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

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.

• miserable function??? – Alexander Grothendieck Nov 30 '13 at 19:16
• Poor little function... :( – Spenser Nov 30 '13 at 20:45

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://www.math.missouri.edu/~stephen/preprints/hardy.html

Here is compactness Is $H(f)={1\over x} \int_0^x f(t)dt$ compact?

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.

• You beat me to this! Ok, i'll just add that $$T^*(u)(x)=\int_x^1 t^{-1}u(t)dt$$$$(TT^*-T^*T)(u)=\left(\int_0^1 u(t)dt\right) \chi_{(0,1)}$$ – Norbert Nov 30 '13 at 20:49
• I had the advantage that I had it all written up beforehand! – Stephen Montgomery-Smith Nov 30 '13 at 20:51
• So what is the final formula for $T^n$ ? I can't understand from that paper. – Norbert Nov 30 '13 at 20:54
• It's on page 5. In this special case $$T^{n+1} f(x) =\frac1x \int_0^x \frac{(\log(x/t))^n}{n!} u(t) \, dt$$ – Stephen Montgomery-Smith Nov 30 '13 at 21:01
• I don't see what your problem is. – Stephen Montgomery-Smith Dec 13 '15 at 18:58