Let $\lbrace P_n(z)\rbrace_{n\in\mathbb N_0}$ be a family of polynomials defined by a generating function $g(t,z)=\sum\limits_{n=0}^\infty P_n(z)t^n$ or by a contour integral $P_n(z)=\frac1{2\pi i}\oint\frac{g(t,z)}{t^{n+1}}dt$. Are there known sufficient conditions on $g$ or on the $P_n$ themselves that guarantee the existence of a weight function $w:I\to \mathbb R^+_0$ (where $I\subset\mathbb R$ is an appropriate interval) such that the $P_n$ are orthogonal w.r.t. $w$?
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2$\begingroup$ Orthogonal polynomials must satisfy a three-term recursion $x P_{n+1}(x) = A_n P_n(x) - B_n P_{n-1}(x)$ and have real interlaced roots. Not clear how to test these necessary conditions from a generating function or contour integral, let alone give sufficient conditions. $\endgroup$– Noam D. ElkiesCommented Jan 27, 2012 at 23:06
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2$\begingroup$ Oops, I started writing my answer before this comment appeared. The three-term recurrence should be $xP_{n}(x) = P_{n+1}(x) + A_n P_n(x) - B_n P_{n-1}(x)$ with $B_n>0$, and then it actually implies that the roots are interlaced. $\endgroup$– Henry CohnCommented Jan 27, 2012 at 23:18
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$\begingroup$ (But you're right that this isn't a particularly useful condition for proving that polynomials are orthogonal. I view Favard's theorem as having mainly psychological value: if you observe a suitable recurrence experimentally, then you really ought to look for an orthogonality proof to explain it. It's sometimes possible to prove the recurrence directly and work from there, but this is generally not the most flexible or illuminating approach.) $\endgroup$– Henry CohnCommented Jan 27, 2012 at 23:21
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$\begingroup$ Henry is right, and I apologize for the typo. $\endgroup$– Noam D. ElkiesCommented Jan 27, 2012 at 23:56
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Favard's theorem characterizes this in terms of the three-term recurrence. Suppose the polynomials $P_n$ are normalized so that they are monic. Then they are orthogonal polynomials with respect to some Borel measure if and only if there are constants $\alpha_n$ and $\beta_n$ such that $P_n(x) = (x+\alpha_n) P_{n-1}(x) + \beta_n P_{n-2}(x)$ and $\beta_n < 0$. (The sign condition on $\beta_n$ is needed to get a positive measure. I think you still get a signed measure if you have a three-term recurrence with $\beta_n \ge 0$, but I'm not certain offhand.)
This is pretty easy to test for in practice if you are given a sequence of polynomials numerically. Strictly speaking, it doesn't guarantee a weight function as specified in your question, since the measure may not be absolutely continuous with respect to Lebesgue measure, but I assume that's not what you really care about. If it is, then I'm not sure offhand how to characterize that case.
answered Jan 27, 2012 at 23:13
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$\begingroup$ Thank you. That's about what I expected: that one of the necessary conditions is essentially sufficient. Like Noam D. Elkies, I didn't really expect there to be any hope of deriving anything directly from a generating function. Let alone the even more hopeless question: If only $g(t,z)$ is given, is there even a way to know, without doing a Taylor expansion, that the coefficients will actually be polynomials in $z$? $\endgroup$– WolfgangCommented Jan 28, 2012 at 10:24