# extending uniformly convergence from the complex plane minus real line to the entire plane

Let $f_n$ be a sequence of entire holomorphic functions which converges uniformly on every compact subset of $\mathbb{C}\setminus\mathbb{R}$. Suppose those $\{f_n\}$ are controlled by following \begin{equation} |f_n(z)|\leqslant\frac{1}{|\mathrm{Im}(z)|^{1/2}},\quad\forall z\in\mathbb{C}\setminus\mathbb{R}. \end{equation} Then how can I show that those $\{f_n\}$ actually converges uniformly on every compact subset of $\mathbb{C}$?

Additionally, what else conditions can replace the given controlling condition?

## 1 Answer

Hint: $f_n(Re^{it})$ converges in $L^1([0,2\pi])$ for each $R > 0.$ Why? That's where the given estimate comes into play. So we can use Cauchy's integral formula to estimate $|f_n(z)-f_m(z)|$ for $|z|\le R/2$ in terms of the integral of $|f_n-f_m|$ on the circle of radius $R.$

• Ok. Why $f_n(Re^{it})$ converges in $L^1$? – S.Gau at Math Jun 9 '17 at 4:39
• @S.GauatMath It follows from the dominated convergence theorem. – zhw. Jun 9 '17 at 6:20
• You mean 1/sin(t)^{1/2} is integrable on [0,2pi] ? – S.Gau at Math Jun 9 '17 at 6:23
• @S.GauatMath Yes, although you want to put absolute values on that. – zhw. Jun 9 '17 at 6:27