# If $f(x + y) = f(x) + f(y)$ showing that $f(cx) = cf(x)$ holds for rational $c$

For $f:\mathbb{R}^n \to \mathbb{R}^m$, if $f(x + y) = f(x) + f(y)$ for then for rational $c$, how would you show that $f(cx) = cf(x)$ holds?

I tried that for $c = \frac{a}{b}$, $a,b \in \mathbb{Z}$ clearly $$f\left(\frac{a}{b}x\right) = f\left(\frac{x}{b}+\dots+\frac{x}{b}\right) = af\left(\frac{x}{b}\right)$$ but I can't seem to finish it, any help?

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See e.g. the answer here. BTW this equation is called Cauchy's functional equation. – Martin Sleziak May 13 '12 at 12:17

Try computing $bf(x/b){}{}{}$.

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When $n$ is an integer we already know that $f(nx) = nf(x)$. Now suppose that $m$ is an integer not equal to zero. Then we have that

$$\begin{eqnarray*}f(x)&=& f\left(\underbrace{\frac{x}{m} + \ldots +\frac{x}{m} }_{\text{m times}} \right) \\ &=& \underbrace{f\left(\frac{x}{m}\right) + \ldots +f\left(\frac{x}{m}\right)}_{\text{m times}}\\ &=& mf\left(\frac{x}{m}\right) \end{eqnarray*}$$

from which it follows that $\frac{1}{m}f(x) = f\left(\frac{x}{m} \right)$. Putting this and the previous observation together we get that for all integers $n,m$ with $m \neq 0$

$$f\left(\frac{nx}{m} \right) = \frac{n}{m}f(x)$$

showing what we wanted to prove.

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For $c \in \mathbb{Z}^*$:

$$f(cx) = cf(x)$$

Let's do a variable substitution: $\hat{x} = cx$:

\begin{align*} f(\hat{x}) &= cf\left(\frac{\hat{x}}{c}\right) \\ \Rightarrow f\left(\frac{\hat{x}}{c}\right) &= \frac{1}{c} f(\hat{x}) \end{align*}

Combine this with your initial findings to get the proof for $c \in \mathbb{Q}$.

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Let's show that $f(x/b) = \frac{f(x)}{b}$. Denote $t= f(x/b)$. Now sum $t$ with itself exactly $b$ times:

$$b\cdot t=t+t+\dots+t=f(x/b)+f(x/b)+\dots + f(x/b)$$ $$=f(x/b+\dots+x/b)=f((bx)/b)=f(x)$$

Divide both sides by $b$ and you have $t=\frac{f(x)}{b}$.

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