# Uniform Convergence

(Lecture 18 of Mathematical Methods II.)

The convergence of a series of functions on a domain~$\mathcal{D}$ can be understood straightforwardly as the convergence of a series of complex numbers:

\begin{equation} \tag{1} (\forall z\in\mathcal{D})(\forall\epsilon>0)(\exists N\in\mathbf{N})(n>N)\Rightarrow(|S_n(z)-f(z)|<\epsilon)\,. \end{equation}

This is aptly called pointwise convergence, as convergence is considered at each point~$z$ independently of what happens at other points, even if they do converge as well.

If we regard convergence of the function as a whole, however, a different (stronger) notion of convergences arises, uniform convergence.

Essentially, uniform convergence means that the function~$S_n$ as a whole converges towards~$f$, that is to say, the particularity of a given point~$z$ does not enter the criterion of convergence. Seen in still another way, this means that the "speed of convergence" of various points is similar. Formally, this merely replaces the statement over~$z$:

\begin{equation} \tag{2} (\forall\epsilon>0)(\exists N\in\mathbf{N})(\forall z\in\mathcal{D}(n>N)\Rightarrow(|S_n(z)-f(z)|<\epsilon)\,. \end{equation}

The concept is important because several properties of the functions~$S_n$, such as continuity and Riemann integrability, are transferred to the limit~$f$ if the convergence is uniform.

It is often convenient to understand a statement to understand what its opposite means. A function is therefore not converging uniformly if there exists $\epsilon$ such that for all~$N$, there are some~$n$ and some~$z_0$ such that $|S_n(z_0)-f(z_0)|>\epsilon$.

Examples of non-converging functions are the geometric series or $z^n$ on the open disk.

One can check the uniform convergence of a series using the Weierstrass $M$-test: If the series $\sum_k u_k(z)$ is such that:

\begin{equation} \tag{3} |u_k(z)|\le M_k \end{equation}

for all~$k\in\mathbf{N}$ and all~$z\in\mathcal{D}$, then if $\sum_k M_k$ converges, then $\sum_k u_k(z)$ converges uniformly.

This is proved first by showing that the series $S_n(z)=\sum_{k=0}^{n-1}u_k(z)$ converges (by showing that it is a Cauchy sequence) and then that it satisfies the definition of uniform convergence.

Proof: first, given the convergence of $\sum_k M_k$, we have:

\begin{equation} \tag{4} (\forall\epsilon>0)(\exists N\in\mathbf{N})(n>N)\Rightarrow(\sum_{k=n}^\infty M_k<\epsilon) \end{equation}

which follows directly from the fact the series converges, say to~$M$, i.e., that $|\sum M_{k=0}^{n-1}-M|<\epsilon$; but the term in the absolute value is precisely $|\sum_{k={n}}^\infty M_k|$. Now, let us compute $|S_n(z)-S_m(z)|$ for~$n>m$. This is $|\sum_{k=m}^{n-1}u_k(z)|\le\sum_{k=m}^{n-1}M_k$ which can be made arbitrarily small for large enough~$m$, so for any~$\epsilon>0$, there exist~$N$ such that~$n,m>N$ implies $|S_n(z)-S_m(z)|<\epsilon$, showing that it is a Cauchy sequence, and thus is convergent since the complex space is complete. Let us call $f(z)=\lim_{n\rightarrow\infty}S_n(z)$. We now show that the convergence of $S_n$ to~$f$ is uniform. We compute $|S_n(z)-f(z)|=|\sum_{k=n}^\infty u_k(z)|\le\sum_{k=n}^\infty M_k$. By Eq.~((4)), for any~$\epsilon>0$, there exists~$N$ such that $|S_n(z)-f(z)|<\epsilon$, QED.

If a power series~$\sum_{k=0}^\infty c_k(z-z_0)^k$ has radius of convergence~$\rho>0$, then for every~$0<r<\rho$, the series converges uniformly on the closed disk~$\{z\,:\,|z-z_0|\le r\}$.

This is a direct consequence of the $M$-test and the fact that a power series converges absolutely, since $|\sum_{k=0}^\infty c_k(z-z_0)^k|\le\sum_k|c_k|r^k$ with $r$ a constant.

We are now in a position to illustrate the importance of the concept of uniform convergence:

Let~$S_k$ a series of functions that are continuous on a domain~$\mathcal{D}$ that contains the contour~$\mathcal{C}$. If $S_k$ converges uniformly to~$f$ on $\mathcal{D}$, then:

1. $f$ is continuous on~$\mathcal{D}$.
2. $\lim_{k\rightarrow\infty}\int_\mathcal{C}S_k(z)\,dz=\int_\mathcal{C}f(z)\,dz$

that is to say, we can interchange the integral sign and the limit. As a corollary, applying this theorem to the case of a power series $\sum c_k(z-z_0)^k$ that converges uniformly to~$f$, we find

\begin{equation} \tag{5} \int f(z)dz=\sum\int c_k(z-z_0)^kdz=\sum\frac{c_k}{k+1}(z-z_0)^{k+1}\,. \end{equation}

We will prove the statement on continuity. We have to show that $\lim_{z\rightarrow z_0}f(z)=f(z_0)$ is true for~$f$ if it is the limit of a uniformly convergent series $S_k(z)$. That is to say, for all $\epsilon>0$, we must show that there exists $\delta$ such that $|z-z_0|<\delta$ implies $|f(z)-f(z_0)|<\epsilon$.

By decomposition and the triangle inequality:

\begin{multline} |f(z)-f(z_0)|=|f(z)-S_k(z)+S_k(z)-S_k(z_0)+S_k(z_0)-f(z_0)|\\ \le|f(z)-S_k(z)|+|S_k(z)-S_k(z_0)|+|S_k(z_0)-f(z_0)|\,. \end{multline}

which can all be made arbitrarily small, by uniform convergence for the first term, by continuity of $S_k$ at $z$ and $z_0$ for the second and third terms. The proof of integration relies on the ML inequality and is left as an exercise.