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Section 8.1 Sequences and Series of Functions

Power series, Fourier series, and many approximation procedures are built from sequences of functions. The basic question is simple: if each function in the sequence is well behaved, what can be said about the limit? Pointwise convergence is often too weak, so the right notion for analysis is usually uniform convergence.

Subsection 8.1.1

Let \(S \subseteq \R\text{,}\) and let \((f_n)\) be a sequence of functions from \(S\) to \(\R\text{.}\) We say that \(f_n\) converges pointwise on \(S\) to a function \(f \colon S \to \R\) if for every \(x \in S\) the numerical sequence \((f_n(x))\) converges to \(f(x)\text{.}\)

Example 8.1.1.

Let \(f_n(x)=x^n\) on \([0,1]\text{.}\) Then \(f_n(x)\to 0\) for every \(x \in [0,1)\text{,}\) while \(f_n(1)=1\) for every \(n\text{.}\) Thus the pointwise limit is
\begin{equation*} f(x)=\begin{cases} 0 & \text{if } 0 \le x \lt 1,\\ 1 & \text{if } x=1. \end{cases} \end{equation*}
Each \(f_n\) is a polynomial, but the limit is not continuous at \(1\text{.}\)
Pointwise convergence looks at one input \(x\) at a time. The index \(N\) that makes \(|f_n(x)-f(x)|\) small is allowed to depend on \(x\text{.}\) Uniform convergence removes that dependence.
We say that \(f_n\) converges uniformly on \(S\) to \(f\) if for every \(\varepsilon \gt 0\) there exists \(N \in \N\) such that
\begin{equation*} |f_n(x)-f(x)| \lt \varepsilon \qquad \text{for all } n \ge N \text{ and all } x \in S. \end{equation*}

Proof.
Fix \(x \in S\text{.}\) If \(f_n \to f\) uniformly on \(S\text{,}\) then for every \(\varepsilon \gt 0\) there exists \(N\) such that \(|f_n(y)-f(y)| \lt \varepsilon\) for all \(y \in S\) and all \(n \ge N\text{.}\) In particular, the same inequality holds when \(y=x\text{.}\) Therefore \(f_n(x)\to f(x)\text{.}\)

Example 8.1.3.

The convergence in Example 8.1.1 is not uniform on \([0,1]\text{.}\) Let \(f\) be the pointwise limit. For \(x_n=n/(n+1)\) we have \(x_n \in [0,1)\text{,}\) so \(f(x_n)=0\text{,}\) while
\begin{equation*} f_n(x_n)=\left(\frac{n}{n+1}\right)^n \to e^{-1}. \end{equation*}
Hence \(|f_n(x_n)-f(x_n)|\) does not tend to \(0\text{.}\) Therefore \(f_n\) cannot converge uniformly to \(f\) on \([0,1]\text{.}\)
If \(f \colon S \to \R\) is bounded, its supremum norm on \(S\) is
\begin{equation*} \|f\|_S := \sup\{|f(x)| : x \in S\}. \end{equation*}
When the set \(S\) is understood, we simply write \(\|f\|\text{.}\)

Proof.
If \(f_n \to f\) uniformly on \(S\text{,}\) then for every \(\varepsilon \gt 0\) there exists \(N\) such that \(|f_n(x)-f(x)| \lt \varepsilon\) for all \(x \in S\) and all \(n \ge N\text{.}\) Taking the supremum over \(x\) gives \(\|f_n-f\|_S \le \varepsilon\) for \(n \ge N\text{.}\)
Conversely, if \(\|f_n-f\|_S \to 0\text{,}\) then for every \(\varepsilon \gt 0\) there exists \(N\) such that \(\|f_n-f\|_S \lt \varepsilon\) whenever \(n \ge N\text{.}\) Hence \(|f_n(x)-f(x)| \le \|f_n-f\|_S \lt \varepsilon\) for every \(x \in S\text{,}\) which is exactly uniform convergence.

Proof.
If \(f_n \to f\) uniformly on \(S\text{,}\) then for every \(\varepsilon \gt 0\) we may choose \(N\) so that \(|f_n(x)-f(x)| \lt \varepsilon/2\) for all \(x \in S\) and all \(n \ge N\text{.}\) Therefore, whenever \(m,n \ge N\text{,}\)
\begin{equation*} |f_m(x)-f_n(x)| \le |f_m(x)-f(x)|+|f_n(x)-f(x)| \lt \varepsilon \end{equation*}
for all \(x \in S\text{.}\)
Conversely, suppose the displayed condition holds. Fix \(x \in S\text{.}\) Then \((f_n(x))\) is a Cauchy sequence in \(\R\text{,}\) so it converges by completeness. Define \(f(x)=\lim_{n\to\infty} f_n(x)\text{.}\) Now let \(\varepsilon \gt 0\) and choose \(N\) as in the hypothesis. For \(n \ge N\) and fixed \(x \in S\text{,}\) the inequality \(|f_m(x)-f_n(x)| \lt \varepsilon\) holds for all \(m \ge N\text{.}\) Letting \(m \to \infty\) gives \(|f(x)-f_n(x)| \le \varepsilon\text{.}\) Since this estimate is valid for every \(x \in S\text{,}\) the convergence is uniform.
A series of functions\(\sum_{n=0}^\infty f_n\) on \(S\) is studied through its partial sums
\begin{equation*} s_N(x)=\sum_{n=0}^N f_n(x). \end{equation*}
We say that the series converges pointwise or uniformly according as the sequence \((s_N)\) does.

Proof.
Let \(s_N(x)=\sum_{n=0}^N f_n(x)\text{.}\) Since \(\sum_{n=0}^\infty M_n\) converges, its tails tend to \(0\text{.}\) Thus, given \(\varepsilon \gt 0\text{,}\) there exists \(N\) such that for all \(m \gt n \ge N\text{,}\)
\begin{equation*} \sum_{k=n+1}^m M_k \lt \varepsilon. \end{equation*}
Hence for every \(x \in S\text{,}\)
\begin{equation*} |s_m(x)-s_n(x)| \le \sum_{k=n+1}^m |f_k(x)| \le \sum_{k=n+1}^m M_k \lt \varepsilon. \end{equation*}
By Proposition 8.1.5, the partial sums converge uniformly on \(S\text{.}\)
For each fixed \(x \in S\text{,}\) the estimate \(|f_n(x)| \le M_n\) also shows that the numerical series \(\sum_{n=0}^\infty |f_n(x)|\) converges by comparison. Therefore the convergence is absolute at every point.

Example 8.1.7.

Let \(0 \lt r \lt 1\text{.}\) On the interval \([-r,r]\) we have \(|x^n| \le r^n\text{,}\) and the geometric series \(\sum_{n=0}^\infty r^n\) converges. Therefore \(\sum_{n=0}^\infty x^n\) converges uniformly on \([-r,r]\) by the Weierstrass M-test.

Proof.
Let \(L=\limsup |a_n|^{1/n}=1/R\text{.}\) Choose a number \(\rho\) such that
\begin{equation*} L \lt \rho \lt \frac{1}{r}. \end{equation*}
By the definition of \(\limsup\text{,}\) there exists \(N\) such that \(|a_n|^{1/n} \le \rho\) for every \(n \ge N\text{,}\) and hence \(|a_n| \le \rho^n\) for all such \(n\text{.}\)
If \(|x-x_0| \le r\) and \(n \ge N\text{,}\) then
\begin{equation*} |a_n(x-x_0)^n| \le \rho^n r^n = (\rho r)^n. \end{equation*}
Since \(\rho r \lt 1\text{,}\) the geometric series \(\sum_{n=N}^\infty (\rho r)^n\) converges. For the finitely many indices \(n \lt N\text{,}\) set \(M_n=|a_n|r^n\text{.}\) Then \(|a_n(x-x_0)^n| \le M_n\) for every \(x \in [x_0-r,x_0+r]\) and every \(n\text{,}\) and the numerical series \(\sum M_n\) converges. The Weierstrass M-test now gives absolute and uniform convergence on the stated interval.

Example 8.1.9.

The geometric series
\begin{equation*} \sum_{n=0}^\infty x^n \end{equation*}
converges to \(1/(1-x)\) on \((-1,1)\text{,}\) but the convergence is not uniform on that open interval. Indeed, if \(s_N(x)=1+x+\dots+x^N\text{,}\) then
\begin{equation*} \frac{1}{1-x}-s_N(x)=\frac{x^{N+1}}{1-x}. \end{equation*}
For each fixed \(N\text{,}\) the right-hand side becomes arbitrarily large as \(x \to 1^-\text{.}\) Hence \(s_N\) cannot converge uniformly to \(1/(1-x)\) on \((-1,1)\text{.}\)
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