I'm having trouble with the proof of the following fact.
Let $E,F,G$ be Banach spaces. Suppose $X$ is open in $E$ and $Y$ is open in $F$. Given functions $f\in C^m(X,F)$, $g\in C^m(Y,G)$ such that $f(X)\subseteq Y$. Then $g\circ f\in C^m(X,G)$.
Here $C^m$ means $m$-times Fréchet differentiable. We write $\mathcal{L}(E,F)$ for the space of bounded linear operators from $E$ to $F$.
My book (Amann & Escher, Analysis II, p.185) says the proof is induction on $m$ and left as an exercise. While I understand that $\partial(g\circ f)(x_0)=\partial g(f(x_0))\partial f(x_0)$, I got stuck for the case $m\geq2$. Here are my thoughts.
So $\partial(g\circ f):X\to\mathcal{L}(E,G)$ is the composition of $(\partial g)\circ f$ and $\partial f$, where $X\xrightarrow{f}F\xrightarrow{\partial g}\mathcal{L}(F,G)$ and $X\xrightarrow{\partial f}\mathcal{L}(E,F)$. However, this does not mean $\partial(g\circ f)=(\partial g)\circ f\circ\partial f$, because the composition of $(\partial g)\circ f$ and $\partial f$ is actually the composition of the values of these functions. So we shall write $\partial(g\circ f)=\big((\partial g)\circ f\big)(\partial f)$. We can decompose the map like this:\begin{matrix}X&\xrightarrow{\varphi}&\mathcal{L}(E,F)\times\mathcal{L}(F,G)&\xrightarrow{\psi}&\mathcal{L}(E,G)\\x_0&\mapsto&(\partial f(x_0),\partial g(f(x_0)))&\mapsto&\partial g(f(x_0))\partial f(x_0)\end{matrix} By assumption the map $x_0\mapsto\partial f(x_0)$ belongs to $C^{m-1}(X,\mathcal{L}(E,F))$. By induction the map $x_0\mapsto\partial g(f(x_0))$, being a composition of $\partial g$ and $f$, both of which are $C^{m-1}$, belongs to $C^{m-1}(X,\mathcal{L}(F,G))$. Hence $\varphi\in C^{m-1}(X,\mathcal{L}(E,F)\times\mathcal{L}(F,G))$. It suffices to show that $\psi\in C^{m-1}(\mathcal{L}(E,F)\times\mathcal{L}(F,G),\mathcal{L}(F,G))$, and then induction completes the proof.
But how do I show that $\psi\in C^{m-1}(\mathcal{L}(E,F)\times\mathcal{L}(F,G),\mathcal{L}(F,G))$? And are there other proofs?
Notice $\psi$ is just the product of linear functions $(A, B) \mapsto A \circ B$ hence is infinitely differentiable (even analytic).
(You probably already know bilinear continuous functions are infinitely differentiable.)
Theorem. Any multilinear continuous function $u:E_1 \times \ldots \times E_r \to F$ is indefinitely differentiable.
Abridged proof. $$u(x_1 + h_1, \ldots, x_r + h_r) - u(x_1, \ldots, x_r) - \sum_{i = 1}^r u(x_1, \ldots, h_i, \ldots, x_r) = \sum u(y_1, \ldots, y_r)$$ where the $(y_1, \ldots, y_r)$ are vectors such that at least two entries are of the form $h_i.$ The right hand side is $o(\|h_1\| + \ldots + \|h_r\|) = o(\|(h_1, \ldots, h_r)\|).$ Hence $$u'(x_1, \ldots, x_r) \cdot (h_1, \ldots, h_r) = \sum_{i = 1}^r u(x_1, \ldots, h_i, \ldots, x_r).$$
Define now $e_i$ to be the canonical inclusion $E_i \mapsto E_1 \times \ldots \times E_r.$ So $u'(x_1, \ldots, x_r) = \sum_{i = 1}^r u(x_1, \ldots, e_i, \ldots, x_r)$ and the function in the right is the sum of $r$ multilinear functions (in $r - 1$ variables). By induction, we are done.