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.

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"Multiplicative function" really means $f(xy)=f(x)f(y)$. Really I would call this linear, but I think most people will read that as "linear and additive", so I'm at a loss for a better suggestion... –  rschwieb Sep 7 '12 at 15:40
@rschwieb: yeah, I'm not sure what the correct name of this restriction is... grateful for suggestions! –  Xodarap Sep 7 '12 at 15:42
@rschwieb: maybe homogeneous of degree 1. –  Willie Wong Sep 7 '12 at 15:50
$f(cx)=cf(x)$ for all $c,x$ means $f(x)=f(1)x$. So all functions that satisfy this equation are of the form $f(x)=ax$ for some $a$ –  Thomas Andrews Sep 7 '12 at 15:58

Let $$f(x,y) = \begin{cases} (x,y) & xy > 0 \\ 0 & xy \leq 0 \end{cases}$$

It is clear that $cf(\vec{z}) = f(c\vec{z})$ for any $c\in \mathbb{R}$. But $f$ is not a linear map.

further counter examples can be constructed in polar coordinates. Let $(r,\omega)\in \mathbb{R}_+ \times \mathbb{S}^{n-1}$ denote the spherical coordinates of $\mathbb{R}^n$. Then a map

$$f(r,\omega) = (\lambda(\omega)r,\omega)$$

satisfies $cf(\vec{v}) = f(c\vec{v})$ for every $\lambda:\mathbb{S}^{n-1}\to \mathbb{R}$ that satisfies $\lambda(\omega) = \lambda(-\omega)$. Clearly many of these are not linear.

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If $f(cx) = cf(x)$ then $f(x) = xf(1)$ and the result follows trivially.
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.