# Uniform Continuity implies Lipschitz Continuity if delta is proportional to epsilon?

I started writing this post wanting to ask a question, but now I think I answered it. Could you please review my proof?

Definition: A continuous function $$f$$ defined on the interval $$I$$ is said to be uniformly continuous if for each $$\epsilon>0,\exists \delta>0$$ s.t. $$\forall x,y\in I, |x−y|\leq\delta \implies |f(x)−f(y)|\leq\epsilon$$.

Definition: A function $$f$$ defined on a set $$I\subseteq\mathbb{R}$$ is said to be Lipschitz continuous on $$I$$ if there exists an $$M$$ so that $$\frac{|f(x)−f(y)|}{|x−y|}\leq M$$ for all $$x$$ and $$y$$ in $$I$$ such that $$x≠y$$.

According to Courant & John's book Introduction to Calculus and Analysis Volume I (Section 1.2, page 43), uniform continuity implies Lipschitz Continuity IF $$\delta$$, (called the "modulus of continuity") is such that $$\delta \leq \epsilon C$$, where C is a constant. How do I prove this?

Trying to prove this, I go back to the definition for uniform continuity on an interval $$I$$, which I give above and I plugged $$\delta \leq \epsilon C$$ into the above definition, and I get the statement that $$\forall \epsilon>0$$ and for all $$x,y$$ is some closed interval $$I$$, $$|x−y|\leq\epsilon C \implies |f(x)−f(y)|\leq \epsilon$$.

Then I choose $$x,y$$ such that $$0<|x−y|=\epsilon C$$, and I divide both sides of this inequality: $$|f(x)−f(y)|\leq \epsilon$$, to get this inequality: $$\frac{|f(x)−f(y)|}{|x−y|}≤\frac{1}{C}$$.

Well? Is it good?

• I've answered the same question asked here: math.stackexchange.com/q/4113043/169085. The proof does not require any specific structure of the interval, as opposed to Kavi Rama Murthy's hint below. May 17, 2022 at 11:30

You have proved the inequality $$|f(x)-f(y)| \leq \frac 1 C |x-y|$$ only for a particular choice of $$x$$ and $$y$$. hat does not prove that $$f$$ is Lipschitz.
For a correct proof let $$x . we can divide the interval $$[x,y]$$ into subinetrvals $$[x_i,x_{i+1}]$$of length at most $$\delta=\epsilon C$$ using at most $$[\frac {y-x} {\delta}]+1$$ intervals. We then get $$|f(y)-f(x)| \leq \sum_i |f(x_{i+1})-f(x_i)| <\epsilon ([\frac {y-x} {\delta}]+1)$$. Can you complete the proof now?