# 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 $$ε>0, ∃ δ>0$$ s.t. $$∀ x,y∈I, |x−y|≤δ ⇒ |f(x)−f(y)|≤ε$$.

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

According to Richard Courant's book in Section 1.2, page 43, Uniform Continuity implies Lipschitz Continuity IF $$\delta$$, (called the "modulus of continuity") is such that $$\delta ≤ \epsilon*C$$, 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 $$\delta ≤ \epsilon*C$$ into the above definition, and I get the statement that $$∀ε>0$$ and for all $$x,y$$ is some closed interval $$I$$, $$|x−y|≤ε*C ⇒ |f(x)−f(y)|≤ε$$.

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

Well? Is it good?

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?