Assume $f$ is a function over $\mathbb{R}$ satisfying $f(x+y)=f(x)+f(y)$ for all $x,y \in \mathbb{R}$. Prove that there is a constant $c$ for which $f(x) = cx$ for all $x \in \mathbb{Q}$.

We know that $f(0) = 0$. Now set $x = n$ and $y=n$ to get $f(2n) = 2f(n)$ where $n$ is a rational number. But doesn't this condition imply all constant functions work? So we must get another condition.

  • $\begingroup$ You're not quite there. Set $f(1) = \alpha$, then what do you conclude about $f(n)$? $\endgroup$ – Cameron Williams Apr 9 '16 at 1:25
  • $\begingroup$ We have $2^n \cdot \alpha = f(2n)$ for positive integers $n$. $\endgroup$ – user19405892 Apr 9 '16 at 1:29
  • $\begingroup$ Also isn't it obvious there is a $c$ since $c(x+y) = cx+cy$ works? $\endgroup$ – user19405892 Apr 9 '16 at 1:42
  • $\begingroup$ Actually the only constant function that works is f(x) = 0. f (x)= c DOESN'T work because c=f (p+c) $\ne $ f (p)+f (q)= 2c. $\endgroup$ – fleablood Apr 9 '16 at 1:53
  • $\begingroup$ No. $f (2^n)=f (2^{n-1} +2^{n-1})=2f (2^{n-1}) =...=2^{n-1}f (2) $ but $f (2n)=2f (n) $. Again, figure out what f (1) equals first. Then note f (q) = qf (1) $\endgroup$ – fleablood Apr 9 '16 at 2:01

Notice: $$ f\left(\frac{m}{n}\right) = m\cdot f\left(\frac{1}{n}\right) = m\cdot \left( \frac{f(1)}{n}\right) = f(1)\cdot \frac{m}{n} $$ where $f\left(\frac{1}{n}\right) = \frac{f(1)}{n}$ because $$ f\left(\frac{1}{n}\right) = f\left( \frac{n}{n} - \frac{n-1}{n}\right) = f(1) - (n-1)f\left(\frac{1}{n}\right) $$ by the linearity assumption.

  • $\begingroup$ Couldn't we just say that since $c(x+y) = cx+cy$ we know there is a $c$? $\endgroup$ – user19405892 Apr 9 '16 at 1:45
  • $\begingroup$ @user19405892 No; you need a constant that depends on a function $f$, where $f$ is any function satisfying the conditions you listed. So, given $f$, find a constant that works. As you've written it, $c$ has not been defined. $\endgroup$ – Derek Allums Apr 9 '16 at 1:48
  • $\begingroup$ Why does $f\left(\frac{m}{n}\right) = m\cdot f\left(\frac{1}{n}\right)$? $\endgroup$ – user19405892 Apr 9 '16 at 1:52
  • $\begingroup$ By assumption on $f$: $f(m/n)=f(1/n+\cdots+1/n)$ with $m$ summands, so you apply your rule $m$ times to see that it equals $f(1/n)+\cdots+f(1/n)=m\cdot f(1/n)$. $\endgroup$ – Derek Allums Apr 9 '16 at 1:56
  • $\begingroup$ Couldn't we also just solve this functional equation to get $f(x) = cx$? Actually, your proof seems to do that since you don't have to assume $m$ and $n$ are integers. $\endgroup$ – user19405892 Apr 9 '16 at 2:02

You can demonstrate that for $n\in\mathbb{Z}$, $f(n)=nf(1)$ and $f(1)=nf(\frac1n)$ for $n\ne0$, so that for $x=\frac{p}{q}\in\mathbb{Q}$, $f(x)=\frac{p}{q}f(1)$, so $c=f(1)$.


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