gen-define-uniform-convergence-and-give-one-short-worked-example-7fda

Define uniform convergence and give one short worked example.

# Uniform Convergence ## Definition A sequence of functions $\{f_n\}$ converges **uniformly** to a function $f$ on a set $D$ if: $$\forall \epsilon > 0, \exists N \in \mathbb{N} \text{ such that } \forall n \geq N, \forall x \in D: |f_n(x) - f(x)| < \epsilon$$ **Key point:** The same $N$ works for *all* $x \in D$ (unlike pointwise convergence where $N$ may depend on $x$). --- ## Worked Example **Show that $f_n(x) = \frac{x}{n}$ converges uniformly to $f(x) = 0$ on $[0, 1]$.** **Solution:** We need to show: $\forall \epsilon > 0, \exists N$ such that $\forall n \geq N, \forall x \in [0,1]: \left|\frac{x}{n} - 0\right| < \epsilon$ For any $x \in [0,1]$: $$\left|f_n(x) - f(x)\right| = \left|\frac{x}{n}\right| = \frac{x}{n} \leq \frac{1}{n}$$ (since $x \leq 1$ on our domain) Given $\epsilon > 0$, choose $N > \frac{1}{\epsilon}$. Then for all $n \geq N$ and all $x \in [0,1]$: $$\left|f_n(x) - f(x)\right| \leq \frac{1}{n} \leq \frac{1}{N} < \epsilon$$ Therefore $f_n \to f$ uniformly on $[0,1]$. ✓