The zero-strip-decreasing conjecture for complex finite difference operators

Counterexample to uniform zero-strip decrease, with the sharp pointwise replacement.

Katkova, Tyaglov, and Vishnyakova, Hermite–Poulain theorems for linear finite difference operators, Constructive Approximation 52 (2020), 357–393, Conjecture 4, ask whether their operator

$$\Delta_{\theta,h}f(z) =\frac{e^{i\theta}f(z+ih)-e^{-i\theta}f(z-ih)}{2i}, \qquad 0\leq\theta<\pi,$$

remains zero-strip decreasing when its domain is enlarged from real polynomials to all complex polynomials.

The term has a crucial uniform quantifier. In the cited original definition, Cardon, Complex zero strip decreasing operators, Definition 1.4(b), if

$$S_r=\{z\in\mathbb C:|\operatorname{Im}z|\leq r\},$$

then, for every fixed $r>0$, there must exist a single $r_1<r$ such that the image of every admissible input with zeros in $S_r$ has all its zeros in $S_{r_1}$. In particular, $r_1$ cannot depend on the input polynomial. The conjectured extension to $\mathbb C[z]$ fails in precisely this uniform sense.

Fix $r>0$, choose the allowed parameters $\theta=\pi/2$ and $h=1$, and for $R>0$ consider

$$f_R(z)=z^2-(R+ir)^2.$$

Its two roots are $R+ir$ and $-R-ir$. They lie on opposite boundaries of the centered strip $S_r$, so the input strip is minimal and has positive width; moreover, their barycenter is zero. The source's exact normalization gives

$$\Delta_{\pi/2,1}f_R(z) =\frac{f_R(z+i)+f_R(z-i)}2 =z^2-(R+ir)^2-1.$$

Write the square root with positive real and imaginary parts as

$$u_R+iv_R=\sqrt{(R+ir)^2+1}, \qquad u_R,v_R>0.$$

The two output roots are $\pm(u_R+iv_R)$. Equating real and imaginary parts yields

$$u_R^2-v_R^2=R^2-r^2+1, \qquad u_Rv_R=Rr.$$

Consequently $t_R=v_R^2$ is the unique positive root of

$$t^2+(R^2-r^2+1)t-R^2r^2=0.$$

Evaluating this polynomial at $0$ and $r^2$ shows $0<v_R<r$. Subtracting its values at $t_R$ and $r^2$ gives the sharper exact identity

$$(r^2-v_R^2)(R^2+v_R^2+1)=r^2,$$

and therefore

$$v_R\longrightarrow r \qquad\text{as }R\longrightarrow\infty.$$

Given any proposed $r_1<r$, choose $R$ sufficiently large that $v_R>r_1$. Then all zeros of $f_R$ lie in $S_r$, while both zeros of $\Delta_{\pi/2,1}f_R$ lie outside $S_{r_1}$. Thus

$$\sup_{\substack{\deg f=2\Z(f)\subseteq S_r}} \ \max_{w\in Z(\Delta_{\pi/2,1}f)} |\operatorname{Im}w| =r,$$

so no uniform strict decrease exists even in fixed degree two, with centered inputs attaining both boundary lines.

In fact, the obstruction applies to every allowed phase and every real $h\neq0$. If $0<\theta<\pi$, the same quadratic has output roots

$$-h\cot\theta \ \pm\sqrt{(R+ir)^2+h^2\csc^2\theta},$$

whose imaginary parts tend to $\pm r$. At the remaining phase $\theta=0$, use instead

$$g_R(z)=\bigl(z^2-(R+ir)^2\bigr)^2.$$

Again its input zeros attain both boundaries, while

$$\Delta_{0,h}g_R(z) =4hz\bigl(z^2-(R+ir)^2-h^2\bigr),$$

whose two nonzero roots have imaginary parts tending to $\pm r$.

There is, however, a valid nonuniform statement that explains the computational intuition behind the conjecture. For arbitrary $h\in\mathbb C\setminus\{0\}$, factor

$$f(z)=c\prod_{j=1}^d(z-\rho_j), \qquad a=\min_j\operatorname{Im}\frac{\rho_j}{h}, \qquad b=\max_j\operatorname{Im}\frac{\rho_j}{h}.$$

If $w$ is an output root and $\zeta=w/h$, $\eta_j=\rho_j/h$, the vanishing equation implies

$$\prod_{j=1}^d|\zeta+i-\eta_j| =\prod_{j=1}^d|\zeta-i-\eta_j|.$$

But each pair of factors satisfies

$$|\zeta+i-\eta_j|^2-|\zeta-i-\eta_j|^2 =4\bigl(\operatorname{Im}\zeta-\operatorname{Im}\eta_j\bigr).$$

If $a<b$ and $\operatorname{Im}\zeta\geq b$, every factor on the left is at least its counterpart on the right, and at least one is strictly larger; zero factors cannot restore the product equality. The same argument excludes $\operatorname{Im}\zeta\leq a$. Hence

$$a<\operatorname{Im}\frac wh<b \qquad\text{for every }w\in Z(\Delta_{\theta,h}f) \quad\text{whenever }a<b.$$

If $a=b$, all output zeros remain on that same line. Thus every individual positive-width minimal strip parallel to $h$ contracts strictly, but the amount of contraction depends on the input and can approach zero even in fixed degree. The pointwise interpretation is true; the standard uniform zero-strip-decreasing conjecture on $\mathbb C[z]$ is false.