I am interested in knowing what we can say in general about when a continuous function $f:\mathbb{R} \to \mathbb{R}$ is differentiable.

To my mind, there are various ways a continuous function can fail to be differentiable. It could have a corner (i.e. its left-derivative is not equal to its right-derivative, but both exist). It could oscillate wildly, like $x\sin \dfrac{1}{x}$ at $x=0$. I'm not really sure if there are other options.

For instance, suppose we eliminate the oscillation by saying that $f$ is monotone on $[a,b]$. Can we then say for instance that its left-derivative exists almost everywhere?

  • $\begingroup$ Another option is that the slopes of the secant lines approach infinity, as is the case for $f(x)=x^{1/3}$ at $x=0$. $\endgroup$ – David Mitra Mar 15 '14 at 10:46

A monotone function is differentiable almost everywhere according to a theorem of Lebesgue. See here for an elementary proof.

  • $\begingroup$ Thanks! Are there more general results about continuous functions, if we look at say, only the left derivative existing? $\endgroup$ – Eric Auld Mar 15 '14 at 14:31
  • $\begingroup$ @EricAuld The Weierstrass function is continuous but has no one-sided derivative anywhere. However, if we assume the stronger property of absolute continuity, then the derivative (ordinary) exists almost everywhere. $\endgroup$ – user127096 Mar 16 '14 at 19:40
  • $\begingroup$ I'm having a hard time getting an intuitive understanding of this proof. I can talk myself through the argument and understand that it works, but I'm not really feeling it. Do you have any tips? $\endgroup$ – Eric Auld Apr 8 '14 at 23:12
  • $\begingroup$ A followup question would be whether the function can fail to be differentiable uncountably often. $\endgroup$ – Solomonoff's Secret Jan 1 '18 at 17:34
  • $\begingroup$ @Solomonoff's Secret: Such a function can fail to be differentiable continuum many times in every nonempty open interval, and even worse. In fact, "most" (in the Baire category sense) strictly increasing continuous functions have even worse behavior --- their continuity sets are first (Baire) category sets (i.e. meager sets). For more than you probably want to know, see Singular continuous functions. $\endgroup$ – Dave L. Renfro Jan 1 '18 at 18:42

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