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Find all continuous functions $f$ defined over $\mathbb{R}^{+*}$ such that $f(x) =\sqrt{2+ f(x^2)}.$

I found the constant function $f(x)=2$ and the function $f(x)=x+1/x$, but are there other functions? i start from the obious fact that $f(x) > 0$ as by definition of $f$...we deduce that $f(x) \geq \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{...+\sqrt{2+f(x^{2^n})}}{}}}} \geq \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{...+\sqrt{2+...}}{}}}} =2$ ... so $f(x) \geq 2$ for every positive real number...

skills foot
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  • Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please [edit] the question. This will help you recognize and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Jun 15 '22 at 08:54
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    What is $\Bbb R^*$? – Martin R Jun 15 '22 at 08:56
  • @MartinR Unless the OP says otherwise I assume it's $\Bbb R\setminus{0}$, so denoted to emphasize it's a group under $\times$. – J.G. Jun 15 '22 at 08:57
  • Martin R : $R^{+}$ is the set of positive real numbers and $R^{}$ is the set of real numbers different from zero – skills foot Jun 15 '22 at 09:05
  • The function $f : x \to x + \frac{1}{x}$ does not work (take $x = 2$).

    Notice that one must have $f(1) = \sqrt{2 + f(1)}$ hence $f(1)$ must be the positive root of $X^2 - X -2$ whence $f(1) = 2$.

     – Olivier Roche Jun 15 '22 at 09:42
  • https://math.stackexchange.com/questions/2819684/can-non-linear-continuous-and-odd-function-fx-satisfy-f2x-2fx j isn't automatically be linear when f(x) is continuous – hih Jun 17 '22 at 14:34

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$$f(x)^2-f(x^2)=2$$ let $x=e^t$ and $g=f\circ \exp$ $$ g(t)^2 - g(2t) = 2 $$ let $g=2\cdot h$ $$ h(2t) = 2\cdot h(t)^2 - 1 $$ let $h=\cosh\circ j$ $$ \cosh( j(2t)) = \cosh(2\cdot j(t)) $$ and you get solutions of the form $$ j(t)=\alpha t,\qquad h(t)=\cosh(\alpha t),\qquad g(t)=2\cosh(\alpha t),\qquad f(x)=2\cosh(\alpha \log x)=x^\alpha+x^{-\alpha}.$$ Then it is not difficult to show that the constant solution and the solutions $x^{\alpha}+x^{-\alpha}$ are the only continuous solutions.

Jack D'Aurizio
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