Given $T: (C[0,1], \|\cdot\|_1) \to (l^\infty,\|\cdot\|_\infty)$ where $(Tf)_n = \int_0^1 x^nf(x)dx$, find $\|T\|$.

Asked Jul 06 '23 at 14:58

Active Jul 06 '23 at 15:04

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Given $T: (C[0,1], \|\cdot\|_1) \to (l^\infty,\|\cdot\|_\infty)$ where $(Tf)_n = \int_0^1 x^nf(x)dx$, find $\|T\|$.

I've shown that $$\|Tf\|_\infty =\sup_{n \in \mathbb{N}}\left|\int_0^1 x^nf(x)dx\right| \le\sup_{n \in \mathbb{N}} \int_0^1x^n|f(x)|dx \le\sup_{n \in \mathbb{N}}\int_0^1 |f(x)|dx =\|f\|_1.$$

If what I've done is correct then $\|T\| \leq 1$. Now I don't know how to continue to show that $\|T\| = 1$. Can I take $f = 1$ for this?

edited Jul 06 '23 at 15:04

Anne Bauval

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asked Jul 06 '23 at 14:58

Peter Sampodiras

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Does $0$ belong to your $\Bbb N?$ – Anne Bauval Jul 06 '23 at 15:15
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It was not specified in the problem, but I guess it does belong to $\Bbb N$ – Peter Sampodiras Jul 06 '23 at 15:18
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Then yes $f=1$ proves your conjecture because $\sup_{n\in\Bbb N}|(Tf)_n|=(Tf)_0=1.$ – Anne Bauval Jul 06 '23 at 15:19
What happens if it doesn't, I dont get it. – Peter Sampodiras Jul 06 '23 at 15:19
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If $0\notin\Bbb N,$ you won't find an $f$ such that $|f|1=1=|Tf|\infty,$ but a sequence $(f_m)$ such that $|f_m|1=1$ and $|Tf_m|\infty\to1:$ take for instance $f_m(x)$ equal to $0$ on $[0,1-1/m],$ and to $2m^2(x-1+1/m)$ on $[1-1/m,1].$ – Anne Bauval Jul 06 '23 at 15:36
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Thanks for your help! – Peter Sampodiras Jul 06 '23 at 15:40

Given $T: (C[0,1], \|\cdot\|_1) \to (l^\infty,\|\cdot\|_\infty)$ where $(Tf)_n = \int_0^1 x^nf(x)dx$, find $\|T\|$.

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