calculus - Let $f:mathbb R rightarrow mathbb R$ continuous function such that $f(x+y)=f(x)+f(y)$. Then there exist a real $r$ such that $f(x) = r x$.

Let $f:\mathbb R \rightarrow \mathbb R$ continuous function such that $f(x+y)=f(x)+f(y)$. Then there exist a real $r$ such that $f(x) = r x$.

My try:

with $f(\frac1q x)$:

$$f(x) = f(q\cdot\frac1q x) = f(\frac1q x+...+\frac1q x) = f(\frac1q x)+...+f(\frac1q x) = qf(\frac1q x)$$
Now $f(\frac pqx)$: Fix $y=\frac xq$ to get
$$f(\frac pqx) = f(p\frac xq) = f(py) = p\,f(y) = p\,f(\frac1q x) = p\cdot\frac1qf(x)$$

and i know too that $\lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h} = \lim_{h \rightarrow 0} \frac{f(x) + f(h) - f(x)}{h} = \lim_{h \rightarrow 0} \frac{f(h)}{h}$

here i stuck how can i conclude and find this $r$, some help please.

Answer

You're thinking too hard. You've shown $f$ is linear for all rationals. Via limits, this implies $f$ is linear in irrationals as well. Specifically any irrational is the limit of rationals and convergent limits commute with continous functions. Most importantly, $f(1):=r$, that's how you define $r$. Then $f(x)=xf(1)=xr$.

Also when you were trying to take derivatives I think you meant $f(x+h)$, not $f(x+y)$.