Uniformly Convergent Sequence of Continuous Functions Converges to Continuous Function
Theorem
Let $S \subseteq \R$.
Let $x \in S$.
Let $\sequence {f_n}$ be a sequence of real functions $S \to \R$ converging uniformly to $f: S \to \R$.
Let $f_n$ be continuous at $x$ for all $n \in \N$.
Then $f$ is continuous at $x$.
Corollary
Let $S \subseteq \R$.
Let $\sequence {f_n}$ be a sequence of real functions $S \to \R$ converging uniformly to $f : S \to \R$.
Let $f_n$ be continuous for all $n \in \N$.
Then $f$ is continuous.
Proof
Let $\epsilon \in \R_{> 0}$.
Since $f_n \to f$ uniformly, there exists some $N \in \N$ such that:
- $\size {\map {f_n} x - \map f x} < \dfrac \epsilon 3$
for all $x \in S$ and $n \ge N$.
Since $f_N$ is continuous at $x$, there exists some $\delta > 0$ such that:
- for all $y$ with $\size {x - y} < \delta$, we have $\size {\map {f_N} x - \map {f_N} y} < \dfrac \epsilon 3$
Then for $y$ with $\size {x - y} < \delta$ we have:
\(\ds \size {\map f x - \map f y}\) | \(=\) | \(\ds \size {\map f x - \map {f_N} x + \map {f_N} x - \map {f_N} y + \map {f_N} y - \map f y}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \size {\paren {\map f x - \map {f_N} x} + \paren {\map {f_N} x - \map {f_N} y} + \paren {\map {f_N} y - \map f y} }\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \size {\map f x - \map {f_N} x} + \size {\map {f_N} x - \map {f_N} y} + \size {\map {f_N} y - \map f y}\) | Triangle Inequality for Real Numbers | |||||||||||
\(\ds \) | \(<\) | \(\ds 3 \times \frac \epsilon 3\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \epsilon\) |
Since $\epsilon$ was arbitrary, $f$ is continuous at $x$.
$\blacksquare$
Sources
- 1973: Tom M. Apostol: Mathematical Analysis (2nd ed.) ... (previous) ... (next): $\S 9.4$: Uniform Convergence and Continuity: Theorem $9.2$