Exercise 1.6.3 (Rolle's theorem fails for almost everywhere differentiability)

Question

Give an example to show that Rolle's theorem can fail if $f$ is continuous, yet only almost everywhere differentiable.

Elu Thingol · Accepted Answer

Consider the absolute value function on the interval negative and positive unit intervals:
$$f:[-1,1]	o[0,1], \quad f(x) = |x| = \left\{\begin{array}{ll} -x &	ext{if } x<0\ 0 &	ext{if } x=0\ x & 	ext{if } x>0\end{array}ight.$$
\begin{figure}
\includegraphics[width=0.33	extwidth]{tao_measure_exercise_1-6-3.png}
\caption{Plot of the function $f(x) = |x|$ which is continuous, differentiable almost everywhere but not differentiable everywhere.}
\end{figure}
Then $f$ is continuous, differentiable on the intervals $x\in [-1,0)$ with $f'(x)=-1$, and on interval $x\in (0,1]$ with $f'(x) = 1$ (cf. \href{../analysis_I}{Analysis I}, Theorem 10.1.13). However, $f$ is not differentiable at $x=0$ since
$$\lim_{x	o 0 -}\frac{f(x)-0}{x-0} = \lim_{x	o 0 -}\frac{-x-0}{x-0} 
eq \lim_{x	o 0 +} \frac{x-0}{x-0} = \lim_{x	o 0 +}\frac{f(x)-0}{x-0}.$$
Thus, $f$ is a continuous function which is almost everywhere differentiable, not everywhere differentiable and its derivative does not become zero on any of the points.

Exercise 1.6.3 (Rolle's theorem fails for almost everywhere differentiability)

Answers

Comments

Exercise 1.6.3 (Rolle's theorem fails for almost everywhere differentiability)

Answers

Comments

Add answer