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Let $u$ be a measurable function in $[0,1]$ and define $T:L^p(0,1)\to L^p(0,1)$ by $Tu(x)=\frac{1}{x}\int_0^x u(t)dt\quad\forall x \in [0,1]$. Let $1<p<\infty$. Prove that $T$ is bounded, non-compact. Determine the spectral radius of $T$ and prove that in the case $p=2$ the operator $TT^*-T^*T$ has range $1$. Can anyone help me? Thank you.

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    $\begingroup$ miserable function??? $\endgroup$ Nov 30, 2013 at 19:16
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    $\begingroup$ Poor little function... :( $\endgroup$
    – Spenser
    Nov 30, 2013 at 20:45

1 Answer 1

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Here is the proof that $T$ is bounded:

Hardy's Inequality for Integrals

Here is the exact calculation of its norm:

Computing the best constant in classical Hardy's inequality

To find its spectral radius, use the formula $\text{radius}(T) = \lim\limits_{n\to\infty}\|T^n\|^{1/n}$

To compute $T^n$ and $T^*$, look at http://faculty.missouri.edu/~stephen/preprints/hardy.html

Here is compactness https://math.stackexchange.com/questions/262221/is-hf-1-over-x-int-0x-ftdt-compact?rq=1

Probably this will get marked as a duplicate, but I don't see anywhere spectral radius was asked before.

Another way to find a lower bound for the spectral radius is to consider $u(t) = t^{-1/r}$ for $r>p$. This will give eigenfunctions.

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    $\begingroup$ You beat me to this! Ok, i'll just add that $$T^*(u)(x)=\int_x^1 t^{-1}u(t)dt$$$$(TT^*-T^*T)(u)=\left(\int_0^1 u(t)dt\right) \chi_{(0,1)}$$ $\endgroup$
    – Norbert
    Nov 30, 2013 at 20:49
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    $\begingroup$ It's on page 5. In this special case $$ T^{n+1} f(x) =\frac1x \int_0^x \frac{(\log(x/t))^n}{n!} u(t) \, dt $$ $\endgroup$ Nov 30, 2013 at 21:01
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    $\begingroup$ @User2313 No, $\int_1^x = - \int_x^1$. $\endgroup$ Oct 25, 2015 at 18:40
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    $\begingroup$ I don't see what your problem is. $\endgroup$ Dec 13, 2015 at 18:58
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    $\begingroup$ @TrostAft I fixed the $T^n$ and $T^*$ link. $\endgroup$ Oct 19, 2020 at 23:15

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