I am trying to show that given a uniformly continuous function $f:\mathbb{R}\to\mathbb{R}$, if we know a specific $\epsilon,\delta>0$ such that

$$|x-y|<\delta \implies |f(x)-f(y)|<\epsilon$$


$$|x-y|<c\delta \implies |f(x)-f(y)|<c\epsilon$$

for any positive real number $c$.

As with most proofs of this form. I am trying to prove this by showing this assumption is true for

  1. Positive integers.
  2. Positive Rational Numbers
  3. Positive Real Numbers

in that order.

The problem

Assume that $f:\mathbb{R}\to \mathbb{R}$ is uniformly continuous.

It can be shown

(Scaling Up $\epsilon$ and $\delta$)

If $$|x-y|<\delta \implies |f(x)-f(y)|<\epsilon $$ $$\forall x,y\in \mathbb{R}$$ then $$|x-y|<k\delta \implies |f(x)-f(y)|<k\epsilon$$ $$\forall x,y\in \mathbb{R}$$ for any positive integer k.

I am wondering if it is possible to show that given

(Scaling Down $\epsilon$ and $\delta$)

If $$|x-y|<\delta \implies |f(x)-f(y)|<\epsilon $$ $$\forall x,y\in \mathbb{R}$$ then $$|x-y|<\frac{1}{k}\delta \implies |f(x)-f(y)|<\frac{1}{k}\epsilon$$ $$\forall x,y\in \mathbb{R}$$ for any positive integer k.

Personal Thoughts

Part of me thinks that I should be able to transform the scaling up argument into a scaling down argument. Possibly by applying a similar trick to the range. I am unsure.

Source material/ Definitions/ Etc

(Uniformly Continuous Definition) Given $\epsilon>0$ we can find $\delta>0$ such that $$|x-y|<\delta \implies |f(x)-f(y)|<\epsilon\text{ }$$ $$\forall x,y\in \mathbb{R}$$

( Scaling Up $\epsilon$ and $\delta$: Proof) Assume without loss of generality that $x<y$ and pick $x_1,...,x_{k-1}\in \mathbb{R}$ such that $$|x-x_1|,|x_1-x_2|,...,|x_{k-1}-y|<\delta$$ then $$|f(x)-f(x_1)|,...,|f(x_{k-1})-f(y)|<\epsilon$$ and thus $$|f(x)-f(y)|\leq |f(x)-f(x_1)|+...+|f(x_{k-1})-f(y)|<k\epsilon$$


2 Answers 2


Unfortunately, your statement is false. If it is true then $|x-y| \leq c \delta$ implies $|f(x)-f(y)| \leq c\epsilon$ by continuity. Now let $x,y \in \mathbb R$ be arbitrary and take $c=\frac {|x-y|} {\delta}$. We get $|f(x)-f(y)| \leq \frac {|x-y|} {\delta}\epsilon$. This proves that $f$ is a Lipschitz function on $\mathbb R$. But there are easy examples of uniformly continuous functions which are not Lipschitz.

One specific example: $f(x)=\sqrt x$ for $0\leq x \leq 1$, $0$ for $x \leq 0$ and $1$ or $x \geq 1$.


Counterexample :

$√:[0, 1]\to \Bbb{R}$ is uniformly continuous.

Choose $\epsilon=\frac{1}{2}$ then $|√x-√y|<\frac{1}{2}$ whenever $|x-y|<\frac{1}{4}$


Now for $\epsilon'=\frac{\epsilon}{2}$

Then $|√x-√y|<\frac{1}{4}$ whenever $|x-y|<\frac{1}{16}$

Here $\delta'=\frac{1}{16}$ But $\delta'<\frac{\delta}{2}=\frac{1}{8}$

We can find two points $x, y$ such that $|x-y|<\frac{\delta}{2}$ but $ |√x-√y|>\frac{\epsilon}{2}$

Choose $x=\frac{1}{12},y=0$ then $|x-y|<\frac{1}{8}$ but $|√x-√y|>\frac{1}{4}$

Conclusion : $|√x-√y|<\frac{1}{2}$ whenever $|x-y|<\frac{1}{4}$ but $|x-y|<\frac{1}{8}$ doesn't imply $|√x-√y|<\frac{1}{4}$

For $\epsilon=\frac{1}{2}, k=\frac{1}{2}$ for the function $√:[0, 1]\to\Bbb{R}$ your result is false.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .