Suppose a continuous function $f:[0,1]\rightarrow [{0,1}]$ satisfies the functional equation $f(x^2)=f(x)^2$. Then I conjecture that we must have $f(x)=0$ or $f(x)=x^r$ for some real number $r\geq 0$. However, I haven't the foggiest idea how to prove (or disprove) that conjecture. Can anyone offer such a proof, or produce a counterexample? (If necessary, it's fine to assume that f is also differentiable.)

One helpful thing: It follows from the original functional equation that $f(x^{2^n})=f(x)^{2^n}$ for every integer $n$.

Thanks in advance!

  • $\begingroup$ Yes, I guess I should just say for all integers n. I'll change that. $\endgroup$ – dcw Dec 13 '13 at 20:48

There are many function $f(x)$ not of the form $x^r$ that satisfy $f(x^2)=f(x)^2$.

We begin with Igor Rivin's suggestion to make a logarithmic substitution. Actually, we can make a double logarithmic substitution to simplify the functional equation even further.

Let's use the extended real numbers $[-\infty,\infty]:=\mathbb{R}\cup\{-\infty,\infty\}$, which have the homeomorphism type of a closed interval. We extend the exponenential and logarithm functions in the usual way (i.e., $2^\infty=\infty$, $2^{-\infty}=0$, $\log 0=-\infty$, $\log \infty=\infty$). The extensions remain continuous.

Let $\varphi:[0,1]\to [-\infty,\infty]$ be given by $$ \varphi(x)=\log_2(-\log_2 x). $$ This is a homeomorphism, with inverse $$ \varphi^{-1}(v)=2^{-(2^v)}. $$

The map $\varphi$ gives a correspondence between continuous $f:[0,1]\to [0,1]$ and continuous $h:[-\infty,\infty]\to[-\infty,\infty]$, given by $$ h=\varphi\circ f\circ\varphi^{-1}. $$ It can be checked that $f(x^2)=f(x)^2$ for all $x\in[0,1]$ if and only if $h(v+1)=h(v)+1$ for all $v\in[-\infty,\infty]$ (with the convention $-\infty+1=-\infty$ and $\infty+1=\infty$).

If $h(v+1)=h(v)+1$, the $h$ is determined by its restriction to any interval of length 1, say $[0,1]$. A continuous function $j:[0,1]\to [-\infty,\infty]$ is the restriction of some $h$ satisfying $h(v+1)=h(v)+1$ if and only if $j$ is either identically $-\infty$, identically infinity, or finite everywhere. When $j$ is identically $\pm\infty$, $f(x)$ is identically $0$ or $1$, so we exclude these cases.

Explicitly: let $h:[0,1]\to (-\infty,\infty)$ be any continuous function satisfying $h(1)=h(0)+1$. We extend $h$ to a continuous function $(-\infty,\infty)\to (-\infty,\infty)$ by $$ h(v)=h\left( v-\lfloor v\rfloor\right)+\lfloor v\rfloor. $$ Finally, we set $h(-\infty)=-\infty$, $h(\infty)=h(\infty)$. Then the function $f:[0,1]\to[0,1]$ given by $f(x)=\varphi^{-1}(h(\varphi(x)))$ satisfies $f(x^2)=f(x)^2$.

As a special case: if $h(v)=v+\lambda$ for some fixed $\lambda\in\mathbb{R}$, then$f(x)=x^{2^\lambda}$, so we get every function $f(x)=x^r$ (for $r>0$) this way.

To get an $f$ not of this form, we need a different $h$. For example, we could take $h(v)=v+\sin(2\pi v)$, which is continuous and satisfies $h(v+1)=h(v)+1$.


Hint : $f(0)=f(0^2)=f(0)^2\iff f(0)\in\{0,1\}$ ; $f(1)=f(1^2)=f(1)^2\iff f(1)\in\{0,1\}$. Also, $f(x^{2n})=f(x^n)^2$. Obviously, $f(x)=x^n$ is one such solution. So is $f(x)=0$. Since $f(x)$ is in $[0,1]$ as well, we also have $f(f(x^2))=f(f^2(x))=f^2(f(x))$, and similarly for greater nesting levels.

  • $\begingroup$ $\large a^{x}$ is $0$ k. too. $\endgroup$ – Felix Marin Dec 13 '13 at 20:44
  • $\begingroup$ @FelixMarin: $a^1=a\not\in\{0,1\}$, unless $a=1$, in which case we have $f(x)=1^x=1=x^0=x^n|_{n=0}$ Also, $f(x)=0$ is another trivial solution. $\endgroup$ – Lucian Dec 13 '13 at 20:47
  • $\begingroup$ Fine. A constant. Thanks. $\endgroup$ – Felix Marin Dec 13 '13 at 20:49
  • $\begingroup$ @Lucian: Yes, that jumps out rather nicely. The problem is showing whether or not $f(x)=x^n$ and $f(x)=0$ are unique solutions. Perhaps I should have stated another useful fact in the post: that if we assume f isn't constant, by continuity we must have $f(0)=0$ and $f(1)=1$. $\endgroup$ – dcw Dec 13 '13 at 20:55

Why not go to log-log scale, so to speak. If $x = \exp(u),$ then $g(2u) = 2 g(u),$ for appropriate choice of $g.$

  • $\begingroup$ One could even use, if you'll pardon the terminology, $\log\log$-$\log\log$ scale: if we normalize appropriately, we get a functional equation like $h(v+1)=h(v)+1$. Or perhaps you meant this by $\log$-$\log$ scale? $\endgroup$ – Julian Rosen Dec 13 '13 at 20:50
  • $\begingroup$ @Igor Rivin: could you please clarify what you mean? I'm unfortunately quite inexperienced with this sort of problem. Do you propose using $g(u)$ as a counterexample, or to produce a new example, or for some other purpose? $\endgroup$ – dcw Dec 13 '13 at 21:02
  • $\begingroup$ @JulianRosen No, I did not mean that, and I agree that an extra log is even better. $\endgroup$ – Igor Rivin Dec 13 '13 at 21:02
  • $\begingroup$ @JulianRosen: How would we normalize to get the equation you described? I'm not very familiar with solution methods for functional equations. $\endgroup$ – dcw Dec 13 '13 at 21:06
  • 1
    $\begingroup$ To be explicit: if we set $h(v)=\log_2\left(-\log_2 f\big(2^{-2^v}\big)\right)$, then $h(v+1)=h(v)+1$. $\endgroup$ – Julian Rosen Dec 13 '13 at 21:11

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.