TITLE: Properties of Hardy operator $T(u)(x)=\frac{1}{x}\int_0^x u(t)dt$
QUESTION [9 upvotes]: 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.

REPLY [11 votes]: 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.