Uniform limit theorem

The uniform limit theorem states: if $(f_n)$ is a sequence of continuous functions $f_n: X \to \mathbb{R}$ on a metric space $X$, and $f_n \to f$ uniformly , then $f$ is continuous.

Statement

For metric spaces $X$ and $Y$: if $f_n: X \to Y$ are continuous and $f_n \rightrightarrows f$ uniformly (meaning $\sup_{x \in X} d(f_n(x), f(x)) \to 0$), then $f$ is continuous.

Proof idea

Given $\varepsilon > 0$ and $x_0 \in X$:

1. Choose $N$ so that $|f_n(x) - f(x)| < \varepsilon/3$ for all $x$ and $n \geq N$.
2. Use continuity of $f_N$ to find $\delta$ such that $|f_N(x) - f_N(x_0)| < \varepsilon/3$ when $d(x, x_0) < \delta$.
3. Triangle inequality: $|f(x) - f(x_0)| < \varepsilon$.

Counterexample for pointwise convergence

$f_n(x) = x^n$ on $[0, 1]$ converges pointwise to a discontinuous limit:

The convergence is not uniform.