Suppose $f(x)$ is linear i.e. $f(x + y) = f(x) +f(y) $ and monotone on $[-\infty, +\infty]$, then $f(x) = ax$, a real.

Question

Suppose $f(x)$ is linear i.e. $f(x + y) = f(x) +f(y) $ and monotone on $[-\infty, +\infty]$, then $f(x) = ax$, $a\in\Bbb R$

We know $f(0)=0$, and if one assumes differentiability it is quite easy.

But how does monotone make up for differentiability?

Answer 1

the functional equation shows that $f$ is $\mathbb{Q}$-linear. since for $n$ a positive integer and any real $x$: $$ f(nx) = f(x) + \dots +f(x) = nf(x) $$ with the consequence that $$ f(x) = f(n.\frac{x}{n})=nf(\frac{x}{n}) $$ i.e. $$ f(\frac{x}{n})= \frac1nf(x) $$ hence for positive integers $p,q$ $$ f(\frac{p}{q}.x)=\frac{p}{q}f(x) $$ also $$ f(0)=f(0+0)=f(0)+f(0) $$ implies $f(0)=0$ and $$ f(0)=f(x+(-x))=f(x) +f(-x) $$ then implies $$ f(-x)=-f(x) $$ and so for any rational $r$ we can write $f(r)=rf(1)$

from the functional equation alone that is as far as we can go.

if, in addition, $f$ is monotonic increasing (say), then for any real $x$ we can choose rationals $r_1, r_2$ such that $r_1 \le x \le r_2$. this implies: $$ f(r_1) \le f(x) \le f(r_2) $$ or $$ r_1f(1) \le f(x) \le r_2f(1) $$ using a squeeze this gives us, for any real $x$: $$ f(x)=xf(1) $$ and from this your required conclusion is an easy consequence

Answer 2

The gist of the argument at https://en.wikipedia.org/wiki/Cauchy%27s_functional_equation is that you can use induction to show that on each "copy" of the rationals i.e. $\mathbb{Q}+x$ for $x$ ranging over the positive irrationals and $0$, an additive function is in fact linear. In principle, it could be linear with different slopes on different copies; indeed using the axiom of choice one can "construct" such functions. But this is impossible if you assume even very slight nice properties of $f$, such as continuity, monotonicity, or even just continuity at one point.

Answer 3

It sufficies to show that $f$ is continuous at zero (and thus continuous because $f$ is linear) since the result is true for the rationals and continuity immediately implies the result for $\mathbb{R}$. To this end, observe that a monontone function has countably many discontinuities and hence is Lebesgue measurable. Now appeal to this MSE answer.