Uniform $f$ is uniformly continuous

Question

$f$ is a map from $f:\mathbb{R} \rightarrow \mathbb{R}$. Let $f$ be continous and assume $f$ is both uniformly contiuous on $(-\infty,0]$ and $[0,\infty)$ . Show that $f$ is uniformly continuous on all of $\mathbb{R}$.

I thought that the function $f$ is actually uniformly continuous over all the real numbers since it is in both the intervals $(-\infty,0]$ and $[0,\infty)$. I guess the "problem" arises on $f(0)$.

I have been trying to prove this for hours upon hours now but I just do not get it even though I get the definitions and that you should assign 3 $\delta$‘s to cover every case (I think) and then a $\delta$ to be the minimum of one of thoes. I can not show how to pick the $\delta$’s as this is hard for me.

Answer 1

Let $\epsilon >0$. There exist $\delta_1$ abnd $\delta_2>0$ such that

1. $|f(x)-f(y)| <\epsilon/2$ if $x, y \leq 0$ and $|x-y| \leq \delta_1$

2. $|f(x)-f(y)| <\epsilon/2$ if $x, y \geq 0$ and $|x-y| \leq \delta_2$

Let $\delta=\min\{\delta_1,\delta_2\}$. Suppose $|x-y| <\delta$. We claim that $|f(x)-f(y)| <\epsilon$. The only case where you face difficulty is when one of $x,y$ is $\leq 0$ and the other is $\geq 0$. Supose $x \leq 0$ and $y \geq 0$. Then $|f(x)-f(y)| \leq |f(x)-f(0)|+|f(0)-f(y)|<\epsilon /2+\epsilon /2=\epsilon$.

Answer 2

Given $\varepsilon>0$, take $\delta>0$ such that $|x-y|\implies\bigl|f(x)-f(y)\bigr|<\frac\varepsilon2$ both on $[0,\infty)$ and on $(-\infty,0]$. Then if $x,y\in\Bbb R$ are such that $|x-y|<\delta$ and if $x\leqslant0\leqslant y$, then$$\bigl|f(x)-f(y)\bigr|\leqslant\bigl|f(x)-f(0)\bigr|+\bigl|f(0)-f(y)\bigr|<\varepsilon$$and you have a similar argument if $x\geqslant0\geqslant y$.