The at-most-two non-real zeros conjecture for normalised polynomial sequences

A stronger localization theorem

Let

$$A(x)=ax+b, \qquad B(x)=cx+d, \qquad a,b,d<0<c,$$

and define

$$W_0(x)=1, \qquad W_1(x)=x, \qquad W_n(x)=A(x)W_{n-1}(x)+B(x)W_{n-2}(x) \quad(n\geq2).$$

In fact, for every $n\geq2$, the polynomial $W_n$ has at least $n-2$ distinct real zeros in

$$\left(-\infty,-\frac dc\right). \tag{1}$$

Because $\deg W_n=n$, it follows that $W_n$ has at most two nonreal zeros. This proves the conjecture, including its full range of real parameters.

A symmetric tridiagonal representation

Put

$$\xi=-\frac dc>0.$$

For $x<\xi$, define

$$t(x)=\sqrt{-B(x)}=\sqrt{-cx-d}>0$$

and consider the real symmetric $n\times n$ matrix

$$J_n(x)= \begin{pmatrix} x&t(x)&0&\cdots&0\\ t(x)&A(x)&t(x)&\ddots&\vdots\\ 0&t(x)&A(x)&\ddots&0\\ \vdots&\ddots&\ddots&\ddots&t(x)\\ 0&\cdots&0&t(x)&A(x) \end{pmatrix}. \tag{2}$$

Write $D_k(x)$ for the determinant of its leading $k\times k$ principal submatrix and set $D_0(x)=1$. Expansion along the last row gives

$$D_1(x)=x, \qquad D_k(x)=A(x)D_{k-1}(x)-t(x)^2D_{k-2}(x) \quad(k\geq2).$$

Since $-t(x)^2=B(x)$, these are exactly the defining initial conditions and recurrence for $W_k$. Therefore,

$$\boxed{\det J_n(x)=W_n(x)} \qquad(x<\xi). \tag{3}$$

Endpoint inertia

Let $\nu_+(M)$ denote the number of strictly positive eigenvalues of a real symmetric matrix $M$, counted with multiplicity. The matrix in (2) extends continuously to $x=\xi$, where $t(\xi)=0$. Since

$$\xi>0, \qquad A(\xi)=a\xi+b<0,$$

we have

$$J_n(\xi) =\operatorname{diag}\bigl(\xi,A(\xi),\ldots,A(\xi)\bigr), \qquad \nu_+\bigl(J_n(\xi)\bigr)=1. \tag{4}$$

At the other end, set $x=-R$ and let $R\to\infty$. Then

$$x=-R, \qquad A(-R)=(-a)R+b, \qquad t(-R)=\sqrt{cR-d}.$$

Because $-a>0$, for every sufficiently large $R$,

$$-R+\sqrt{cR-d}<0, \qquad (-a)R+b-2\sqrt{cR-d}>0. \tag{5}$$

The Gershgorin interval corresponding to the first row of $J_n(-R)$ lies entirely in $( -\infty,0)$, whereas all the other Gershgorin intervals lie entirely in $(0,\infty)$. Since these two collections are disjoint, their eigenvalue counts equal the numbers of corresponding rows. Consequently,

$$\nu_+\bigl(J_n(-R)\bigr)=n-1 \tag{6}$$

for all sufficiently large $R$.

Counting the crossings

For every $x<\xi$, all off-diagonal entries of $J_n(x)$ are strictly positive. Hence

$$\dim\ker J_n(x)\leq1. \tag{7}$$

Indeed, the first component of a vector in the kernel determines its second component from the first row, and each subsequent row determines the next component uniquely. If the first component is zero, this recursion forces every component to vanish.

The eigenvalues of $J_n(x)$ vary continuously with $x$. Its positive inertia can change only when $J_n(x)$ is singular, which by (3) is equivalent to

$$W_n(x)=0.$$

At any such point, (7) shows that at most one eigenvalue can cross zero. Therefore each distinct real zero in $(-R,\xi)$ changes $\nu_+$ by at most one. Equations (4) and (6) give the total change

$$\left| \nu_+\bigl(J_n(-R)\bigr) -\nu_+\bigl(J_n(\xi)\bigr) \right| =n-2.$$

Thus $W_n$ has at least $n-2$ distinct real zeros in $(-R,\xi)$, and therefore in the interval (1).

Finally, the recurrence shows that

$$\deg W_n=n, \qquad [x^n]W_n(x)=a^{n-1}\neq0 \quad(n\geq1).$$

At most two zeros can remain outside the $n-2$ distinct real zeros already located. Since $W_n$ has real coefficients, any nonreal zeros form one conjugate pair. Hence every $W_n$ has at most two nonreal zeros, as asserted.

D. G. L. Wang and Jiarui Zhang, Piecewise interlacing zeros of polynomials, Conjecture 5.2.