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 $x,y$?
All functions $\frac{1}{f\left(y^2f(x)\right)} = \big(f(x)\big)^2\left(\frac{1}{f\left(x^2-y^2\right)} + \frac{2x^2}{f(y)}\right)$
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8-1 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? – 2011-06-13
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0Dear 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 $f(x)=\frac{1}{x^2+1}$. – 2011-06-13
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2You should have mentioned these before. – 2011-06-13
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0Dear Amir, thank you very much! I have now upvoted your answer. – 2011-06-13
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0Hint: did you try putting y=0? – 2011-06-13
1 Answers
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 $$f(x) = \frac{1}{x^2 + 1}.$$
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0@J.J.: +1. The range of $g$ need not be $\mathbb{R}^+$ even if $f(x) = g(x^2)$ – 2011-06-13
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0@J.J. very nice! Thank you for this nice solution. :) please see [here](http://math.stackexchange.com/q/45260/6715), too. – 2011-06-14
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2@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? – 2011-06-14
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0I just spotted a possible mistake: $1/f(0)=f(0)$ does not implies $f(0)=1$. It only implies, that $f(0)\in\{-1,1\}$. – 2011-06-15
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0@FUZxxl: The range of $f$ was required to be $\mathbb{R}^+$, so $-1$ is not an option. – 2011-06-15
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0@J. J. Ah, thanks. I didn't read the ${}^+$ in $\mathbb R^+$. – 2011-06-15
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0@J.J. Instead of proving the solution for $f(x)$, suppose you do not know the solution a priori, what would you do to find the solution? Any intuition and motivation would be appreciated. – 2013-03-09