# Functional equation which only has constant functions: $f(x)+2f^2(x^2)-1=0,\forall x\in (1,+\infty)$

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.

• 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 $f(x^2)\times f(x^2)$.
– lulu
Dec 27, 2018 at 11:55
• I think he/she mean $$f^2(x^2)=(f(x^2))^2$$ Dec 27, 2018 at 12:02
• @henrik the usual multiplication Dec 27, 2018 at 12:02
• 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. Dec 27, 2018 at 12:08
• I would write $(f(x^2))^2$ in the equation and save everyone the trouble of explaining what $f^2(x^2)$ means. Dec 27, 2018 at 13:10

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

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.