> [!NOTE] Theorem > Let $f_n: \Omega \rightarrow \mathbb{C}$ be a sequence of [[Complex Differentiability|analytic]] functions on an open set $\Omega$. If $f_n$ [[Uniform Convergence of Sequence of Real Functions|converges uniformly]] to $f$, then $f$ is also analytic. Recall that for a function to be analytic at one point we require that the function be differentiable in a neighbourhood of the point, and therefore the assumption on $\Omega$ being open is natural. Being analytic is a local property, and requiring that the uniform convergence holds only on compact sets would suffice. ###### Proof Let $z \in \Omega$. Choose $r>0$ sufficiently small so that $B_r(z) \subset \Omega$. Since $f_n$ is analytic in $\Omega$ we can apply [[Cauchy's Integral Formula]] to obtain $f_n(z)=\frac{1}{2 \pi \mathrm{i}} \int_{\partial B_r(z)} \frac{f_n(w)}{w-z} \mathrm{~d} w$ Taking limits as $n$ goes to infinity, and assuming that we can move the limit inside the integral we would obtain $f(z)=\frac{1}{2 \pi \mathrm{i}} \int_{\partial B_r(z)} \frac{f(w)}{w-z} \mathrm{~d} w$ We have seen before that this implies that $f$ is differentiable (in fact infinitely differentiable) and obtained an expression for its derivative (see [[Extension of Cauchy's Integral Formula]]). So the only thing left is to justify moving the limit inside the integral. Notice that this is really a one dimensional integral and we can apply the results learnt earlier in the year. Taking $\gamma(t)=z+r \mathrm{e}^{\mathrm{i} t}$ for $t \in[0,2 \pi)$, we have $\gamma^{\prime}(t)=\mathrm{i} r \mathrm{e}^{\mathrm{i} t}$ and so $\int_{\partial B_r(z)} \frac{f_n(w)}{w-z} \mathrm{~d} w=\int_0^{2 \pi} \frac{f_n\left(z+r \mathrm{e}^{\mathrm{i} t}\right)}{r \mathrm{e}^{\mathrm{i} t}} \mathrm{i} r \mathrm{e}^{\mathrm{i} t} \mathrm{~d} t=\mathrm{i} \int_0^{2 \pi} f_n\left(z+r \mathrm{e}^{\mathrm{i} t}\right) \mathrm{d} t \tag{1}$ For fixed $z$, as a function of $t$ we have that $f_n\left(z+r \mathrm{e}^{\mathrm{i} t}\right)$ converges uniformly to $f\left(z+r \mathrm{e}^{\mathrm{i} t}\right)$ and applying [[Integral of Uniformly Convergent Series of Real Functions equals Series of Their Integrals]] we can move the limit inside the integral, obtaining $\lim _{n \rightarrow \infty} \int_{\partial B_r(z)} \frac{f_n(w)}{w-z} \mathrm{~d} w=\lim _{n \rightarrow \infty} \mathrm{i} \int_0^{2 \pi} f_n\left(z+r \mathrm{e}^{\mathrm{i} t}\right) \mathrm{d} t=\mathrm{i} \int_0^{2 \pi} f\left(z+r \mathrm{e}^{\mathrm{i} t}\right) \mathrm{d} t$ Notice that we have (reading the expression $(1)$ backwards, now for $f$ instead of for $f_n$ ) $\mathrm{i} \int_0^{2 \pi} f\left(z+r \mathrm{e}^{\mathrm{i} t}\right) \mathrm{d} t=\int_{\partial B_r(z)} \frac{f(w)}{w-z} \mathrm{~d} w$ obtaining the result.