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Here is the question:

Suppose $P_0, P_1, P_2, \dots$ are polynomials orthonormal with respect to the inner product $$(f,g)=\int_a^b f(x)g(x)W(x)dx,$$ where $W(x) > 0$ is a weight function and $P_n$ is of degree $n$. Is it true that $P_n$ has $n$ distinct roots in $(a,b)$?

Clearly $P_0$ has no roots and since $(P_0,P_1)=0$, I know $P_1$ must cross the $x$-axis at least once otherwise the integral would not equal $0$, so $P_1$ has one root in $(a,b)$. However, I'm not sure how to prove this for an arbitrary $n$-value (if it is true). I would appreciate any advice.

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The fact that $(P_0, P_k) = 0$ when $k\neq 0$ shows that all $P_k$ have a root in $(a,b)$, not just for $k = 1$. – Jason DeVito Oct 4 '11 at 3:04
That's true but how can I show that Pn has n distinct roots in (a,b)? – Nick Oct 4 '11 at 3:10

This is true. The proof I know is beautiful but quite clever in my opinion.

Because $\deg P_k = k$ for all $k$, you get $\mathbb{R}_{n-1}[X] = \textrm{Span}(P_0, \ldots, P_{n-1})$. And because of orthogonality, if $\deg f < n$ then $(f, P_{n}) = 0$.

Now, let $r_1, \ldots, r_k$ be the roots of $P_n$ that are in $[a,b]$ and have odd multiplicity (if there are some). Denote $Q = (X-r_1) \ldots (X-r_n)$, and write $P_n = Q R$. All the roots of $R$ in $[a,b]$ have even multiplicity, so $R$ does not change sign in $[a,b]$. In order to conclude, we need to show $\deg Q = \deg P_n$. If we assumed $\deg Q < \deg P_n$, then we would have

$$(Q, P_n) = 0$$

So we'd get

$$\int_a^b Q^2 R w = 0$$

But the function $Q^2 R w$ does not change sign on $[a,b]$, so $Q^2 R w= 0$. This is obviously impossible, so $\deg Q = \deg P_n$, which concludes.

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Thank you this is very helpful!! – Nick Oct 4 '11 at 3:30
Very neat :) ${}$ – Mariano Suárez-Alvarez Oct 4 '11 at 5:56
also, why do all the roots have to be distinct? – Nick Oct 4 '11 at 6:21
@Nick : The roots of $Q$ are distinct by construction (you take each once each root that has odd multiplicity). – Joel Cohen Oct 4 '11 at 7:24
I used a similar approach in the second part of this answer. – J. M. Oct 4 '11 at 7:55

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