# Linearity of Uniformly Continuous functions

### Context

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

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

then

$$|x-y|

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 $$|x-y| $$\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 $$x and pick $$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)|

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 $$|x-y|<\frac{1}{4}$$

$$\delta=\frac{1}{4}$$

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.