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I was wondering if there is a continuous function $f:[0,1]\to\mathbb R$ satisfying $$f(x)^2=f\left(x^2\right)\text,$$ for all $x\in[0,1]$, $f(0)=1$ and $ f(1)=0$.

Clearly, some easy functions like polynomials are not satisfied. I guess there is a way to construct an example since it only needs a continuous function.

Mohsen Shahriari
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Luke Chen
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    Out of curiosity, what makes it obvious for polynomials? –  Jul 20 '19 at 00:17
  • @Gae.S. $f(x^2)$ only has even degrees – Luke Chen Jul 20 '19 at 10:35
  • @LukeChen ... and so may $(f(x))^2$. –  Jul 20 '19 at 12:02
  • @Gae.S. Actually if $[f]^2=f(x^2)]$, then $f$ only has even degrees. Let $f=a_0+a_1x^2+...+a_nx^(2n)$ and assume $a_0$ not equal to 0, when $[f]^2=f(x^2)$, consider the coefficient of $x^2$ in $[f]^2$ and it implies $a_1=0$. Repeat the operation and we get all coefficients are zero and it contradicts. – Luke Chen Jul 21 '19 at 02:19
  • @Gae.S. Also we can do derivation to both sides and calculate the degrees to get a contradiction. – Luke Chen Jul 21 '19 at 02:21

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There is no such function. Let $g(x)=|f(x)|$. Note that $g(x^{2})=(g(x))^{2}$. By iteration we get $g(x^{2^{n}})=(g(x))^{2^{n}}$ for every positive integer $n$. By continuity $g(x^{2^{n}}) \to g(0)=1$ for $|x| <1$. So $(g(x))^{2^{n}} \to 1$ for such $x$. But this implies $g(x)=1$. [Indeed $g(x) >1$ implies $(g(x))^{2^{n}} \to \infty$ and $g(x)<1$ implies $(g(x))^{2^{n}} \to 0$]. We have proved that $|f(x)|=1$ for $|x|<1$. By continuity $|f(1)|=1$ contradicting the hypothesis.

Kavi Rama Murthy
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