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I was reading through Akhiezer's book Lectures on Integral Transforms and in chapter nine, he states that the Hankel transform is unitary for $\nu > -1$, so that for a suitable function, $f$,

$f(y) = \int_0^{\infty}dr \sqrt{ry}J_{\nu}(ry) \int_0^{\infty} dx f(x) \sqrt{xr} J_{\nu}(xr)$.

He leaves the proof that it is unitary to the reader, but I cannot for the life of me even begin to know what the angle of attack should be. I've done some pretty extensive Googling, but have come up completely short. Can anyone point me in the right direction or if you know where to find the proof itself, can link me to it? Much appreciated.

I have found a couple of proofs for the case where $\nu > -1/2$, but none for the range $-1 < \nu < -1/2$.

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http://en.wikipedia.org/wiki/Hankel_transform#Orthogonality

This ensures orthogonality. You should think in terms of linear algebra - transition matrix to any orthogonal basis in orthogonal. So you need to prove orhogonality of Bessels. This is stated in Wiki Link above.

PS

Actually I do not quite understand where this orthogonality comes from ? Can one comment ?

Because usually orthogonality comes easily - the eigenfunctions of some symmetric matrix (operator) are orthogonal. But Bessels are eigenfucntions in $\nu$ , not in "x". So I am puzzled.

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  • $\begingroup$ That is basically what I want to prove, however the proofs I have seen all deal with the $\nu > -1/2$ and I'm not sure how to prove it for the more general case $\nu > -1$. The fact that it is true (at least according to Akhiezer and the NIST Handbook of Mathematical Functions and I checked it with Mathematica for some functions) is great, but I'm interested in the proof itself as I need to adapt it for research purposes. $\endgroup$
    – Cameron Williams
    Jan 10, 2012 at 18:25
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    $\begingroup$ I see your point now. Hankel can be seen as FT for functions in R^3 depending on radius only. Would it be helpful ? What is the meaning of $\nu=-1/2$ ? $\endgroup$
    – Alexander Chervov
    Jan 11, 2012 at 5:59
  • $\begingroup$ Sorry for the delayed response, but I didn't actually need to use the proof itself. I just needed to use the result. Apparently I had goofed when doing some Dirac delta algebra, but it all worked out. Thanks for your input anyway! $\endgroup$
    – Cameron Williams
    Mar 7, 2012 at 0:37
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you can see the book:

Integral Transforms for Engineers By Larry C. Andrews, Bhimsen K.

best regard, Ramin

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  • $\begingroup$ Hi Ramim, it turned out that I did not need the proof after all (as I went a more measure-theoretic route). Thanks for the reference though. :) $\endgroup$
    – Cameron Williams
    Jan 14, 2014 at 16:44
  • $\begingroup$ Was a missing prove really found somewhere? $\endgroup$
    – Sergei
    Oct 3, 2014 at 17:41

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