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

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

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    I suspect it might be easier to look at $A$ in terms of the basis $e_n(t) = e^{i n 2\pi t}$.2012-06-09

3 Answers 3

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

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    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?2017-11-19
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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 $\cos(\pi x/2)$.

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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: http://math.stackexchange.com/questions/151425/spectrum-of-indefinite-integral-operators/151444#1514442012-06-10