I am following Smoothness and renorming by Deville et.el. On page no. 5 there is an example as given below. Let $K$ be a locally compact space. Then $\|.\|_{\infty}$ the cannonical sup norm on $C_0(K)$ the space of all real valued functions on $K$ that vanish at infinity is Frechet differentiable if and only if the set of points where the norm is attained is a singleton consisting of an isolated point. My problem starts in the converse part. It is mentioned compactness. But compactness of what? How the norm is equal to a continuous linear functional? Any help is appreciated.
“From the compactness”: I would suggest the fact that $S=\{y,\,2|x(y)| \geq |x(t_0)|\}$ is compact, so, since $t_0$ is an isolated point in $K$, $S’=S \backslash \{t_0\}$ is compact so $|x|$ has a maximum on $S’$, which is smaller than $|x(t_0)|$.
In this situation, note that as stated, there is $\alpha > 0$ such that for all $y \neq t_0$, $|x(y)| +\alpha < |x(t_0)|$.
Thus, if $\|f\|_{\infty} < \alpha/3$, it is easy to see that $\|x+f\|_{\infty}=|x(t_0)+f(t_0)|=|x(t_0)| + sf(t_0)$, $s$ being the sign of $x(t_0)$.