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I need to solve the following functional equation: $$f(x)+2f^2(x^2)-1=0,\forall x\in (1,+\infty)$$

where $f^2(x^2)$ means the multiplication of the real number $f(x^2)$ with itself.

I sense that any solution of the above functional equation belongs to the class of constant functions on $(1,+\infty)$.

Can anyone suggest something that would help me? I wish happy new year with peace along the world. May only love be in our hearts.

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  • $\begingroup$ Well...a non-constant example would be $f(x)=-1$ if $x$ is algebraic and $\frac 12$ if $x$ is transcendental. You could exclude things like that requiring continuity (say). Note: I am reading $f^2(x^2)$ as $f(x^2)\times f(x^2)$. $\endgroup$
    – lulu
    Dec 27, 2018 at 11:55
  • $\begingroup$ I think he/she mean $$f^2(x^2)=(f(x^2))^2$$ $\endgroup$ Dec 27, 2018 at 12:02
  • $\begingroup$ @henrik the usual multiplication $\endgroup$ Dec 27, 2018 at 12:02
  • $\begingroup$ How do you get the sense that any solution will be constant? Even if we add a constraint of continuity (like @lulu demonstrates is needed), it's far from obvious to me. $\endgroup$ Dec 27, 2018 at 12:08
  • $\begingroup$ I would write $(f(x^2))^2$ in the equation and save everyone the trouble of explaining what $f^2(x^2)$ means. $\endgroup$
    – David K
    Dec 27, 2018 at 13:10

2 Answers 2

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The conjecture is false even if we require the function to be continuous on $(1,+\infty)$. Take $f(x)=-\cos\left(\frac{1}{\ln x}\right)$.

To see that this is a solution, use the double angle formula $\cos x=2\cos^2\frac x2-1$:

$$-f(x)=\cos\left(\frac{1}{\ln x}\right)=2\cos^2\left(\frac{1}{2\ln x}\right)-1=2[f(x^2)]^2-1$$

Hence, $f(x)+2[f(x^2)]^2-1=0$.

Edit: added a proof of its correctness

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Let me solve this functional equation under some reasonable assumptions. First note that $f(x)=\dfrac{1}{2}, -1$ are the only constant function solutions. Moreover, always $f(x)\le 1$ and equality occures iff $f(x^2)=0$ iff $f(x^4)=\pm\dfrac{\sqrt{2}}{2}.$

Next, note that the given domain can be replace by $(0,\infty)$ with the order reversing bijection $t\mapsto e^{1/2t}$ to transform the given functional equation to $$g(2t)=2(g(t))^2-1$$ for all $t\in (0,\infty),$ where $g(t)=-f(e^{1/2t}).$ Clearly, cosine (on a restricted domain) satisfices the given equation. In fact, for any $a\neq0$ $$g(t)=\cos\left(\dfrac{\pi t}{a}\right)$$ is the only solution under assumptions mentioned in this answer.

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