, so it follows that $[D\phi (a)](t_{1}) = [D\psi (a)](t_{1})$ for all $t_{1}\in E$. But it can be shown that
$$[D\phi (a)](t_{1}) = [D^{n+1}f(a)](t_{1}, h_{2}, \ldots , h_{n+1})$$
and
$$[D\psi (a)](t_{1}) = [D^{n+1}f(a)](t_{1}, h_{\sigma (2)}, \ldots , h_{\sigma (n+1)})$$
holds for all $t_{1}\in E$, so the second statement follows by taking $t_{1}$ as $h_{1}$.