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


  • $\begingroup$ $\left\|A\right\|_{2}\le 1$. 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. $\endgroup$
    – user14717
    Commented Jun 9, 2012 at 5:18
  • $\begingroup$ I suspect it might be easier to look at $A$ in terms of the basis $e_n(t) = e^{i n 2\pi t}$. $\endgroup$
    – copper.hat
    Commented Jun 9, 2012 at 7:00

4 Answers 4


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 $x(s) = \cos \frac{\pi}{2}s$.

  • 11
    $\begingroup$ This is reverse engineering :) When you know the answer, it is always much easier to get it. $\endgroup$
    – Norbert
    Commented Jun 9, 2012 at 14:10
  • $\begingroup$ @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 :) $\endgroup$
    – Math1000
    Commented May 28, 2016 at 22:36
  • $\begingroup$ 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? $\endgroup$
    – user110320
    Commented Nov 19, 2017 at 20:53

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 $\lambda$, $$\|V\|=\frac2\pi.$$

  • 1
    $\begingroup$ This is my favorite answer!!! $\endgroup$
    – Landau
    Commented Dec 11, 2021 at 14:03
  • 1
    $\begingroup$ Very nice derivation (intended pun). $\endgroup$
    – Jean Marie
    Commented Oct 24, 2022 at 6:04
  • $\begingroup$ How do you know that $\lambda >0$? $\endgroup$ Commented Nov 1, 2022 at 3:34
  • 1
    $\begingroup$ $$ \lambda\|f\|^2=\langle \lambda f,f\rangle=\langle V^*Vf,f\rangle=\langle Vf,Vf\rangle=\|Vf\|^2.$$ If you mean why nonzero, $V$ is injective so $0$ is not an eigenvalue of $V^*V$. $\endgroup$ Commented Nov 1, 2022 at 3:50

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.


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 $\cos(\pi x/2)$.


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