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$$f(\tfrac{1}{2}+x)+f(\tfrac{1}{2}-x)=8xf\big(4(\tfrac{1}{2}+x)(\tfrac{1}{2}-x)\big)\qquad\text{for}\qquad x\in(0,\tfrac{1}{2})$$

I have no idea about how to tackle this equation. The original problem asks me to verify that a function indeed fits in, but I want to know how to tell the solution directly from the equation. Thanks to the comments, I have to add that $f$ could be defined on $(0,1)$. Also if no one knows how to proceed for a while, I will post the answer the book gives (perhaps as well as the motivation for this problem), and maybe some people then know how to tackle this equation with the given answer.

Surb
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Apocalypse
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  • This is a very odd functional equation. If you let $x=1/4,$ e.g., which is supposedly in the domain, you are forced to evaluate, on the LHS, $f(1/2+1/4)=f(3/4),$ which is supposedly not in the domain! Are you sure it isn't $f(x)+f(1/2-x)=8x f(4x(1/2-x))?$ – Adrian Keister Aug 02 '19 at 16:27
  • It might not be a domain, instead it might be a set for which the functional equation is true. In any way, the domain of $f(x)$ should be clarified. – Sil Aug 02 '19 at 16:33
  • @AdrianKeister Oh! Sorry I forget to mention the domain for $f$. Indeed the $f$ is defined on $(0,1)$ – Apocalypse Aug 02 '19 at 16:35
  • Do we know anything else about $f$? i.e is it continuous on $(0, 1)$, or differentiable? or maybe we don't know anything about it? – Jihoon Kang Aug 02 '19 at 17:14
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    @JihoonKang The background of this problem doesn’t seem to force any condition on $f$, though the result the book gives is actually smooth. – Apocalypse Aug 02 '19 at 17:16
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    Let $f(x) = g(4x)$ and substitute $x = z/4$ to get the easier to read equation $g(2+x) + g(2-x) = 2x g((2+x)(2-x))$. – eyeballfrog Aug 02 '19 at 17:26

1 Answers1

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If you have the values of a continuous $f$ given on $(\frac12,1)$, then you get values on $(\frac14,\frac12)$ from the functional equation isolated for the first term, $$ f(\tfrac12-x)=8xf(1-4x^2)-f(\tfrac12+x), \\ f(a)=4(1-2a)f(4a(1-a))-f(1-a). $$ To ensure continuity in $a=\frac12$, we need $f(\frac12)=0$.

From the values in $(\frac14,1)$ you can determine the values of $f(a)$ for all $a$ such that $$ a_1=\tfrac14<4a(1-a)\iff (a-\tfrac12)^2<\tfrac3{16}\iff a_2=\tfrac12-\tfrac{\sqrt3}4<a<\tfrac12+\tfrac{\sqrt3}4 $$

From these you get values for the interval $(\frac12-\frac{\sqrt3}4,\frac14)$ for $x\in (0,\frac14,\frac12)$. Continuity in $a=\frac14$ follows from continuity in $a=\frac34$. With these values we can extend to $(a_3,1)$ etc. with $$ a_{k+1}=\frac12(1-\sqrt{1-a_k})=\frac12\frac{a_k}{1+\sqrt{1-a_k}} $$ This iteration of the lower interval bounds $a_0=\frac12, a_1=\frac14,a_2=\frac1{4(2+\sqrt3)},...$ converges to zero, so that all values of $f$ are determined by fixing the values on $[\frac12,1]$ to an arbitrary continuous function with $f(\frac12)=0$.

Lutz Lehmann
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    Thank you for the answer! It seems that without further constraints $f$ can’t be uniquely determined up to a coefficient. You actually miss something in your proof which is really important. When $a=1/2$, $f(1/2)$ is not necessarily zero because $f(1)$ could blow up and make the RHS an undetermined $0\cdot\infty$ limit, which could be arbitrary. The result the book gives is $\frac{1}{\sqrt{(x(1-x))}}$, and the motivation for this problem is to calculate the invariant measure under the logistic map $4\mu x(1-x)$ for $\mu=1$. The result is the $f$ above with a coefficient $1/\pi$ to normalize it – Apocalypse Aug 02 '19 at 17:51