# 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 $\|T\|\geq 2$?

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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 $a\to0$.
I did a proof using your idea @Din $$f_n(t)=\left\{\begin{array}{ccl} -1&\textrm{if}&x\in [-1, -1/n]\\ nx&\textrm{if}&x\in (-1/n, 1/n)\\ 1&\textrm{if}&x\in [1/n, 1] \end{array}\right..$$ Note $\|f_n\|_\infty=1$ for all $n\in\mathbb N$ so that,
\begin{align}|Tf_n|&=\left|\int_{-1}^0f_n(t)\ dt-\int_{0}^1f_n(t)\ dt\right|\\ &=\left|\int_{-1}^{-1/n}f_n(t)\ dt+\int_{-1/n}^0f_n(t)\ dt- \int_{0}^{1/n}f_n(t)\ dt-\int_{1/n}^1f_n(t)\ dt\right|\\ &=\left|-\int_{-1}^{-1/n}\ dt+n\int_{-1/n}^{0}t\ dt-n\int_{0}^{1/n}t dt -\int_{1/n}^1\ dt\right|\\ &=\left|\frac{1}{n}-1+n\frac{1}{2n^2}-n\frac{1}{2n^2}+\frac{1}{n}-1\right|\\ &=\left|\frac{2}{n}-2\right|.\end{align}
So, $$\|T\|\geq \left|\frac{2}{n}-2\right|,$$ for all $n\in\mathbb N$. Taking the limit as $n\to \infty$ we have, $$\|T\|\geq |-2|=2,$$ hence $\|T\|=2$. Thanks for the tip @Din..
I've just fixed that..I meant $T$.. – PtF Apr 12 '13 at 20:38