Montel’s theorem gives compactness for locally uniformly bounded families of holomorphic functions. In the proof of the Riemann mapping theorem it produces a locally uniformly convergent subsequence of injective maps, and Hurwitz's theorem shows that the nonconstant limit is injective.
Throughout, “locally uniformly” means “uniformly on every compact subset”.
Equicontinuity
Equicontinuous family (Tag D-TIHRR)
Let \(S\subseteq\CC\) and let \((f_n)_{n\ge1}\) be a sequence of functions \(f_n\colon S\to\CC\). The family \(\{f_n\}\) is equicontinuous on \(S\) if for every \(\varepsilon>0\) there exists \(\delta>0\) such that for all \(x,y\in S\), \[ \abs{x-y}<\delta\implies\abs{f_n(x)-f_n(y)}<\varepsilon\qquad\text{for all } n\ge1. \]
The number \(\delta\) depends only on \(\varepsilon\), not on \(n\), \(x\), or \(y\). [@Rud76] states this definition for families of complex functions on a subset of a metric space.
An equicontinuous family is uniformly continuous with one \(\delta(\varepsilon)\) for every member.
Arzelà--Ascoli theorem (Tag T-6F3GO)
Let \(X\) be a compact Hausdorff space, and give \(C(X;\RR)\) the uniform norm \[ \norm{f}_{\infty,X}\coloneqq\sup_{x\in X}\abs{f(x)}, \] under which it is a Banach space. A subset \(A\subseteq C(X;\RR)\) is compact if and only if \(A\) is closed, uniformly bounded, and equicontinuous at every point of \(X\).
In particular, for \(X=[a,b]\subseteq\RR\), every uniformly bounded and uniformly equicontinuous sequence in \(C([a,b];\RR)\) has a uniformly convergent subsequence.
Applied on each compact set, the Arzelà–Ascoli theorem shows that a pointwise bounded, equicontinuous sequence of continuous functions on a compact metric space has a uniformly convergent subsequence, whose limit is continuous.
(Negation of equicontinuity)
A family \(\mathcal F\) of functions on a metric space \(X\) is not equicontinuous if and only if there exist \(\varepsilon>0\), points \(x_k, y_k\in X\), and \(f_k\in \mathcal F\) with \(\abs{x_k-y_k}\to 0\) and \(\abs{f_k(x_k) - f_k(y_k)} \geq \varepsilon\) for all \(k\).
Normal families
Normal families (Tag D-IJMPJ)
Let \(\Omega\subseteq\CC\) be open. A family \(\mathcal F=\{f_j\}_{j\in J}\) of holomorphic functions on \(\Omega\) is normal if every sequence in \(\mathcal F\) has a subsequence that converges locally uniformly on \(\Omega\).
Families uniformly bounded on compact subsets (Tag D-MBDTR)
Let \(\Omega\subseteq\CC\) be open and let \(\mathcal F\) be a family of holomorphic functions on \(\Omega\). The family \(\mathcal F\) is uniformly bounded on compact subsets of \(\Omega\) if for every compact set \(K\subseteq\Omega\) there exists \(M>0\) such that \[ \abs{f(z)}\le M\qquad\text{for all } z\in K \text{ and all } f\in\mathcal F. \]
(Local boundedness implies equicontinuity)
Let \(\mathcal F\) be a family of holomorphic functions on \(\Omega\) that is uniformly bounded on compact subsets. Then \(\mathcal F\) is equicontinuous on every compact \(K\subseteq\Omega\).
Choose \(r>0\) such that \(K_{2r}\coloneqq\ts{z : \operatorname{dist}(z,K)\leq 2r}\subseteq\Omega\), and let \(M\) bound every \(f\in\mathcal F\) on the compact set \(K_{2r}\). For \(\xi\in K_r\) the closed disc \(\overline{D_r(\xi)}\) lies in \(K_{2r}\), so Cauchy’s estimate gives \(\abs{f'(\xi)}\leq M/r\). For \(z,w\in K\) with \(\abs{z-w}<r\), the segment \([z,w]\) lies in \(D_r(z)\subseteq K_r\), so \(\abs{f(z)-f(w)}\leq (M/r)\abs{z-w}\) for every \(f\in\mathcal F\).
Montel’s theorem follows by applying the Arzelà–Ascoli theorem on each set of an exhaustion of \(\Omega\) by compact sets and taking a diagonal subsequence.
Locally uniformly bounded holomorphic families are equicontinuous and normal (Tag T-MCB7V)
Let \(\Omega\subseteq\CC\) be open and let \(\mcf\) be a family of holomorphic functions on \(\Omega\) that is uniformly bounded on compact subsets of $\Omega$. Then \(\mcf\) is equicontinuous on every compact subset of \(\Omega\), and \(\mcf\) is a normal family.
Univalent functions (Tag D-HL4KE)
Let \(U\subseteq\CC\) be open. A holomorphic function \(f\colon U\to\CC\) is univalent if \(f\) is injective.
(Univalent maps)
If \(f\colon \Omega \to \Omega'\) is holomorphic, injective, and surjective, then \(f'\) does not vanish, and \(\inverseof{f}\) is holomorphic. So a univalent surjection produced by the Riemann mapping argument is a biholomorphism.
(The real analogue fails)
The map \(f(x) = x^3\) is a bijection \(\RR\to\RR\) with \(f'(0)=0\), and \(\inverseof{f}(x) = x^{1/3}\) is not differentiable at \(0\).
A sequence with uniformly bounded derivatives on $[0,1]$ is equicontinuous (Tag E-ISFYB)
Let \(f_k:[0,1]\to\CC\) be differentiable and suppose \[ |f_k'(x)|\le M \] for every \(k\) and every \(x\in[0,1]\), where \(M\) is independent of \(k\). Show that \(\ts{f_k}\) is equicontinuous.
Hint: MVT.
Solution
For all \(x,y\in[0,1]\) and all \(k\), the mean-value estimate gives \[ |f_k(x)-f_k(y)|\le M|x-y|. \] Hence the sequence is uniformly Lipschitz. Given \(\varepsilon>0\), take \(\delta=\varepsilon/M\) when \(M>0\) (and arbitrary \(\delta>0\) when \(M=0\)). Then \(|x-y|<\delta\) implies \[ |f_k(x)-f_k(y)|<\varepsilon \] for every \(k\), so \(\ts{f_k}\) is equicontinuous.
A non-equicontinuous sequence (Tag E-LXY7N)
Exhibit a sequence of functions that is not equicontinuous.
Solution
The family \(f_k(x) = x^k\) is not equicontinuous, since fixing \(x_0 \in (0, 1)\) we have \[ \abs{f_k(x_0) - f_k(1)} \convergesto{k\to \infty} 1 > \eps .\]
Uniformly bounded but not equicontinuous (Tag E-YFL4K)
Produce a sequence that is uniformly bounded but not equicontinuous.
Solution
\(f_k(x) = (\sin k x)\) is uniformly bounded but not equicontinuous on \((0, 1)\) since it has no convergent subsequence on any compact subset.
Montel’s theorem
Montel's theorem (Tag T-4ALS2)
Let \(\Omega\subseteq\CC\) be open and let \(\mcf\) be a family of holomorphic functions on \(\Omega\) that is uniformly bounded on compact subsets of \(\Omega\). Then
\(\mcf\) is equicontinuous on every compact subset of \(\Omega\), and
\(\mcf\) is a normal family.
Let \(K\subseteq\Omega\) be compact and choose \(r>0\) with \(3r<\operatorname{dist}(K,\CC\sm\Omega)\) (any \(r>0\) if \(\Omega=\CC\)). The set \(K'\coloneqq\ts{z\st\operatorname{dist}(z,K)\le2r}\) is a compact subset of \(\Omega\), so there is \(M\) with \(\abs f\le M\) on \(K'\) for all \(f\in\mcf\). For \(z,w\in K\) with \(\abs{z-w}<r\), the circle \(C=\ts{\abs{\zeta-w}=2r}\) lies in \(K'\), and the Cauchy integral formula gives \[ \abs{f(z)-f(w)}=\abs{\frac{1}{2\pi i}\int_Cf(\zeta)\Big(\frac{1}{\zeta-z}-\frac{1}{\zeta-w}\Big)d\zeta}\le\frac{1}{2\pi}\cdot2\pi(2r)\cdot M\cdot\frac{\abs{z-w}}{r\cdot2r}=\frac{M}{r}\abs{z-w}, \] which is independent of \(f\); this is equicontinuity on \(K\). Normality follows from the Arzelà–Ascoli theorem applied on an exhaustion of \(\Omega\) by compact sets, with a diagonal argument.
Montel’s second theorem uses the extended convention for normality in the remark on D-QTJ7T, where a subsequence may instead tend to \(\infty\) uniformly on compact subsets: a family of meromorphic functions on a region \(\Omega\) that omits the same three values of the Riemann sphere is normal in that sense.
Locally uniformly bounded families of holomorphic functions are normal.
(Pointwise convergence)
If a locally uniformly bounded sequence of holomorphic functions on \(\Omega\) converges pointwise on \(\Omega\), then it converges locally uniformly. By Montel’s theorem every subsequence has a further subsequence converging locally uniformly, and pointwise convergence forces all of these limits to equal the pointwise limit. Consequently, if \(f_n \to f\) pointwise on \(\Omega\) and \(f\) is discontinuous or fails to be holomorphic at some point, then \(\ts{f_n}\) is not uniformly bounded on compact subsets of \(\Omega\).
A sequence of holomorphic functions on \(\Omega\) with values in a fixed disc is uniformly bounded, so it has a locally uniformly convergent subsequence, and the limit is holomorphic.
Exercises
Equicontinuity + pointwise convergence implies uniform convergence (Tag E-UJAF4)
Prove the following: if \(\ts{f_n}\) is equicontinuous on \(K\) a compact set and \(f_n\to f\) pointwise, then \(f_n\to f\) uniformly.
Solution
Fix \(\eps\), it suffices to find an \(n= n(\eps)\) to bound \(\norm{f_n - f}_{\infty, K } < \eps\). A standard \(\eps/3\) argument works: write \[ \abs{f_n(x) - f(x) } \leq \abs{f_n(x) - f_n(y)} + \abs{f_n(y) - f(y)} + \abs{f(y) - f_n(y)} .\]
Use equicontinuity to bound \(\abs{f_n(x) - f_n(y)}\) for all \(n\geq N_0 = N_0(\eps)\), for all \(x,y\in K\). This takes care of the 1st and 3rd terms.
For the 2nd term, cover \(K\) by \(\delta\dash\)balls and by compactness obtain a finite cover \(B_{\delta}(y_k)\covers K\). Then \(x\in B_\delta(y)\) for \(y=y_j\) for some \(j\), and in this ball use pointwise convergence of \(f_n\to f\).
Uniformly bounded derivatives imply equicontinuity (Tag E-C5QHZ)
Let \(\mcf\) be a family of differentiable functions \(f:[0,1]\to\CC\) such that \[ |f'(x)|\le M \] for every \(f\in\mcf\) and every \(x\in[0,1]\), for one constant \(M\) independent of \(f\). Show that \(\mcf\) is equicontinuous.
Hint: apply the MVT.
Solution
For every \(f\in\mcf\) and \(x,y\in[0,1]\), the mean-value estimate gives \[ |f(x)-f(y)|\le M|x-y|. \] Thus the whole family is uniformly Lipschitz with the same constant \(M\). Given \(\varepsilon>0\), if \(M>0\) take \(\delta=\varepsilon/M\); if \(M=0\) any \(\delta>0\) works. Then \(|x-y|<\delta\) implies \[ |f(x)-f(y)|<\varepsilon \] for every \(f\in\mcf\). Hence \(\mcf\) is equicontinuous.
The family $\{z^k\}$ on $[0,1]$ is not equicontinuous (Tag E-GFNDF)
Give an example of a non-equicontinuous family.
Solution
Take \(f_k(x)=x^k\) on \([0,1]\) and test equicontinuity at the endpoint \(1\). Set \(\varepsilon=1/2\). Given any \(\delta>0\), choose \(x\in[0,1)\) with \(|1-x|<\delta\). Since \(x<1\), choose \(k\) large enough that \(x^k<1/2\). Then \[ |f_k(1)-f_k(x)|=1-x^k>{1\over2}=\varepsilon. \] Thus no single \(\delta\) works for the whole family at \(1\), so \(\{f_k\}\) is not equicontinuous.