I'd like to know if the following exercise is correct. I'm not completely sure about the last point but also I don't know what more I'd say. I really appreciate corrections or any suggestion you can provide.

Let $f, f_1 , f_2 ,\ldots$ be elements of $H(G)$ and show that $f_n\to f$ iff for each closed rectifiable curve $\gamma$ in $G$, $f_n (z) \to f (z)$ uniformly for $z$ in $\{\gamma\}$.

Since $\{\gamma\}$ is compact the sufficiently is clear. Conversely, let $a\in G$ and $r>0$ such that $\overline B(a,r)\subset G$, then there is a number $R>r$ such that $\overline B(a,R)\subset G$. Define $\gamma (t)= a + R\exp (2\pi it)$ (for $0\le t \le 1$) and $\vert z-a \lvert \le r $,

$$\lvert f(z) -f_n(z)\rvert=\frac{1}{2\pi }\bigg \rvert\int_{\gamma}\frac{f(\zeta)-f_n(\zeta)}{\zeta-z}d\zeta\bigg \lvert\le 2\sup\{ \rvert f(\zeta)-f_n(\zeta)\lvert: \zeta \in \{\gamma\} \} $$

Given $\epsilon>0$, $\sup\{ \rvert f(\zeta)- f_n(\zeta)\lvert: \zeta \in \{\gamma\} \}< \epsilon/3$ whenever $n\ge N$ and so

$$\lvert f(z) -f_n(z)\lvert<2\epsilon/3$$

Then, $\sup \{ \lvert f(z)-f_n(z) \lvert: z\in \overline B(a,r),n\ge N\}<\epsilon$. It follows that $f_n$ converges uniformly on $\overline B(a,r)$. Now if $K\subset G$ is an arbitrary compact and $0<r<\operatorname{dist}(\operatorname{fr}(G),K)$, then we can cover by $r$-disk the compact $K$

$$K\subset \bigcup \{B(z_n:r):1\le n\le N \text{ and } z_n\in K\}$$

Since the convergence is uniformly on each disk $B(z_n,r)$, hence the convergence must be uniformly on $K$ as desired.

  • $\begingroup$ By $f_n \mapsto f$ do you mean uniformly on compact subsets of $G$? $\endgroup$ – copper.hat May 21 '15 at 2:39
  • $\begingroup$ @copper.hat Yes exactly (or the metric given a $C(G,\bf{C})$) $\endgroup$ – Jose Antonio May 21 '15 at 2:41
  • $\begingroup$ It would seem that you could establish from the above (using $f_n(z)-f(z)$, instead) that $f_n \to f$ uniformly on $\overline{B}(a,{r \over 2})$, and then use compactness to conclude that $f_n \to f$ uniformly on any compact set? $\endgroup$ – copper.hat May 21 '15 at 3:10
  • $\begingroup$ I think I got it. I made and edition with your suggestions. Thanks @copper.hat $\endgroup$ – Jose Antonio May 21 '15 at 3:50
  • $\begingroup$ Looks good! ${}{}$ $\endgroup$ – copper.hat May 21 '15 at 4:02

$\forall z$ and $r$ s.t. $\overline{B(z,r)}\subseteq G, d(\overline{B(z,r)},\mathbb{C}\setminus G)>0$ since if that's not the case, i.e. $d(\overline{B(z,r)},\mathbb{C}\setminus G)=0$, (since $\overline{B(z,r)}$ and $\mathbb{C}\setminus G$ are both closed,) $\exists a\in \overline{B(z,r)}\cap \mathbb{C}\setminus G$, contradicting the fact that $\overline{B(z,r)}\subseteq G$.

$\epsilon:=d(\overline{B(z,r)},\mathbb{C}\setminus G)$, let $R$ be s.t. $r<R<r+\epsilon$. Obviously $\overline{B(z,R)}\subseteq G$.

$\forall z\in \overline B(z,r)$, Cauchy's theorem tells us $$\big|f(z)-f_n(z)\big|= \dfrac{1}{2\pi} \bigg|\oint_{|\xi-z|=R}\dfrac{f(\xi)-f_n(\xi)}{z-\xi}d\xi\bigg|\leqslant\dfrac{R}{R-r}||f-f_n||_{|\xi-a|=R}$$

Notice that since $||f-f_n||_{|\xi-a|=R}$ tends to $0$,previous equation dictates that $f_n\rightrightarrows f$ (convergence uniformly) for any $\overline{B(z,r)}\subseteq G$.

Now claim that $f_n\rightrightarrows f$ for any compact $K\subseteq G$.

$$\bigcup_{z\in K} B(z,d(z,\mathbb{C}\setminus G))$$ is an open cover of $K$ that's in $G$, which can be reduced to a finite cover, say $$\bigcup_{i=1}^n B(z_i,r_i)$$...

(what's coming next is obvious. Hint: $\forall\epsilon, \exists N_i$ for each $\overline{z_i, r_i)}$,take $N:=\max\{N_i\}<\infty$


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