Find $\Vert f \Vert$ for a functional $f$ over $C[-1,1]$. 
Let , $\displaystyle f:C[-1,1]\to C[-1,1]$ be defined by $\displaystyle f(x)=\int_{-1}^0 x(t)\,dt -\int_0^1x(t)\,dt$. Find $\Vert f\Vert$ with respect to sup norm defined on $C[-1,1]$.

Firstly , $\displaystyle |f(x)|\le \int_{-1}^0\sup_{-1\le t\le 1}|x(t)|\,dt+\int_0^1 \sup_{_1\le t\le 1}|x(t)|\,dt=2\Vert x\Vert$.
Then we get , $\Vert f\Vert \le 2$.
I couldn't find the reverse inequalty. I'm unable to construct $\Vert f\Vert \ge 2$.
I saw this but that answer is not clear to me.

Consider the sequence of functions 
$$x_n(t) = \left\{ \begin{array} 
.1 & \forall t \in [-1,-\frac{1}{n}] \\
- nt & \forall t \in [-\frac{1}{n},\frac{1}{n}] \\
-1 & \forall t \in [\frac{1}{n},1] 
\end{array} \right.$$
The $x_n$ are continuous, $\|x_n\| = 1$, and $f(x_n) \to 2$ when $n\to\infty$
