# Suppose that each function in a sequence $\{f_n\}$ is continuous in an open set $U$ and that the sequence converges normally in $U$ to the limit…

I am reading the book of Palka, An Introduction to Complex Function Theory, and the author on page 248 states two theorems, which does not prove and I would like to know what the demonstration is. The topics are the following:

Theorem $1.4.$ Suppose that each function in a sequence $\{f_n\}$ is continuous in an open set $U$ and that the sequence converges normally in $U$ to the limit function $f$. Then $f$ is continuous in $U$. Furthermore $\int_{\gamma}f(z)dz=\lim_{n\rightarrow \infty }\int_{\gamma}f_n(z)dz$ for every piecewise smooth path $\gamma$ in $U$.

Theorem $1.5.$ A sequence $\{f_n\}$ of functions defined in on open set $U$ converges normally in $U$ if and only if it is a uniform Cauchy sequence on each closed disk that is contained in $U$.

I do not understand why Palka does not show these results, could someone explain to me please, I think that to solve this one can use one of the following results:

• How is normal convergence defined? – amsmath Oct 30 '17 at 22:16
• @amsmath $\{f_n\}$ converges normally i $U$ to the limit function $f$ if $\{f_n\}$ is pointwise convergent to $f$ in $U$ and if, in addition, the convergence is uniform on each compact set in $U$. – user424241 Oct 30 '17 at 22:24
• What is the conditions of $A$ indicated in Theorem 1.1, if no any restrictions to A, why not just put $U=A$ in Theorem 1.4 if you are allowed to apply Theorem 1.1? – user284331 Oct 31 '17 at 0:04
• @user284331 yes, $A\subseteq \mathbb{C}$ – user424241 Oct 31 '17 at 0:29
• So the Theorem 1.5 is just a matter of Theorem 1.2, so do you accept, or have you seen the proof of Theorem 1.2? If so, then nothing needs to be proved. – user284331 Oct 31 '17 at 0:31

Theorem 1.4: Let $z_0\in U$ and consider a $r>0$ such that the closed disk $\overline{D(z_0,r)}\subset U$. Then $\overline{D(z_0,r)}$ is a compact set and therefore the restriction of $(f_n)_{n\in\mathbb N}$ converges uniformly to $f$ on $\overline{D(z_0,r)}$. Therefore, by theorem 1.1, $f$ is continuous at $z_0$.
Besides, since the image $\Gamma$ of $\gamma$ is compact and $(f_n)_{n\in\mathbb N}$ converges uniformly to $f$ on $\Gamma$, theorem 1.1 tells us that$$\lim_{n\to\infty}\int_\gamma f_n(z)\,\mathrm dz=\int_\gamma f(z)\,\mathrm dz.$$
On the other hand, if a sequence $(f_n)_{n\in\mathbb N}$ of functions is a uniformly Cauchy sequence on every closed disk and if $K$ is a arbitrary compact subset of U, one can use the compacity of $K$ to find a finite set $z_1,z_2,\ldots,z_n$ of elements of $U$ and a finite set $r_1,r_2,\ldots,r_n$ of numbers greater than $0$ such that each closed disk $\overline{D(z_k,r_k)}$ is a subset of $U$ and that$$K\subset\bigcup_{k=1}^n\overline{D(z_k,r_k)}.$$Since $(f_n)_{n\in\mathbb N}$ is a uniformly Cauchy sequence on each such disk, it is a uniformly Cauchy sequence on $K$.