0

I have a continuous $f:\mathbb{R}\rightarrow \mathbb{R}$ which is such that $f(x+y)=f(x)+f(y)$ $\forall x, y \in \mathbb{R}$. I am trying to prove that $f(n)=nf(1)$ for all $n \in \mathbb{Z}$.

The claim is trivial for $n \in \mathbb{N}$. Below I attempt to prove it for $n \in \mathbb{Z}_{<0}$, and all rational numbers. Is my attempt correct?

First, negative integers. Consider arbitrary negative integer, $n$. Then, $f(n)=f(n-(n-1))-f(-n+1)=f(1)-f(1-n)=f(1)-(1-n)f(1)=nf(1)$ (where the second last equality follows from the truth of the claim for all natural numbers).

Second, rational numbers. Consider arbitrary rational number $\frac{a}{b}$ (where both $a,b \in \mathbb{Z}). $ $f(\frac{a}{b})=f(\sum_{i=1}^{a}\frac{1}{b})=af(\frac{1}{b})$. I now show that $f(\frac{1}{b})=\frac{1}{b}f(1)$.

$f(\frac{1}{b})=f(\frac{1}{b}+\frac{b-1}{b})-f(\frac{b-1}{b})=f(1)-(b-1)f(\frac{1}{b})$. So, we have that $f(\frac{1}{b})=f(1)-(b-1)f(\frac{1}{b}).$ Rearranging, we get that $bf(\frac{1}{b})=f(1)$, and so $f(\frac{1}{b})=\frac{1}{b}f(1). $

Charles
  • 705
  • It seem to be correct, but what is your question? – Khosrotash Sep 22 '22 at 06:26
  • @Khosrotash Just to receive verification that my proof is correct (hence the tag proof verification) – Charles Sep 22 '22 at 06:27
  • 1
    If $f$ is continuous then it's true for all values. – Brian Tung Sep 22 '22 at 06:38
  • Analogously, to prove the statement only for rational numbers $\frac ab$, the continuity assumption is not necessary (note that you never used it!). – Greg Martin Sep 22 '22 at 07:10
  • $f(x+1)=f(x)+f(1)$. Consider case $x=1$ to prove for $n=2$, then use induction with $x=n$ to prove for all integer $n>2$ and backward induction with $x=n-1$ to prove for all integer $n < 2$. – Ivan Kaznacheyeu Sep 22 '22 at 14:35
  • Using induction one can show that to show $f(na)=nf(a)$ for any $a\in \mathbb{R}, n\in \mathbb{Z}$, then one can show that $f(\frac1{n})=\frac1{n} f(1)$ for any $n\in \mathbb{N}$. Combining of these gives proof of $f(\frac{a}{b})=\frac{a}{b}f(1)$ for any $a\in\mathbb{Z},b\in\mathbb{N}$. – Ivan Kaznacheyeu Sep 22 '22 at 14:39
  • As for negative $n$, maybe simplest method is to show $f(0)=0$ first and then use $f(0)=f(na)+f(-na) \Rightarrow f(-na)=-f(na)=-n f(a)$. – Ivan Kaznacheyeu Sep 22 '22 at 14:41

0 Answers0