Let $X=C[0,1]$ with the norm $\|x\|_1=\int_0^1 |x(t)|\,dt$, $x\in C[0,1]$ and $\Omega=\left\{f\in X':\|f\|=1\right\}$, where $X'$ denotes the dual space of $X$. Let $C(\Omega)$ be the linear space of continuous functions on $\Omega$ with the norm $\|u\|=\sup_{s\in\Omega}|u(s)|$, $u\in C(\Omega)$. Let $\Phi:X\rightarrow C(\Omega)$ be the natural linear isometry defined by $\Phi_x(f)=f(x)$ for $f\in\Omega$ and $x\in X$.
Problem: Show that $\Phi:X\rightarrow C\left(\Omega\right)$ is not onto.
Thanks in advance for any help.
If $f\in \Omega$ then $(-f)\in\Omega$. We have $$\Phi_x(-f)=-\Phi_x(f)\ \forall\, x\in X, f\in\Omega.\quad (*)$$ Let $x_0\in X$ such that $x_0\neq0$. Then $$\Phi_{x_0}(-f_0)\neq0\neq \Phi_{x_0}(f_0)$$ for some $f_0\in\Omega$. Define $\psi:\Omega\to\mathbb{R}$ by the formula $$\psi(f)=\big(\Phi_{x_0}(f)\big)^2.$$ Function $\psi$ is continuous as a composition of continuous maps, i.e. $\psi\in C(\Omega)$. But $\psi$ does not satisfy $(*)$: $$\psi(f_0)=\big(\Phi_{x_0}(f_0)\big)^2=\big(\Phi_{x_0}(-f_0)\big)^2=\psi(-f_0)$$ So, $\psi\notin \Phi(X)$. In the last chain of equalities, it is important that $\psi(f_0)≠0$.