11

Let $f :\mathbb{R}→ \mathbb{R}$ be a function such that $f^2$ and $f^3$ are differentiable. Is $f$ differentiable?

Similarly, let $f :\mathbb{C}→ \mathbb{C}$ be a function such that $f^2$ and $f^3$ are analytic. Is $f$ analytic?

Julien
  • 44,791
ROBINSON
  • 2,269
  • 3
    What do you mean by $f^p$? Is it the $p^{\text{th}}$ power or the $p^{\text{th}}$ iterate? If it's just the $p^{\text{th}}$ power, then $|x|$ is a counter example for the first question, which is not so exciting. – Mark McClure May 20 '13 at 02:02
  • @julien Yes, I like your answer. I make the point mainly because I assumed the question refers to iterates, rather than powers. It's a natural question as well. – Mark McClure May 20 '13 at 02:09
  • @MarkMcClure Right. But only ROBINSON knows. I took a chance. – Julien May 20 '13 at 02:11
  • @julien Exactly - which is why asked! In the meantime, I'll just sit on my fabulous example. :) – Mark McClure May 20 '13 at 02:13
  • @MarkMcClure He, he. Or you can answer the "iterate" version and we'll see what the OP says when he/she comes back. – Julien May 20 '13 at 02:14
  • Very related: http://mathoverflow.net/questions/125861/f3-f2-are-the-cube-and-quadratic-of-f-respectively-and-both-infinite-differen – Andrés E. Caicedo May 20 '13 at 02:24
  • @AndresCaicedo I like the fact that you also link to a related question in that link. We have the beginning of an iteration here. – Julien May 20 '13 at 02:36
  • The analytic version of this was asked on this site not too long ago —see here http://math.stackexchange.com/q/348627/274 — and that was already a duplicate of an older question. Please do search for old questions before asking! – Mariano Suárez-Álvarez May 20 '13 at 15:43

2 Answers2

14

In short: not necessarily in the real case, but yes in the complex case.

In the real case like in the complex one, $f$ is differentiable whenever it does not vanish and $$ f'=\left(\frac{f^3}{f^2}\right)'=\frac{(f^3)'f^2-f^3(f^2)'}{f^4}. $$

Real case: if $f$ vanishes, it no longer needs to be differentiable. For an alternative example to $f(x)=|x|$, observe that $$ f(x)=x\sin\left(\frac{1}{x}\right)\quad \forall x\neq 0\qquad f(0)=0 $$ is not differentiable at $0$, while $f^2$ and $f^3$ are differentiable everywhere.

Complex case: the zeros of $f^2$ and those of $f$ coincide. If $f$ is constant equal to $0$, the result is clear. Otherwise, its zeros are isolated since $f^2$ is holomorphic non constant equal to $0$. On the open set which is the complement of these zeros, the argument above shows that $f$ is holomorphic. And $f^2$, whence $f$, is bounded near each zero. So these are removable singularities. Hence $f$ is holomorphic on its domain.

Edit: for a different argument, see this thread.

Note: if you replace the assumptions by $f^2$ (or $f^n$, $n\geq 2$, more generally) differentiable and $f$ continuous, you get the same conclusions in both cases. Indeed, where $f(x_0)\neq 0$, we have $$ \frac{f(x)-f(x_0)}{x-x_0}=\frac{f^2(x)-f^2(x_0)}{x-x_0}\cdot\frac{1}{f(x)+f(x_0)}\longrightarrow \frac{(f^2)'(x_0)}{2f(x_0)}. $$ But if you remove the continuity assumption on $f$, the complex case can fail as well. Just pick a random square root of $z$.

Community
  • 1
Julien
  • 44,791
4

If you do want $f^{p}$ to represent the $p^{\text{th}}$ iterate, then you can let $f$ denote the characteristic function of the irrational numbers. Then $f^2$ and $f^3$ are both identically zero, yet $f$ is nowhere continuous, let alone differentiable.

Mark McClure
  • 30,510
  • Nicely done, +1. You'll notice that my wonderful complex example was false...So after all $|x|$ is somehow the best. Yes, in the complex case, we must have $f$ holmorphic no matter what. – Julien May 20 '13 at 03:43
  • @julien Thanks! Judging from the real/complex distinction, though, I'd say that the power interpretation is the more probable. – Mark McClure May 20 '13 at 03:59
  • I really like your example. Just for the record, we can find a better behaved counterexample, namely $f$ continuous. For instance, we can glue $|x+1|$ on $[-\infty,-1/2]$ and $x^2$ on $[-1/4,+\infty)$ smoothly with $f\geq 0$ on $\mathbb{R}$. Then $f$ is not differentiable at $-1$, but $f^2(x)$ and $f^3$ are smooth. – Julien May 20 '13 at 16:32