Let $f : \mathbb{R} \rightarrow \mathbb{R}$.
- We say that $f$ is additive if $f(a+b) = f(a) + f(b)$.
- We say that $f$ is bounded from below on any open interval $(a,b)$ if there is an $M \in \mathbb{R}$ depending on the interval so that $M \leq f(x)$ for all $x \in (a,b)$.
- We say that $f$ is linear if for any $a, b,c \in \mathbb{R}$, we have $f(ab + c) = af(b) + f(c).$
Step 1: The only linear functions on $\mathbb{R}$ are of the form $f(x) = \alpha x$ for some $\alpha \in \mathbb{R}$.
Proof: Consider the function $g(x) := f(x)/x$. We claim that this function is constant outside of zero. Observe that for any non-zero $q \in \mathbb{R}$, we have
$$ g(qx) = \frac{f(qx)}{qx} = \frac{qf(x)}{qx} = \frac{f(x)}{x} = g(x).$$
For any non-zero $x, y \in \mathbb{R}$, let $q = (y/x)$. Then $g(x) = g(qx) = g(y)$, so $g$ is constant on $\mathbb{R} \setminus \{0\}$. Denote this constant by $\alpha$. Then for all $x$ non-zero, we have $f(x) = \alpha x$, and this also holds at $x=0$. $\blacksquare$
Remark: Note that this also works if we replace $\mathbb{R}$ with $\mathbb{Q}$.
Step 2: Let $f$ be additive. We claim that, restricting $f$ to the rationals, we have $f : \mathbb{Q} \rightarrow \mathbb{Q}$ is of the form $f(x) = \alpha x$.
Proof: First, observe that $f(0) = f(0+0) = 2f(0)$, so $f(0) = 0$. Next, notice that for $n \in \mathbb{Z}_+$, we have
$$ f(n) = f \left( \sum_{i=1}^n 1 \right) = n f(1).$$
Now, notice that if $n \in \mathbb{Z}$ is less than zero, then $f(n) = n f(1)$. Indeed,
$$ 0 = f(0) = f(n + (-n)) = f(n) + f(-n) = f(n) + (-n) f(1).$$
Now we need to deal with rational numbers. Notice that for $n \in \mathbb{Z}$ positive, we have
$$ f(1) = f \left( \sum_{i=1}^n \frac{1}{n} \right) = n f(1/n).$$
This shows that $f(1/n) = (1/n) f(1)$. The same trick as above can be used to show that this works for negative $n$. Finally, the same tricks can be used to show that $f$ is linear on the rationals. Now use Step 1. $\blacksquare$
Step 3: Assume that $f : \mathbb{R} \rightarrow \mathbb{R}$ is additive and bounded from below (on any closed interval). We claim that $f(x) = \alpha x$ for all $x \in \mathbb{R}$.
Proof: Let $\alpha$ be from Step 2. Let $M$ be such that for $x \in [0,1]$, we have $f(x) \geq M$. Notice that for $x \in [0,T]$, where $T > 0$ is an rational number, we have
$$ 0 \leq x \leq T \implies 0 \leq \frac{x}{T} \leq 1 \implies M \leq f \left( \frac{x}{T}\right) \implies MT \leq f(x).$$
If we have $[-T, 0]$ instead, then
$$-T \leq x \leq 0 \implies 0 \leq -\frac{x}{T} \leq 1 \implies M \leq f \left( - \frac{x}{T}\right) \implies f(x) \leq - TM.$$
In particular, this shows that $f$ is bounded above on $[-T, 0]$. But by assumption, it is also bounded below on $[-T, 0]$. This shows us that $f$ is bounded on $[-T,0]$ (and we can use a symmetric argument to show that $f$ is bounded on $[0,1]$ as well).
Let's now assume that $|f(x)| \leq M$ for $x \in [-1,1]$ (one may need to make $M$ larger). Notice that if $x \in \mathbb{R}$, then there is a sequence of rational numbers $r_n$ so that $|x - r_n| \leq 1/n$. Rearranging, we get $ -1 \leq n(x - r_n) \leq 1$. Using additivity, $f(n(x- r_n)) = nf(x) - nf(r_n) = f(x) - \alpha n r_n$, and thus
$$ -M \leq f(n(x - r_n)) = nf(x) - n\alpha r_n \leq M \implies \alpha r_n - \frac{M}{n} \leq f(x) \leq \alpha r_n + \frac{M}{n}.$$
Take $n \rightarrow \infty$ to get $\alpha x \leq f(x) \leq \alpha x$, as desired. $\blacksquare$
