Uniform limit theorem

Statement

Let $S \subseteq \mathbb{R}$, and let $f_n \colon S \to \mathbb{R}$ be continuous for each $n$. If $f_n \to f$ uniformly on $S$, then $f$ is continuous on $S$.

Remarks

Fix $c \in S$ and $\varepsilon > 0$. Choose $n$ with $|f_n - f| < \varepsilon/3$ on $S$, and $\delta$ for the continuity of $f_n$ at $c$. Then $|f(x) - f(c)| \le |f(x) - f_n(x)| + |f_n(x) - f_n(c)| + |f_n(c) - f(c)| < \varepsilon$ for $|x - c| < \delta$. The [[sequence $x^n$ on $[0, 1]$|0YSRR1JTKZ]] shows that pointwise convergence is not enough. Mathlib states the theorem as TendstoUniformly.continuous.

History

Augustin-Louis Cauchy claimed in 1821 that a series of continuous functions has a continuous sum, whenever it converges. In 1826, Niels Henrik Abel found counterexamples among Fourier series. In modern terms, Cauchy's argument needs uniform convergence.

Explanations

L1 human

If continuous graphs approach a limit graph uniformly, the limit graph has no jumps. A jump in the limit would force some $f_n$ to jump as well.