Suppose $\mathscr F \subset H(\Omega$) for some region (i.e. open connected) $\Omega$. ($H(\Omega)$ means the set of all holomorphic function in $\Omega$)

We call $\mathscr F$a normal family if every sequence of members of $\mathscr F$ contains a subsequence which converges uniformly on compact subsets of $\Omega$.

Now, can I ask for an explicit example of a normal family?

Using the first montel theorem, I thought of $\displaystyle \mathscr F = \left\{f_m: f_m(z) = \frac m{m+1}\frac 1z, m \in \mathbb N\right\}$ on $\Omega = \{z: 0 < |z| < 1\}$

The functions $f_m$ are not bounded but they are uniformly bounded on compact subsets of $\Omega$, hence they form a normal family. But it feels like cheating because the $f_m$ are actually converging uniformly on compact subsets to $\displaystyle \frac 1z$!

  • $\begingroup$ The family is not uniformly bounded but rather uniformly bounded on compact subsets of $\Omega.$ $\endgroup$ – zhw. May 17 '15 at 16:04
  • $\begingroup$ @zhw. Of course! That's what I meant, thanks for the correction $\endgroup$ – Ant May 17 '15 at 16:07

Of course, any family of bounded functions would do. If you want an unbounded example, take $$ \mathscr F = \left\{\frac{z}{z-a},\ |a|\ge 1\right\},\qquad \Omega=\{z:|z|<1\} $$

The normality can be verified directly: for any sequence of such functions, either $a_n\to \infty$ (and then the functions converge to $0$), or $\{a_n\}$ has a convergent subsequence. And if $a_n\to a$, then convergence is uniform on compact subsets of $\Omega$ because $$ \left|\frac{z}{z-a_n}-\frac{z}{z-a }\right| = |a_n-a|\frac{|z|}{|z-a_n||z-a|}\le |a_n-a|\frac{|z|}{(1-|z |^2)} $$

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  • $\begingroup$ thanks for the answer! But why if it has a convergent subsequence then the convergence is uniform on compact subsets? $\endgroup$ – Ant May 19 '15 at 15:02
  • $\begingroup$ I think I got it. Thank you! :) $\endgroup$ – Ant May 19 '15 at 16:34

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