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What are all functions with $x>1$ and $y>1$ $\rho$ that follows $$\rho(xy)=\frac{1}{\frac{y}{\rho(x)}+\frac{x}{\rho(y)}}$$ and is continuous

If this doesn't have any solutions then prove no such solution exists.

I would like to understand how to solve a problem like this

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2 Answers 2

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Rewrite the property as \begin{equation*} \rho(xy) (\rho(x)x + \rho(y)y) = \rho(x)\rho(y) \end{equation*} If $\rho$ has that property, it follows that $\rho(1) = 0$. See this by setting $x = y = 1$ and getting $2\rho^2(1) = \rho^2(1)$.

Now look at $y = 1/x$ for any $x$. Using the above we get $\rho(x)\rho(1/x) = 0$, so $\rho = 0$ for the formulation above.

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    $\begingroup$ Good answer @Chrystomath, just need to exclude $x = 1$. $\endgroup$
    – A rural reader
    Commented Jan 17, 2021 at 3:14
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    $\begingroup$ $\rho=0$ is not a valid function either, unless one accepts non-real equations such as $0=\frac{1}{\frac{1}{0}+\frac{1}{0}}$. $\endgroup$
    – Chrystomath
    Commented Jan 17, 2021 at 9:11
  • $\begingroup$ @crowja okay I changed the problem to be x and y bigger then 1 because of Chrystomath $0=\frac{1}{\frac{1}{0}+\frac{1}{0}}$ $\endgroup$
    – user872441
    Commented Jan 17, 2021 at 15:21
  • $\begingroup$ @SuperWarm the limit x->0 of $0=\frac{1}{\frac{1}{x}+\frac{1}{x}}$ then we still get zero so it should be fine. $\endgroup$
    – user808403
    Commented Jan 17, 2021 at 15:25
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To solve $$\frac{1}{xy\rho(xy)}=\frac{1}{x\rho(x)}+\frac{1}{y\rho(y)}$$ consider $f(t):=1/(t\rho(t))$. Then it is well known that the only continuous functions satisfying $$f(xy)=f(x)+f(y)$$ are $$f(x)=A\log x$$ Hence $$\rho(x)=\frac{a}{x\log x}$$ where $a\ne0$.

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