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本帖最后由 hbghlyj 于 2023-1-11 18:48 编辑 complex.pdf page49
Proposition 7.21. Suppose that $U$ is a domain and the sequence of holomorphic functions $f_n: U \rightarrow \mathbb{C}$ converges to $f: U \rightarrow \mathbb{C}$ uniformly on compacts in $U$. Then $f$ is holomorphic.
Proof. Note by the above that $f$ is continuous on $U$. Since the property of being holomorphic is local, it suffices to show for each $w \in U$ that there is a ball $B(w, r) \subseteq U$ within which $f$ is holomorphic. Since $U$ is open, for any such $w$ we may certainly find $r>0$ such that $B(w, r) \subseteq U$. Then as $B(w, r)$ is convex, Cauchy's theorem shows that for every closed path $\gamma:[a, b] \rightarrow B(w, r)$ whose image lies in $B(w, r)$ we have $\int_\gamma f_n(z) d z=0$ for all $n \in \mathbb{N}$.
But $\gamma^*=\gamma([a, b])$ is a compact subset of $U$, hence $f_n \rightarrow f$ uniformly on $\gamma^*$. By Proposition 5.16.
\[
0=\int_\gamma f_n(z) d z \rightarrow \int_\gamma f(z) d z,
\]
so that the integral of $f$ around any closed path in $B(w, r)$ is zero. But then Theorem 5.22 shows that $f$ has a primitive $F$ on $B(w, r)$. By Theorem 7.18. (Morera’s theorem), $f$ is holomorphic. $\square$
Remark. This is false for real functions, for example here we have a sequence of real differentiable functions converging uniformly to $f(x)=|x|$ which is not differentiable at 0. |
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