Restricted Cauchy equation on a non-dense domain

Question

It is really well known that if $f: \mathbf{R}\to \mathbf{R}$ is continuous and $$ \forall x,y \in \mathbf{R},\,\,\,\,f(x+y)=f(x)+f(y) $$ then $f$ is linear, i.e., there exists $a \in \mathbf{R}$ such that $f(x)=ax$ for all $x \in \mathbf{R}$.

More generally, it is known that if $f: K \to \mathbf{R}$ is continuous, where $K\subseteq \mathbf{R}$ is a non-empty open connected set with $K+K\subseteq K$, then the functional equation $$ \forall (x,y) \in K^2,\,\,\,\,f(x+y)=f(x)+f(y)\,\,\,\,\,\,\,\,\,\,\,(\star) $$ implies that $f$ is linear.

Here a related question:

Question. Does there exist a set $S\subseteq \mathbf{R}^2$ which is not dense in $\mathbf{R}^2$, not containing an open connected set, and such that if $f: \mathbf{R}\to \mathbf{R}$ is continuous function satisfying ($\star$) for all $(x,y) \in S$ then $f$ is linear?

Easy observation: $S$ cannot be bounded (see also the comment of TheoBandit below). Indeed, in the opposite, we could set $f(x)=0$ for all $x$ in a ball with sufficiently large radius; then, any continuous extension will work.

@TheoBendit it contains an open connected set; well, let me add it at the question, thank you — Paolo Leonetti, Jun 04 '19 at 05:59
I think it suffices to require the measure of $S$ is greater than $0$, then by Steinhaus theorem, we can reduce the case to what you have mentioned. — Bach, Jun 04 '19 at 06:00
@PaoloLeonetti Why to avoid the previous and what is your previous case? And are you aware that the measure of $\mathbb R\times\mathbb N$ is $0$? — Bach, Jun 04 '19 at 06:06
@Bach Just to understand what happens in the other cases. Indeed, $\mathbf{R}\times \mathbf{N}$ has Lebesgue measure $0$ and it does not contain an open connected set, that's why I was asking that — Paolo Leonetti, Jun 04 '19 at 06:13
Also, on the topic of this result, does it not contradict your own easy observation? For example, take $S$ to be a (bounded and connected) open ball. — Theo Bendit, Jun 04 '19 at 06:36
@KaviRamaMurthy Here it is how connectedness plays a role in the question: suppose that $A$ and $B$ are two open balls in $\mathbf{R}^2$ with radius $1$ and their centers are sufficiently far from each other, let us say $(0,0)$ and $(10,0)$. Then you can define $f: \mathbf{R}\to \mathbf{R}$ as any continuous extension of $f(x)=x$ if $|x| \le 2$ and $f(x)=0$ if $|x-10|\le 2$. Theb $f$ satisfies the equation and it is not linear. About the reference on the open connected case (in truth, with an additional hypothesis), you can have a look here: https://arxiv.org/abs/1905.13541 — Paolo Leonetti, Jun 04 '19 at 09:37
@TheoBendit You are correct. The proper result is: "Let $K\subseteq \mathbf{R}$ be a non-empty open connected set such that $K+K\subseteq K$ and assume that equation ($\star$) holds for all $(x,y) \in K^2$, with $f: K \to \mathbf{R}$ continuous. Then $f$ is linear." (Hence, your observation is related to the domain of $f$.) — Paolo Leonetti, Jun 04 '19 at 09:44

score 3 · Accepted Answer · edited Jun 04 '19 at 10:09

Yes, there is such a domain. Consider, $$S = \bigcup_{n \in \Bbb{Z}} \{(x, nx) : x \in \Bbb{R}\}.$$ Note that $S$ is not dense and has empty interior. I claim that, if $f : \Bbb{R} \to \Bbb{R}$ is continuous and satisfies $$f(x + y) = f(x) + f(y)$$ for all $(x, y) \in S$, then $f$ is scalar homogeneous. First, note that $(0, 0) \in S$, hence $$f(0 + 0) = f(0) + f(0) \implies f(0) = 0.$$ Next, suppose $n \ge 1$ is an integer. I wish to show that $f(nx) = nf(x)$ for all $x$. We proceed by induction.

The base case is clear. Suppose $f(nx) = nf(x)$ for some $n$. Then, because $(x, nx) \in S$, we have $$f((n + 1)x) = f(x + nx) = f(x) + f(nx) = f(x) + nf(x) = (n + 1)f(x).$$ The claim holds by induction.

Similarly, this holds for $n < 0$. If we assume $n \le 0$ is such that $f(nx) = nf(x)$, then $$f(x) + f((n - 1)x) = f(x + (n - 1)x) = f(nx) = nf(x),$$ which implies $f((n - 1)x) = (n - 1)f(x)$ as required. By induction, $$f(nx) = nf(x) \quad \forall n \in \Bbb{Z}, x \in \Bbb{R}.$$

Next, as you might guess, we establish rational scalar homogeneity. However, this follows directly from the integer scalar homogeneity. We have $$qf\left(\frac{p}{q}x\right) = f\left(q \frac{p}{q}x\right) = f(px) = pf(x) \implies f\left(\frac{p}{q}x\right) = \frac{p}{q}f(x)$$ where $p, q \in \Bbb{Z}$ and $q \neq 0$.

Since the rationals are dense in $\Bbb{R}$, we must have $f(\lambda x) = \lambda f(x)$ for any real $\lambda$, hence $f$ is linear.

Restricted Cauchy equation on a non-dense domain

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