# $f^{(n)}(x)$ exist $\forall x \in I$, an interval in $\mathbb{R}$ but $f^{(n+1)}(x)$ does not exist for any $x \in I$

$f'(x)$, $f''(x)$, $f^{(3)}(x)$, ..., $f^{(n)}(x)$ all exist $\forall x \in I$, an interval in $\mathbb{R}$ but $f^{(n+1)}(x)$ does not exist for any $x \in I$

Can such a function exist? I think it can't but can't think of the reason why. I can construct a function with countably infinite points in I where the function is only differentiable n times, but, it seems impossible to have every point in the interval, I, have this property.

• Have you heard of Weierstrass' nowhere differentiable, continuous function? (Integrate it.) – David Mitra Apr 18 '18 at 11:23
• Yes. Is there a way to smooth it out so it can be differentiated a finite number of times before failing at every point? – futurebird Apr 18 '18 at 11:24

## 2 Answers

Yes. Take the Weierstrass function $f$. It is differentiable nowhere. If $F$ is a primitive of $f$ (which exists, since $f$ is continuous), then $F'$ exists, but $F''$ exists nowhere.

Take any continuous, nowhere differentiable function (e.g. Weierstrass function). Then integrate it $n$ times. First time you can do it, as the function is continuous. Later, you are integrating differentiable, so continuous, functions.

• How do I know the first anti-derivative of the Weierstrass function is also continuous? I understand why I can integrate it once, but am still trying to understand doing it more than once. – futurebird Apr 18 '18 at 11:35
• The integral of a continuous function is differentiable, so is continuous and hence integrable. – B. Mehta Apr 18 '18 at 11:44
• If $F(x) = \int_0^x f(t)\;dt$, then $F'(x) = f(x)$ provided $f$ is continuous at the point $x$. Perhaps your calculus textbook contains this fundamental theorem. – GEdgar Apr 18 '18 at 11:49