Is the differential of a functional norm-dependent? A functional $J: (\text{normed function space}) \to \mathbb{R}$  is said to be differentiable at $f$ if there exists a linear functional $\phi$ such that
$$\forall h: J(f + h) - J(f) = \phi(h) + r(h) ||h||$$
where $r(h)$ is a functional which goes to $0$ as $||h|| \to 0$. My questions is: does the existence, and value of, the derivative depend on the norm $||\cdot||$?
My intuition is that both the existence and value are norm-dependent. As a special case though, I think that if two norms are equivalent, then the existence and value of the derivatives match for both norms. That is, suppose $||\cdot||_0$ and $|| \cdot ||_1$ are two norms with
$$c|| \cdot ||_1 \le || \cdot ||_0 \le C || \cdot ||_1$$
for some positive constants $c, C$ and suppose that $J$ is a functional differentiable at $f$ w.r.t. $|| \cdot ||_1$. Then we can write
$$\forall h: J(f + h) - J(f) = \phi(h) + r(h) ||h||_1$$
with $\phi$ linear and $r(h) \to 0$ as $||h||_1 \to 0$.
Fix $\epsilon > 0$. There exists a $\delta > 0$ such that $||h||_1 < \delta \implies ||r(h)||_1 < \frac{\epsilon}{C}$. But then $\frac 1C ||r(h)||_0 \le \frac {\epsilon}{C}$, so $||r(h)||_0 \le \epsilon$. Thus $J$ is differentiable at $f$ w.r.t. $|| \cdot ||_0$ and has the same derivative.
Another special case is that a linear functional $J$ has derivative $J$ at every point, w.r.t. any norm.
 A: The value of the derivative does not depend on the norm. Suppose that $J$ is differentiable at $f$ w.r.t. both $\|\cdot\|_0$ and $\|\cdot\|_1$ with derivatives $\phi_0$ and $\phi_1$. For arbitrary $h$ and $\tau > 0$ we have
\begin{align}
J(f+\tau h) &= J(f) + \tau \phi_0(h) + r_0(\tau h) \tau \|h\|_0, \\
J(f+\tau h) &= J(f) + \tau \phi_1(h) + r_1(\tau h) \tau \|h\|_1.
\end{align}
Next, we take the difference, divide by $\tau > 0$ and let $\tau \searrow 0$. This results in
$$
\phi_0(h) - \phi_1(h) = 0.
$$
Since $h$ was arbitrary, this implies $\phi_0 = \phi_1$.
Concerning the existence,
let us take the Lebesgue space $L^2(0,1)$ and
$$J(f) := \int_0^1 \sin(f(t)) \, \mathrm dt.$$
Then, it can be checked that $J$ is differentiable w.r.t. the norms of $L^p(0,1)$ for all $p > 1$, but it is not differentiable w.r.t. the norm of $L^1(0,1)$, although the derivative (w.r.t. $L^p(0,1)$) is
$$
J'(f) h = \int_0^1 \cos(f(t)) h(t) \, \mathrm{d} t
$$
and this is even continuous (for fixed $h$) w.r.t. $h \in L^1(0,1)$.
