Properties of Hardy operator $T(u)(x)=\frac{1}{x}\int_0^x u(t)dt$

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.