Proving that $ψ \in X^*$, but there is no $f \in B_X$ such that $|ψ(f(x))|=\|ψ\|$, where $ψ: X= (C[-1,1], \| \cdot \|_\infty) \rightarrow \mathbb{R}$. We consider $X= (C[-1,1], \| \cdot \|_\infty)$ and $ψ :X \rightarrow \mathbb{R}$ a linear functional that is defined by $$ψ(f) = \int_{-1}^{0}f(t)dt- \int_{0}^{1}f(t)dt. $$
We need to prove that $ψ \in X^*$, which means that we need to prove that $ψ$ is linear and bounded (equvalently, there is an $M >0$ such that $|ψ(f)| \leq M \|f\|_\infty$),but there is no $f \in B_X = \{ f \in X \ \ : \ \ \|f\|_\infty \leq 1 \} $ such that $|ψ(f(x))|=\|ψ\|$. 
Any help would be greatly appreciated.
 A: \begin{align*}
|\psi(f)|\leq\int_{-1}^{0}|f(t)|dt+\int_{0}^{1}|f(t)|dt\leq\|f\|_{\infty}\left(\int_{-1}^{0}dt+\int_{0}^{1}dt\right)=2\|f\|_{\infty}.
\end{align*}
Next we claim that $\|\psi\|=2$. Take $f_{n}(t)=1$ for $t\in[-1,-1/n]$, $f_{n}(t)=-nt$ for $t\in[-1/n,1/n]$ and $f_{n}(t)=-1$ for $t\in[1/n,1]$, then $\|f_{n}\|_{\infty}=1$ and that $0\leq\phi(f_{n})\rightarrow 2$.
Now assume that such an $f$ exists for which $|\psi(f)|=2$.
If there were some $-1<t_{0}<0$ such that $|f(t_{0})|<1$, say, $|f(t_{0})|<\epsilon_{0}<1$, then we can choose some $\delta>0$ such that $|f(t)|<\epsilon_{0}$ for all $t\in[t_{0}-\delta,t_{0}+\delta]\subseteq[-1,0]$, for then
\begin{align*}
\left|\int_{-1}^{0}f(t)dt\right|&\leq\int_{-1}^{t_{0}-\delta}|f(t)|dt+\int_{t_{0}-\delta}^{t_{0}+\delta}|f(t)|dt+\int_{t_{0}+\delta}^{0}|f(t)|dt\\
&\leq\int_{-1}^{t_{0}-\delta}dt+\epsilon_{0}\int_{t_{0}-\delta}^{t_{0}+\delta}dt+\int_{t_{0}+\delta}^{0}dt\\
&=1-2\delta+\epsilon_{0}\cdot 2\delta\\
&<1-2\delta+2\delta\\
&=1.
\end{align*}
By triangle inequality, we have $|\psi(f)|<2$, a contradiction.
We conclude that $|f(t)|=1$ for $t\in(-1,0)$. Of course we extend by continuity that $|f(t)|=1$ on $[-1,0]$.
By symmetry, one argues that $|f(t)|=1$ on $[0,1]$.
We conclude that $|f(t)|=1$ on whole $[-1,1]$. For $f(0)=1$, then $f(t)=1$ for any other $t\in[-1,1]$, if not, we can choose by Intermediate Value Theorem that $f(c)<1$ for some $c\in[-1,1]$.
Similarly, if $f(0)=-1$, then $f(t)=-1$. In either case, one can compute that $\psi(f)=0$, a contradiction.
