Equivalence of additivity and homogeneity for $\mathbb{R} \rightarrow \mathbb{R}$ strictly increasing functions

Question

In my (sort of) intro to proofs course, we were given the following theorem without proof (which was then used in other proofs):

If $f : \mathbb{R} \rightarrow \mathbb{R}$ is a strictly increasing function, the following statements are equivalent:

(i) $\forall n \in \mathbb{Z} : \forall x \in \mathbb{R} : f(nx) = nf(x)$
(ii) $\forall x \in \mathbb{R} : f(x) = ax$, where $a = f(1) > 0$ *
(iii) $\forall x_1, x_2 \in \mathbb{R} : f(x_1 + x_2) = f(x_1) + f(x_2)$

*Since f goes from $\mathbb{R}$ to $\mathbb{R}$, I believe (ii) is equivalent to degree 1 homogeneity.

I managed to show two implications:

(ii) $\implies$ (iii)

$f(x_1 + x_2) = a(x_1 + x_2) = ax_1 + ax_2 = f(x_1) + f(x_2)$

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(iii) $\implies$ (i)

if $n \in \mathbb{Z}$ and $n > 0$:

$f(nx) = f(x + ... + x)$ [$n$ times] $= f(x) + ... + f(x)$ [$n$ times] $= nf(x)$

This is easy to extend to $n <= 0$ if you observe that additivity implies

$f(0) = 0$, because if it didn't, $f(x) = f(x + 0) = f(x) + f(0) \neq f(x)$
$f(-x) = -f(x)$, since $f(x) + f(-x) = f(x-x) = f(0) = 0$

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Now I need help figuring out how to show that (i) $\implies$ (ii). I'm sure I could use (i) to prove that $\forall q \in \mathbb{Q} : f(qx) = qf(x)$, but I have no idea how to extend that to the reals. I suppose that's where the fact $f$ is strictly increasing becomes relevant?

Thanks in advance!

Answer 1

As you said, one can show that for $q\in\mathbb{Q}$, $f(q)=q f(1)$. Indeed, taking $x=1$ and $x=m^{-1}$, $f(n)=n f(1)$ and $f(1=m/m)=m f(1/m)$, so $f(n/m)=n f(1/m)=n/m f(1)$.

As the rational numbers are dense, take $q^1_n\nearrow x\in\mathbb{R}$ (increasing sequence converging to $x$). Then, $$ f(q^1_n)=q^1_n f(1)\to xf(1). $$ Similarly for a decreasing sequence $q^2_n\to x$, $$ f(q^2_n)=q^2_n f(1)\to xf(1). $$

Using that $f$ is increasing, $$ \lim_{n\to \infty}f(q^2_n)\ge f(x)\ge \lim_{n\to \infty}f(q^1_n), $$ but $\lim_{n\to \infty}f(q^i_n)=xf(1)$ for $i=1,2$, so $f(x)$ is sandwiched to $xf(1)$ and we are done. Obviously, as $f$ is strictly increasing, $f(1)>0$.