I am new in this forum. My question: Suppose a real valued function $f: \mathbb{R} \rightarrow \mathbb{R}$ is continuous everywhere. Is it possible to construct $f$ that is differentiable at only one point? If possible, please give an example.

Note: I am aware that there is a function which is differentiable at a single point but discontinuous elsewhere. I also know about Weierstrass function that continuous everywhere but nowhere differentiable. But is there a function which is continuous but only differentiable in one point?

In fact, I found this discussion but unfortunately it still does not give a definitive answer. Moreover they consider only in an interval, whereas my problem is for the entire domain. Thank you very much

  • 1
    $\begingroup$ Another answer can be found in the last sentence here. $\endgroup$ – Jonas Meyer Feb 12 '12 at 4:39

It is certainly possible. Fix a nowhere-differentiable function $f$ such that $0\leq f(x)\leq 1$ for all $x$. Now consider $x^2f(x)$. This is differentiable at $0$ but nowhere else. You can verify it is differentiable at $0$ using the limit definition of derivative. $$\lim_{h\to 0} \frac{h^2f(h)-0^2f(0)}{h}=\lim_{h\to 0}hf(h)$$ and $0\leq f(h)\leq 1$ implies $0\leq hf(h)\leq h$. So the limit goes to $0$ by the squeeze theorem.

To see it is not differentiable elsewhere is a slightly harder exercise. Suppose $x^2f(x)$ is differentiable at $x\neq 0$. Then $$\lim_{h\to 0} \frac{(x+h)^2f(x+h)-x^2f(x)}{h}=L.$$ Adding and subtracting a mixed term $x^2f(x+h)$ in the middle, this becomes $$ \lim_{h\to 0} \frac{(x+h)^2f(x+h)-x^2f(x+h)}{h}+\frac{x^2f(x+h)-x^2f(x)}{h}=L $$ The left-hand term limits to $2x f(x).$ The right-hand term limits to $x^2f'(x)$. This implies that $f'(x)$ exists, since it is equal to $x^{-2}(2xf(x)-L)$. (This fails for $x=0$.)

  • 2
    $\begingroup$ Presumably you mean to take $f$ continuous as well? In that case $xf(x)$ works. $\endgroup$ – Jonas Meyer Feb 12 '12 at 4:36
  • 1
    $\begingroup$ Thanks @Jonas for the improvement. I was fixated on getting a derivative of $0$... $\endgroup$ – Cheerful Parsnip Feb 12 '12 at 4:43
  • 11
    $\begingroup$ To show that $x^2 f(x)$ is not differentiable anywhere else, just use the fact that $g(x) h(x)$ is differentiable where $g$ and $h$ are, with $g(x) = x^2 f(x)$ and $h(x) = 1/x^2$. $\endgroup$ – Robert Israel Feb 12 '12 at 4:55
  • $\begingroup$ Thanks for your answers. Could you please give an example what $f$ is? If it is assumed to be continuous as @Jonas said, should it be Weierstrass function? Or can it be discontinuous function? $\endgroup$ – netsurfer Feb 12 '12 at 4:56
  • 1
    $\begingroup$ @netsurfer: If you take $x^2$ as Jim did, it could be any nowhere differentiable function that is bounded in a neighborhood of $0$. If you want the result to be continuous, then $f$ should be taken to be a continuous nowhere differentiable function, of which the Weierstrass function is only one example while any would do. And if $f$ is continuous, then $xf(x)$ is differentiable at $0$ with derivative $f(0)$, but not differentiable elsewhere because $f(x)=\frac{1}{x}\cdot xf(x)$ when $x\neq 0$ and $\frac{1}{x}$ is differentiable (see Robert's comment). $\endgroup$ – Jonas Meyer Feb 12 '12 at 5:01

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.