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
The proof is divided in three parts:


 * Some preliminary results;
 * Showing that the result holds for $\Q$, that is, $\forall r \in \Q: f\left({r}\right) = a r$;
 * Showing that the result holds for $\R$, that is, $\forall x \in \R: f\left({x}\right) = a x$.

Preliminaries
As $f$ is additive, we have:


 * $f \left({1}\right) = f \left({0 + 1}\right) = f\left({0}\right) + f \left({1}\right)$, that is:


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

Thus, for all $x\in\R$, we have:


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

It follows that the function $f$ is odd:


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

Proof for $\Q$
Denote $a = f \left({1}\right)$.

Trivially, we have:


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

Next, suppose that:


 * $f \left({n x}\right) = n f \left({x}\right)$

By additivity of $f$, we have:


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

Now induction gives us:


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

By oddness of $f$, we conclude:


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

Given $q \in \Z, q \ne 0$, we have:


 * $a = f \left({1}\right) = f \left({\dfrac q q}\right) = f \left({q \dfrac 1 q}\right) = q f \left({\dfrac 1 q}\right)$

It follows that:


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

Given $p, q \in \Z, q \neq 0$, we have:


 * $f \left({\dfrac p q}\right) = f \left({p \dfrac 1 q}\right) = p f \left({\dfrac 1 q}\right) = p \dfrac a q = a \dfrac p q$

Therefore we conclude:


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

Proof for $\R$
Let $x \in \R \setminus \Q$.

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

Likewise, let $\left \langle{s_n}\right \rangle$ be decreasing, with $s_n \in \Q$ for each $n \in \N$, 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, by passing to $g = -f$ if necessary, we can assume 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 $n \in \N$.

As we have $f \left({r_n}\right) = a r_n$ and $f \left({s_n}\right) = a s_n$, it follows that:


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

Since both $a r_n$ and $a s_n$ converge to $a x$, we have $f \left({x}\right) = a x$ by the Squeeze theorem.