T(f)=$\int_0^12xf(x)dx$. Then prove that $\|T\| =1$.

Question

Let $T:C[0,1]\to R$ be defined by $T(f)=\int_0^12xf(x) \, dx$. Then prove that $\|T\| =1$. Here $C[0,1]$ is equipped with supremum norm.

Answer 1

$||T(f)||=||\int_0^12xf(x)dx||\le \int_0^1||2xf(x)||dx \le2.||f||.\int_0^1xdx $

$\frac{||T(f)||}{||f||}\le 2.\int_0^1xdx$

${\sup_f{\in C[0,1]}}{\frac{||T(f)||}{||f||}}\le1$

$||T||\le 1$

The other inequality can be shown easily.

Answer 2

You may write $$ Tf = \int_{0}^{1}f(x)d\mu(x), \;\;\; \mu(S)=\int_{S}2xdx. $$ Therefore, $$ \|T\| = \|\mu\|= \int_{0}^{1}2xdx=1. $$