$f : U \to Y$ where $U\subset X$ is open and $X, Y$ are normed spaces is called Gâteaux differetiable at $u\in U$ if there exists a bounded linear operator $T$ from $U$ to $Y$ such that for $h\to 0$ we have $$ \frac{f(u+hv)-f(u)}{h} \to Tv $$ for all $v\in U$. $f$ is called Fréchet differentiable if the convergence is uniform in $v\in B_X$ (the closed unit ball in $X$).
Here, let $f=\|\|$ be the supremum norm on the space $C[0,1]$ (the space of continuous functions on $[0,1]$) and $c_0$ (the space of null convergent sequences), respectively.
Unfortunately, I haven't come very far. It's easy for see that in neither space $\|.\|$ can be differentiable at $0$. Also for both spaces, the triangle inequality yields $$ \frac{\|f_0+h g\|-\|f_0\|}{h} \leq \frac{\|f_0\|+\|h g\|-\|f_0\|}{h} = \|g\|\, . $$ So the derivative's norm, if it exists, must be bounded by $1$.
Moreover, as the norm is homogeneous, $\lambda f_0$ is differentiable for some $\lambda > 0$, then so is $f_0$.
From here, I don't know how to proceed. One idea I had: Since we have the upper bound for $T$ which has to be a linear functional, it might be possible to invoke something like of the Hahn-Banach theorem. But I don't quite see how to make a connection.
Can anyone help me with this? (And also with the $c_0$ case?)
It's a lemma by Banach that in $C(K)$ for $K$ compact and metrizable, $\|\cdot \|$ is Gâteaux differentiable at $f\in C(K)$ iff $|f|$ attains a strict maximum.
Reference: Lemma 1.2.1 in Isometries on Banach Spaces: Function Spaces by James E. Jamison and Richard J. Fleming.