# Prove ax+b is uniformly continuous on R

Let f(x) = ax+ b Let any x,y ∈ R. Let ε > 0 be abritraty

|f(x) – f(y)| = |ax+b – (ay+b)|

= |ax-ay| < ε

= |a(x-y)| < ε

≤ |a||x-y| < ε

If |a||x-y| < ε then |x-y| < ε / |a|


Let δ = ε / |a| > 0 |f(x)-f(y) < ε for |x-y|< δ where δ = ε / |a|

• You should distinguish the case when $a=0$, but in that case it is obvious. – egreg Dec 10 '16 at 20:26
• You can just proves that $f$ is Lipschitz continuous, which implies uniform continuity. – Henricus V. Dec 11 '16 at 2:48

I agree with your answer. Here is how I would have formulated it.

We need to prove that $$\forall \epsilon > 0 \ \exists \delta : \forall x,y \in \mathbb{R} \ |x-y| < \delta \Rightarrow |f(x) - f(y)|< \epsilon$$

Now let $\epsilon > 0$ and assume $a \not = 0$, choose $\delta = \frac{\epsilon}{|a|}$, then

$$|x-y| < \delta = \frac{\epsilon}{|a|} \Leftrightarrow |a| |x - y | = | ax - ay| = |(ax - b) -(ay - b)| < \epsilon$$

Thus $\delta$ does not depend on $x,y$.

As pointed out in the comment under this answer, for $a = 0$, you can choose $\delta$, for example, to be $1$.

• For $a\ne0$, of course. For $a=0$, just take $\delta=1$ or whatever. – egreg Dec 10 '16 at 20:27
• You are correct, Thanks for pointing it out @egreg. – Olba12 Dec 10 '16 at 20:33
• Doesnt for a=0, any value of δ work? So could I just show the equation and say "any value works" or should I show an actual value for δ? – Aggrawal Puja Dec 10 '16 at 20:40
• Yes, indeed, any works. But if you read the defintion, it says $\exists \delta$ which means that there is atleast one. Hence it is sufficient to show the uniform continuity for only one $\delta$ of your choosing. @AggrawalPuja – Olba12 Dec 10 '16 at 20:42