Calculating a Fréchet derivative Let $I$ be a set, and let $B(I)$ be the space of bounded, real-valued functions on $I$ equipped with the sup-norm. Let $\phi: \mathbb R \to \mathbb R$ be bounded and continuously differentiable everywhere. Finally, let $S: B(I) \to \mathbb R$ be linear and continuous.
Define $\Phi: B(I) \to \mathbb R$ by $\Phi(x) = S(\phi \circ x)$.

Is $\Phi$ Fréchet differentiable at every $x \in B(I)$, and if so is it the case that $\Phi'(x) = S(\phi' \circ x)$?

I can show that the function $x \mapsto S(\phi' \circ x)$ is linear and bounded, using the corresponding facts about $\phi'$ and $S$, but I'm not sure I can show that this function satisfies the definition of the Fréchet derivative.
I have to show, for every $x \in B(I)$, that
$$\lim_{\| h \|_\infty \to 0} \frac{| \Phi (x + h) - \Phi(x) -  S(\phi' \circ h) |}{\| h \|_\infty} = 0,\tag{1}$$
where $\| \cdot \|_\infty$ is the sup-norm on $B(I)$. Now, by the definition of $\Phi$ and the linearity of $S$
$$ \frac{\Phi (x + h) - \Phi(x) -  S(\phi' \circ h)}{\|h\|_\infty} =  \frac{S(\phi \circ (x+h)) - S(\phi \circ x) -  S(\phi' \circ h)}{\|h\|_\infty}  = S\Big(\frac{[\phi \circ(x+h)] - [\phi \circ x] - [\phi' \circ h]}{\|h\|_\infty} \Big).$$
From here I would like to argue that as $\|h\|_\infty \to 0$,
$$\frac{[\phi \circ(x+h)] - [\phi \circ x] - [\phi' \circ h]}{\|h\|_\infty} \to 0, \tag{2}$$
and then use the continuity of $S$ to conclude. I haven't convinced myself that (2) holds however.
 A: For every $t\in I$, by the differentiability of $\phi$, $$(\phi (x+h))(t) - \phi(x(t))=\phi(x(t) + h(t)) - \phi(x(t)) =\phi'(x(t))h(t) + o(h(t)).$$
That is to say, at least formally,
$$ \phi\circ (x+h) - \phi\circ x = (\phi'\circ x) h + o(h).$$
To justify this, we need to use $\phi\in C^1$ to write for $x,h\in \mathbb R$,
$$ \phi(x+h) - \phi(x) - \phi'(x)h= R(x,h)h$$
and the boundedness of $\phi'$ gives the boundedness of $R$ on bounded subsets on $\mathbb R^2$. Letting $x,h\in B(I)$ gives the result.
By chain rule and linearity, $d (S\circ f)(x)h = dS(f(x))[df(x)h]=S(df(x)h)$, so
$$ d\Phi(x)h = S((\phi'\circ x) h). $$
Note carefully the difference between $h\mapsto (\phi'\circ x) h$ and $x\mapsto \phi'\circ x$. The Frechet derivative at each point is supposed to be a linear map, and I don't think you could have shown $x\mapsto S(\phi'\circ x)$ is linear. Well, actually you could, because the only convex, bounded functions $\phi$ are constants, and so $\Phi$ is a constant function, with 0 derivative, and abusing notation with 0 would give you linearity. But ignoring either boundedness or convexity(as you have edited the question), your formula doesn't give a linear map in the general case. The map $x\mapsto S((\phi'\circ x)\cdot)$ maps from $B(I)$ to the space of linear maps on $B(I)$ taking values in $\mathbb R$.
