TITLE: Does Peetre's theorem hold in complex analysis?
QUESTION [5 upvotes]: Let $E, F$ be two smooth vector bundles over a smooth manifold $M$. Peetre's theorem states that any $\mathbb{R}$-linear morphism $D: \mathcal{E} \to \mathcal{F}$ of the sheaves of sections of $E$ and $F$ is a differential operator, i.e. there exist trivializing charts $U_i$ such that on $D|_{U_i}$ is a differential operator in the sense of real analysis.
Question 1: Does this result also hold for complex manifolds?
By this I mean that $M$ is a complex manifold, $E$ and $F$ are holomorphic vector bundles and $D$ is $\mathbb{C}$-linear. We then also restrict to sheaves of holomorphic sections.
I don't think the proof of the real version can be adapted directly, since it relies on bump functions.
This is related to my m.SE question in which I showed that the result does not hold for the sheaf of rational functions on $\mathbb{C}$.
However, on any locally ringed space $(X, \mathcal{O})$ one can define differential operators of order $0$ as multiplication with a global section of $\mathcal{O}$ and differential operators of order $n$ inductively as $\mathbb{C}$-linear morphisms of sheaves $D: \mathcal{E} \to \mathcal{F}$ satisfying that $f D - D f$ is a differential operator of order $n-1$. General differential operators can then be defined as $\mathbb{C}$-linear morphisms of sheaves which locally are differential operators of finite order.
This gives the expected result for the differential operators on a Scheme (EGA IV Prop 16.8.8) (i. e. on $n$-dimensional affine space they are polynomials in the $n$ partial derivatives). By Peetre's theorem this definition also coincides with the classical notion of differential operators on a smooth manifold.
Question 2: If $\mathcal{E}$ and $\mathcal{F}$ are sheaves of sections of holomorphic vector bundles over a complex manifold, does the above definition of differential operators produce the classical differential operators between holomorphic vector bundles?

REPLY [3 votes]: I think Peetre's theorem is false in the holomorphic category. Namely, one can construct a counter example (a differential operator of infinite order) already in the case of one complex variable. We want to exhibit a sheaf morphism $D\colon \mathcal{O} \to \mathcal{O}$, where $\mathcal{O}$ is the sheaf of holomorphic functions in one complex variable $z$ of the form
$$
  D[f](z) = \sum_{k=0}^\infty d_k \frac{f^{(k)}(z)}{k!} ,
$$
with infinitely many non-zero $d_k$. Since this is a purely local question, we can just assume that we are only dealing with functions $f$ defined on some neighborhood of $z=0 \in \mathbb{C}$.
First, recall that the Taylor coefficients of any holomorphic function $f(z)$ grow no faster than exponentially (as a consequence of the Cauchy integral formula), that is
$$
  \left|\frac{f^{(k)}(0)}{k!}\right| < \frac{C}{r^k} ,
$$
for some $C, r > 0$, where $r$ is at least as large as the radius of a disk that fits into the domain to which $f(z)$ has a unique analytic continuation. Since the same argument works also about any point of the domain of $f(z)$, we can always find $0 < R \le r$ such that
$$
  \left|\frac{f^{(k)}(z)}{k!}\right| < \frac{C}{R^k} ,
$$
uniformly on some (possibly small) neighborhood of $z=0$. The point is that such a neighborhood and corresponding constants $C, R > 0$ exist for any function $f(z)$ holomorphic at $z=0$.
Now, let $d_k$ be the coefficients of some non-polynomial entire function $d(z) = \sum_{k=0}^\infty d_k z^k$, meaning that infinitely many $d_k\ne 0$ and the sums $\sum_{k=0}^\infty |d_k|/R^k$ converge for any $R>0$. For example $d_k = 1/k!$. The question to answer is the following: given $f(z)$ holomorphic on some neighborhood of $z=0$, does $D[f](z)$ define a holomorphic function on some (possibly smaller) neighborhood of $z=0$? By combining the above estimates, we see that the answer is Yes, since the inequalities
$$
  \left|\sum_{k=0}^N d_k \frac{f^{(k)}(z)}{k!} \right|
  \le C \sum_{k=0}^\infty \frac{d_k}{R^k}
  < \infty
$$
for arbitrarily large $N$ mean that the series defining $D[f](z)$ converges uniformly (and hence to a holomorphic function) on the same neighborhood of $z=0$ on which $|f^{(k)}(z)/k!| < C/R^k$ uniformly.
So $D$ maps germs of holomorphic functions to germs of holomorphic functions and, given how it was defined, it also clearly satisfies all the other properties of a morphism of sheaves. And yet, $D$ is not a differential operator of finite order. By construction, $D$ is a differential operator of infinite order with constant coefficients. But clearly, we can take the coefficients $d_k(z)$ as holomorphic functions, provided that they satisfy similar locally uniform bounds. So there are lots of possibilities for constructing sheaf morphisms that are not finite order differential operators.