[[Noteworthy Uses]]:: *[[Noteworthy Uses]]* [[Proved By]]:: *[[Proved By|Crucial Dependencies]]* ---- - Let $X$ be a [[topological space]] and let $(Y,d)$ be a [[metric space]]; - Let $f_{n}: X \to Y$ be a [[sequence]] of functions [[uniform convergence|converging uniformly]] to a function $f: X \to Y$. > [!theorem] Theorem. ([[uniform limit theorem]]) > If each $f_{n}$ is [[continuous]], then so is $f$. > [!proof]- Proof. ([[uniform limit theorem]]) > Recall that in order to prove $f$ [[continuous]], we must show that given $\varepsilon>0$, around any $x \in X$ we can find an [[epsilon-ball]] $U$ s.t. $d\big(f(x), f(y)\big) < \varepsilon \ \ \fa y \in U.$ > Fix $\varepsilon > 0$. Since $f_{n} \to f$ [[uniform convergence|uniformly]] there exists $N \in \nn$ such that $d_{Y}\big(f_{N}(t), f(t)\big) < \frac{\varepsilon}{3}$ for all $t \in X$. Moreover, since $f_{N}$ is [[continuous]] on $X$ by assumption, about every $x$ there is a [[neighborhood]] $U$ s.t. $d\big(f_{N}(x), f_{N}(y)\big) \leq \frac{\varepsilon}{3}$ for all $y \in U$. Now apply the [[triangle inequality]] like so: $d\big(f(x), f(y) \big)\leq d\big(f(x), f_{N}(x)\big) + d\big(f_{N}(x), f_{N}(y) \big) + d\big(f_{N}(y), f(y)\big) < \frac{\varepsilon}{3}+ \frac{\varepsilon}{3}+ \frac{\varepsilon}{3} = \varepsilon, \ \ \ \forall y \in U.$ > > > ---- #### ----- #### References > [!backlink] > ```dataview TABLE rows.file.link as "Further Reading" FROM [[]] FLATTEN file.tags GROUP BY file.tags as 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 > ```