Timeline for Product of two cuspforms is not a cuspform

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May 4, 2012 at 22:01	history	edited	paul garrett	CC BY-SA 3.0	added 521 characters in body
May 4, 2012 at 14:58	comment	added	Matt Young		@Paul Yes there is a potential issue with reversing the limit as $y\rightarrow \infty$, and the sum over $j$. If $f$ is sufficiently smooth then using the fact that the Laplacian is self-adjoint, one can show that the spectral coefficients decay as the reciprocal of some polynomial of the Laplace eigenvalue. The sup norm of a Maass form grows like a fixed polynomial in the Laplace eigenvalue, and also $u_j(x+iy)$ is exponentially small once $y$ is a little larger than the square-root of the Laplace eigenvalue. Maybe I'll go back and edit my answer when I get some time...
May 4, 2012 at 12:57	history	edited	paul garrett	CC BY-SA 3.0	added 1590 characters in body
May 4, 2012 at 1:41	comment	added	Eren Mehmet Kiral		What are the important principles that you say? In the case of vanishing at cusps but not being cuspidal case, the function I had found had good properties in terms of its spectral decomposition. (I know this doesn't violate any theorem, but did violate my intuition about cuspidal functions).
May 3, 2012 at 13:42	history	edited	paul garrett	CC BY-SA 3.0	added 87 characters in body
May 2, 2012 at 22:38	history	answered	paul garrett	CC BY-SA 3.0