Sinofine Lotusie

About Uniform

All numbers discussed here is in ℝ\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:

f(x) is continuous at x0⇔∀ϵ∈ℝ∃δ∈ℝ:|x−x0|<δ⇒|f(x)−f(x0)|<ϵ\begin{aligned}&\text{$f(x)$ is continuous at $x_0$}\Leftrightarrow\\ &\forall{\epsilon\in \mathbb R}\exists{\delta\in \mathbb R} : |x-x_0|<\delta \Rightarrow |f(x)-f(x_0)|<\epsilon\end{aligned}

As is seen above, δ\delta is somewhere dependent towards x0x_0. What if we genernalize that idea such that δ\delta shall be dependent not to x0x_0 but to an interval? Just modify the concept as below:

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

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

Proof.

If f(x)f(x) is derivable in 𝐈\mathbf{I}, according to Lagrange's mean value theorem, we can conclude that f(x1)−f(x2)x1−x2=f′(ξ)≤max{f′(x)}<M\frac{f(x_1)-f(x_2)}{x_1-x_2}=f'(\xi)\leq \max\{f'(x)\}< M, thus denoting |x1−x2|≤δ|x_1-x_2|\leq \delta and we can say

|f(x1)−f(x2)|<Mδ<ϵ⇒δ=ϵM|f(x_1)-f(x_2)|<M\delta<\epsilon\Rightarrow \delta=\frac{\epsilon}{M}

Just let δ\delta be ϵM\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.

f(x) is continuous at x0⇔limx→x0f(x)=f(x0)\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:
∀a,b∈𝐈, f(x) is uniform continuous at 𝐈⇔lim||T||→0|f(a)−f(b)|=0(||T|| denotes |a−b|)\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}