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 \(g:{\mathbf{C}}\to{\mathbf{C}}\).
  • On every line segment in \({\mathbf{C}}\), \(f_n \to g\) uniformly.

Show that

  • \(g\) is entire, and
  • \(f_n\to g\) uniformly on every compact subset of \({\mathbf{C}}\).

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 \(\Omega \subset {\mathbf{C}}\) be a connected open subset. Let \(\left\{f_{n}: \Omega \rightarrow {\mathbf{C}}\right\}_{n=1}^{\infty}\) be a sequence of holomorphic functions uniformly bounded on compact subsets of \(\Omega\). Let \(f: \Omega \rightarrow {\mathbf{C}}\) 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.


  • 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*}


  • 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 \(\overline{{\mathbb{D}}}\) 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, \(f(z) \coloneqq\overline{z}\) is continuous on \(\overline{{\mathbb{D}}}\) 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*}

#complex/exercise/completed #complex/qual/work #complex/qual/completed