---- - Suppose $X$ is a set and $(Y, d)$ is a [[metric space]]. > [!definition] Definition. ([[uniform convergence]]) > A [[sequence]] of functions $(f_{n}:X \to Y)$ is said to **converge uniformly** to $f:X \to Y$, often written as $\lim_{ n \to \infty } f_{n} = f \text{ uniformly},$ > if for all $\varepsilon > 0$ there exists $N \in \nn$ s.t. for all $n \geq N$ we have $d\big(f_{n}(x), f(x)\big) < \varepsilon$ for all $x \in X$. > [[Uniform convergence]] is a manifestly stronger notion than [[pointwise converge|pointwise convergence]]. When a sequence converges uniformly, it also converges pointwise to the same ([[limits in Hausdorff spaces are unique|unique]]) limit. > [!equivalence] > One has $f_{n} \to f \text{ uniformly} \iff \sup_{x \in X} d\big( f_{n}(x), f(x) \big) \xrightarrow[n \to \infty]{}0$ > where the [[supremum]] is taken in $[0, \infty]$. > > > [!proof]- Proof. > > The proof is obvious: > > > > Suppose $f_{n} \to f$ uniformly. Then for all $\varepsilon>0$ there exists $N \in \mathbb{N}$ such that for all $n \geq N$ $d\big( f_{n}(x), f(x) \big)\leq \varepsilon \text{ for all }x \in X.$ > > Taking the [[supremum]] preserves this inequality: $\sup_{x \in X} d\big( f_{n}(x), f(x) \big) \leq \varepsilon \text{ for all } x \in X.$ > > Hence $(\implies)$ is shown. Conversely, assume $\sup_{x \in X} d\big( f_{n}(x), f(x) \big) \xrightarrow[n \to \infty]{}0$. Fix $\varepsilon>0$. Obtain $N \in \mathbb{N}$ such that for all $n \geq N$, $\sup_{x \in X} d\big( f_{n}(x), f(x)\big)<\varepsilon$. Then $d\big( f_{n}(x), f(x) \big)<\varepsilon$ for all $x \in X$. Uniform convergence, and thus $(\impliedby)$, follow. > > > > [!equivalence] Equivalence. (Convergence in [[uniform topology]]) > One has $f_{n} \to f \text{ uniformly } \iff f_{n} \to f \text{ in the uniform topology on }Y^{X}.$ > > > > > > [!proof]- Proof. > > First recall the terminology in [[uniform metric]]. To say a [[sequence]] $(f_{n})_{n \in \mathbb{N}} \in (Y^{X})^{\mathbb{N}}$ converges to $f \in Y^{X}$ in the [[uniform metric]] is to say that for all $\varepsilon>0$ there exists $N \in \mathbb{N}$ such that for any $n \geq N$ one has $\sup_{x \in X} \min\{1, d\big( f_{n}(x), f(x) \big)\}< \varepsilon .$ > > > > Now suppose $f_{n} \to f$ uniformly. Then for all $\varepsilon>0$ there exists $N \in \mathbb{N}$ such that for all $n \geq N$ one has $d\big( f_{n}(x), f(x) \big) \leq \varepsilon \text{ for all } x \in X.$ > > Passing to the [[supremum]] over $x \in X$ preserves this inequality, as does taking the $\min$ against $1$. Thus $f_{n} \to f$ in the uniform metric on $Y^{X}$. > > > > Conversely, suppose $f_{n} \to f$ in the uniform metric on $Y^{X}$. Fix $\varepsilon>0$, and obtain $N \in \mathbb{N}$ such that for all $n \geq N$ one has $\sup_{x \in X} \min\{1, d\big( f_{n}(x), f(x) \big)\}< \varepsilon .$ > > Clearly then $\sup_{x \in X} \min \{ 1, d\big(f_{n}(x), f(x)\big) \} \xrightarrow[n \to \infty]{}0,$ which implies $\sup_{x \in X}d(f_{n}(x), f(x))\to 0$. Thus $f_{n} \to f$ uniformly by the first equivalent definition of uniform convergence. > ^equivalence-uniform-topology > [!basicproperties] > - [[uniform limit theorem|uniform limits respect continuity (uniform limit theorem)]] ^properties > [!basicnonexample] > Recall this example: > ![[pointwise converge#^c6bab8]] > > This sequence converges pointwise, but not uniformly. Indeed, if it did converge uniformly, it would have to converge to $f(x)=1_{\{ x=0 \}}+1$ as it does pointwise. Let $\varepsilon=\frac{1}{59}$. We see that, given any choice of $k \in \mathbb{N}$, we can pick $x=-\frac{1}{59k} \in (-\frac{1}{k}, 0)$ and observe > > $|f_{k}(x)-f(x)|=|\overbrace{ kx }^{ =-1/59 }+1|>\varepsilon.$ > > ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```