About Uniform[WIP]

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 below: \[\begin{aligned}&\mbox{$f(x)$ is continuous at $x_0$}\Leftrightarrow\\ &\forall{\varepsilon\in \mathbb R}\exists{\delta\in \mathbb R} : \vert x-x_0\vert <\delta \Rightarrow \vert f(x)-f(x_0)\vert <\varepsilon\end{aligned} \]

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: \[\begin{aligned}&\mbox{$f(x)$ is \textbf{uniform} continuous at interval $\mathbf{I}$}\Leftrightarrow\\ &\forall{\varepsilon\in\mathbb R}\exists{\delta\in\mathbb R, x_1,x_2\in\mathbf{I}}: \vert x_1-x_2\vert <\delta \Rightarrow \vert f(x_1)-f(x_2)<\varepsilon\vert \end{aligned}\]

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 (\(\forall x \in\mathbf{I}\exists A \in\mathbb{R}, \vert f'(x)\vert <A\)), that function is then uniform continuous.


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 \(\vert x_1-x_2\vert \leq \delta\) and we can say \[\vert f(x_1)-f(x_2)\vert <M\delta<\varepsilon\Rightarrow \delta=\frac{\varepsilon}{M}\]

Just let \(\delta\) be \(\frac{\varepsilon}{M}\) 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. \[\mbox{$f(x)$ is continuous at $x_0$}\Leftrightarrow\lim_{x\to x_0}f(x)=f(x_0) \] The same way the uniform continuity can be as: \[\begin{aligned}\mbox{$\forall{a,b}\in\mathbf{I}$, $f(x)$ is uniform continuous at $\mathbf{I}$}\Leftrightarrow\lim_{\Vert T\Vert \to 0}\vert f(a)-f(b)\vert =0 \\(\mbox{$\Vert T\Vert $ denotes $\vert a-b\vert $}) \end{aligned}\]