TITLE: Are all multiplicative functions additive?
QUESTION [10 upvotes]: Suppose $cf(x)=f(cx)$ and $f:\mathbb{R}\to\mathbb{R}$. I believe it follows that $f(x+y)=f(x)+f(y)$. 
Proof: There is some $c$ such that $y=cx$. Then 
$$f(x+y)=f\left((1+c)x\right)=(1+c)f(x)=f(x)+cf(x)=f(x)+f(cx)=f(x)+f(y)$$
QED.
I wonder if the same thing holds for when $f:\mathbb{R}^n\to\mathbb{R}^n$? I can't use the same trick, because all vectors are not scalar multiples.
I tried thinking of it in terms of basis units, but didn't get anywhere.

REPLY [7 votes]: If $f(cx) = cf(x)$ then $f(x) = xf(1)$ and the result follows trivially.
There are two ways to extend this to multiple variables.
In the case $f(cx, y) = f(x, cy) = cf(x, y)$ we have that $f(x, y) = xyf(1, 1)$ and again we have linearity.
In case $f(cx, cy) = cf(x, y)$ we have $f(x, y) = xf(1, \frac{y}{x})$. Notice that now $f(1, \circ)$ could be any function.