Let $E$ be a Banach space. $\mathcal{L}(E)$ is the space of continuous linear mappings from $E$ to $E$. Let $f: \mathcal{L}(E) \rightarrow \mathcal{L}(E)$ such that $$f(u) = u \circ u \circ u $$ Show that $f$ is differentiable.
I am pretty unsure on how to approach this. I know this is a dumb question, but I find it hard to work with differentiability of functions that map to other functions.
Can someone tell me how to approach such a question?
I will give you some hints. I am going to use a sufficient condition for differentiability, based on the definition of Gâteaux differentiability.
Recall that $u$ is linear. Hence, for $t \in \mathbb{R}$ and $v \in \mathcal{L}(E)$, we can write $$ (u+tv) \circ (u+tv) = u \circ u + t u \circ v + t v \circ u + v \circ v. $$ It follows immediately that $$ \frac{d}{dt} (u+tv) \circ (u+tv) = u \circ v + v \circ u. \tag{1} $$ Now you must show that the right-hand side of (1) depends continuously on $u \in \mathcal{L}(E)$. This should be rather easy.
Finally, iterate twice to show that $f$ is differentiable.