Let $D= \{z\in \mathbb{C}:|z| < 1\}$ be the unit disk. And consider the sheaf of holomorphic functions $\mathcal{O}_{D}$.

Question (?) : Is there a sheaf endomorphisms $\phi : \mathcal{O}_D \to \mathcal{O}_D$ which is not a (possibly infinite order) differential operator. I.e. not of the form:

$$\phi=\Sigma_{n=0}^{\infty} b_n(z) \partial^n$$

Where $\partial =\frac{d}{d z}$ and $b_n \in \mathcal{O}_D$ .

EDIT: Suppose I require that $\phi$ be continuous w.r.t. to the natural frechet topology on $\mathcal{O}_D$ coming from uniform convergence on compact subsets, does the answer change?

  • 1
    $\begingroup$ Such an operator cannot be support non-increasing, by Peetre's theorem ( en.wikipedia.org/wiki/Peetre_theorem ). $\endgroup$ Commented Feb 19, 2018 at 11:23
  • $\begingroup$ @Rafael Mrđen: what about $\bar{\partial}$ ? $\endgroup$
    – Qfwfq
    Commented Feb 19, 2018 at 12:03
  • $\begingroup$ @RafaelMrđen I'm not sure how Peetre's theorem can be applied here, $\mathcal{O}_D$ can't be in any obvious sense a sheaf of smooth sections of a vector bundle as smooth functions don't act on it by multiplication. $\endgroup$ Commented Feb 19, 2018 at 12:07
  • $\begingroup$ Do you have any continuity constraint on sheaf endomorphisms? $\endgroup$ Commented Feb 19, 2018 at 12:56
  • $\begingroup$ Oh, I see I was wrong, thanks. Of course, $\phi$ need not a priori be extendable to continuous functions. $\endgroup$ Commented Feb 19, 2018 at 12:59

1 Answer 1


Probably this paper answers your question negatively.

In particular it shows that if $X$ is an open subset of $\mathbb{C}^n$ and the sheaf $\mathcal{O}$ of holomorphic functions on $X$ is given the topology of compact convergence then every continous endomorphism of $\mathcal{O}$ is given by a convergent $\mathcal{O}$-linear sum of operators $\partial_1^{\alpha_{1}}\cdots \partial_n^{\alpha_n}$ with $\alpha_i\in \mathbb{N}$.


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