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.
T(f)=$\int_0^12xf(x)dx$. Then prove that $\|T\| =1$.
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real-analysis
functional-analysis
normed-spaces
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1What are your thoughts on the problem? What have you tried so far? Do you understand what's being asked? – 2017-02-02
2 Answers
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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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0Hint: 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 $f(x)=1$, $||f(x)||=1$. – 2017-02-02
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0@Cuoredicervo that's not really a hint so much as a completion of this answer – 2017-02-02
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0My idea what to give a method to find in the general situation – 2017-02-02