I have a short question, related to the ongoing search of mathematics instructors for counter-examples to common undergraduate mistakes.

The classical example of a function that is differentiable everywhere but has discontinuous derivative is \begin{equation} f(x)=\left\{ \begin{array}{cc} x^2\sin(1/x) &(x\neq0), \\ 0 &(x=0), \end{array}\right. \end{equation} which has derivative \begin{equation} f'(x)=\left\{ \begin{array}{cc} 2x\sin(1/x)-\cos(1/x) &(x\neq0), \\ 0 &(x=0). \end{array}\right. \end{equation} $f'$ fails to be continuous at $0$ purely because its left- and right-hand limits do not even exist at $0$.

However, suppose that we have found a function $g$ whose derivative $g'$ has finite but unequal left- and right-hand limits at some cluster point $x_0$ in its domain. May we conclude that $g$ is not differentiable at $x_0$?

If this is not the case, is there a simple counter-example? (I'm guessing such a counter-example ought to be more complicated than the $f$ I have given above, as $f$ is sometimes claimed to be the simplest example of a differentiable function with discontinuous derivative.)

Thanks in advance!

  • $\begingroup$ This would produce a $g$ with a cusp, i.e., a point at which it is not differentiable. $\endgroup$ – Stromael Apr 24 '15 at 11:11
  • $\begingroup$ I have to say, deleting comments is very annoying. $\endgroup$ – Stromael Apr 24 '15 at 13:29
  • $\begingroup$ It's desirable, actually, when they do not help at all. $\endgroup$ – Git Gud Apr 24 '15 at 16:44

Suppose $f$ is differentiable over $(x_0-k,x_0+k)$, for some $k>0$, and that $$ \lim_{x\to x_0^-}f'(x_0)=l \qquad\text{and}\qquad \lim_{x\to x_0^+}f'(x_0)=r $$ with $l\ne r$. It is not restrictive to assume $l<r$ (otherwise take $-f$).

Take $\varepsilon=(r-l)/4$. Then there exists $\delta$ with $0<\delta<k$ such that \begin{align} f'(x)&<l+\varepsilon &&\text{for every $x\in(x_0-\delta,x_0)$}\\ f'(x)&>r-\varepsilon &&\text{for every $x\in(x_0,x_0+\delta)$} \end{align}

This contradicts Darboux’s theorem, because $f'(x)$ should assume every value in the interval $(f'(x_0-\delta/2),f'(x_0+\delta/2))$, for some $x\in(x_0-\delta/2,x_2+\delta/2)$. But we see that infinitely many values are missed, as $f'(x_0)$ can only provide one of them.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.