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Let $X = SL_2(\mathbb{Z}) \backslash \mathbb{H}$ be the modular surface. Consider a basis of $L^2$-normalized Hecke-Maass cusps forms $\phi_j$ on $X$ with $-\Delta$-eigenvalue $\lambda_j$. Hejhal-Rackner (1991, on the Topography of Maass forms) seems to claim that for any fixed set $A \subset X$ with finite measure, $$\int_A \phi_j (z) d\mu z \to 0$$ as $\lambda_j \to \infty$.

I am probably overlooking something, but is there a simple proof of this? One of the even/odd Maass forms case should be trivial by sign considerations, but unless I am missing something the other case does not seem as apparent.

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This is an analog of the Riemann-Lebesgue lemma. Let $f$ be a smooth function on $X$. Then by self-adjointness of the Laplacian $$ \lambda_j \langle \phi_j, f \rangle = \langle \Delta \phi_j ,f \rangle = \langle \phi_j, \Delta f\rangle , $$ and by Cauchy-Schwarz this is bounded in size by $\Vert \phi_j \Vert \Vert \Delta f\Vert$. Therefore $\langle \phi_j, f\rangle$ goes to zero as $\lambda_j$ gets large. Now approximate the characteristic function $\chi_A$ of the set $A$ by suitable smooth functions $f$, and bound $\langle \phi_j, f-\chi_A \rangle$ by $\Vert \phi_j \Vert \Vert f-\chi_A \Vert$.

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  • $\begingroup$ I'm not seeing why the $L^2$ norm of $\Delta f$ might not blow up as $f$ better-and-better approximates a characteristic function... $\endgroup$
    – paul garrett
    Commented Jan 20, 2015 at 15:10
  • $\begingroup$ @paul garrett: Once you approximate the characteristic function by $f$, just use Cauchy's inequality to bound $\langle \phi_j, f-\chi_A \rangle$ by $\le \Vert f-\chi_A\Vert$. (I edited that in to the answer) $\endgroup$
    – Lucia
    Commented Jan 20, 2015 at 15:14
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    $\begingroup$ For clarity I would add after your last sentence the following. "This way one can see that for any $\epsilon>0$ we have $\limsup_{j\to\infty}|\langle\phi_j,\chi_A\rangle|\leq\epsilon$, whence $\lim_{j\to\infty}\langle\phi_j,\chi_A\rangle=0$." $\endgroup$
    – GH from MO
    Commented Jan 20, 2015 at 16:09
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    $\begingroup$ Although interaction with eigenvalues and such is certainly interesting, isn't the literal question a consequence of Bessel's inequality? That is, $\sum_j |\langle \varphi_j,f\rangle|^2\le |f|^2$, so the sum converges, so the terms go to $0$, ... $\endgroup$
    – paul garrett
    Commented Jan 20, 2015 at 16:41
  • $\begingroup$ @paulgarrett: Sure, that works too! (I wrote the first thing that came to mind.) $\endgroup$
    – Lucia
    Commented Jan 20, 2015 at 16:45

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