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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.

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    $\begingroup$ What are your thoughts on the problem? What have you tried so far? Do you understand what's being asked? $\endgroup$ – Omnomnomnom Feb 2 '17 at 5:33
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You may write $$ Tf = \int_{0}^{1}f(x)d\mu(x), \;\;\; \mu(S)=\int_{S}2xdx. $$ Therefore, $$ \|T\| = \|\mu\|= \int_{0}^{1}2xdx=1. $$

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$||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.

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  • $\begingroup$ 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 $f(x)=1$, $||f(x)||=1$. $\endgroup$ – Cuoredicervo Feb 2 '17 at 6:51
  • $\begingroup$ @Cuoredicervo that's not really a hint so much as a completion of this answer $\endgroup$ – Aweygan Feb 2 '17 at 7:44
  • $\begingroup$ My idea what to give a method to find in the general situation $\endgroup$ – Cuoredicervo Feb 2 '17 at 7:47

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