I understand that differentiability implies continuity, whereas the converse isn't true. But must a continuous function have both a left and right derivative, not necessarily equal to one another?

  • 6
    $\begingroup$ $f(x) = x \sin(1/x),$ with $f(0)$ defined to be $0,$ is continuous and has neither a left nor a right derivative at $x=0.$ $\endgroup$ – Dave L. Renfro Feb 21 '14 at 22:12
  • $\begingroup$ There is a nice graph of the function I gave in the StackExchange question Calculating Dini derivatives (which I found by doing a google image search for "right derivative" AND "sin(1/x)". $\endgroup$ – Dave L. Renfro Feb 21 '14 at 22:28

For $f(x)=x \sin{\frac{1}{x}}$, $f(0)=0$ (Just like Dave commented) you have: $$\frac{f(0 + \Delta x) - f(0)}{ \Delta x} = \frac{(0 + \Delta x) \sin (\frac{1}{0 +\Delta x}) - 0}{\Delta x} = \sin (\frac{1}{\Delta x})$$ As $\Delta x$ approaches $0$ from the left/right you will just get $- \sin(x)$ / $\sin(x)$ as $x$ tends to $\infty$, i.e, no limit.

But, let $\epsilon > 0$ and $ \delta = \epsilon$: $$|x|=|x-0| < \delta \implies |x \sin (\frac{1}{x}) - 0| = |x \sin (\frac{1}{x})| \leq |x| < \delta = \epsilon $$ Hence $f$ is continous at $0$.

  • $\begingroup$ This does not address the OP's question at all. $\endgroup$ – Mark Viola Mar 29 at 22:17

The famous Weierstrass function, which is continuous but nowhere differentiable, fails everywhere to have left- and right-derivatives, as well. This takes a bit of work to check.

  • $\begingroup$ I wouldn't have believed it was possible at the time $\endgroup$ – user10389 Feb 22 '14 at 13:42

If $f$ has a left and right derivative at every point, then it certainly has finite Dini derivatives at every point, and then it follows immediately from the Denjoy-Young-Saks Theorem that $f$ is differentiable away from a set of measure $0$. But as pointed out in Ted Shifrin's answer, there are certainly continuous functions which are nowhere differentiable, e.g. Weierstrass's function.

[Or in other words: it is relatively common to describe Weierstrass-like nowhere differentiable functions as "having corners at every point". The above paragraph shows that this is very loose language: taken literally, there are no such functions!]

Thus the Weierstrass function is continuous everywhere and certainly has left and right hand derivatives only at a set of measure zero. This checks a weak form of Ted's answer, but enough to use the Weierstrass function to answer the OP's question...lazily.

Really though this answer is almost a joke: I believe I am applying too big a theorem to deduce too small a result. When I first read the question I was tempted to say that a function which has (finite!) left and right hand derivatives at every point must be differentiable on the complement of a countable set, just like a function which has left and right hand limits at every point must be continuous except at countably many points. Is that actually true? (Added: Hmm, I think I've changed my guess; I now doubt that's true. But I'd still like to know.)


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.