How to prove that the following functional does not attain a maximum over the unit sphere in $C^0[0,1]$ with the supremum norm? This is an extension of a previously asked question 
Can we show that the supremum of the following functional over the unit sphere in $C^0[0,1]$ is $1$?
In the above link, we have shown that, for the following functional defined on $C^0[0,1]$, $f(x)=\int_0^{1/2}x(t)dt-\int_{1/2}^1x(t)dt$, its supremum over the unit sphere $S(0,1)\doteq \big\{x \in C^0[0,1]: ||x||=1\big\}$, is $1$. 
I am wondering how to show that this supremum can never be achieved over this $S(0,1)$? Thanks in advance!
 A: We have :
$$|f(x)| = \left| \int_0^{1/2}x(t)dt-\int_{1/2}^1x(t)dt \right| \le \left| \int_0^{1/2}x(t)dt\right|+\left|\int_{1/2}^1x(t)dt \right| \le \int_0^{1}|x(t)|dt \le1.$$
Suppose $x$ is such that $|f(x)|=1$, since $|x(t)| \le 1$ and by the above inequality, we must have $|x(t)| =1$ for every $t$ (see the edit for details).
Then $x(t) = \delta(t)$ with $\delta :[0,1] \to \{-1,1\}$.
Since we require $x$ to be continuous, either $\delta \equiv 1$ or $-1$. In both cases, $|f(x)|=0$, then the supremum cannot be achieved.
EDIT :
Suppose $|x(t_0)|<\alpha$ for some $t_0$ and $0<\alpha<1$, by continuity, we cand find some interval $I\subset [0,1], t_0 \in I$ such that $|x(t)|<\alpha$ for every $t \in I$. Then :
$$\int_0^{1}|x(t)|dt \le \int_{I}|x(t)|dt +\int_{[0,1] \setminus I}|x(t)|dt \le m(I)\alpha+(1-m(I)) < 1.$$
$m(I)$ being the length of the interval I.
A: Notice that $C[0,1] \subseteq L^2[0,1]$ and $\chi_{\left[0,\frac12\right]} -\chi_{\left[\frac12, 1\right]} \in L^2[0,1]$.
Assume $x \in C[0,1]$ is the maximizer, i.e. $\|x\|_\infty = 1$ and $|f(x)| = 1$. We have
$$1 = |f(x)| = \left|\left\langle x, \chi_{\left[0,\frac12\right]} -\chi_{\left[\frac12, 1\right]}\right\rangle\right| \stackrel{CS}{\le}\|x\|_2\left\|\chi_{\left[0,\frac12\right]} -\chi_{\left[\frac12, 1\right]}\right\|_2 = \|x\|_2 \le \|x\|_\infty = 1$$
The equality condition in Cauchy-Schwarz implies that $x$ is proportional to $\chi_{\left[0,\frac12\right]} -\chi_{\left[\frac12, 1\right]}$ so either $x \equiv 0$ (which contradicts $\|x\|_\infty = 1$), or $x \notin C[0,1]$.
Hence the maximum of $f$ is not attained on the unit sphere of $C[0,1]$.
