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We see that we can find a function nowhere differentiable or finitely not differentiable. But I want to understand, can a continuous function on $[0,1]$ be constructed which is differentiable exactly at two points in $[0,1]$? How can I construct such a function which is finitely differentiable?

Jonas Meyer
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Sayantan Koley
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    Just multiply the Weierstrass function by a function that tends to zero quickly enough at exactly two points, say $x^2(x-1)^2$ and use that Weierstrass' is Holder. – Pp.. Jan 21 '15 at 03:40
  • @Pp..: That works but there is no need to square, or to use Holder. – Jonas Meyer Jan 21 '15 at 04:30

1 Answers1

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Take a function that is continuous everywhere and differentiable nowhere, call it $g$.

Take $2$ points, $a$ and $b$. Then $f(x) = (x-a)(x-b)g(x)$ is continuous everywhere, differentiable only at $a$ and $b$.

You can use continuity of $g$ and the definition of the derivative to verify very quickly that $f$ is differentiable at $a$ and $b$. To see that $f$ is differentiable nowhere else, you could note that if $f$ were differentiable at $c\not\in \{a,b\}$, then so would be $g(x) =\dfrac{f(x)}{(x-a)(x-b)}$ by the quotient rule.

Jonas Meyer
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  • how can a function be continuous everywhere and differentiable nowhere? Does such a function exist? – Jason Jan 21 '15 at 18:15
  • @Jason: The most famous example was brought to us by Weierstrass, but there are many others. They seem rare from ordinary experience in calculus, and yet in some sense, differentiability is such a strong condition that it is the rare special case (in some precise senses like Baire category). – Jonas Meyer Jan 21 '15 at 18:25