Norm of the operator $Tf=\int_{-1}^0f(t)\ dt-\int_{0}^1f(t)\ dt$ Consider the operator $T:(C[-1, 1], \|\cdot\|_\infty)\rightarrow \mathbb R$ given by, $$Tf=\int_{-1}^0f(t)\ dt-\int_{0}^1f(t)\ dt,$$ is $\|T\|=2$. How to show $\|T\|=2$? On the one hand it is easy,
$$\begin{align}
|Tf|&=\left|\int_{-1}^0f(t)\ dt-\int_{0}^1f(t)\ dt\right|\\&\leq \int_{-1}^0|f(t)|\ dt+\int_{0}^1|f(t)|\ dt\\ &\leq \|f\|_\infty \left(\int_{-1}^0\ dt+\int_{0}^1\ dt\right)\\ &=2.
\end{align}$$ How to show $\|T\|\geq 2$?

Try $f$ piecewise linear with $f(x)=-1$ if $x\leqslant -a$, $f(x)=x/a$ if $-a\leqslant x\leqslant a$ and $f(x)=+1$ if $x\geqslant a$, when $a\to0$.
