The Question: Prove that if a function f defined on S⊆R is Lipschitz continuous then f is uniformly continuous on S.

Definition. A function f defined on a set S⊆R is said to beLipschitz continuouson S if there exists an M so that |f(x)−f(c)||x−c|≤M for all x and c in S such that x≠c.My Heuristic Interpretation: if f is Lipschitz continuous then the “absolute slope” of f is never unbounded i.e. no asymptotes.

Definition. A continuous function f defined on Dom(f) is said to beuniformly continuousif for each ε>0 ∃ δ>0 s.t. ∀ x,c∈Dom(f) |x−c|≤δ ⇒ |f(x)−f(c)|≤ε

Proof:f Lipschitz continuous ⇒ |f(x)−f(c)|≤M|x−c|. Since we suppose |x−c|≤δ for uniform continuity, we have x within δ of c, so |x|≤|c|+δ. So taking δ=ε/M

|f(x)−f(c)|≤M|x−c|≤Mδ=ε

My Question: Is my proof valid with the assumptions taken?

**Answer**

It’s not very well organized, and it has some extraneous clutter, but it also has the core of the argument. You want to show that for each ϵ>0 there is a δ>0 such that |f(x)−f(c)|<ϵ whenever x,c∈domf and |x−c|<δ, so in a polished version of the argument your first step should be:

Suppose that f is Lipschitz continuous on some set S with Lipschitz constant M, and fix ϵ>0.

You’ve already worked out that ϵ/M will work for δ, so you can even start out with:

Fix ϵ>0 and let δ=ϵM.

Now you want to show that this choice of δ does the job.

Clearly δ>0. Suppose that x,c∈S and |x−c|<δ. Then by the Lipschitz continuity of f we have |f(x)−f(c)|≤M|x−c|<Mδ=ϵ, so f is uniformly continuous on S. ⊣

**Added:** Your heuristic interpretation of Lipschitz continuity is inaccurate enough that it may well lead you astray at some point. Consider the function

f(x)={xsin1x,if x≠00,if x=0.

This function has no vertical asymptotes, but it’s not Lipschitz continuous:

f(12nπ)−f(12nπ+π2)12nπ−12nπ+π2=12nπ+π212nπ−12nπ+π2=12nπ+π22nπ−1=2nππ/2=4n,

which can be made as large as you want. This function has very, very steep bits, but they’re also very, very short.

**Attribution***Source : Link , Question Author : Moderat , Answer Author : Brian M. Scott*