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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!

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  • $\begingroup$ Yes, I guess I should just say for all integers n. I'll change that. $\endgroup$ – dcw Dec 13 '13 at 20:48
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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 $h(v+1)=h(v)+1$.

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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.

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  • $\begingroup$ $\large a^{x}$ 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
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Why not go to log-log scale, so to speak. If $x = \exp(u),$ then $g(2u) = 2 g(u),$ for appropriate choice of $g.$

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  • $\begingroup$ 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
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    $\begingroup$ 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

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