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I accidentally stumbled upon a problem of complex analysis in several variables, and I have a hard time understanding what I read, it might be related to the Cousin II problem but I cannot say for sure. Let me state the problem, maybe you can tell me if you can think of a relevant tool.

Let me start with a simple instance of my problem, I state the general problem later. Is it possible to have an entire function $\tilde \psi(u_1,u_2)$ on $\mathbb{R}^2$ such that $|\psi(u_1,u_2)|\leq (u_1-u_2)^2$ on $B(0,1)^c$ and $\psi(0)\neq 0$? I insist that the uniqueness theorem does not work here, for instance $(u_1-u_2)^2$ is an entire function with non-isolated zeros.

(no need to bother further if the answer is no).

General version: I have an entire function $\psi(u)$ on $\mathbb{R}^n$, of exponential type $1$ (when seen as a function of $n$ complex variables). I want to build another entire function $\tilde \psi(u)$ of exponential type $1$ which satisfies:

  • $|\tilde\psi(u)|\leq c|\psi(u)|,u\in \mathbb{R}^n$ outside some small ball $B(0,\varepsilon)$
  • $\tilde\psi(0)\neq 0$.

In dimension $n=1$ it is easy, if $\psi(u)=cu^k(1+o(1))$ as $u\to 0$ for some $k\in\mathbb{N},c\neq 0$, simply define $$\tilde \psi(u)=\frac{u^{-k}}{1+u^{2k}}\psi(u)$$.

From what I understand to the "Cousin II problem", it seems to give conditions to glue two meromorphic functions together, such as $\frac{1}{u_1-u_2}$ on an open domain not containing $0$, and $1$ (or another non-vanishing function) on an open domain containing $0$, but I am not sure it applies.

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    $\begingroup$ If I understand your simple version correctly, your function vanishes on the diagonal $u_1=u_2$. Then it vanishes at $(0,0)$ just by continuity. $\endgroup$
    – Oleg Eroshkin
    Commented Jun 20 at 21:33
  • $\begingroup$ It is supposed to vanish on the diagonal only away from zero, the question is wether it implies vanishing close to 0 also $\endgroup$
    – kaleidoscop
    Commented Jun 21 at 4:34
  • $\begingroup$ Apply the uniqueness theorem for the restriction to the diagonal to conclude that the function vanishes on the full diagonal. $\endgroup$
    – Jochen Wengenroth
    Commented Jun 21 at 4:50
  • $\begingroup$ The uniqueness theorem works for functions of two complex variables? How is it stated? $\endgroup$
    – kaleidoscop
    Commented Jun 21 at 6:36
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    $\begingroup$ Just define the function $f(z):= \psi(z, z)$, then $f$ is entire function of one variable, and it is zero if $z\in \mathbb{R}, |z| > 1$. Hence, it is identically zero by the uniqueness theorem, in particular $0=f(0) =\psi(0, 0)$. $\endgroup$
    – Aleksei Kulikov
    Commented Jun 21 at 11:03

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