Uniformly convergent sequence of holomorphic function in every compact subset converges to holomorphic function

$$\{f_n\}^\infty_{n=1}$$ 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?

• What do you mean by "don't we just need a continuity"? Nov 15, 2020 at 17:07
• 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? Nov 15, 2020 at 17:09

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.
• 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 $\int f_ndz\to\int fdz$? Nov 15, 2020 at 17:37
• Of course we need uniform convergence in $a)$ but I want to know why uniform convergence of $f_n$ is crucial here. Nov 15, 2020 at 17:39
• 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 $f_n$. 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)