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Let $\langle\cdot,\cdot\rangle$ be any inner product on the set of functions $[a,b]\rightarrow\mathbb{R}$, and let $(p_n)_{n\in\mathbb{N}}$ be a sequence of polynomials defined by:

$p_{-1}(x)=0$, $p_0(x)=1$, $p_{n+1}=(x-a_n)p_n(x)-b_np_{n-1}(x)$ for $n\geq0$

where $a_n$ and $b_n$ are constants chosen so that all $p_n$ are orthogonal with respect to $\langle\cdot,\cdot\rangle$. Now I've proved that $p_n$ is a monic polynomial of degree $n$ for all $n$, and found what the constants $a_n$ and $b_n$ must be, but why does every $p_n$ have to have $n$ distinct roots in $[a,b]$?

If $\langle\cdot,\cdot\rangle$ was the standard inner product, or any Sturm-Liouville inner product with a weight function, we could prove this by letting $x_1,\dots,x_k$ be the simple roots of $p_n$ in $[a,b]$ (where $k$ may be 0) and $q(x):=(x-x_1)\dots(x-x_k)$. Then $p_n(x)q(x)$ has constant sign, so $\langle p_n,q\rangle\neq0$, so the degree of $q$ must be $n$. But for a general inner product, I don't think this argument works. Can we do something else with $q$, e.g. write it as a linear combination of $p_0,\dots,p_k$? Or is there another method?

Many thanks for any help with this!

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  • $\begingroup$ I tend to prefer the linear-algebraic route myself; there is always a symmetric tridiagonal matrix whose characteristic polynomial is an orthogonal polynomial, and whose entries are made up of the recursion coefficients. Then, you can show that a symmetric tridiagonal matrix without any zero off-diagonal entries necessarily has simple eigenvalues. $\endgroup$
    – J. M. ain't a mathematician
    Apr 21, 2013 at 16:18
  • $\begingroup$ Yes, there are restrictions on the inner product. I don't have my copy with me now, but if memory serves there's a description of this route in Chihara's book on orthogonal polynomials. If it's in your library, you should try to get a hold of it. $\endgroup$
    – J. M. ain't a mathematician
    Apr 21, 2013 at 22:52

1 Answer 1

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As noted, you cannot use just any inner product.

Suppose $q(t)$ is a factor of $p_n$ such that $q(t)\ge0$ for $t$ in $[a,b]$. Let $r(t)=p(t)/q(t)$. Then the product $rp_n$ is non-negative on $[a,b]$ and is not identically zero, and therefore $\langle 1,rp_n\rangle>0$.

On the other hand, $p_n$ is orthogonal to all polynomials of degree less than $n$, and therefore if the degree of $q$ is greater than zero, $\langle r,p_n\rangle=0$. (Remark: this shows we cannot use just any inner product on polynomials, we need $\langle tp,q\rangle=\langle p,tq\rangle$ for all $p$ and $q$, that is, multiplication by $t$ must be self adjoint.)

Since $\langle1,rp_n\rangle = \langle r,p_n\rangle$, this is a contradiction. We conclude that $p_n$ has no non-constant non-negative factor. In particular $(t-\lambda)^2$ cannot be a factor and therefore $p_n$ has only simple zeros.

A minor variant of this argument also shows that the zeros of $p_n$ are all real.

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  • $\begingroup$ Thanks for your answer. Just two questions: why is $\langle1,rp_n\rangle>0$ in the second para, and why does $\langle1,rp_n\rangle=\langle r,p_n\rangle$ in the fourth? $\endgroup$
    – Harry Macpherson
    Apr 21, 2013 at 18:47
  • $\begingroup$ Any non-negative polynomial $q(t)$ in one real variable is a sum of squares of polynomials, and therefore $\langle 1,q\rangle>0$ if $q$ is not identically zero. Given the condition in the remark in the third paragraph, $\langle t^kp,q\rangle=\langle p,t^k,q\rangle$. This yields the claim in the fourth paragraph. $\endgroup$
    – Chris Godsil
    Apr 22, 2013 at 0:28

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