This weird question was given by my professor as a part of my assignment :
$\textbf{Question :}$
“Let $f: \mathbb{R} \to \mathbb{R}$ be a function such that it satisfies $f(x + y) = x f(\frac{1}{y}) + y f(\frac{1}{x})$, whenever $x$ and $y$ are both irrational numbers. Then prove that $f(0)$ is always $0$.”
$\textbf{My attempt :}$
Since $0$ is rational, we can’t let any of the variables to be $0$. So I tried the substitution $y = -x$ to get $f(0) = x f(\frac{1}{-x}) – x f(\frac{1}{x})$.
I tried proving f to be an even function after this, so that $f(0)$ becomes $0$. But I couldn’t do so or proceed further anyhow.
Can somebody kindly provide me hints or solutions for this problem ?
Best Answer
With $x = y = t$ (any irrational) you get $$ f(2t) = 2 t f(1/t) $$ and with $x = y = 1/(2t)$ you get $$f(1/t) = f(2t)/t$$ Substitute the first into the second and it says $$ f(1/t) = 2 f(1/t)$$ from which you conclude $f(1/t) = 0$, i.e. $f$ is $0$ on all irrationals. Then use $y = -x$ to get $f(0) = 0$.