Directional derivative of $f:H^1 \to H^1$ Let $f:\mathbb{R} \to \mathbb{R}$ be a smooth function. We think of $f:H^1(\Omega) \to H^1(\Omega)$ in the sense that $f(u)(x) = f(u(x))$.
I want to know what the directional derivative is of $f:H^1(\Omega) \to H^1(\Omega)$. 
Is it simply (the derivative of $f$ at $u$ in the direction $d$)
$$Df(u;d)(x) = f'(u(x))d(x)?$$
where $f'$ is the standard derivative of $f:\mathbb{R} \to \mathbb{R}$?
 A: First of all, a clarification: since the elements of $H^1 (\Omega)$ are not functions, but equivalence classes of functions modulo equality on null sets, it doesn't make sense to "apply" $u$ at $x \in \Bbb R$, therefore the writing $u(x)$ doesn't make sense a priori. Of course, in order for it to become meaningful you would have to embed $H^1 (\Omega)$ in a space of functions. This can be done, assuming $\Omega$ to be regular enough. If $\dim \Omega = n$ and $r \ge 0$ and $\alpha \in (0,1]$ are such that $r + \alpha = 1 - \frac n 2$, then $H^1 (\Omega)$ may be embedded into the Hölder space $C^{r,\alpha} (\Omega)$. In order for this to happen, the only possibility is $n=1$, $r=0$ and $\alpha = \frac 1 2$.
Let us note the action of $f$ on $H^1 (\Omega)$ as $\tilde f$, in order to avoid confusion.
Now, working on the image of $H^1 (\Omega)$ as a Banach subspace of $C^{0,\frac 1 2} (\Omega)$, the definition of the directional derivative is quite clear: the directional derivative of $\tilde f$ at the point $\phi$ in the direction $\psi$ is
$$D \tilde f(\phi, \psi) = \lim _{t \to 0} \frac {\tilde f (\phi + t \psi) - \tilde f (\phi)} {t} ,$$
whence, using $\tilde f (u) = f \circ u$ and viewing $D \tilde f (\phi, \psi)$ as a function of $x \in \Omega$, it follows that
$$D \tilde f(\phi, \psi) (x) = \lim _{t \to 0} \frac {\tilde f \circ (\phi + t \psi) - \tilde f \circ (\phi)} {t} (x) = \lim _{t \to 0} \frac {f \big( (\phi(x) + t \psi(x) \big) - f \big( \phi(x) \big)} {t} = \\
\frac {\Bbb d} {\Bbb d t} [t \mapsto f \big( (\phi(x) + t \psi(x) \big)] \Big| _{t = 0} = f' \big( \phi(x) \big) \psi (x) = \widetilde {f'} (\phi) (x) \psi (x) ,$$
whence $D \tilde f(\phi, \psi) = \widetilde {f'} (\phi) \psi = (f' \circ \phi) \psi$, so yes, your intuition is correct.
