I've been trying to find the norm of the linear operator
$(Tf)(x) = \int_{-1}^1 xyf(y)dy$, where
$T:L_{\infty}(-1,1) \rightarrow L_{1}(-1,1)$ and $f\in L_{\infty}(-1,1)$.
From definition the norm $||T||$ is defined as
$\sup_{f\neq 0}\frac{||Tf||}{||f||_{\infty}}$, where $||f||_{\infty}$ is the essential supremum of function $f$ over $(-1,1)$.
I've already shown that $||T||$ is bounded by $1$.
In order to find the lower bound of $||T||$ I've been trying to find some sequence of functions that show
$\sup_{f\neq 0}\frac{||Tf||}{||f||_{\infty}} \geq \sup_{n}\frac{||Tf_{n}||}{||f{n}||_{\infty}}=1$
but to no avail. Thanks in advance for all the answers.
Best Answer
Since $|Tf|\le T|f|$ we can assume for convenience that $f\ge0$. Now $$||Tf||_1=\int\left|\int_{-1}^1 xyf(y)\,dy\right|\,dx\le \int_{-1}^1\int_{-1}^1|xyf(y)|\,dxdy\le||f||_\infty\int_{-1}^1\int_{-1}^1|xy|\,dydx=||f||_\infty,$$so $||T||\le1$, as you said. Let $f=\chi_{[0,1]}-\chi_{[-1,0]}$. Then $$Tf(x)=\int_0^1xy\,dy-\int_{-1}^0 xy\,dxdy=x.$$So $||f||_\infty=1$ and $||Tf||_1=1$, hence $||T||\ge1$.