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Let $f_n\colon \mathbb{C} \to \mathbb{C}$ be a sequence of entire functions, such that $f_n$ converges to the zero function on an open dense subset $U$ of $\mathbb{C}$ pointwise (or equivalently normally). Then, does $f_n$ genuinely converge to the zero function on $\mathbb{C}$?

It seems to me that it does converge, but I am not sure. If the assumption is a convergence on a dense subset of $\mathbb{C}$,then $f_n$ does not necessarily converge (https://math.stackexchange.com/questions/3651442/pointwise-convergence-of-holomorphic-functions-on-a-dense-set/3651462#3651462).

I would appreciate any comments! Thanks.

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It is not true. Consider the compact sets $$K_n=\{ z:|z|\leq n, |\arg z|\geq 1/n\}\cup\{0\}.$$ By Runge's approximation theorem, there exist polynomials $f_n$, such that $|f_n(z)|<1/n,\; z\in K_n,$ and $f_n(1)=n.$ This sequence of polynomials evidently converges uniformly on compact subsets of the dense open set $C\backslash[0,+\infty)$ but does not converge at the point $1$.

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Due to Weierstrass there exists an entire function $F$ such that: $F(0)\neq 0$ and $F(re^{it})\to 0$ if $r$ tends to $\infty$, for any real $t$. Then $g_n(z):=F(nz)$ gives a counterexample. See the book by Remmert "Funktionentheorie II", chapter 12, §3. I am sure that there is an English translation.

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  • $\begingroup$ Such a function is described here: mathoverflow.net/questions/264651/… There is no evidence that Weierstrass knew it. BTW, the sequence converges to 0 at EVERY point of the plane. $\endgroup$ Commented Aug 21, 2021 at 12:59

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