Can You Solve This Challenging Functional Equation?

[stevendaryl]

Let $c$ be a rational number (possibly irrational), then there exist a sequence $\{q_n\}_{n\in\mathbb{N}}$ of purely rational numbers that converges to $c$. Therefore (assuming $f$ is continuous)
$$
\begin{align*}
f(cx) &= \lim_{n\rightarrow\infty}f(q_nx) \\
&= \lim_{n\rightarrow\infty}q_nf(x) \\
&= cf(x)
\end{align*}
$$
and thus $f(ax+by) = af(x) + bf(y)$ is true even when $a$ and $b$ are irrational.

I guess I didn’t make it explicit, but I was asking about the case where we drop the assumption of continuity.

 

[haruspex]

I guess I didn’t make it explicit, but I was asking about the case where we drop the assumption of continuity.

If we plug f(x)+f(y)=f(x+y) back into the original equation,
$f(xf(y))+f(f(x))+yf(y)=f(x)+yf(x)+yf(y)$
$f(xf(y))=yf(x)$
We know that f(1) is either 0 or 1. If it is 1, setting x=1:
$f(f(y))=y$
But we already know f(f(y))=f(y), so f(y)=y.
If f(1)=0, f(f(y))=0, so f(y)=0.

So either f(x)=x for all x or f(x)=0 for all x,