Monotone Additive Function is Linear

Theorem
Let $f: \R \to \R$ be a monotonic function which is additive, i.e.:
 * $\forall x, y \in \R: f \left({x + y}\right) = f \left({x}\right) + f \left({y}\right)$

Then:
 * $\exists a \in \R: \forall x \in \R: f\left({x}\right) = a x$

Proof
Let $a = f \left({1}\right)$.

Then $f \left({1}\right) = a \times 1$.

Supposing, by induction, that $f \left({n}\right) = a n$ for some $n \in \N$.

Then we have $f \left ({n+1}\right) = f \left({n}\right) + f\left({1}\right)$ which is $a n + a = a \left({n+1}\right)$.

So:


 * $\forall n \in \N: f\left({n}\right) = a n$

As $f$ is additive, we also have:
 * $f \left({1}\right) = f \left({0 + 1}\right) = f\left({0}\right) + f \left({1}\right)$, that is:


 * $f\left({0}\right) = 0$

Another milestone: for all $x\in\R$, we have:
 * $0 = f \left({0}\right) = f \left({x + \left({-x}\right)}\right) = f \left({x}\right) + f \left({-x}\right)$

therefore the function $f$ is odd:


 * $\forall x \in \R: f\left({-x}\right) = - f \left({x}\right)$

Then it follows that:
 * $f \left({-n}\right) = -\left({a n}\right) = a \left({-n}\right)$

Since we have already proved to the naturals, to zero, and to the negative integers, we see that is holds for all integers:


 * $\forall p \in \Z: f \left({p}\right) = a p$

Now, we have:
 * $\forall x \in \R: f \left({1 x}\right) = 1 f \left({x}\right)$

If we suppose, by induction, that:
 * $f \left({n x}\right) = n f \left({x}\right)$

we have:
 * $f \left({\left({n + 1}\right) x}\right) = f \left({n x + x}\right) = f \left({n x}\right) + f \left({x}\right) = n f \left({x}\right) + f \left({x}\right) = \left({n+1}\right) f \left({x}\right)$

It also holds for $n=0$:
 * $\forall x \in \R: f \left ({0 x}\right) = f \left({0}\right) = 0 = 0 f \left({x}\right)$

For the negative integers, we have:
 * $f \left({\left({-n}\right) x}\right) = - f \left({n x}\right) = - \left({n f \left({x}\right)}\right) = \left({- n}\right) f \left({x}\right)$

Therefore:
 * $\forall p \in \Z, x \in \R: f \left({p x}\right) = p f \left({x}\right)$

Given $q \in \Z, q \ne 0$, we have:
 * $a = f \left({1}\right) = f \left({\dfrac q q}\right) = f \left({q \dfrac 1 q}\right) = q f \left({\dfrac 1 q}\right)$

i.e.:
 * $\forall q \in \Z, q \ne 0: f \left({\dfrac 1 q}\right) = \dfrac a q$

Given $p, q \in \Z, q \neq 0$, we have:
 * $f \left({\dfrac p q}\right) = f \left({p \dfrac 1 q}\right) = p f \left({\dfrac 1 q}\right) = p \dfrac a q = a \dfrac p q$

and then:
 * $ \forall r \in \Q: f \left({r}\right) = a r$

Let $x \in \R - \Q$.

Let $\left \langle {r_n}\right \rangle$ be an increasing sequence, with $r_n \in \Q$ for each $n \in \N$, and with $\displaystyle \lim_{n \to \infty} r_n = x$.

Likewise, let $\left \langle{s_n}\right \rangle$ be decreasing, with each term a rational number, such that $\displaystyle \lim_{n \to \infty} s_n = x$.

From the Peak Point Lemma, it is always possible to construct sequences like these, for $\Q$ is dense in $\R$.

Now, we will assume (without loss of generality) that $f$ is increasing.

Then we have $f \left({r_n}\right) \leq f \left ({x}\right) \leq f \left({s_n}\right)$ for all $n \in \N$.

But $f \left({r_n}\right) = a r_n$, and $f \left({s_n}\right) = a s_n$, and then:


 * $a r_n \leq f \left({x}\right) \leq a s_n$

Since both sequences $a r_n$ and $a s_n$ converge to $a x$, $f \left({x}\right)$ must be $a x$ (because $f \left({x}\right)$ must be $\ge \displaystyle \lim_{n \to \infty} a r_n$, and $\le \displaystyle \lim_{n \to \infty} a s_n$).