I was going through a handout on functional equations (I am very new to this),and there is a theorem which says the following - Suppose $ f : \mathbb{R} \to \mathbb{R}$ satisfies $f(x + y) = f(x) + f(y) $ . Then $ f(qx) = qf(x)$ for any $q \in \mathbb{Q} $. Moreover, f is linear if any of the following are true:
• $f$ is continuous in any interval.
• $f$ is bounded (either above or below) in any nontrivial interval.
• There exists $(a, b)$ and $\epsilon > 0$ such that $(x − a)^2 + (f(x) − b)^2 > \epsilon $ for every $x$. (i.e. the graph of $f$ omits some disk, however small).
Can you please explain the intuition behind this ?
[Edit: Changed tags]
The general solution $f\colon\mathbb{R}\to\mathbb{R}$ of $f(x+y)=f(x)+f(y)$ is given by all $\mathbb{Q}$-linear functions from $\mathbb{R}$ to itself where $\mathbb{R}$ is considered as a linear space ober $\mathbb{Q}$. Then, using a basis of this vector space, it can bei shown that the graph of $f$ is dense in $\mathbb{R}\times\mathbb{R}$ provided that $f$ is not $\mathbb{R}$-linear. All conditions in the question imply that the graph of $ f$ is not dense in $\mathbb{R}\times\mathbb{R}$.