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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.

Rosa Maria Gtz.
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  • You could write $t=\dfrac{p}{q}$. Note, however, that you're proving this only for rational $r$. How about irrational $r$? –  May 08 '14 at 02:22
  • can i take a sequence of irrationals that converges to rational and apply this? im confused with this exercise , more hints if its possible – Rosa Maria Gtz. May 08 '14 at 02:26

2 Answers2

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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 $f(x+y)$.

Alex R.
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First, you have to prove that $f$ is $\Bbb R$-linear. You have yet proved that it is $\Bbb Q$-linear. Now take a real number $k$ and consider a sequence of rationals that converges towards $k$. Use the continuity of $f$ to prove that $f(kx)=kf(x)$.

Now, you have to look for that $r$, but $f(x)=f(x\cdot1)=xf(1)$.

ajotatxe
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