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

Question

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?

Answer 1

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 $x <y$. 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?