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


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Where is $B$?${}{}$ – martini Jun 9 '12 at 4:41
Sorry, it was a mistake, there's only one operator, $A$. – Frank Duque Jun 9 '12 at 4:47
Then you should edit your question :) – martini Jun 9 '12 at 4:50
$\left\|A\right\|_{2}\le 1$. 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. – Nick Thompson Jun 9 '12 at 5:18
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 9 '12 at 7:00
up vote 7 down vote accepted

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

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This is reverse engineering :) When you know the answer, it is always much easier to get it. – Norbert Jun 9 '12 at 14:10
@Norbert :D it's also quite common exercise ;) – qoqosz Jun 9 '12 at 14:13
@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 at 22:36

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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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.

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If the book is not handy, I also recently wrote up the proof here:… – user12014 Jun 10 '12 at 8:02

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