About Uniform

All numbers discussed here is in $\mathbb{R}$

What I'd like to discuss here is all about uniform concepts I've learnt since. And I wanna summarize them.

About uniform continuity

In calculus, there exists a concept called Continuity as expressed: $f(x)$ is continuous at $x_0 \iff$ $$\forall \varepsilon \in \mathbb{R} \exists \delta \in \mathbb{R} \text{s.t.} |x-x_{0}|<\delta\implies |f(x)-f(x_{0})|<\varepsilon$$

As is seen above, $\delta$ is somewhere dependent towards $x_0$. What if we genernalize that idea such that $\delta$ shall be dependent not to $x_0$ but to an interval? Just modify the concept as below: $f(x)$ is uniform continuous at interval $\mathbf{I}\iff$ $$\forall\varepsilon\in\mathbb{R}\exists \delta\in\mathbb{R},\forall x_1,x_2\in\mathbf{I} : |x_1-x_2|<\delta \implies |f(x_1)-f(x_2)<\varepsilon|$$

And with Lagrange's mean value theorem, we can distinguish that in an open interval, if every point is derivable and its value is bounded ($\forall x \in\mathbf{I}\exists A \in\mathbb{R}, |f'(x)|<A$), that function is then uniform continuous.

Proof.

If $f(x)$ is derivable in $\mathbf{I}$, according to Lagrange's mean value theorem, we can conclude that $\frac{f(x_1)-f(x_2)}{x_1-x_2}=f'(\xi)\leq \max\{f'(x)\}< M$, thus denoting $|x_1-x_2|\leq \delta$ and we can say $$|f(x_1)-f(x_2)|<M\delta<\epsilon\Rightarrow \delta=\frac{\epsilon}{M}$$

Just let $\delta$ be $\frac{\epsilon}{M}$ and then the requirements for uniform continuity are met.

That type of continuity is particularly called Lipschitz continuity.

in a prettier writing style

If we combine the limit's symbols with the continuity concept, the statement can be easily expressed as follows. $$\text{$f(x)$ is continuous at $x_0$}\Leftrightarrow\lim_{x\to x_0}f(x)=f(x_0)$$ The same way the uniform continuity can be as: $$\begin{aligned}\text{$\forall{a,b}\in\mathbf{I}$, $f(x)$ is uniform continuous at $\mathbf{I}$}\Leftrightarrow\lim_{||T||\to 0}|f(a)-f(b)|=0 \\(\text{$||T||$ denotes $|a-b|$}) \end{aligned}$$