Six regularity conditions each force an additive $f : \mathbb{R} \to \mathbb{R}$ to be $x \mapsto f(1)x$: continuity at a single point, monotonicity on a nondegenerate interval, boundedness above on one, boundedness below on one, constancy of sign on one, and a graph that is not dense in $\mathbb{R}^{2}$

Assume at least one of the six conditions holds. The six steps below treat the six conditions in turn and are exhaustive for that assumption; in each the conclusion reached is $f(x) = cx$ for every real $x$.

Condition 3. If $f$ is bounded above on a nondegenerate interval, that interval contains a closed $[p,r]$ with $p < r$ on which $f$ is bounded above, and the boundedness lemma gives $f(x) = f(1)x = cx$ for every real $x$.

Condition 6, in the contrapositive: if $f$ is not $x \mapsto cx$ then $\Gamma$ is dense in $\mathbb{R}^{2}$. Suppose $f(x_{2}) \ne c\,x_{2}$ for some real $x_{2}$. Then $x_{2} \ne 0$, since $f(0) = 0$. Put $x_{1} := 1$, $v_{1} := (x_{1}, f(x_{1})) = (1, c)$ and $v_{2} := (x_{2}, f(x_{2}))$, and put $\Delta := x_{1}f(x_{2}) - x_{2}f(x_{1}) = f(x_{2}) - c\,x_{2}$, which is nonzero by assumption.

Condition 4. If $f$ is bounded below on a nondegenerate interval $I$, say $f(z) \ge m$ for $z \in I$, then $-f$ is additive and satisfies $-f(z) \le -m$ on $I$, so $-f$ is bounded above on $I$; by step 2.1 applied to $-f$ we get $-f(x) = (-f)(1)\,x = -cx$, hence $f(x) = cx$.

Condition 2. Let $f$ be monotone on a nondegenerate interval, which contains $[p,r]$ with $p < r$. If $f$ is nondecreasing there then $f(z) \le f(r)$ for every $z \in [p,r]$, and if $f$ is nonincreasing there then $f(z) \le f(p)$; either way $f$ is bounded above on $[p,r]$ and step 2.1 applies.

Condition 1. Let $f$ be continuous at a point $c_{0}$. Taking $\varepsilon := 1$ gives a real $\delta > 0$ with $|f(x) - f(c_{0})| < 1$, hence $f(x) < f(c_{0}) + 1$, for every $x$ with $|x - c_{0}| < \delta$. The set of such $x$ is the nondegenerate interval $(c_{0}-\delta,\ c_{0}+\delta)$, so $f$ is bounded above on a nondegenerate interval and step 2.1 applies.

Let $(a,b) \in \mathbb{R}^{2}$ and let $\varepsilon > 0$ be real. Put $\alpha := (a\,f(x_{2}) - b\,x_{2})/\Delta$ and $\beta := (b\,x_{1} - a\,f(x_{1}))/\Delta$. Then $\alpha x_{1} + \beta x_{2} = a$ and $\alpha f(x_{1}) + \beta f(x_{2}) = b$, as multiplying out and cancelling $\Delta$ shows in each case.

Condition 5. If $f(z) \ge 0$ for every $z$ in a nondegenerate interval $I$ then $f$ is bounded below on $I$ by $0$ and step 3.1 applies; if $f(z) \le 0$ for every $z \in I$ then $f$ is bounded above on $I$ by $0$ and step 2.1 applies. So sign-constancy is a special case of the two preceding conditions and needs no separate argument.

Choose rationals $q_{1}, q_{2}$ with $|q_{1} - \alpha| < \eta$ and $|q_{2} - \beta| < \eta$, where $\eta > 0$ is a real chosen with $\eta\,(|x_{1}| + |x_{2}|) < \varepsilon$ and $\eta\,(|f(x_{1})| + |f(x_{2})|) < \varepsilon$; such rationals exist because a rational lies strictly between any two distinct reals, and such an $\eta$ exists because for a real $K \ge 0$ the inequality $\eta K < \varepsilon$ holds for all small enough $\eta > 0$.

Put $x := q_{1}x_{1} + q_{2}x_{2}$. Then $f(x) = q_{1}f(x_{1}) + q_{2}f(x_{2})$ by additivity and rational homogeneity, so $(x, f(x)) \in \Gamma$. Moreover $|x - a| = |(q_{1}-\alpha)x_{1} + (q_{2}-\beta)x_{2}| \le \eta(|x_{1}| + |x_{2}|) < \varepsilon$ and likewise $|f(x) - b| \le \eta(|f(x_{1})| + |f(x_{2})|) < \varepsilon$.

So every open ball of $\mathbb{R}^{2}$ meets $\Gamma$, that is, $\Gamma$ is dense in $\mathbb{R}^{2}$. Reading this contrapositively: if $\Gamma$ is not dense in $\mathbb{R}^{2}$ then $f(x) = cx$ for every real $x$, which is condition 6.

Each of the six conditions has now been shown to force $f(x) = cx$ for every real $x$: condition 1 at step 3.3, condition 2 at step 3.2, condition 3 at step 2.1, condition 4 at step 3.1, condition 5 at step 4.1 and condition 6 at step 6.1.