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The finite-difference stability conjecture for stable polynomials

From papers

Let pp be a stable polynomial, meaning that all of its zeros lie in the open left half-plane. For parameters θ\theta and hh as in the operator ]Δθ,h]]\Delta_{\theta,h}], and for integers m=1,…,deg⁡p−1m=1,\ldots,\deg p-1, consider the iterates

Δθ,hm(p)(z).\Delta_{\theta,h}^m(p)(z).

Finite-difference stability conjecture. For any stable polynomial pp, the polynomials

Δθ,hm(p)(z),m=1,…,deg⁡p−1,\Delta_{\theta,h}^m(p)(z),\qquad m=1,\ldots,\deg p-1,

have no nonreal zeroes in the closed right half-plane. The conjecture concerns the action of the complex zero-decreasing operator on stable polynomials; calculations suggest it, but the source gives no resolution.

Progress summary

Open

The conjecture remains open: its source gives only supporting calculations, and no verified proof or counterexample was found.

The conjecture asks whether every permitted iterate Δθ,hm(p)\Delta_{\theta,h}^{m}(p) of a polynomial whose zeros lie in the open left half-plane has no nonreal zeros in the closed right half-plane. It is stated as Conjecture 1 in a 2019 paper on linear finite-difference operators; the retrieved source does not identify a resolution.

Known results

  • The operator Δθ,h\Delta_{\theta,h} is described as complex zero-decreasing, with related root-location results, but these do not settle the conjecture.

Current status (as of August 2026): The conjecture is unproved, with no verified counterexample or claimed resolution in the retrieved record.

Sources
Sources & referencesView supporting material

Primary source

Olga Katkova, Mikhail Tyaglov and Anna Vishnyakova, “Hermite-Poulain theorems for linear finite difference operators”, arXiv:1901.06398 (2019).

complex-analysisfinite-difference-operatorsstable-polynomials

Solutions 1

Counterexample

Use the finite-difference operator in the normalization of the conjecture:

Δθ,hp(z)=eiθp(z+ih)−e−iθp(z−ih)2i.\Delta_{\theta,h}p(z) =\frac{e^{i\theta}p(z+ih)-e^{-i\theta}p(z-ih)}{2i}.

Take

p(z)=z2+2z+10=(z+1)2+9,h=2,θ=3π4.p(z)=z^2+2z+10=(z+1)^2+9,\qquad h=2,\qquad \theta=\frac{3\pi}{4}.

The zeros of pp are −1±3i-1\pm3i, so pp is strictly Hurwitz stable.

Since

p(z+2i)=z2+2z+6+i(4z+4),p(z+2i)=z^2+2z+6+i(4z+4),

direct substitution gives

Δ3π/4,2p(z)=z2−2z+22.\Delta_{3\pi/4,2}p(z) =\frac{z^2-2z+2}{\sqrt2}.

The zeros of the resulting polynomial are

z=1+i,z=1−i,z=1+i,\qquad z=1-i,

both in the open right half-plane.

Thus the first permitted iterate m=1=deg⁡(p)−1m=1=\deg(p)-1 already violates the conjectured stability. This provides a counterexample of the minimum possible degree.

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Shivam Patel ·