How can I find all continuous functions $f:\mathbb{R} \rightarrow \mathbb{R}^+$ such that $$\frac{1}{f\left(y^2f(x)\right)} = \big(f(x)\big)^2\left(\frac{1}{f\left(x^2-y^2\right)} + \frac{2x^2}{f(y)}\right)$$ for all reals $x,y$?

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    $\begingroup$ -1 Dear Amir, you have not shown any research effort. Could you please explain your attempt(s) at the solution? It would also be helpful to have some motivation, i.e., what lead you to this problem, or if this is a homework (or olympiad) problem, what techniques do you think are applicable? $\endgroup$ – Amitesh Datta Jun 13 '11 at 9:57
  • $\begingroup$ Dear Amitesh, I was looking for some functional equations problems to make a PDF file, and suddenly found this with no solutions. And all I know is that the only solution is $f(x)=\frac{1}{x^2+1}$. $\endgroup$ – Amir Hossein Jun 13 '11 at 10:44
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    $\begingroup$ You should have mentioned these before. $\endgroup$ – user9413 Jun 13 '11 at 11:44
  • $\begingroup$ Dear Amir, thank you very much! I have now upvoted your answer. $\endgroup$ – Amitesh Datta Jun 13 '11 at 12:15
  • $\begingroup$ Hint: did you try putting y=0? $\endgroup$ – leonbloy Jun 13 '11 at 13:57

Let's first show that $f(x) = f(-x)$ for all $x \in \mathbb{R}$. Put $x=y=0$, and we get $$\frac{1}{f(0)} = f(0),$$ which implies that $f(0) = 1$. Now letting $x=0$ we get $$\frac{1}{f(y^2)} = \frac{1}{f(-y^2)},$$ which implies that $f(y^2) = f(-y^2)$, proving our first claim.

Now since $f(x) = f(-x)$ there exists a function $g\colon [0,\infty) \to \mathbb{R}^+$ such that $f(x) = g(x^2)$. The condition on $f$ becomes $$\frac{1}{g(y^4g^2(x^2))} = (g(x^2))^2 \left( \frac{1}{g((x^2 - y^2)^2)} + \frac{2x^2}{g(y^2)} \right).$$ By changing $x^2 \mapsto x$ and $y^2 \mapsto y$, this turns to $$\frac{1}{g(y^2 g^2(x))} = (g(x))^2 \left( \frac{1}{g((x-y)^2)} + \frac{2x}{g(y)}\right).$$ Now notice that we already know that $g(0) = f(0) = 1$. Also, by setting $x=1$, $y=0$ we get $$1 = (g(1))^2 \left( \frac{1}{g(1)} + 2 \right) \Leftrightarrow g(1) = (g(1))^2 + 2 (g(1))^3 \Leftrightarrow$$ $$1 = g(1) + 2 (g(1))^2 \Leftrightarrow 2(g(1) - \frac{1}{2})(g(1) + 1) = 0.$$ Hence $g(1) = \frac{1}{2}$.

We will show by induction that $g(n) = \frac{1}{1+n}$ for all $n \in \mathbb{N}$. Suppose that the claim holds for some $n \in \mathbb{N}$, then $$\frac{1}{g((n+1)^2 g^2(n))} = (g(n))^2 \left( \frac{1}{g((n-(n+1))^2)} + \frac{2n}{g(n+1)}\right) \Leftrightarrow$$ $$\frac{1}{g(1)} = \frac{1}{(n+1)^2} \left( \frac{1}{g(1)} + \frac{2n}{g(n+1)}\right) \Leftrightarrow$$ $$2 (n+1)^2 = 2 + \frac{2n}{g(n+1)} \Leftrightarrow g(n+1) = \frac{1}{n+2},$$ and by induction we have that $$g(n) = \frac{1}{n+1}$$ for all $n \in \mathbb{N}$.

Consider now the original condition on $g$ and let $x$ and $y$ be natural numbers. We get that $$\frac{1}{g(\frac{y^2}{(x+1)^2})} = \frac{1}{(x+1)^2} \left( (x-y)^2 + 1 + 2x(y+1) \right) \Leftrightarrow$$ $$\frac{1}{g(\frac{y^2}{(x+1)^2})} = \frac{1}{(x+1)^2} \left( (x+1)^2 + y^2 \right) \Leftrightarrow$$ $$g(\frac{y^2}{(x+1)^2}) = \frac{(x+1)^2}{(x+1)^2 + y^2} = \frac{1}{\frac{y^2}{(x+1)^2} + 1},$$ and hence the formula $$g(x) = \frac{1}{x+1}$$ holds for all squares of rational numbers. But they are dense in $[0,\infty)$ and since $g$ was continuous, we get that the only solution is $$g(x) = \frac{1}{x+1}.$$ (Checking that this is indeed a solution is straightforward.) Thus $$f(x) = \frac{1}{x^2 + 1}.$$

  • $\begingroup$ @J.J.: +1. The range of $g$ need not be $\mathbb{R}^+$ even if $f(x) = g(x^2)$ $\endgroup$ – user17762 Jun 13 '11 at 21:14
  • $\begingroup$ @J.J. very nice! Thank you for this nice solution. :) please see here, too. $\endgroup$ – Amir Hossein Jun 14 '11 at 6:04
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    $\begingroup$ @Sivaram: In this case the domain of $g$ is $[0,\infty)$ and $x \mapsto x^2$ is a bijection on that interval, so $g$ can't take non-positive values. Or am I missing something? $\endgroup$ – J. J. Jun 14 '11 at 6:19
  • $\begingroup$ I just spotted a possible mistake: $1/f(0)=f(0)$ does not implies $f(0)=1$. It only implies, that $f(0)\in\{-1,1\}$. $\endgroup$ – FUZxxl Jun 15 '11 at 17:52
  • $\begingroup$ @FUZxxl: The range of $f$ was required to be $\mathbb{R}^+$, so $-1$ is not an option. $\endgroup$ – J. J. Jun 15 '11 at 19:24

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