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Let $J_1$ be the Bessel function of the first kind and let $H_1(x) = \frac{J_1(|x|)}{|x|}$ for $n = 1$. Define the operator $Tf(x) = (f * H_1)(x)$ from $L^2$ to $L^2$.

Since the $H_1$-function is the Fourier transform of something it must be in $L^2$, so we have a Hilbert-Schmidt operator which is in this case self-adjoint and compact, so the spectral theorem applies which for example says all the eigenvalues form a countable set, are real and go to zero.

What I am now interested in is the largest eigenvalue of $T$. What theorem or method could I try to obtain this? (I don't need a full solution, just a hint would suffice).

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For large $x>0$, $J_1(x)$ is asymptotic to $\sqrt{2/(\pi x)}\cos(x-3\pi/4)$, so that it is not in $L^p$ for $1\le p\le 2$. The Fourier transform of $J_1(x)$ is unbounded with support in $[-1,1]$, so that $T$ does not map $L^2$ into itself. – Julián Aguirre Sep 16 '10 at 20:23
@Julián Aguirre: I modified it a bit, $H_1(x)$ is the Fourier transform of the $\textrm{circ}$-function. – Jonas Teuwen Sep 16 '10 at 20:42
up vote 1 down vote accepted

In Fourier space, the operator $T$ is given by $$\hat{(T f)}(\xi)=c\sqrt{1-\xi^2}\hat f(\xi)$$ for some constant $c\ne0$ ($c$ dependson the normalization chosen for thr Fourier transform.) If $\lambda$ is an eigenvalue and $f$ is a corresponding eigenfunction, then for almost all $\xi$ $$\lambda \hat f(\xi)=c\sqrt{1-\xi^2}\hat f(\xi).$$ It follows that the only eigenvalue is $\lambda=0$, and the coresponding eigenfunctions are all $L^2$ functions whose Fourier transform vanish almost everywhere on $[-1,1]$.

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The easy way to deal with convolution operators $f \mapsto g\ast f$ is to Fourier-transform. Then the operator is just the multiplication by $\hat{g}$ and the spectrum is the essential range of $\hat{g}$. Of course this is not an eigenvalue, these all live on the constancy intervals of $\hat{g}$, so just determine these. In your edit these are $1$ and $0$ for the circ function.

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