Consider the following function (I think it has a name, but I don't remember it): $$ f(x) = \cases{-1 & $x < 0$ \\ 0 & $x = 0$ \\ 1 & $x > 0$} $$

$f'(x)$ is zero everywhere except at $x=0$, where $f$ is not continuous. But suppose we ignore the right half of the real line and define $f(0)$ to be $-1$. Then $f$ has a left derivative at $x=0$, and it is zero. We can do the same thing from the right, so in a way it could make a little bit of sense to say that $f'(0) =0$.

Of course, I understand that going by the definition $f$ isn't differentiable at $x=0$. But one could imagine an alternative definition of derivative for discontinous functions, in which one calculates lateral derivatives by redefining the function to be continuous, and then we see if the lateral derivatives match. This doesn't always work; for example it's hard to meaningfully assign a derivative to $x \mapsto |x|$ at $x=0$.

Are there other functions with this property? Does it have a name?

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    $\begingroup$ Your $f$ is generally called $\operatorname{sgn}$ or the sign function. $\endgroup$ – Hagen von Eitzen Sep 24 '12 at 16:01
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    $\begingroup$ You might be interested in Dini derivatives and removable singularities, which is what your derivative function has. $\endgroup$ – Robert Mastragostino Sep 24 '12 at 16:17
  • $\begingroup$ Although this does not apply to your particular function, there is a notion of differentiability that does not imply continuity: approximate differentiability. (It does imply approximate continuity, though, and your function is not approximately continuous at zero.) $\endgroup$ – user31373 Sep 24 '12 at 17:37
  • $\begingroup$ @RobinSaunders Four years later, I don't really recall what I was trying to get at with the comment. Going to remove it. $\endgroup$ – rschwieb Aug 15 '16 at 17:07
  • $\begingroup$ Yes, sorry, I only noticed the date after responding! This question came up as related to a recent one, and I didn't think to check the date. $\endgroup$ – Robin Saunders Aug 15 '16 at 17:16

This "lateral derivative" approach is already taken in many analysis texts. Royden's analysis text does this, for example. (I though that I have seen them referred to as left and right "derivates", but I can't seem to come up with a source...)

By the way, differentiability of a function on its domain always implies it is continuous there. This works for "left differentiability implies left continuity" and on the right hand too. That is not contradicting your example, however!

If you "ignore the right half of the plane", you are effectively changing the domain to be only the half you are focused on, and indeed, you defined it to be continuous there.

  • $\begingroup$ Royden defines 4 derivates. 2 are on the left and if both exist and agree, then the left-hand derivative exists. Similarly the other 2 are used for right-hand derivative. See en.wikipedia.org/wiki/Dini_derivative $\endgroup$ – Graphth Sep 24 '12 at 16:18
  • $\begingroup$ @Graphth That must have been where I saw it, then :) Thanks! $\endgroup$ – rschwieb Sep 24 '12 at 16:34
  • $\begingroup$ As for his last example , the absolute value function $x \mapsto |x|$, that is a convex function, and for those there is the concept of subderivative. $\endgroup$ – kjetil b halvorsen Sep 24 '12 at 16:50

Interestingly, you can assign a derivative of the function $\operatorname{abs}$ at $0$ by using the following definition: $$\frac{\mathrm df(x)}{\mathrm dx}=\lim_{h\to0}\frac{f(x+h)-f(x-h)}{2h}.$$

Thus, taking the limit, $$\operatorname{abs}'(x)=\frac{\mathrm d}{\mathrm dx}|x|=\operatorname{sgn}(x)=\cases{-1, & $x < 0;$ \\ \phantom{-}0, & $x = 0;$ \\\phantom{-}1, & $x > 0.$}$$.

  • $\begingroup$ Well, but it's also true that for any constant $c$ in $(0,1)$, $\lim_{h\to0}\frac{f(x+ch)-f(x-(1-c)h)}{h}$ has the value $f'(x^{-}) + c(f'(x^{+}) - f'(x^{-}))$. So just as with any function with a jump discontinuity at 0, you can assign any intermediate value to $f'(0)$ (and still be locally monotone). $\endgroup$ – Robin Saunders Aug 15 '16 at 17:13

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