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Find all continuous functions $f:\mathbb R^{+}\to\mathbb R^{+}$ such that $f(x)^2=f(x^2)$ for all positive reals $x$.

If $f$ is a constant function, then $f(x)=1$. If $f$ is non-constant, I'm suspecting the only solutions are of the form $f(x)=x^k$, where $k$ is a constant, but I have no idea how to prove this. Thanks in advance.

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    $\begingroup$ I think $f(x)=x^{\sin(2\pi\log_2|\ln x|)}$ also works (with $f(1)=1$). $\endgroup$ – stewbasic Oct 7 '16 at 2:13
  • $\begingroup$ $f(x)=x^{\sin..}$ does not work. $\endgroup$ – Takahiro Waki Oct 7 '16 at 6:51
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Why, the solutions are plenty, even the continuous ones. Let $f(2)=a$; then $f(4)=a^2$. Now draw a freehand curve from the point $(2,a)$ to $(4,a^2)$. That would be your function. Using $f(x^2)=f(x)^2$, extend it to $[4,16]$, then to $[16,256]$, and so on. Then use the equation in reverse and extend to $[\sqrt2,2]$ and so on. Then maybe we'll have to repeat the entire trick for $x<1$.

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    $\begingroup$ Yes it does work. $f(\sqrt2)=\sqrt{a}$, then $f(\sqrt[4]2)=\sqrt[4]{a}$ and so on, hence $\lim\limits_{x\to1+}f(x)=1$. If we start from any other $x$ than 2, we'll get the same limit. True, the intervals get smaller and smaller, but so does the variation of function along the interval. $\endgroup$ – Ivan Neretin Oct 7 '16 at 7:26
  • $\begingroup$ Perhaps I'm misunderstanding something, but how does this answer account for all positive reals $x$? $\endgroup$ – John Smith Oct 7 '16 at 13:50
  • $\begingroup$ You define the function arbitrarily on $[2,4]$ and then grow this interval both ways, thus covering $[1,\infty]$. Then you define it arbitrarily on $\left[{1\over2},{1\over4}\right]$ and extend that to cover $[0,1]$. $\endgroup$ – Ivan Neretin Oct 7 '16 at 13:57

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