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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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    Hint: did you try putting y=0?2011-06-13

1 Answers 1

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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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    @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