What is the norm of $T$ : $L^1$ (($-1$,$1$) , $\mathbb{R}$) $\rightarrow$ $\mathbb{R}$

Question

I have the following question

What is the norm of $T$ : $L^1(-1,1)\to \mathbb{R}$, where $T(f)= \int_{-1}^{1} t f(t) dt$?

We know that $$\|T\|= \sup_{f \ne 0} \dfrac{|T(f)|}{\|f\|_{1}}$$

I was able to show that $\|T\|\leq 1$. How I can show that $\|T\|=1$?

I know I must bring a function $f$ with unit norm and $|T(f)|=1$ to say that $\|T\|= 1$ but I couldn't. Any help is welcome.

Answer 1

The functions $f_{\epsilon}\in L^1([-1,1])$ defined by $$ f_{\epsilon}(x) = \begin{cases} 0, & t < 1-\epsilon\\ 1/\epsilon, & t \ge 1-\epsilon \end{cases} $$

have norm 1 and satisfy $|Tf_{\epsilon}| = 1-\frac{\epsilon}{2}$. This means that, for arbitrarily small $\epsilon$, we have that $$ \|T\| = \sup_{\|f\|=1}|Tf|\ge |Tf_{\epsilon}|=1-\frac{\epsilon}{2}. $$

As a consequence, you have that $\|T\| \ge 1$ and, since you already know that $\|T\|\leq 1$, you conclude that $\|T\|=1$.