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

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.

• What are your thoughts on the problem? What have you tried so far? Do you understand what's being asked? – Omnomnomnom Feb 2 '17 at 5:33

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

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

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

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

$||T||\le 1$

The other inequality can be shown easily.

• Hint: to prove the equality you can use a particular $f(x)$, in general it is helpful use a function such that the norm is equal to 1. Then $f(x)=1$, $||f(x)||=1$. – Cuoredicervo Feb 2 '17 at 6:51
• @Cuoredicervo that's not really a hint so much as a completion of this answer – Aweygan Feb 2 '17 at 7:44
• My idea what to give a method to find in the general situation – Cuoredicervo Feb 2 '17 at 7:47