Answer to Question #242618 in Real Analysis for saduni

Question

discuss the continuity/uniform continuity/Lipchitz continuity and differentiability of the functions |x| on [−1,1].

Answer 1

ANSWER.Function $f(x)=|x|$ is Lipschitz continuous, therefore, uniformly continuous, hence continuous on [-1,1]. The function is not differentiable at all points of the [-1,1] (At the point x=0 the function is not differentiable).

EXPLANATION.

Function $f(x)=|x|$ is Lipschitz continuous, because for all ${ x }_{ 1 },{ x }_{ 2 }\in \left[ -1,1 \right] :\ \left| f({ x }_{ 1\ })-f\ ({ x }_{ 2 }) \right| =\left| \left| { x }_{ 1\ } \right| -\left| { x }_{ 2\ } \right| \right| \ \le \left| { x }_{ 1 }-{ x }_{ 2 } \right|$ (1).

Since for any $ε>0$ and $δ=ε$ for any ${ x }_{ 1 },{ x }_{ 2 }\in \left[ -1,1 \right]$ such that $\left| { x }_{ 1 }-{ x }_{ 2 } \right| <\delta \quad$from (1) it follows $\left| f({ x }_{ 1\ })-f\ ({ x }_{ 2 }) \right| =\left| \left| { x }_{ 1\ } \right| -\left| { x }_{ 2\ } \right| \right| < ε$ , then $f$ is uniformly continuous.

$f$ is differentiable on $[-1,0)$ and $(0,1]$ , because

$f(x)=\begin{cases} -x,-1\le x<0 \\ 0,\quad x=0 \\ x,\quad 0<x\le 1 \end{cases}, f'(x)=\begin{cases} -1,-1\le x<0 \\ 1\ ,\quad 0<x\le 1 \end{cases}$ .

Since ${ f }_{ - }^{ ' }(0)=\lim _{ x\rightarrow { 0 }^{ - } }{ \frac { f(x)-f(0) }{ x } } =\lim _{ x\rightarrow { 0 }^{ - } }{ \frac { -x-0 }{ x } =-1,\quad { f }_{ + }^{ ' }(0)= } \lim _{ x\rightarrow { 0 }^{ + } }{ \frac { f(x)-f(0) }{ x } } =\lim _{ x\rightarrow { 0 }^{ + } }{ \frac { \ x-0 }{ x } =\ 1 }$

then the left derivative is not equal right derivative. So , the function is not differentiable at the point $x=0.$