Douglas question for functions of the form z̄+h

Let D={z∈C:∣z∣<1}\mathbb{D}=\{z\in\mathbb{C}:|z|<1\}, let A(D)A(\mathbb{D}) denote the disk algebra, and let A2(D)A^2(\mathbb{D}) be the Bergman space. For h∈A(D)h\in A(\mathbb{D}), define the Bergman-space Toeplitz operator Tz‾+h=PA2(D)Mz‾+hT_{\overline{z}+h}=P_{A^2(\mathbb{D})}M_{\overline{z}+h}. Is it true that Tz‾+hT_{\overline{z}+h} is invertible on A2(D)A^2(\mathbb{D}) whenever inf⁡z∈D∣z‾+h(z)∣>0\inf_{z\in\mathbb{D}}|\overline{z}+h(z)|>0?

References

Progress summary

Refreshed
Claimed solved

A new preprint claims to settle the invertibility question for this explicit family, while the broader Douglas question remains open.

The problem asks whether the relevant operator is invertible for symbols of the form zˉ+h\bar z+h, with hh in the disk algebra, under a positive lower-bound hypothesis. The result concerns this explicit class rather than every version of Douglas's question.

Known results

  • McDonald and Sundberg proved the positive result for real harmonic symbols.
  • Analytic symbols reduce to pointwise multipliers.
  • Zhao and Zheng produced a counterexample to the general question using a second-order real polynomial in ∣z∣|z|.
  • Earlier work established partial results for harmonic symbols under quantitative lower-bound assumptions.

August 2026 preprint

The preprint claims an affirmative answer for symbols zˉ+h\bar z+h satisfying the stated positive lower-bound condition. This gives a claimed solution of the tracked case, but the result remains unverified; the source explicitly cautions that it does not cover every formulation of Douglas's question.

Current status (as of August 2026): The stated positive-lower-bound case for symbols zˉ+h\bar z+h is claimed solved by the August 2026 preprint, but broader formulations remain open.

Sources

Solutions 0

No solutions have been posted yet.