Uniform Convergence

Section 7.1 Uniform Convergence

Complex functions are the key to unlocking many of the mysteries encountered when power series are first introduced in a calculus course. We begin by discussing an important property associated with power series–uniform convergence.

Recall that, for a function $f$ defined on a set $T\text{,}$ the sequence o f functions $\{S_n\}$ converges to $f$ at the point $z_0 \in T\text{,}$ provided $\lim\limits_{n \to \infty}S_n(z_0)=f(z_0)\text{.}$ Thus, for the particular point $z_0\text{,}$ we know that for each $\varepsilon>0\text{,}$ there exists a positive integer $N_{\varepsilon,z_0}$ (depending on both $\varepsilon$ and $z_0$) such that

\begin{equation} \text{ if } n \ge N_{\varepsilon,z_0}, \text{ then } |S_n(z_0) -f(z_0)|\lt \varepsilon\text{.}\tag{7.1.1} \end{equation}

If $S_n(z)$ is the $n$th partial sum of the series $\sum\limits_{k=0}^{\infty}c_k(z-\alpha)^k\text{,}$ Statement (7.1.1) becomes

\begin{equation*} If n \ge N_{\varepsilon,z_0}, \text{ then } \left|\sum_{k=0}^{n-1}c_k(z_0-\alpha)^k-f(z_0)\right| \lt \varepsilon\text{.} \end{equation*}

For a given value of $\varepsilon\text{,}$ the integer $N_{\varepsilon,z_0}$ needed to satisfy Statement (7.1.1) often depends on our choice of $z_0\text{.}$ This is not the case if the sequence $\{S_n\}$ converges uniformly. For a uniformly convergent sequence, it is possible to find an integer $N_{\varepsilon}$ (depending only on $\varepsilon$) that guarantees Statement (7.1.1) no matter what value for $z_0 \in T$ we pick. In other words, if $n$ is large enough, the function $S_n$ is uniformly close to the function $f$ for all $z \in T\text{.}$ Formally, we have the following definition.

Definition 7.1.1. Uniform convergence.

The sequence $\{S_n(z)\}$ converges uniformly to $f(z)$ on the set $T$ if for every $\varepsilon>0\text{,}$ there exists a positive integer $N_{\varepsilon}$ (depending only on $\varepsilon$) such that

\begin{equation} \text{ if } n \ge N_{\varepsilon}, \text{ then } |S_n(z) -f(z)| \lt \varepsilon \text{ for all } z \in T\text{.}\tag{7.1.2} \end{equation}

If $S_n(z)$ is the $n$th partial sum of the series $\sum\limits_{k=0}^{\infty}c_k(z-\alpha)^k\text{,}$ we say that the series $\sum\limits_{k=0}^{\infty}c_k(z-\alpha)^k$ converges uniformly to $f(z)$ on the set $T\text{.}$

Example 7.1.2.

The sequence $\{S_n(z)\}=\{e^z+\frac{{1}}{n}\}$ converges uniformly to the function $f(z) =e^z$ on the entire complex plane because for any $\varepsilon>0\text{,}$ Statement (7.1.2) is satisfied for all $z$ for $n \ge N_{\varepsilon}\text{,}$ where $N_{\varepsilon}$ is any integer greater than $\frac{{1}}{{\varepsilon}}\text{.}$ We leave the details of showing this result as an exercise.

A good example of a sequence of functions that does not converge uniformly is the sequence of partial sums forming the geometric series. Recall that the geometric series has $S_n(z) =\sum\limits_{k=0}^{n-1}z^k$ converging to $f(z) =\frac{1}{1-z}$ for $z \in D_1(0)\text{.}$ Because the real numbers are a subset of the complex numbers, we can show that Statement (7.1.2) is not satisfied by demonstrating that it does not hold when we restrict our attention to the real numbers. In that context, $D_1(0)$ becomes the open interval $(-1,1)\text{,}$ and the inequality $|S_n(z) -f(z)|\lt \varepsilon$ becomes $| S_n(x) -f(x)|\lt \varepsilon\text{,}$ which for real variables is equivalent to the inequality $f(x)-\varepsilon\lt S_n(x)\lt f(x)+\varepsilon\text{.}$ If Statement (7.1.2) were to be satisfied, then given $\varepsilon >0, \, S_n(x)$ would be within an $\varepsilon$-bandwidth of $f(x)$ for all $x$ in the interval $(-1,1)$ provided $n$ were large enough. Figure 7.1.3 illustrates that there is an $\varepsilon$ such that, no matter how large $n$ is, we can find $x_0 \in (-1,1)$ with the property that $S_n(x_0)$ lies outside this bandwidth. In other words, Figure 7.1.3 illustrates the negation of Statement (7.1.2), which in technical terms we state as:

\begin{align} \amp \text{ There exists } \varepsilon>0 \text{ such that, for all positive integers } N,\notag\\ \amp \text{ there is some } n \ge N \text{ and some } z_0 \in T\notag\\ \amp \text{ such that } |S_n(z_0)-f(z_0)| \ge \varepsilon\text{.}\tag{7.1.3} \end{align}

Figure 7.1.3. The geometric series does not converge uniformly on $(-1,1)$

In the exercises, we ask you to use Statement (7.1.3) to show that the partial sums of the geometric series do not converge uniformly to $f(z) =\frac{1}{1-z}$ for $z \in D_1(0)\text{.}$

A useful procedure known as the Weierstrass $M$-test can help determine whether an infinite series is uniformly convergent.

Theorem 7.1.4. Weierstrass $M$-test.

Suppose that the infinite series $\sum\limits_{k=0}^{\infty}u_k(z)$ has the property that for each $k\text{,}$ $|u_k(z)| \le M_k$ for all $z \in T\text{.}$ If $\sum\limits_{k=0}^{\infty}M_k$ converges, then $\sum\limits_{k=0}^{\infty}u_k(z)$ converges uniformly on $T\text{.}$

Proof.

Let $S_n(z) =\sum\limits_{k=0}^{n-1}u_k(z)$ be the $n$th partial sum of the series. Note that, If $n > M\text{,}$ then

\begin{equation*} \big|S_n(z) -S_{m}(z)\big| = \big|u_{m}(z)+u_{m+1}(z) +\cdots +u_{n-1}(z)\big| \le \sum_{k=m}^{n-1}M_k\text{.} \end{equation*}

Because the series $\sum\limits_{k=0}^{\infty}M_k$ converges, we can make the last expression as small as we want to by choosing a large enough $m\text{.}$ Thus, for $\varepsilon>0\text{,}$ there is a positive integer $N_{\varepsilon}$ such that if $n\text{,}$ $m>N_{\varepsilon}\text{,}$ then $|S_n(z)-S_{m}(z)|\lt \varepsilon\text{.}$ But this means that for all $z \in T, \, \{S_n(z)\:$ is a Cauchy sequence. According to Theorem 4.1.8, this sequence must converge to a number, which we might as well designate by $f(z)\text{.}$ That is, $f(z)=\lim\limits_{n \to \infty}S_n(z) =\sum\limits_{k=0}^{\infty}u_k(z)\text{.}$ This observation gives us a function to which the series $\sum\limits_{k=0}^{\infty}u_k(z)$ converges. However, we still must show that the convergence is uniform. Let $\varepsilon>0$ be given. Again, since $\sum\limits_{k=0}^{\infty}M_k$ converges, there exists $N_{\varepsilon}$ such that if $n \ge N_{\varepsilon}\text{,}$ then $\sum\limits_{k=n}^{\infty}M_k\lt \varepsilon\text{.}$ Thus, if $n \ge N_{\varepsilon}$ and $z \in T\text{,}$ then

\begin{align*} \big|f(z) -S_n(z)\big| \amp = \left|\sum\limits_{k=0}^{\infty}u_k(z) - \sum\limits_{k=0}^{n-1}u_k(z)\right|\\ \amp =\left|\sum\limits_{k=n}^{\infty}u_k(z)\right|\\ \amp \le \sum\limits_{k=n}^{\infty}M_k\\ \amp \lt \varepsilon\text{,} \end{align*}

which completes the argument.

Theorem 7.1.5 gives an interesting application of the Weierstrass $M$-test.

Theorem 7.1.5.

Suppose that the power series $\sum\limits_{k=0}^{\infty}c_k(z-\alpha)^k$ has radius of convergence $\rho>0\text{.}$ Then for each $r\text{,}$ $0\lt r\lt \rho\text{,}$ the series converges uniformly on the closed disk $\overline{D}_r(\alpha)\text{,}$ where (by way of reminder) $\overline{D}_r(\alpha) =\{z:|z-\alpha| \le r\}\text{.}$

Proof.

Given $r\text{,}$ with $0\lt r\lt \rho\text{,}$ choose $z_0 \in D_{\rho}(\alpha)$ such that $|z_0-\alpha| =r\text{.}$ The proof of Theorem 4.4.1 part (ii) reveals that $\sum\limits_{k=0}^{\infty}c_k(z-\alpha)^k$ converges absolutely for $z \in D_{\rho }(\alpha)\text{,}$ from which it follows that $\sum\limits_{k=0}^{\infty}|c_k(z_0-\alpha)^k| = \sum \limits_{k=0}^{\infty}|c_k| r^k$ converges. Moreover, for all $z \in\overline{D}_r(\alpha)\text{,}$

\begin{equation*} |c_k(z-\alpha)^k| =|c_k| |z-\alpha|^k \le |c_k|r^k\text{.} \end{equation*}

The conclusion follows from the Weierstrass $M$-test with $M_k=|c_k|r^k\text{.}$

An immediate consequence of Theorem 7.1.5 is Corollary 7.1.6.

Corollary 7.1.6.

For each $r, \, 0\lt r\lt 1\text{,}$ the geometric series converges uniformly on the closed disk $\overline{D}_r(0)\text{.}$

The following theorem gives important properties of uniformly convergent sequences.

Theorem 7.1.7.

Suppose that $\{S_k\}$ is a sequence of continuous functions defined on a set $T$ containing the contour $C\text{.}$ If $\{S_k\}$ converges uniformly to $f$ on the set $T\text{,}$ then

i.
$f$ is continuous on $T\text{,}$ and
ii.
$\lim\limits_{k \to \infty} \int_CS_k(z)\,dz = \int_{C}\lim\limits_{k \to \infty}S_k(z)\,dz= \int_{C}f(z)\,dz\text{.}$

Proof.

Given $z_0 \in T\text{,}$ we must prove $\lim\limits_{z \to z_0}f(z) = f(z_0)\text{.}$ Let $\varepsilon>0$ be given. Since $\{S_k\}$ converges uniformly to $f$ on $T\text{,}$ there exists a positive integer $N_{\varepsilon}$ such that for all $z \in T, \, |f(z)-S_k(z)|\lt \frac{\varepsilon}{3}$ whenever $k \ge N_{\varepsilon}\text{.}$ And, as $S_{N_{\varepsilon}}$ is continuous at $z_0\text{,}$ there exists $\delta >0$ such that if $|z-z_0| \lt \delta\text{,}$ then $|S_{N_{\varepsilon}}(z)-S_{N_{\varepsilon}}(z_0)|\lt \frac{\varepsilon}{3}\text{.}$ Hence, if $|z-z_0|\lt \delta\text{,}$ we have

\begin{align*} |f(z) -f(z_0)| \amp = \big|f(z) -S_{N_{\varepsilon}}(z)+S_{N_{\varepsilon}}(z) - S_{N_{\varepsilon}}(z_0)+S_{N_{\varepsilon}}(z_0) -f(z_0)\big|\\ \amp \le \big|f(z) -S_{N_{\varepsilon}}(z)\big| + \big|S_{N_{\varepsilon}}(z)-S_{N_{\varepsilon}}(z_0)\big| + \big|S_{N_{\varepsilon}}(z_0) -f(z_0)\big|\\ \amp \lt \frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3}\\ \amp =\varepsilon\text{,} \end{align*}

which completes part (i).

To prove part (ii), let $\varepsilon>0$ be given and let $L$ be the length of the contour $C\text{.}$ Because $\{S_k\}$ converges uniformly to $f$ on $T\text{,}$ there exists a positive integer $N_{\varepsilon}$ such that, if $k \ge N_{\varepsilon}\text{,}$ then $|S_k(z) -f(z)| \lt \frac{\varepsilon}{L}$ for all $z \in T\text{.}$ Because $C$ is contained in $T\text{,}$ $\max\limits_{z \in C} \big|S_k(z) -f(z)\big| \lt \frac{\varepsilon}{L}$ if $k \ge N_{\varepsilon}\text{,}$ and we can use the ML inequality (Theorem 6.2.19) to get

\begin{align*} \left|\int_CS_k(z)\,dz- \int_{C}f(z)\,dz\right| \amp = \left|\int_C[S_k(z)-f(z)]\,dz\right|\\ \amp \le \max_{z \in C}\big|S_k(z) -f(z)\big| L\\ \amp \lt \left(\frac{\varepsilon}{L}\right) L\\ \amp = \varepsilon\text{.} \end{align*}

Corollary 7.1.8.

If the series $\sum\limits_{n=0}^{\infty}c_n(z-\alpha)^n$ converges uniformly to $f(z)$ on the set $T$ and $C$ is a contour contained in $T\text{,}$ then

\begin{equation*} \sum_{n=0}^{\infty}\int_Cc_n(z-\alpha)^n\,dz = \int_{C}\sum\limits_{n=0}^{\infty}c_n(z-\alpha)^n\,dz = \int_{C}f(z)\,dz \end{equation*}

Example 7.1.9.

Show that $\mathrm{Log}(1-z)=\sum\limits_{n=1}^{\infty}\frac{1}{n}z^n\text{,}$ for all $z \in D_1(0)\text{.}$

Solution.

For $z_0 \in D_1(0)\text{,}$ we choose $r$ and $R$ so that $0 \le |z_0| \lt r \lt R \lt 1\text{,}$ thus ensuring that $z_0 \in \overline{D}_r(0)$ and that $\overline{D}_r(0) \subset D_{R}(0)\text{.}$ By Corollary 7.1.6, the geometric series $\sum\limits_{n=0}^{\infty}z^n$ converges uniformly to $\frac{1}{1-z}$ on $\overline{D}_r(0)\text{.}$ If $C$ is any contour contained in $\overline{D}_r(0)\text{,}$ Corollary 7.1.8 gives

\begin{equation} \int_C\frac{1}{1-z}\,dz = \sum_{n=0}^{\infty} \int_{C}z^n\,dz\text{.}\tag{7.1.4} \end{equation}

Clearly, the function $f(z) =\frac{1}{1-z}$ is analytic in the simply connected domain $D_{R}(0)\text{,}$ and $F(z) = -$Log $(1-z)$ is an antiderivative of $f(z)$ for all $z \in D_{R}(0)\text{,}$ where Log is the principal branch of the logarithm. Likewise, $g(z)=z^n$ is analytic in the simply connected domain $D_{R}(0)\text{,}$ and $G(z)=\frac{1}{n+1 }z^{n+1}$ is an antiderivative of $g(z)$ for all $z \in D_{R}(0)\text{.}$ Hence, if $C$ is the straight-line segment joining $0$ to $z_0\text{,}$ we can apply Theorem 6.4.5 to Equation (7.1.4) to get

\begin{equation*} -\mathrm{Log}(1-z)\bigg|_{z=0}^{z=z_0}= \left.\sum\limits_{n=0}^{\infty} \left(\frac{1}{n+1}z^{n+1}\right)\right|_{z=0}^{z=z_0}\text{,} \end{equation*}

which becomes

\begin{equation*} -\mathrm{Log}(1-z_0) =\sum\limits_{n=0}^{\infty}\frac{1}{n+1}z_0^{n+1}=\sum\limits_{n=1}^{\infty}\frac{1}{n}z_0^n\text{.} \end{equation*}

The point $z_0 \in D_1(0)$ was arbitrary, so the solution is complete.

Exercises Exercises

1.

This exercise relates to Figure 7.1.3.

(a)

For $x$ near $-1\text{,}$ is the graph of $S_n(x)$ above or below $f(x)\text{?}$ Explain.

Solution.

By definition, $f(-1) =\frac{1}{1-(-1)}=\frac{1}{2}\text{.}$ It appears from the graph that the value of the upper function is approximately 1 (certainly larger than $\frac{1}{2}$), so the graph of $S_n$ must be above the graph of $f\text{.}$

(b)

Is the index $n$ in $S_n(x)$ odd or even? Explain.

(c)

Assuming that the graph is accurate to scale, what is the value of $n$ in $S_n(x)\text{?}$ Explain.

Solution.

From the graph, we approximate $S_n(1)=5\text{.}$ As $S_n(x)=\sum\limits_{k=0}^{n-1}x^k\text{,}$ we deduce that $n=5\text{.}$ Explain.

2.

Complete the details to verify the claim of Example. Example 7.1.2.

3.

Prove that the following series converge uniformly on the sets indicated.

(a)

$\sum\limits_{k=1}^{\infty}\frac{1}{k^2}z^k$ on $\overline{D}_1(0) = \{z:|z| \le 1\}$

Solution.

We see that $|\frac{1}{k^2}z^k| \le \frac{1}{k^2}$ for $z\in \bar{D}_1(0)\text{.}$ By the Weierstrass M-test, the series $\sum\limits_{k=1}^{\infty}\frac{1}{k^2} z^k$ converges uniformly on $\bar{D}_1(0)=\{z:|z| \le 1 \}\text{,}$ because the series $\sum\limits_{k=1}^{\infty}\frac{1}{k^2}$ converges.

(b)

$\sum\limits_{k=0}^{\infty}\frac{1}{(z^2-1)^k}$ on $\{z:|z| \ge 2\}$

(c)

$\sum\limits_{k=0}^{\infty}\frac{z^k}{z^{2k}+1}$ on $\overline{D}_r(0)\text{,}$ where $0\lt r\lt 1\text{.}$

4.

Show that $S_n(z)=\sum\limits_{k=0}^{n-1}z^k=\frac{1-z^n}{1-z}$ does not converge uniformly to $f(z)=\frac{1}{1-z}$ on the set $T=D_1(0)$ by appealing to Statement (7.1.3). \hint{Given $\varepsilon>0$ and a positive integer $n\text{,}$ let $z_n=\varepsilon^{\frac{1}{n}}\text{.}$}

5.

Why can’t we use the arguments of Theorem 7.1.5 to prove that the geometric series converges uniformly on all of $D_1(0)\text{?}$

Solution.

The crucial step in the theorem is the statement, “Moreover, for all $z\in \bar{D}_{r}(\alpha)$ it is clear that $|c_k(z-\alpha)^k| = |c_k||z-\alpha|^k \le |c_k|r^k\text{.}$” If we allowed $r=1\text{,}$ we would not be able to claim that $\sum\limits_{k=0}^{\infty }|c_k|r^k$ converges. Explain.

6.

By starting with the series for the complex cosine given in Section 5.4, choose an appropriate contour and use the method in Example 7.1.9 to obtain the series for the complex sine.

7.

Suppose that the sequences of functions $\{f_n\}$ and $\{g_n\}$ converge uniformly on the set $T\text{.}$

(a)

Show that the sequence $\{f_n+g_n\}$ converges uniformly on $T\text{.}$

Solution.

Let us say that $\{f_n\}$ and $\{g_n\}$ converge uniformly on $T$ to $f$ and $g$ respectively. Let $\varepsilon>0$ be given. The uniform convergence of $\{f_n\}$ means there exists an integer $N_{\varepsilon}$ such that $n \ge N_{\varepsilon}$ implies $|f_n(z) -f(z)| \lt \frac{\varepsilon}{2}$ for all $z\in T\text{.}$ Likewise, there exists an integer $M_{\varepsilon}$ such that $n \ge M_{\varepsilon}$ implies $|g_n(z) -g(z)| \lt \frac{\varepsilon}{2}$ for all $z\in T\text{.}$ If we set $L_{\varepsilon} = Max\{N_{\varepsilon}, \, M_{\varepsilon}\}\text{,}$ then for $n \ge L_{\varepsilon}, \, \Big|\big(f_n(z)+g_n(z)\big) - \big(f(z)+g(z)\big)\Big| \le |f_n(z)-f(z)| + |g_n(z)-g(z)| \lt \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$ for all $z\in T\text{.}$

(b)

Show by example that it is not necessarily the case that $\{f_n\,g_n\}$ converges uniformly on $T\text{.}$

Solution.

For all $n\text{,}$ let $f_n(x)=x\text{,}$ and $g_n(x)=\frac{1}{n}\text{,}$ for all $x\in T\text{,}$ where $T$ is the real numbers. Then $f_n(x)$ converges uniformly to $x\text{,}$ and $g_n(x)$ converges uniformly to $0$ (verify). However, even though $f_n(x) g_n(x)$ converges to $0$ (explain), the convergence is not uniform (verify). Can you come up with a different example?

8.

On what portion of $D_1(0)$ does the sequence $\{nz^n\}_{n=1}^{\infty}$ converge, and on what portion does it converge uniformly?

9.

Consider the function $\zeta (z)=\sum\limits_{n=1}^{\infty}n^{-z}\text{,}$ where $n^{-z}=\exp(-z \ln n)\text{.}$

(a)

Show that $\zeta (z)$ converges uniformly on $A=\{z:\mathrm{Re}(z) \ge 2\}\text{.}$

Solution.

For $z \in A, \, |n^{-z}| = |\exp[-(x+iy) \ln n]| = |\exp(-iy)|\,|\exp(-x\ln n)| = n^{-x}\text{.}$ Since $z \in A\text{,}$ we know $\mathrm{Re}(z) = x \ge 2\text{,}$ so $n^{-x} \le \frac{1}{n^2}\text{.}$ Thus, with $M_n=\frac{1}{n^2}\text{,}$ we see that $\zeta (z)$ converges uniformly on $A$ by the Weierstrass $M$-test.

(b)

Let $D$ be a closed disk contained in $\{z:\mathrm{Re}(z)>1\}\text{.}$ Show that $\zeta(z)$ converges uniformly on $D\text{.}$

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