Proving uniform continuity of absolute value

Question

Prove that the function $f(x) = |x-a| - |x-b|$ is uniformly continuous on $\mathbb{R}$.

Answer 1

Fix $x,x_0 \in \mathbb{R}$. Let $\delta=\frac{\epsilon}{2}$, for some $\epsilon > 0$. If $|x-x_0| < \delta$, then \begin{align} |f(x)-f(x_0)|&=|(|x-a|-|x-b|)-(|x_0-a|-|x_0-b|)| \\ &= ||x-a|-|x-b|-|x_0-a|+|x_0-b)| \\ &= |(|x-a|-|x_0-a|)-(|x-b|-|x_0-b|)| \\ &\le ||x-a|-|x_0-a||+||x-b|-|x_0-b|| &\text{Triangle Inequality} \\ &\le |(x-a)-(x_0-a)|+|(x-b)-(x_0-b)| &\text{Reverse Triangle Inequality} \\ &= |x-x_0-a+a|+|x-x_0-b+b| \\ &=|x-x_0|+|x-x_0| \\ &=2|x-x_0| \\ &<2\delta \\ &=2\frac{\epsilon}{2} \\ &=\epsilon \end{align} Thus, by definition, $f(x)$ is uniformly continuous on $\mathbb{R}$.

Answer 2

Hints:

• $|f(x)-f(y)|\leq ||x-a|-|y-a||+||x-b|-|y-b||$.

• For every pair of real numbers $c$ and $d$ it is true that $||c|-|d||\leq |c-d|$.

Using this, you should be able to arrive at the expression $|f(x)-f(y)|\leq 2|x-y|$. This is sufficient to show that your function is uniformly continuous.