# Prove an additive function has property f(x)=x

So I am given a function $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f(x+y)=f(x)+f(y)$ for all real $x$ and $y$, is continuous at $x=0$, and $f(1)=1$. I need to show that $f(x)=x$ for all real $x$.

I have proven that $f$ is continuous for all real $x$ from the fact that it is continuous at one point, and I am not sure where to go next. (My prof gave a hint that I should prove $f(r)=r$ for all rational $r$, but I'm not sure how to do that.)

• Your prof's hint is a good one. If you don't see how to do it, start by computing $f(1/2)$. (Then $f(1/3)$ or $f(1/4)$.) After that, I think it will be clear how to proceed. – mrf Nov 3 '14 at 22:00
• I know that I should somehow use the additive property to compute f, but I'm not sure how to use it exactly – DennisKRQ Nov 3 '14 at 22:04
• From what you have, you know that every irrational $x$ can be approximated by rationals. Consider a sequence $x_n$ of rationals, and $x_n \to x$. What will happen with $f(x_n) \to$? – FormerMath Nov 3 '14 at 22:05
• $f(1/2) + f(1/2) = f(1) = 1$, so $f(1/2) = 1/2$. Do you see how to continue? – mrf Nov 3 '14 at 22:07
• Ah yes, everything makes sense now! – DennisKRQ Nov 3 '14 at 22:11

Because is continuous in $x=0$ $$\lim_{y\to 0}f(x+y)=f(0)+f(x)$$ so, $f(x)=f(0)+f(x)$ and from here $f(0)=0$
Now if $n \in \Bbb Z, f(n)=nf(1)=n$
For $n \in \Bbb Q,f(\frac1n+\frac1n+...\frac1n)=nf(\frac1n).$
Here you see that $f(1)=nf(\frac1n)\Rightarrow f(\frac1n)=\frac1n \Rightarrow f(q)=q,q \in \Bbb Q^*$
For $x \in \Bbb I$ you know that x is an infinite sum of $q \in \Bbb Q$
Now the conclusion that $f(x)=x$ for all real.