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Let $ \Omega $ be a sufficiently smooth planar region in $ \mathbb{R}^2 $ with spectrum $ \Gamma $ (the set of eigenvalues of the Laplace operator on functions which vanish on the boundary $ \partial \, \Omega $, i.e. acceptable parameters in the Dirichlet problem) and associated orthonormal basis (normalized eigenfunctions) $ \{ e_i(x)\}_{i=1}^\infty $, $x \in \Omega $. Then define the function $$ \phi(s,x) = \sum_{n=1}^\infty \lambda_n^{-s} e_n(x) $$ for $ s \in \mathbb{C} , x \in \Omega $. This series doesn't converge for $ \mathrm{Re} (s) \le l(x) \in \mathbb{R}$, but it can be analytically continued to all of $\mathbb{C}$ minus singularities (I don't recall my source for this fact, but do remember that I read it somewhere.) So I was wondering, if we use $ \langle g , h \rangle_{\Omega} = \int_{\Omega} f(x) \overline{g(x)} d\mu(x) $ as the natural inner product, and define $ \zeta_{\Omega}(s) = \sum \lambda^{-s} $ wherever it converges and analytically continued otherwise, is the following necessarily true? $$ \langle \phi(\alpha,\cdot) , \phi(\beta,\cdot) \rangle_{\Omega} = \zeta_{\Omega}(\alpha + \beta^*).$$

EDIT: It might help to reconsider this in a more general setting. Holding $\beta$ fixed above, the inner product can be seen as a functional $B:L^2(\Omega)\to\mathbb{C}$ acting on $\phi(\alpha,x)$ with respect to $x$. Similarly, analytic continuation may be seen as an operator $A:\Omega\to\Gamma\supset\Omega$ which, for $f\in C^\infty(\Gamma)$, sends the restriction $f|_\Omega$ to $f$ - in this case acting on $\phi(\alpha,x)$ with respect to $\alpha$. The question above is therefore a special case of a more generalized query: do $A$ and $B$ commute wherever their composition makes sense? If I had more background in category theory, I might express this in a morphism-flavored way lending itself to some elementary identity or other I'm not aware of, but I don't.

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Dear anon, Regarding your question on tags, it can happen if one tag has been set (by the moderators) to be synonomous with another. That seems to be the case here. I have posted a meta question about this:… Regards, – Matt E Jun 17 '11 at 11:29
Dear anon, Your tag problem has been resolved! Regards, – Matt E Jun 17 '11 at 11:40

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