# Montel and Function Convergence

## Convergence of holomorphic functions on line segments #complex/exercise/completed

Suppose $$\left\{{f_n}\right\}_{n\in {\mathbb{N}}}$$ is a sequence of entire functions where

• $$f_n \to g$$ pointwise for some $$g:{\mathbb{C}}\to{\mathbb{C}}$$.
• On every line segment in $${\mathbb{C}}$$, $$f_n \to g$$ uniformly.

Show that

• $$g$$ is entire, and
• $$f_n\to g$$ uniformly on every compact subset of $${\mathbb{C}}$$.

Note that $$g$$ is entire by Morera’s theorem, since $$0 = \int_T f_n \to \int_T g$$ by uniform convergence and the $$f_n$$ are holomorphic. By Cauchy’s theorem, up to a constant we have \begin{align*} f_n(z) = \oint_T {f_n(\xi) \over \xi - z}\,d\xi g(z) = \oint_T {g(\xi) \over \xi - z}\,d\xi ,\end{align*} Thus fixing $$K$$ and $${\varepsilon}$$, for any $$T \subseteq K$$ containing $$z$$,  \begin{align*} {\left\lvert {f_n(z) - g(z)} \right\rvert} &= {\left\lvert { \oint_T {f_n(\xi) \over \xi - z},d\xi

• \oint_T {g(\xi) \over \xi - z},d\xi } \right\rvert} \ &\leq \oint_T {{\left\lvert {f_n(\xi) - g(\xi)} \right\rvert} \over \xi - z },d\xi\ &\leq \oint_T { \sup_{\xi\in T}{\left\lvert {f_n(\xi) - g(\xi)} \right\rvert} \over \xi - z },d\xi\ &\leq \oint_T { {\varepsilon}\over \xi - z },d\xi\ &= {\varepsilon}C \to 0 ,\end{align*} {=html} where $$n = n({\varepsilon}, T)$$ can be chosen to produce this $${\varepsilon}$$ using that $$f_n\to g$$ uniformly on $$T$$. Taking a sup over the $$z$$ enclosed by $$T$$ on the LHS yields a bound on the open region enclosed by $$T$$. Taking a union of all such $$T$$ in $$K$$ yields an open cover of $$K$$, which by compactness has a finite subcover. This yields a finite collection $$\left\{{n = n({\varepsilon}, T_k)}\right\}_{k\leq N}$$, and taking the maximum such $$n$$ yields a uniform bound for all of $$K$$.

## Tie’s Extra Questions: Spring 2015 #complex/qual/work

Assume $$f_n \in H(\Omega)$$ is a sequence of holomorphic functions on the region $$\Omega$$ that are uniformly bounded on compact subsets and $$f \in H(\Omega)$$ is such that the set $$\displaystyle \{z \in \Omega: \lim_{n \rightarrow \infty} f_n(z) = f(z) \}$$ has a limit point in $$\Omega$$. Show that $$f_n$$ converges to $$f$$ uniformly on compact subsets of $$\Omega$$.

## Spring 2019.7 #complex/qual/completed

Let $$\Omega \subset {\mathbb{C}}$$ be a connected open subset. Let $$\left\{f_{n}: \Omega \rightarrow {\mathbb{C}}\right\}_{n=1}^{\infty}$$ be a sequence of holomorphic functions uniformly bounded on compact subsets of $$\Omega$$. Let $$f: \Omega \rightarrow {\mathbb{C}}$$ be a holomorphic function such that the set \begin{align*} \left\{z \in \Omega \mathrel{\Big|}\lim _{n \rightarrow \infty} f_{n}(z)=f(z)\right\} \end{align*} has a limit point in $$\Omega$$. Show that $$f_{n}$$ converges to $$f$$ uniformly on compact subsets of $$\Omega$$.

Write $$g(z) \coloneqq\lim_{n\to\infty}f_n(z)$$ for the pointwise limit, then $$g(z) = f(z)$$ on a set with a limit point. By the identity principle, $$g\equiv f$$ on $$\Omega$$, making $$f$$ the pointwise limit of the $$f_n$$.

By Montel, locally uniformly bounded implies normal and locally equicontinuous. So $$\left\{{f_n}\right\}$$ is normal, and thus has a locally uniformly convergent subsequence $$\left\{{f_{n_k}}\right\}$$. Since singletons $$\left\{{z}\right\}$$ are compact, $$f_{n_k}(z) \to g(z)$$ pointwise, and by uniqueness of limits, $$\lim_{k\to\infty } f_{n_k} = g = f$$ on any compact $$K \subseteq \Omega$$.

It remains to show that the original sequence $$\left\{{f_n}\right\}$$ converges locally uniformly to $$f$$, not just the subsequence. Suppose not, then there exists a compact $$K \subseteq \Omega$$ and $${\varepsilon}>0$$ so that $${\left\lVert {f_n - f} \right\rVert}_{K, \infty} > {\varepsilon}$$ for infinitely many $$n$$. This produces a subsequence $$\left\{{f_{n_j}}\right\}$$ with $${\left\lVert {f_{n_j} - f} \right\rVert} > {\varepsilon}$$ for all $$j$$. However, since $${\mathcal{F}}$$ was normal, every subsequence has a locally uniformly convergent subsequence, so this has a further subsequence $$f_{n_{j'}}$$ uniformly converging to $$f$$, a contradiction.

# Function Convergence

## Fall 2021.4 #complex/qual/completed

Prove that the sequence $$\left(1+\frac{z}{n}\right)^{n}$$ converges uniformly to $$e^{z}$$ on compact subsets of $$\mathbb{C}$$.

Hint: $$e^{n \log w_{n}}=w_{n}^{n}$$ and $$e^{z}$$ is uniform continuous on compact subsets of $$\mathbb{C}$$.

Let $$K$$ be compact, where $$z\in K\implies {\left\lvert {z} \right\rvert} \leq R$$ for some constant $$R$$. For the remainder of the problem, we only work in $$K$$.

$$f_n(z) \coloneqq n\log(1 + {z\over n}) \to z$$ uniformly.

$$f_n$$ are uniformly bounded on $$K$$.

$$e^z$$ is uniformly continuous on $$K$$.

If $$g_n\to g$$ uniformly and $$F$$ is uniformly continuous, then $$F \circ g_n \to F\circ g$$ uniformly.

Why these claims imply the result:

If $$f_n(z)\to z$$ uniformly, both are uniformly bounded, and $$e^z$$ is uniformly continuous, then $$e^{f(z)}\to e^z$$ uniformly.

We’ll first show that for $$w$$ in a neighborhood of zero avoiding 1, there exists a constant $$C$$ such that \begin{align*} {\left\lvert { 1 - {\log(1+w) \over w} } \right\rvert} \leq C{\left\lvert {w} \right\rvert} .\end{align*} This follows from estimating the series expansion about $$w=0$$: \begin{align*} {\left\lvert { 1 - {\log(1+w) \over w} } \right\rvert} &= {\left\lvert {w^{-1}\sum_{k\geq 1} { (-w)^k \over k} } \right\rvert} \\ &= {\left\lvert {\sum_{k\geq 2} {(-w)^{k-1} \over k} } \right\rvert} \\ &\leq {\sum_{k\geq 2} {{\left\lvert {w} \right\rvert}^{k-1} \over k} } \\ &= {\sum_{k\geq 1} {{\left\lvert {w} \right\rvert}^{k} \over k+1} } \\ &\leq {\sum_{k\geq 1} {{\left\lvert {w} \right\rvert}^{k} \over 2} } \\ &= {1\over 2}\qty{{1\over 1 - {\left\lvert {w} \right\rvert}} - 1 } \\ &= {1\over 2}{\left\lvert {2} \right\rvert} \qty{1\over 1 - {\left\lvert {w} \right\rvert}} \\ &\leq C {\left\lvert {w} \right\rvert} ,\end{align*} using that $${1\over 1-x}$$ is bounded in compact sets avoiding $$x=1$$.

We can now apply the $$M{\hbox{-}}$$test: \begin{align*} {\left\lvert {n\log\qty{ 1 + {z\over n} } - z } \right\rvert} &= {\left\lvert {z} \right\rvert}\cdot {\left\lvert { {{ \log\qty{1 + {z\over n}} \over {z\over n}} - 1} } \right\rvert} \\ &\leq {\left\lvert {z} \right\rvert} \cdot C{\left\lvert {z\over n} \right\rvert} \\ &\leq M\cdot C\qty{M\over n} \\ &= {CM^2 \over n}\\ &\overset{n\to\infty}\longrightarrow 0 .\end{align*}

## Spring 2021.6, Spring 2015, Extras #complex/qual/completed

Let $$\left\{{f_n}\right\}_{n=1}^\infty$$ is a sequence of holomorphic functions on $${\mathbb{D}}$$ and $$f$$ is also holomorphic on $${\mathbb{D}}$$. Show that the following are equivalent:

• $$f_n\to f$$ uniformly on compact subsets of $${\mathbb{D}}$$.
• For $$0 < r < 1$$, \begin{align*} \int_{{\left\lvert {z} \right\rvert} = r} {\left\lvert {f_n(z) - f(z)} \right\rvert} {\left\lvert {dz} \right\rvert} \overset{n\to\infty}\longrightarrow 0 .\end{align*}

Note: $${\left\lvert {\,dz} \right\rvert} = {\left\lvert {\gamma'(t)} \right\rvert}\,dt$$ for $$\gamma$$ a parameterization of the curve.

$$\implies$$:

• Fix $$r \in (0, 1)$$ and let $$\gamma = \left\{{{\left\lvert {z} \right\rvert} = r}\right\}$$. This is compact, so $$f_n\to f$$ uniformly on $$\gamma$$: \begin{align*} \int_\gamma {\left\lvert {f_n(z) - f(z) } \right\rvert} \,dz &\leq\int_\gamma \sup_{w\in \gamma } {\left\lvert {f_n(w) - f(w) } \right\rvert} \,dz\\ &\leq\int_\gamma {\left\lVert {f_n(w) - f(w) } \right\rVert}_{\infty} \,dz\\ &= {\left\lVert {f_n(w) - f(w) } \right\rVert}_{\infty} \int_\gamma \,dz\\ &= {\left\lVert {f_n(w) - f(w) } \right\rVert}_{\infty} \mathop{\mathrm{length}}(\gamma) \\ &\overset{n\to\infty}\longrightarrow 0 .\end{align*}

$$\impliedby$$:

• Let $$K$$ be compact, then choose $$\gamma$$ enclosing but not intersecting $$K$$.

• Since $$\gamma, K$$ are disjoint compact sets, define $$M \coloneqq\inf \left\{{{\left\lvert {z-\xi} \right\rvert} {~\mathrel{\Big\vert}~}z\in K, \xi\in \gamma}\right\}$$, the $$0<M<\infty$$.

• Apply Cauchy’s formula to the function $$F_n(z) \coloneqq f_n(z) - f(z)$$, where we want to show $${\left\lvert {F_n(z)} \right\rvert} < {\varepsilon}$$: \begin{align*} F_n(z) &= {1\over 2\pi i} \int_\gamma { F_n(\xi) \over z-\xi} \,d\xi\\ \implies {\left\lvert {f_n(z) - f(z) } \right\rvert} &\leq {1\over 2\pi }\int_\gamma {\left\lvert {f_n(\xi) - f(\xi) \over z-\xi} \right\rvert} \,d\xi\\ &\leq {1\over 2\pi} \int_\gamma {{\left\lvert { f_n(\xi) - f(\xi) } \right\rvert} \over M} \,d\xi\\ &\leq {1\over 2\pi M} \int_\gamma {{\left\lvert { f_n(\xi) - f(\xi) } \right\rvert} } {\left\lvert {\,d\xi} \right\rvert} \\ ,\end{align*} where by hypothesis we can bound this integral by an $${\varepsilon}$$. So given $${\varepsilon}$$, choose $$n$$ large enough to bound the integral as above by some $${\varepsilon}$$ depending only on $$n$$ and not on $$z$$. Taking $$\sup$$ of both sides yields $${\left\lVert {f_n - f} \right\rVert}_{\infty, K} \leq {{\varepsilon}\over 2\pi M}$$, so $$f_n\to f$$ uniformly on $$K$$.

## Spring 2020 HW 2, SS 2.6.10 #complex/qual/completed

Can every continuous function on $$\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{D}}\mkern-1.5mu}\mkern 1.5mu$$ be uniformly approximated by polynomials in the variable $$z$$?

Hint: compare to Weierstrass for the real interval.

No: polynomials are holomorphic and the uniform limit of holomorphic functions is holomorphic. However, $$f(z) \coloneqq\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu$$ is continuous on $$\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{D}}\mkern-1.5mu}\mkern 1.5mu$$ but not holomorphic, so can not be uniformly approximated by any sequence of polynomials.

## Spring 2020 HW 2.5 #complex/qual/completed

Assume $$f$$ is continuous in the region $$\left\{{x+iy {~\mathrel{\Big\vert}~}x\geq x_0, ~ 0\leq y \leq b}\right\}$$, and the following limit exists independent of $$y$$: \begin{align*} \lim_{x\to +\infty}f(x+iy) = A .\end{align*}

Show that if $$\gamma_x \coloneqq\left\{{z = x+it {~\mathrel{\Big\vert}~}0 \leq t \leq b}\right\}$$, then \begin{align*} \lim_{x\to +\infty} \int_{\gamma_x} f(z) \,dz = iAb .\end{align*}

The key insight: \begin{align*} \int_\gamma A \,dz &= \int_0^b A \cdot i \,dt&& z=x+it,\, \,dz= i\,dt\\ &=iA \int_0^b \,dt\\ &= iAb .\end{align*}

So now estimate the difference: \begin{align*} {\left\lvert { \int_{\gamma} f(z) \,dz- iAb } \right\rvert} &= {\left\lvert { \int_\gamma f(z) \,dz- \int_\gamma A \,dz} \right\rvert} \\ &= {\left\lvert { \int_\gamma \qty{ f(z) - A } \,dz} \right\rvert} \\ &\leq\int_\gamma {\left\lvert { f(z) - A } \right\rvert} \,dz\\ &\leq \sup_{z = x+iy\in \gamma} {\left\lvert {f(x+iy) - A} \right\rvert} \cdot \mathop{\mathrm{length}}(\gamma_x) \\ &\overset{x\to \infty}\longrightarrow 0 ,\end{align*} using that $$\mathop{\mathrm{length}}(\gamma_x) = b$$ is constant.

## Limiting curve variant #complex/qual/completed

Let $$0\leq \alpha \leq 2\pi$$ be a fixed angle. Suppose $$f$$ is continuous on the region $$\Omega = \left\{{{\left\lvert {z} \right\rvert} \geq R, \operatorname{Arg}(z) \in [0, \alpha]}\right\}$$ and $$\lim_{z\to \infty} zf(z) = A$$. Show that \begin{align*} \lim_{z\to \infty} \int_{\gamma_R} f(z) \,dz= iA\alpha ,\end{align*} where $$\gamma_R \coloneqq\left\{{ {\left\lvert {z} \right\rvert} = R, \operatorname{Arg}(z) \in [0, \alpha]}\right\}$$ is an arc.

Key observation: \begin{align*} iA\alpha = \int_\gamma {A\over z}\,dz .\end{align*} Why this is true: \begin{align*} \int_\gamma {A\over z}\,dz= \int_0^\alpha {1\over Re^{it}} iRe^{it}dt = \int_0^\alpha iA \,dt= iA\alpha .\end{align*}

Now estimate the difference:

\begin{align*} {\left\lvert { \int_\gamma f(z) \,dz- iA\alpha } \right\rvert} &= {\left\lvert { \int_\gamma f(z) \,dz- \int_\gamma {A\over z} \,dz} \right\rvert}\\ &= {\left\lvert {\int_\gamma f(z) - {A\over z} \,dz} \right\rvert} \\ &= {\left\lvert {\int_\gamma{zf(z) - A \over z} \,dz} \right\rvert} \\ &\leq \int_\gamma {\left\lvert {zf(z) - A \over z} \right\rvert} \,dz\\ &= \int_\gamma { {\left\lvert {zf(z) - A} \right\rvert} \over R} \,dz\\ &\leq {1\over R } \int_\gamma {\left\lVert {zf(z) - A} \right\rVert}_{\infty, \gamma} \,dz\\ &= {{\varepsilon}\over R}\cdot \mathop{\mathrm{length}}(\gamma) \\ &= {{\varepsilon}\over R} \cdot R\alpha \\ &= {\varepsilon}\alpha \\ &\overset{R\to\infty}\longrightarrow 0 .\end{align*}