Let $C[-1,1]$ denote the normed space of all (real or complex-valued) functions defined and continuous on the closed interval $[-1,1]$ on the real line, with the norm given by$$\Vert x \Vert_{C[-1,1]} \colon= \max_{0 \leq t \leq 1} \vert x(t) \vert \ \ \ \mbox{ for all } \ \ x \in C[-1,1].$$
Let the functional $f$ on $C[-1,1]$ be defined by $$f(x) \colon= \int_{-1}^0 x(t) \ \mathrm{d} t \ - \ \int_0^1 x(t) \ \mathrm{d} t \ \ \ \mbox{ for all } \ \ x \in C[-1,1].$$
Then $f$ is linear and bounded with norm $\Vert f \Vert \leq 2$. How to determine the exact value of $\Vert f \Vert$?
My work:
For any $x \in C[-1, 1]$, we have \begin{eqnarray*}\vert f(x) \vert &=& \left\vert \int_{-1}^0 x(t) \ \mathrm{d} t \ - \ \int_0^1 x(t) \ \mathrm{d} t \ \right\vert \\ &\leq& \left\vert \int_{-1}^0 x(t) \ \mathrm{d} t \ \right\vert \ + \ \left\vert \int_0^1 x(t) \ \mathrm{d} t \ \right\vert \\ &\leq& \int_{-1}^0 \vert x(t) \vert \ \mathrm{d} t + \int_0^1 \vert x(t) \vert \ \mathrm{d} t \\ &=& \int_{-1}^1 \vert x(t) \vert \ \mathrm{d} t \\&\leq& \int_{-1}^1 \max_{0\leq s \leq 1} \vert x(s) \vert \ \mathrm{d} t \\&=& \int_{-1}^1 \Vert x \Vert_{C[-1, 1]} \ \mathrm{d} t \\&=& \Vert x \Vert_{C[-1, 1]} \ \int_{-1}^1 \mathrm{d} t \\ &=& 2 \Vert x \Vert_{C[-1,1]}.\end{eqnarray*}Thus, $f$ is bounded and taking the supremum over all $x \in C[-1, 1]$ such that $\Vert x \Vert_{C[-1, 1]} = 1$, we get $$\Vert f \Vert \leq 2.$$
Now we require an element $x_0 \in C[-1, 1]$ such that $\Vert x_0 \Vert_{C[-1, 1]} = 1$ and $\vert f(x_0) \vert = 2$.
However, I haven't been able to locate such an $x_0$. Can anybody please be of any help?