8
$\begingroup$

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 $x,y$?

$\endgroup$
5
  • 9
    $\begingroup$ -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? $\endgroup$ Jun 13, 2011 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 $f(x)=\frac{1}{x^2+1}$. $\endgroup$ Jun 13, 2011 at 10:44
  • 3
    $\begingroup$ You should have mentioned these before. $\endgroup$
    – user9413
    Jun 13, 2011 at 11:44
  • $\begingroup$ Dear Amir, thank you very much! I have now upvoted your answer. $\endgroup$ Jun 13, 2011 at 12:15
  • $\begingroup$ Hint: did you try putting y=0? $\endgroup$
    – leonbloy
    Jun 13, 2011 at 13:57

1 Answer 1

16
$\begingroup$

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 $$f(x) = \frac{1}{x^2 + 1}.$$

$\endgroup$
7
  • $\begingroup$ @J.J.: +1. The range of $g$ need not be $\mathbb{R}^+$ even if $f(x) = g(x^2)$ $\endgroup$
    – user17762
    Jun 13, 2011 at 21:14
  • $\begingroup$ @J.J. very nice! Thank you for this nice solution. :) please see here, too. $\endgroup$ Jun 14, 2011 at 6:04
  • 2
    $\begingroup$ @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, 2011 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, 2011 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, 2011 at 19:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .