“Locally” means “on all compact subsets”.
Equicontinuity
A family of functions \(f_n\) is equicontinuous iff for every \({\varepsilon}\) there exists a \(\delta = \delta({\varepsilon})\) (not depending on \(n\) or \(f_n\)) such that \begin{align*} {\left\lvert {x-y} \right\rvert}<{\varepsilon}\implies {\left\lvert {f_n(x) - f_n(y)} \right\rvert} < {\varepsilon} && \forall n .\end{align*}
Equicontinuity is uniform continuity which is also uniform across the family.
For \(X\) compact Hausdorff, consider the Banach space \(C(X; {\mathbb{R}})\) equipped with the uniform norm \begin{align*} {\left\lVert {f} \right\rVert}_{\infty, X} \coloneqq\sup_{x\in X} {\left\lvert {f(x)} \right\rvert} .\end{align*}
A subset \(A \subseteq C(X; {\mathbb{R}})\) is compact iff \(A\) is closed, uniformly bounded, and equicontinuous.
For \(X = [a,b]\subseteq {\mathbb{R}}\), if a sequence is uniformly bounded and uniformly equicontinuous, then there exists a uniformly convergent subsequence.
If \(A\) is a sequence of continuous functions, it contains a subsequence converging uniformly and the limit is continuous. The proof is an \({\varepsilon}/3\) argument.
Normal Families
A family of functions \({\mathcal{F}}\coloneqq\left\{{f_j}\right\}_{j\in J}\) is normal iff every sequence \(\left\{{f_k}\right\}\) has a subsequence that converges locally uniformly, i.e. \(\left\{{f_{k_i}}\right\}\) converges uniformly on every compact subset.
Suppose \({\mathcal{F}}\) is locally uniformly bounded. Then \({\mathcal{F}}\) is locally equicontinuous and a normal family.
A function \(f\in \mathop{\mathrm{Hol}}(U; {\mathbb{C}})\) is called univalent if \(f\) is injective.
If \(f: \Omega \to \Omega'\) is a univalent surjection, \(f\) is invertible on \(\Omega\) and \(f^{-1}\) is holomorphic. Compare to real functions: \(f(x) = x^3\) is injective on \((-c, c)\) for any \(c\) but \(f'(0) = 0\) and \(f^{-1}(x) \coloneqq x^{1/3}\) is not differentiable at zero.
A family \({\mathcal{F}}= \left\{{f_j}\right\}_{j\in J}\) of holomorphic functions on \(\Omega\) is normal if every sequence of functions from \({\mathcal{F}}\) has a locally uniformly convergent subsequence (so they converge on every compact subset of \(\Omega\)).
A family \({\mathcal{F}}\) of holomorphic functions is uniformly bounded on compact subsets of \(\Omega\) iff for each compact \(K \subseteq \Omega\) if \begin{align*} \exists M>0 \text{ such that } {\left\lvert {f(z)} \right\rvert} < M \qquad \forall z\in K,\,\forall f\in {\mathcal{F}} .\end{align*}
A family \({\mathcal{F}}\) of holomorphic functions is equicontinuous on \(K\) if \begin{align*} \forall {\varepsilon}>0,\, \exists \delta = \delta({\varepsilon}) \text{ such that } z,w\in K,\, {\left\lvert {z-w} \right\rvert}< \delta \implies {\left\lvert {f(z) - f(w)} \right\rvert} < {\varepsilon}\quad \forall f\in {\mathcal{F}} .\end{align*}
Equicontinuity is uniform continuity, where the uniformity extends across all \(f\in {\mathcal{F}}\). The following is a stark difference between holomorphic and smooth functions, and is used in the Riemann mapping theorem:
To negate equicontinuity, show that there exists \({\varepsilon}>0\) and a bad tuple \((x, y, f\in {\mathcal{F}})\) such that for any \(\delta\), we can arrange \({\left\lvert {x-y} \right\rvert} < \delta\) to be small but \({\left\lvert {f(x) - f(y)} \right\rvert} > {\varepsilon}\) is large. This produces sequences \(x_k, y_k, f_k\) with \({\left\lvert {x_k-y_k} \right\rvert}\to 0\) but \({\left\lvert {f_k(x_k) - f_k(y_k)} \right\rvert} > {\varepsilon}\).
Show that if \(\left\{{f_k}\right\}\) is a differentiable sequence with \({\left\lvert {f_k'(x)} \right\rvert} \leq M\) uniformly bounded, then \(\left\{{f_k}\right\}\) is equicontinuous.
Hint: MVT.
Exhibit a sequence of functions that is not equicontinuous.
The family \(f_k(x) = x^k\) is not equicontinuous, since fixing \(x_0 \in (0, 1)\) we have \begin{align*} {\left\lvert {f_k(x_0) - f_k(1)} \right\rvert} \overset{k\to \infty}\longrightarrow 1 > {\varepsilon} .\end{align*}
Produce a sequence that is uniformly bounded but not equicontinuous.
\(f_k(x) = (\sin k x)\) is uniformly bounded but not equicontinuous on \((0, 1)\) since it has no convergent subsequence on any compact subset.
Montel’s Theorem
If \({\mathcal{F}}\) is a family of locally uniformly bounded holomorphic functions on \(\Omega\), then
- \({\mathcal{F}}\) is a normal family by Arzela-Ascoli, and
- \(f\) is locally equicontinuous (so equicontinuous on every compact subset).
Equivalently: a family \({\mathcal{F}}\) of meromorphic functions on a domain \(\Omega\) that omits three values is normal.
Locally uniformly bounded families are normal. For bounded sequences of holomorphic functions, pointwise convergence is the same as uniform convergence on bounded sets.
This says that a sequence of holomorphic functions avoiding the exterior of a disc contains a locally uniformly convergent subsequence. In particular, the limit is holomorphic.
Moreover, if \(f_n\to f\) pointwise where \(f\) fails continuity or differentiability at a single point, then \(\left\{{f_n}\right\}\) can not be uniformly bounded on all compact subsets.
Exercise
Prove the following: if \(\left\{{f_n}\right\}\) is equicontinuous on \(K\) a compact set and \(f_n\to f\) pointwise, then \(f_n\to f\) uniformly.
Fix \({\varepsilon}\), it suffices to find an \(n= n({\varepsilon})\) to bound \({\left\lVert {f_n - f} \right\rVert}_{\infty, K } < {\varepsilon}\). A standard \({\varepsilon}/3\) argument works: write \begin{align*} {\left\lvert {f_n(x) - f(x) } \right\rvert} \leq {\left\lvert {f_n(x) - f_n(y)} \right\rvert} + {\left\lvert {f_n(y) - f(y)} \right\rvert} + {\left\lvert {f(y) - f_n(y)} \right\rvert} .\end{align*}
Use equicontinuity to bound \({\left\lvert {f_n(x) - f_n(y)} \right\rvert}\) for all \(n\geq N_0 = N_0({\varepsilon})\), for all \(x,y\in K\). This takes care of the 1st and 3rd terms.
For the 2nd term, cover \(K\) by \(\delta{\hbox{-}}\)balls and by compactness obtain a finite cover \(B_{\delta}(y_k)\rightrightarrows K\). Then \(x\in B_\delta(y)\) for \(y=y_j\) for some \(j\), and in this ball use pointwise convergence of \(f_n\to f\).
Show that if \({\mathcal{F}}\) is a family of differentiable functions with uniformly bounded derivatives, then \({\mathcal{F}}\) is equicontinuous.
Hint: apply the MVT.
Give an example of a non-equicontinuous family.
Take \(f_k(z) \coloneqq z^k\) on \([0, 1]\) – fix any \(z_0\in [0, 1)\), then \({\left\lvert {f_k(1) - f_k(x_0)} \right\rvert} \overset{k\to\infty}\longrightarrow 1\).