Let us define the basis of polynomials given by: $$ \begin{array}\ P_0=1, \\ P_1=x, \\ P_2=x(x1), \\ P_3=x(x1)(x2), \\ P_4=x(x1)(x2)(x3), \ldots\\ \end{array} $$ I would like to know if this basis is orthogonal with respect to some measure. Thank you very much!
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2$\begingroup$ Well, you could define an ad hoc inner product by saying that if $p(x) = \sum a_ip_i$ and $q(x)=\sum b_ip_i$, then $\langle p,q\rangle = \sum a_ib_i$, which would make it an orthogonal (even orthonormal) basis. But presumably you are looking for more than just "some" measure? $\endgroup$ – Arturo Magidin Oct 25 '18 at 21:50

$\begingroup$ Ps: I was thinking about an inner product of the form $\int dx P_i(x) P_j(x) \mu(x)$ for some measure $\mu(x)$ $\endgroup$ – fernando Oct 25 '18 at 22:22

$\begingroup$ You should explain more clearly what kind of measure you are asking? A real measure on the real line? A complex measure on a subset of the complex plane? $\endgroup$ – Alexandre Eremenko Oct 25 '18 at 23:50
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If a sequence of monic polynomials is orthogonal with respect a measure, it satisfies a threeterm recurrence \[ p_{n+1}(t) = (ta_n)p_n(t)  b_n p_{n1}(t) \] where $b_n>0$. From this it follows that consecutive terms in the sequence cannot have a common zero. Your sequence fails badly on this test.

1$\begingroup$ Thank you very much! Indeed you are right. But can we define something similar to a orthogonality condition? e.g. we can also consider the family of polynomials P_n(x) = x^n, which is a complete basis, but in this case we can define, for instance $<P_n P_m> = \int \frac{dz}{z} P_n(x) P_{m}(x)$, where the integral is a contour integral around zero. $\endgroup$ – fernando Oct 25 '18 at 22:35

2$\begingroup$ Very nice answer! Do you know of a reference elaborating on this link between orthogonality and recurrences? $\endgroup$ – Emanuele Tron Oct 25 '18 at 23:04

1$\begingroup$ Any text book on orthogonal polynomials should treat this, it’s quite basic. My “Algebraic Combinatorics” discusses it at lengh too ( if you’ll forgive the plug). $\endgroup$ – Chris Godsil Oct 25 '18 at 23:28