# If $f(x+y)=f(x)+f(y)$ and $f$ is monotone, prove that $f(x)=ax$

Suppose $$f:\Bbb{R}\to\Bbb{R}$$ is a monotone function satisfying $$f(x+y)=f(x)+f(y) \quad \forall \ x,y\in\Bbb{R}$$ Prove that $$\exists a\in \Bbb R,\forall x\in\Bbb{R}, f(x)=ax$$

I proved that $$f(x)=ax \quad \forall x\in \Bbb{Q}$$. Assume it is monotonically increasing. Now let $$\alpha$$ be any irrational number. Then, choose sequences $$\{x_n\}$$ increasing to $$\alpha$$ and $$\{y_n\}$$ decreasing to $$\alpha$$. Thus, $$f(x_n)\le f(\alpha)\le f(y_n)$$ i.e. $$ax_n\le f(\alpha)\le ay_n$$ Now, letting $$n\to \infty$$, we get $$f(\alpha)=a\cdot \alpha$$

Is my proof correct for irrationals?

• is your function continuous in a point ? – idm Aug 31 '14 at 14:22
• @idm: That is not necessary here, because the function is monotone. (It turns out that the function is always continuous, but you don't need to assume it.) – TonyK Aug 31 '14 at 14:24
• More or less. The function might be monotone decreasing in which case the inequalities would be the other way round. You have established uniqueness. That does not establish existence, but in this case that is obvious. – almagest Aug 31 '14 at 14:26
• Minor point. don't worry about it. A not bad name is $\alpha$. – André Nicolas Aug 31 '14 at 14:41
• You also want to point out that $x_n$ and $y_n$ are sequences of rational numbers as this is needed to go from $f(x_n)$ to $ax_n$. If this is for a course you also might want to add "we get $f(\alpha) = a\alpha$ by the squeeze theorem". – Winther Aug 31 '14 at 14:54

Yes,you can add that $\{x_n\},\{y_n\}$ are rational sequences.And letting $n\to\infty$ we get $a\cdot\alpha\leq f(\alpha)\leq a\cdot\alpha$ this implies $f(\alpha)=a\cdot\alpha$ but those are just small things(for clearness of proof)
• For completeness you can add that the number $a$ that you are after is simply $f(1)$. – Arnaud Mortier Nov 19 at 7:07