Norm of powers of the Volterra Operator

**Insensus** · June 17th 2011, 12:27 PM

Hey there,

For a course in Functional Analysis we were asked to show that the norm of the n-th power of the Volterra operator is less than or equal to 1 over n!.

$$V: C[0,1] \rightarrow C[0,1]$\\(Vf)(s) = \displaystyle{\int_0^s f(t)dt}$
Show $\left|\left|V^n\right|\right| \leq \frac{1}{n!}$

Now I know how the operator norm is defined
$$\left|\left|T\right|\right| = \displaystyle{\sup_{\left|\left|x\right|\right|=1} } \left|\left|Tx\right|\right|$
and I've been throwing it at this problem for at least 3 hours, but I can't seem to crack it.

I've so far assumed that induction would be the path to choose but I'm starting to doubt that, yet I see no other method.

I hope someone is able to nudge me in the right direction.

Thanks,
Insensus

**Also sprach Zarathustra** · June 17th 2011, 12:47 PM

Originally Posted by Insensus

Hey there,

For a course in Functional Analysis we were asked to show that the norm of the n-th power of the Volterra operator is less than or equal to 1 over n!.

$$V: C[0,1] \rightarrow C[0,1]$\\(Vf)(s) = \displaystyle{\int_0^s f(t)dt}$
Show $\left|\left|V^n\right|\right| \leq \frac{1}{n!}$

Now I know how the operator norm is defined
$$\left|\left|T\right|\right| = \displaystyle{\sup_{\left|\left|x\right|\right|=1} } \left|\left|Tx\right|\right|$
and I've been throwing it at this problem for at least 3 hours, but I can't seem to crack it.

I've so far assumed that induction would be the path to choose but I'm starting to doubt that, yet I see no other method.

I hope someone is able to nudge me in the right direction.

Thanks,
Insensus

Maybe this will help you:

SpringerLink - Integral Equations and Operator Theory, Volume 27, Number 4

Hope that is free...

**Jose27** · June 17th 2011, 03:04 PM

How about this:

Consider $f_n(t)=t^n$ for $n\geq 0$ then it's not difficult to see that $V^k(f_n)(t) = \frac{t^{n+k}}{(n+k)...(n+1)}t^n$ from which we immediately deduce that $\| V^k(p)(t)\| \leq \frac{1}{k!}$ where $p$ is any polynomial. Now use Weierstrass' approximation theorem.

**Insensus** · June 18th 2011, 02:49 AM

That's very nice, thanks.

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