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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]

Anonymous_original
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Tanya
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  • Do you see why the first assertion is true? – Anonymous_original Jun 10 '20 at 10:13
  • The intuition comes from finding an example which is non-linear. This starts by splitting the non-zero reals into an uncountable number of sets where $x \in A \implies qx \in A$ for $q \in \mathbb Q \backslash {0}$ and having different sets having different linear factors. You then find that none of the bullet points can be true – Henry Jun 10 '20 at 10:20
  • Well, in that same handout it's explained how to solve $f(x+y)=f(x)+f(y)$ over $Q$, ($f(x)=kx$) so if $f$ has a definite solution over ALL RATIONALS, and it is a continuous function, then for the graph to be completed it must hold for irrationals as well. – Anas A. Ibrahim Jun 10 '20 at 10:34

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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 $\mathbb{R}\times\mathbb{R}$.

Jens Schwaiger
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