I have a trouble in handling a following question.
Given a composition operator $ F :C^1([0,1]) \to C^1([0,1]) ; F(f) = f(f(x))$, I have a Gâteaux derivative of $F$.
\begin{align}
\lim_{t \to 0} \frac{F(f+th) – F(f) }{t} &= \frac{(f+th)(f(x)+th(x)) – f(f(x))}{t} \\
&= \frac{f(f(x)+th(x)) + th(f(x)+th(x))- f(f(x))}{t} \\
&= \frac{f(f(x))+ tf'(f(x))h(x)+ o(t) +th(f+th(x))- f(f(x)) }{t} \\
&= \lim_{t \to 0} f'(f(x))h(x)+h(f+th(x))+\frac{o(t)}{t} \\
&= f'(f(x))h(x) + h(f(x)).
\end{align}
for given $h \in C^1([0,1])$.
Therefore, a Gâteaux derivative of $F$ at $f$ is $DF_f(h) = f'(f(x))h(x) + h(f(x))$.
Since it is a linear in terms of $h$, and continous in terms of $f$, can we say a composition operator $F$ is actually Fréchet differentiable?
I am curious that I am right.
Edit: I changed $C^\infty(\mathbb{R})$ to $C^1([0,1])$ with the norm $\|f\|_{C^1([0,1])} = \sup |f(x)| + \sup|f'(x)|$.
Best Answer
Continuity respect to the main argument is not sufficient for Frechet differentiability. You can see this for the particular choice $f(x)=2x$ and $h(x)=e^\frac{1}{2-x}$: $$ \frac{\delta F}{\delta f}[h](x) = 2e^\frac{1}{2-x}+ e^\frac{1}{2(1-x)}\implies \left\|\frac{\delta F}{\delta f}[h](x)\right\|_{\,C^1[0,1]}=+\infty. $$ The norm of the functional derivative depends on the functional argument $h$ in an essential way, thus the functional derivative $\frac{\delta F}{\delta f}[h](x)$ of the functional $f\mapsto F(f)=f\big(f(x)\big)$ is not Frechet.