Monotone Additive Function is Linear

Theorem
Let $f: \R \to \R$ be a monotone real function which is additive, that is:
 * $\forall x, y \in \R: f \paren {x + y} = f \paren x + f \paren y$

Then:
 * $\exists a \in \R: \forall x \in \R: f \paren x = a x$

Proof 1
Denote $a = f \paren 1$.

As Additive Function is Linear for Rational Factors:
 * $ \forall r \in \Q: f \paren r = r f \paren 1 = a r$

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

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

Likewise, let $\sequence {s_n}$ be decreasing, with $s_n \in \Q$ for each $n \in \N$, such that $\displaystyle \lim_{n \mathop \to \infty} s_n = x$.

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:
 * $f \paren {r_n} \le f \paren x \le f \paren {s_n}$

for all $n \in \N$.

As we have $f \paren {r_n} = a r_n$ and $f \paren {s_n} = a s_n$, it follows that:


 * $a r_n \le f \paren x \le a s_n$

Since both $a r_n$ and $a s_n$ converge to $a x$, we have $f \paren x = a x$ by the squeeze theorem.

Proof 2
We use a Proof by Contraposition.

To that end, suppose $f$ is not linear.

We know that Graph of Nonlinear Additive Function is Dense in the Plane.

Therefore $f$ is not bounded on any nonempty open interval.

But then $f$ is certainly not monotone.

Hence, by Rule of Transposition, if $f$ is monotone, then it is linear.