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What is the norm of integral operators $A$ in $L_2(0,1)$?

$Ax(t)=\int_0^tx(s)ds$

Rudy the Reindeer
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Frank Duque
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  • $\left|A\right|_{2}\le 1$. This can be shown using some integral inequalities and Holder's inequality. I have not been able to show that this bound is tight, unfortunately. – user14717 Jun 09 '12 at 05:18
  • I suspect it might be easier to look at $A$ in terms of the basis $e_n(t) = e^{i n 2\pi t}$. – copper.hat Jun 09 '12 at 07:00

4 Answers4

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It's enough to use Schwarz inequality in the following manner:

$$ \| A x \|^2 = \int_0^1 \left| \int_0^t x(s) \, ds \right|^2 dt = \int_0^1 \left| \int_0^t \sqrt{\cos \frac{\pi}{2}s} \cdot \frac{x(s)}{\sqrt{\cos \frac{\pi}{2}s}} \,ds \right|^2 dt \le \int_0^1 \left( \int_0^t \cos \frac{\pi}{2}s \, ds \int_0^t \frac{|x(s)|^2}{\cos \frac{\pi}{2}s}\right) dt = \frac{2}{\pi} \int_0^1 \int_0^t \sin \frac{\pi}{2}t \, \frac{|x(s)|^2}{\cos \frac{\pi}{2}s} \, ds\,dt = \frac{2}{\pi}\int_0^1 \left( \int_s^1 \sin \frac{\pi}{2} t \, dt \right) \frac{|x(s)|^2}{\cos \frac{\pi}{2}s} \,ds = \left( \frac{2}{\pi} \right)^2 \| x \|^2 $$
Equality holds for $x(s) = \cos \frac{\pi}{2}s$.

qoqosz
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    This is reverse engineering :) When you know the answer, it is always much easier to get it. – Norbert Jun 09 '12 at 14:10
  • @qoqosz It should be $x(s)=\sqrt2\cos\frac\pi 2s$. I am only pointing this out because I spent the past hour trying to figure out why I wasn't getting the right result :) – Math1000 May 28 '16 at 22:36
  • How is it that you are able to obtain that $|\int_0^t \sqrt{\cos(\frac{\pi s}{2})} \frac{x(s)}{\sqrt{\cos(\frac{\pi s}{2})}}|^2 \leq \int_0^t \cos(\frac{\pi s}{2})ds \int_0^t \frac{|x(s)|^2}{\cos(\frac{\pi s}{2})}ds$? Is this always true or why can you do this in this case? – user110320 Nov 19 '17 at 20:53
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Here is a rather direct way to obtain the norm of the Volterra operator, without previous knowledge of what it should be. We take advantage of the C$^*$-identity $$\tag1\|V\|=\|V^*V\|^{1/2}.$$ Because $V^*V$ is compact and positive, its norm is equal to its greatest eigenvalue.

The eigenvectors will necessarily be C$^\infty$.

We have $$\tag2 V^*Vf(x)=\int_x^1\int_0^tf(s)\,ds\,dt. $$ If we differentiate the equality $V^*Vf(x)=\lambda f(x)$ twice, we get the differential equation $-f(x)=\lambda f''(x)$. From $(2)$ we get the initial conditions $f(1)=0$, $f'(0)=0$. Since we know that $\lambda>0$ (because $V^*V$ is an injective positive operator) this gives, up to a multiple, $$ f(x)=\cos\frac x{\sqrt\lambda},\qquad\text{subject to $f(1)=0$.} $$ Thus $$ \frac1{\sqrt\lambda}=\frac{(2k+1)\pi}2,\qquad k\in\mathbb N\cup\{0\}, $$ so $$ \sqrt{\lambda}=\frac2{(2k+1)\pi}. $$ Using $k=0$ to get the largest $\lambda$, $$\|V\|=\frac2\pi.$$

Martin Argerami
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It is Problem 188 in the book by P. Halmos, "A Hilbert space problem book". In the solution, the author writes that "A direct approach seems to lead nowhere." The norm is indeed $2/\pi$, and is computed through the adjoint $A^*$ and a suitable kernel. It is a rather long proof, so please try to read it on Halmos' book.

Siminore
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  • If the book is not handy, I also recently wrote up the proof here: http://math.stackexchange.com/questions/151425/spectrum-of-indefinite-integral-operators/151444#151444 –  Jun 10 '12 at 08:02
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The norm of the Volterra operator is $2/\pi$. I will try to recall the proof; the bound suggests that the optimum occurs for some trigonometric polynomial, say $\cos(\pi x/2)$.

Erick Wong
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