Normal Families and Montel's Theorem — Examples

On every compact subset of $\mathbb{D}$, all points satisfy $\mathrm{\mid }z\mathrm{\mid }\le r<1$, so $\mathrm{\mid }{z}^n\mathrm{\mid }\le 1$ there for every $n$. Hence the family is locally bounded on $\mathbb{D}$, and [L1] makes it normal.

On the plane, the closed disc $\bar{D}(2,1\mathrm{/}2)$ lies in $\mathbb{C}$ and every point of it has modulus at least $3\mathrm{/}2$, so $\mathrm{\mid }{z}^n\mathrm{\mid }\ge (3\mathrm{/}2{)}^n$ there. Thus the family is not locally bounded near $2$, and [L1] shows it is not normal on $\mathbb{C}$.

Around each point of $\mathrm{\Omega }$, choose a closed disc still contained in $\mathrm{\Omega }$. The uniform bound $\mathrm{\mid }f\mathrm{\mid }<1$ holds on that disc for every $f\in \mathcal{B}$, so the family is locally bounded.

Applying [L1] to the local boundedness from step 1.1 shows that $\mathcal{B}$ is normal.

Apply [L1] to choose a subsequence converging uniformly on the first compact disc, then a further subsequence converging uniformly on the second compact disc, and continue stage by stage.

The diagonal term at stage $n$ lies in every earlier chosen subsequence, so for each fixed compact stage the diagonal sequence eventually belongs to the corresponding uniformly convergent subsequence. Hence the diagonal sequence converges locally uniformly on the whole disc.

In $\mathbb{D}$, the boundary distance is $1-\mathrm{\mid }z\mathrm{\mid }$, so the canonical condition $\mathrm{dist}(z,\mathrm{\partial }\mathbb{D})\ge 1\mathrm{/}n$ is exactly $\mathrm{\mid }z\mathrm{\mid }\le 1-1\mathrm{/}n$. Thus $K_n=\{\mathrm{\mid }z\mathrm{\mid }\le 1-1\mathrm{/}n\}$ and in particular $K_1=\{0\}$.

On $K_n$, the difference $\mathrm{\mid }f(z)-g(z)\mathrm{\mid }=\mathrm{\mid }z\mathrm{\mid }$ has supremum $1-1\mathrm{/}n$, which is already at most $1$. Substituting into the definition from [L1] gives $d_K(f,g)={\sum }_{n\ge 1}{2}^{-n}(1-1\mathrm{/}n)$.

Choosing a closed disc $\bar{D}(0,r)\subseteq \mathrm{\Omega }$, the point $r\mathrm{/}2$ satisfies $\mathrm{\mid }{f}_n(r\mathrm{/}2)\mathrm{\mid }=nr\mathrm{/}2\to \mathrm{\infty }$. So the family is not locally bounded near the zero point.

Fact [L1] makes local boundedness necessary for normality, so this family is not normal on any domain containing $0$.

If $K\subseteq H$ is compact, then some $\delta >0$ satisfies $\mathrm{Re}z\ge \delta$ on $K$, and [L1] gives $\mathrm{\mid }{f}_n(z)\mathrm{\mid }\ge e^{n\delta }$. Therefore $\chi (f_n(z),\mathrm{\infty })=2\mathrm{/}\sqrt{1+\mathrm{\mid }{f}_n(z){\mathrm{\mid }}^2}\le 2e^{-n\delta }$ on $K$, so $f_n\to \mathrm{\infty }$ chordally locally uniformly on $H$.

Every subsequence has the same chordal limit $\mathrm{\infty }$, so no subsequence can converge locally uniformly to a finite holomorphic function. By [L2], the family is not normal in the finite-valued holomorphic sense.

Fact [L1] gives a normal family whose defining sequence has local uniform limit $0$.

The zero function is not one of the nonzero powers $z^n$, so the limit does not lie in the family. Hence the statement is false.

Local boundedness gives pointwise relative compactness on each compact stage, but it is not itself the equicontinuity hypothesis required by [L1]. Thus Arzelà–Ascoli cannot yet be applied.

Fact [L2] supplies the missing complex-analytic step by converting local boundedness into local equicontinuity. Montel's proof uses that step before applying Arzelà–Ascoli, so Arzelà–Ascoli alone does not prove the theorem.

Fact [L1] provides an explicit holomorphic sequence whose chordal local uniform limit is the constant map $\mathrm{\infty }$.

That witness is exactly contrary to the statement, so the statement is false.