Fix a function $f \in L^2(\Bbb R)$. In fact, let $f$ be continuously differentiable.
Now define a map $\Phi \colon \Bbb R \rightarrow L^2(\Bbb R)$ given by $$ \Phi(a) = f(a \bullet),$$ meaning the image of $a$ by $\Phi$ is the function $$x \mapsto f(ax).$$
I know the Fréchet derivative at $a \in \Bbb R$ is a linear map $A \colon \Bbb R \rightarrow L^2(\Bbb R)$ satisfying $$ \lim_{h \to 0} \frac{\Vert f\left(( a + h) \bullet \right) - f(a \bullet) - Ah \Vert_{L^2}}{|h|} = 0,$$ and I had a suspicion that it would be $h \mapsto hf'(h\bullet)$, but the problem is that this map does not seem to be linear, and I am not sure how it would satisfy the above limit.
Can anyone offer some assistance?
The derivative $\Phi'$ is a mapping from $\mathbb R$ to $L^2$ itself, the directional derivative $h\mapsto \Phi'(a)h$ has to be a $L^2$ function.
If $f$ is $C^1$, then the difference quotient $f( (a+h)x-f(ax))/h$ converges pointwise to $f'(ax)x$. Hence if $\Phi$ is Fréchet differentiable, then the directional derivative is given by $$ (\Phi'(a)h)(x) = f'(ax)hx. $$ This implies that $\int_{\mathbb R} x^2 f'(x)^2 dx<\infty$ is a necessary condition for Fréchet differentiability (and for Gâteaux as well).
If one assumes that $f$ has compact support (and $f(0)=0$ such that $\Phi(0)$ is well-defined) then it should be possible to prove Fréchet differentiability (use dominated convergence theorem).