Sequence of Functions is Uniformly Cauchy iff Uniformly Convergent
Theorem
Let $S \subseteq \R$.
Let $\sequence {f_n}$ be a sequence of real functions $S \to \R$.
Then $\sequence {f_n}$ is uniformly Cauchy on $S$ if and only if $\sequence {f_n}$ converges uniformly on $S$.
Proof
Sufficient Condition
Let $\sequence {f_n}$ be uniformly Cauchy on $S$.
Let $\epsilon \in \R_{> 0}$ be arbitrary.
As $\sequence {f_n}$ is uniformly Cauchy on $S$, there exists $N \in \N$ such that:
- $\size {\map {f_n} x - \map {f_m} x} < \dfrac \epsilon 2$
for all $n, m > N$ and $x \in S$.
Let $x \in S$ be fixed.
Then:
- $\size {\map {f_n} x - \map {f_m} x} < \epsilon$
for all $n, m > N$.
As $\epsilon$ was arbitrary, the sequence $\sequence {\map {f_n} x}$ is therefore Cauchy.
By Cauchy's Convergence Criterion on Real Numbers, it follows that $\sequence {\map {f_n} x}$ is convergent.
Define a function $f : S \to \R$ by:
- $\ds \map f x = \lim_{n \mathop \to \infty} \map {f_n} x$
for all $x \in S$.
We aim to show that $f_n \to f$ uniformly.
For all $x \in S$, we have:
- $\ds \lim_{m \mathop \to \infty} \size {\map {f_n} x - \map {f_m} x} = \size {\map {f_n} x - \map f x}$
We established that for all $\epsilon \in \R_{> 0}$ we can find $N \in \N$ such that:
- $\size {\map {f_n} x - \map {f_m} x} < \dfrac \epsilon 2$
for $x \in S$ and $n, m > N$.
We therefore have:
- $\size {\map {f_n} x - \map f x} \le \dfrac \epsilon 2 < \epsilon$
for all $x \in S$ and $n > N$.
So $\sequence {f_n}$ converges uniformly to $f$ on $S$.
$\Box$
Necessary Condition
Let $\sequence {f_n}$ converge uniformly on $S$ to $f$.
Let $\epsilon \in \R_{> 0}$ be arbitrary.
Since $f_n \to f$ uniformly, there exists some $N \in \N$ such that:
- $\size {\map {f_n} x - \map f x} < \dfrac \epsilon 2$
for all $x \in S$ and $n > N$.
Then if $x \in S$ and $n, m > N$, we have:
\(\ds \size {\map {f_n} x - \map {f_m} x}\) | \(=\) | \(\ds \size {\map {f_n} x - \map f x - \paren {\map {f_m} x - \map f x} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \size {\map {f_n} x - \map f x} + \size {\map {f_m} x - \map f x}\) | Triangle Inequality for Real Numbers | |||||||||||
\(\ds \) | \(<\) | \(\ds \epsilon\) |
Since $\epsilon$ was arbitrary, $\sequence {f_n}$ is uniformly Cauchy on $S$.
$\blacksquare$
Sources
- 1973: Tom M. Apostol: Mathematical Analysis (2nd ed.) ... (previous) ... (next): $\S 9.5$: The Cauchy Condition for Uniform Convergence: Theorem $9.3$