# Define $f(x)=mx+b$ for all $x$. Prove that the function $f:\mathbb{R}\rightarrow\mathbb{R}$ is uniformly continuous.

Define $f(x)=mx+b$ for all $x$. Prove that the function $f:\mathbb{R}\rightarrow\mathbb{R}$ is uniformly continuous.

Let $\{u_n\}$ and $\{v_n\}$ be sequences of $\mathbb{R}$ and converges to $x_0$, then we have $$\lim\limits_{n\rightarrow\infty} (u_n-v_n)=\lim\limits_{n\rightarrow\infty} u_n-\lim\limits_{n\rightarrow\infty} v_n = x_0-x_0=0$$ next we have: $$\lim\limits_{n\rightarrow\infty} (f(u_n)-f(v_n))=\lim\limits_{n\rightarrow\infty} f(u_n)-\lim\limits_{n\rightarrow\infty} f(v_n)=(mx_0+b)-(mx_0+b)=0$$

Hence, $f(x)$ is continuous.

Can anyone check my solution? I am not sure that I have show enough information, I just follow the definition. Thanks

• The proof of continuity looks ok. But uniform continuity is a much stronger notion, which you haven't yet proved. Oct 26 '15 at 0:57
• Use the $\varepsilon$-$\delta$-definition you got in your lecture for sure. It is a straightforward calculation and a good opportunity to really understand the definition. Oct 26 '15 at 0:59
• @NateRiver Can you give me a hit about how to show $f(x)$ is uniformly continuous? Oct 26 '15 at 0:59
• @JohnSmith I haven't covered $\epsilon-\delta$ definition yet Oct 26 '15 at 1:00
• Ok, how did you define uniform continuity? Oct 26 '15 at 1:01

## 2 Answers

You've already shown continuity. I'll stick to your definition. Let $(u_n)$ and $(v_n)$ be real sequences with $\lim_{n\to\infty} \vert u_n-v_n \vert=0$. We have: $$\vert f(u_n)-f(v_n)\vert=\vert mu_n+b-mv_n-b\vert = \vert m(u_n-v_n) \vert \to 0$$ as $n\to\infty$.

• Dear @John I've made some necessary edits in your answer,hope you will not mind it. Oct 26 '15 at 1:15
• @ArpitKansal Sure, whatever is needed for clarification should be added! Oct 26 '15 at 1:45

Hint: For any $x$ and $y$ in $\mathbb R$,note that $\vert f(x)-f(y) \vert \leq m \vert x-y \vert$.Hence $f$ is Lipschitz continuous which is stronger than uniform Continuity.