Suppose $g:[0,1)→R$ is a continuous function satisfying $g(x^2)=x−g(x)$
for every x on interval $[0,1)$.
How does the function g(x) behave as x tends to 1?
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$\begingroup$ Emm Tends to 0.5? $\endgroup$– EugeneCommented Mar 15, 2017 at 5:39
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1$\begingroup$ Obviously, if there is a limit, then it has to be 1/2. One can prove that $g(x)=\sum_k(-1)^kx^{2^k}$. I think that this is nondecreasing and tends to $1/2$ as $x\to 1$, but it does not seem to be trivial to prove that. $\endgroup$– Neil StricklandCommented Mar 15, 2017 at 7:44
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1$\begingroup$ @NeilStrickland That can't be the only solution, given that the homogeneous form $g(x^2) = -g(x)$ has nontrivial solutions, of the form $\sum_n c_n cos((2n + 1) \pi log_2(-ln(x)) + d_n sin((2n + 1) \pi log_2(-ln(x))$. $\endgroup$– user44191Commented Mar 15, 2017 at 8:18
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1$\begingroup$ @user44191 Your functions are not continuous at $x=0$. For example, if $g(x)=\cos(\pi\log_2(-\ln(x)))$ then we can consider $a_n=\exp(-2^n)$ and $a_n\to 0$ with $g(a_n)=(-1)^n$. If $g(x)=-g(x^2)$ then $g(x)=g(x^{2^{2k}})$ and $x^{2^{2k}}\to 0$, so if $g$ is continuous then $g(x)=g(0)$. Also, the identity $g(x)=-g(x^2)$ gives $g(0)=0$. $\endgroup$– Neil StricklandCommented Mar 15, 2017 at 8:38
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$\begingroup$ @NeilStrickland Ah, of course, was so focused on 1 that I forgot about 0. The periodic function I derived my homogeneous solution must converge as it goes to $\infty$, so it must be constant, and the only solution is $g(x) = 0$. $\endgroup$– user44191Commented Mar 15, 2017 at 8:53
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1 Answer
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This appears as Problem 4 of an MSRI Emissary publication, and indeed this problem has been discussed by Noam Elkies. See also his web page (problem 8 here), where he links to a solution.
The conclusion is that $g(x)$ has no limit as $x$ approaches $1$; it starts off monotone increasing until you get pretty close to $1$, but one finds $g(.995) \approx .50088 > 1/2$. Note that $g(x) = x - x^2 + g(x^4) > g(x^4)$, so one concludes there are infinitely many points where $g(x) > 1/2$. But as already observed, if a limit exists, it would have to be $1/2$; therefore no limit exists.
In fact one finds that $g$ oscillates more and more quickly as $x$ approaches $1$, each oscillation occurring about 4 times more quickly than the previous.
answered Mar 15, 2017 at 11:00