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Let $E = L^p(0,1)$ with $1 ≤ p < ∞$. Given $u ∈ E$, set

$$Tu(x):=\int_0^x u(t)dt$$

Prove that $T$ is compact on $E$.

I would like to use Ascoli-Arzela', but I need to prove:

$$|T u(x) − T u(y)| ≤ |x − y|^{1/p′}||u||_p\ \ (1)\ .$$

I have problems proving (1):

$$|T u(x) − T u(y)|=|\int_x^y u(t)dt\ |\leq\int_x^y |u(t)|dt$$

but I do not see how to continue to get to the desired result.

Moreover, applying Ascoli-Arzela' to $T(B_E)$ (where $B_E$ is the unit ball in $E$), we get that the closure of $T(B_E)$ is compact in $C[0,1]$. Why does this imply that $T(B_E)$ has compact closure in $L^p(0,1)$ as well? Any help?

mastro
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1 Answers1

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Use Holder's inequality: $$\int_x^y |u(t)| \, dt \le \left( \int_x^y 1 \, dt \right)^{1/p'} \left( \int_x^y |u(t)|^p \, dt \right)^{1/p} \le |x-y|^{1/p'} \|u\|_p.$$

Umberto P.
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  • applying Ascoli-Arzela' to $T(B_E)$ (where $B_E$ is the unit ball in $E$), we get that the closure of $T(B_E)$ is compact in $C[0,1]$. Why does this imply that $T(B_E)$ has compact closure in $L^p(0,1)$ as well? – mastro Jan 20 '15 at 20:50
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    Filippo: Sorry but I prefer to wait that you post your own answer and accept it one minute later. – Did Jan 20 '15 at 20:54