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The Volterra operator acting on $L^2[0,1]$ is defined by $$A(f)(x)=\int_0^x f(t) dt$$ How can I calculate the spectral radius of $A$ using the spectral radius formula for bounded linear operators: $$\rho(A)=\lim_{n\rightarrow \infty} \|A^n\|^{1/n} \text{?}$$ This was given as an exercise in a book right after introducing the spectral radius formula, so it should be simple exercise, but I don't see immediately how to do the calculations. Any hint is appreciated.

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Here is how to find $A^n(f)(x).$ –  Mhenni Benghorbal Mar 4 '13 at 14:15
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2 Answers

up vote 2 down vote accepted

Hint 1. Prove by induction that $$ A^n(f)(x)=\int\limits_0^x f(s)\frac{(x-s)^{n-1}}{(n-1)!}ds\tag{1} $$ Hint 2. Given $(1)$ show that $$ \Vert A^n\Vert\leq\frac{1}{(n-1)!} $$ Hint 3. To compute $\rho(A)$ recall the following Stirling's approximation $$ N!\approx\left(\frac{N}{e}\right)^N\sqrt{2\pi N} $$

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So is it correct that the spectral radius is $0$? –  Cantor Mar 4 '13 at 13:57
    
Yes${}{}{}{}{}$ –  Norbert Mar 4 '13 at 14:04
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Below I will try to follow Norbert's hints:

Hint 1.

Let's go from $n=2$ to $n=3$ to check the formula.

$$A^2f(x)=\int_0^x f(s)(x-s) ds$$

Then \begin{eqnarray*} A^3f(x) &=& \int_0^x\int_0^tf(s)(t-s)ds dt\\ &=&\int_0^1 \int_0^1 \chi_{[0,x]}(t)\chi_{[0,t]}(s)f(s)(t-s) ds dt \\ &\stackrel{*}{=}& \int_0^1f(s)\chi_{[0,x]}(s)\int_0^1\chi_{[s,x]}(t)(t-s)dt ds\\ &=& \int_0^xf(s)(\frac{x^2}{2}-sx+\frac{s^2}{2})ds \\ &=&\int_0^xf(s)\frac{(x-s)^2}{2}ds \end{eqnarray*} where $\chi_{B}(\cdot)$ is the characteristic function of a set $B$ and at * I have used $\chi_{[0,x]}(t)\chi_{[0,t]}(s)=\chi_{[0,x]}(s)\chi_{[s,x]}(t)$ and Fubini to exchange order of integrals.

Hint 2.

\begin{eqnarray*} \|A^nf\|_2^2 &=& \int_0^1 \left| \int_0^x f(s) \frac{(x-s)^n}{(n-1)!}ds \right|^2 dx \\ &\leq& \left( \frac{1}{(n-1)!} \right)^2\|f\|_2^2 \end{eqnarray*} since $|(x-s)^n|\leq 1$ for $s \in [0,x]$, $x \leq1$.

Hint 3.

$$\|A^n\|^{1/n}\leq \left( \frac{1}{(n-1)!} \right)^{1/n} \rightarrow 0$$ since $$(n-1)!^{1/n} \geq (\sqrt{2\pi})^{1/n}(n-1)^{1-\frac{1}{2n}}e^{-1+\frac{1}{n}} \rightarrow \infty \text{.}$$ In this last step the following lower bound is used: $$n! \geq \sqrt{2\pi}n^{n+\frac{1}{2}}e^{-n} $$

Conclusion

The spectral radius of the Volterra operator is $0$.

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