$\{f_n\}^\infty_{n=1}$ is a sequence of holomorphic functions that converges uniformly to a function $f$ in every compact subset of $\Omega$, then $f$ is holomorphic in $\Omega$.

We let $D$ be any disc whose closure is contained in $\Omega$. Then for any triangle $T$ contained in $D$, by Goursat's theorem, we have $\int _T f_n(z)dz=0$. It then asserts that

$$\int_T f_n(z)dz\to \int_T f(z)dz\text{ as }z\to \infty$$

in the closure of $D$, because of the uniform convergence of $f_n$. This seems a basic question, but can anybody please elaborate what is happening here?

  • $\begingroup$ What do you mean by "don't we just need a continuity"? $\endgroup$
    – angryavian
    Nov 15, 2020 at 17:07
  • $\begingroup$ Oh, I think I mixed it up. Sequence of function needs to be either point-wise convergent or uniformly convergent, if they were to be convergent to some function, is that correct? Then why doesn't it suffice to have only point-wise convergence? $\endgroup$
    – able20
    Nov 15, 2020 at 17:09

2 Answers 2


The idea of the proof is that according to Morera's theorem, if the integral of a continuous function $f$ along the boundary of all triangles contained in the domain vanishes, then $f$ is holomorphic. So it has to be proved that a) $f$ is continuous, and b) $\oint_T f(z)\mathrm dz=0$ for all triangles $T$ in $D$. Because then $f$ is holomorphic on $D$.

For a), we can use the usual result that uniform convergence preserves continuity. So since $f_n$ are all continuous and converge to $f$ uniformly on $\overline D$ (since that's a compact set), $f$ is also continuous.

For b), they use the fact that if $f_n\to f$ uniformly on the contour along which we integrate (which they do, since the contour is compact), then $\int f_n\mathrm dz\to\int f\mathrm dz$. Since the integrals over $f_n$ along triangles are all $0$ according to Cauchy-Goursat, the sequence converges to $0$, so we have $$\oint_T f(z)\mathrm dz=0$$ as desired.

  • $\begingroup$ Thank you so much. "continuity and Integrality preserved under uniform convergence" was one thing I wasn't able to think of. This is kind of dumb question, but do we need to have a uniform convergence when $\int f_ndz\to\int fdz$? $\endgroup$
    – able20
    Nov 15, 2020 at 17:37
  • $\begingroup$ Of course we need uniform convergence in $a)$ but I want to know why uniform convergence of $f_n$ is crucial here. $\endgroup$
    – able20
    Nov 15, 2020 at 17:39
  • 1
    $\begingroup$ Take the sequence of functions which is $n$ on the interval $(0,1/n)$ and $0$ elsewhere. This sequence converges to $0$ pointwise, but not uniformly. The integral over this sequence is $1$ for all $n$, which does not converge to $0$, which would be the integral of the limit of the sequence. So pointwise convergence is not sufficient. But if convergence is uniform, then $\vert\int f-\int f_n\vert=\vert\int f-f_n\vert\leq L\sup\vert f-f_n\vert$, where $L$ is the length of the contour and the supremum is taken over the contour. This supremum converges to $0$ due to uniform convergence of $f_n$. $\endgroup$ Nov 15, 2020 at 18:50

A comment that got too long that discuses (locally) uniform convergence vs pointwise convergence. One can use Runge theorem to construct a sequence of polynomials $P_n$ that converge to zero pointwise except at the origin, where $P_n(0)=1$; then although the integral of $P_n$ converges on every closed curve to the integral of $f$ as $f(z)=0, z \ne 0, f(0)=1$, $f$ is not continuous at zero hence not holomorphic there; in general with pointwise convergence, Osgood's theorem ($f_n \to f$ pointwise on $U$ and $f_n$ analytic, then here is an open dense subset $U_0$ of $U$, where $f_n \to f$ locally uniform so $f$ is analytic at least on $U_0$ is the best one can get; a good (fairly elementary) discussion of what happens is in the American Math Monthly paper by Beardon and Minda (pdf LINK)


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