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A function $f: \mathbb{R} \to \mathbb{R}$ is nowhere continuous but still additive, that is, $f(x+y) = f(x) + f(y)$. The existence of such functions is discussed on this site (e.g. here and here). Given that they exist, what do we know about them?

Claim 1: $f(0) = 0, f(-x) = -f(x), f(qx) = qf(x) \text{ for } x \in \mathbb{R}, q \in \mathbb{Q}.$

Proof: $f(x) = f(x+0) = f(x) + f(0),$ so $f(0) = 0$. And $0 = f(0) = f(x + (-x)) = f(x) + f(-x),$ so $f(-x) = -f(x)$. Finally, by induction, for all $n \in \mathbb{N}, f(nx)=nf(x), f(\frac 1 n x) = \frac 1 n f(x),$ so $f(qx) = qf(x)$ for $q \in \mathbb{Q}$.

Definition: A set $X \subset \mathbb{R}$ is rationally independent if for any function $b: X \to \mathbb{Q},$ then $\sum_{x \in X}b(x)x = 0$ implies $b(x) = 0$ for all $x \in X$. A rationally independent set $X$ is maximal rationally independent if adding any $r \in \mathbb{R}$ to it would make it rationally dependent.

Claim 2: An additive function $f$ is fully specified by its values on any maximal rationally independent set.

Proof: For any $x \in \mathbb{R}$, there is exactly one way to write $x = q_1x_1 + q_2x_2 + q_3x_3 + ...$ for $q_n \in \mathbb{Q}, x_n \in X$. This specifies the value of $f(x)$ by Claim 1.

Claim 3: A rationally independent set contains at most one rational non-zero element.

Proof: If $q_1, q_2 \in X$ and $q_1 \neq q_2$, then let $b(x) = \frac{-q_2}{q_1} \text{ for } x = q_1, 1 \text{ for } x = q_2, \text { otherwise } 0.$

Conjecture 4: An additive nowhere continuous function $f$ is not bounded on any open interval.

Argument: Assume there exists open interval $(a,b), a < b$ such that $x \in (a,b) \implies |f(x)| < K$. I conjecture that this allows for any $\varepsilon > 0$ constructing a punctured interval around $0$ where $|f(x)| < \varepsilon$. If correct, this would imply $f$ is continuous at $0$, a contradiction.

Conjecture 5: A rationally independent set is countable.

Conjecture 6: Let $X$ be a rationally independent set not containing $0$. Let $\hat f(x): X \to \mathbb{R}$. Then there exists an additive nowhere continuous function $f$ such that for all $x \in X, f(x) = \hat f(x)$.

That is, $f$ is uniquely specified for any maximal rationally independent set not containing $0$.

Questions:

  • Are these claims (and their proposed proofs) correct?
  • Can you prove or disprove the conjectures?
  • Are there other ways to characterize additive, nowhere continuous functions?
SRobertJames
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  • If $ |f|$ is bounded on $(a,b),$ say by M, then for $x_0={a+b\over 2}$ and $|x|<{b-a\over 2}$ we have $x_0+x\in (a,b)$ and $$f(x)=f(x+x_0)-f(x_0)$$ Hence $|f(x)|\le M+|f(x_0)|.$ By the way in definition of rationally independent set you should consider finite sums. – Ryszard Szwarc Nov 29 '22 at 21:37
  • The set $\mathbb{R}$ is a linear space over $\mathbb{Q}.$ Then it contains a basis $X.$ Then $X$ is a maximal rationally independent set by definition. Every real number is a finite linear combination of elements in $X,$ with rational coefficients. Therefore the set $X$ has cardinality the continuum. – Ryszard Szwarc Nov 29 '22 at 21:46
  • @RyszardSzwarc Thanks. 1. Can you clarify your remark "in definition of rationally independent set you should consider finite sums"? 2. I understand your first comment to be "If $f$ is bounded on any open interval, $f$ is bounded on an open interval around zero" but you are not saying "for any $\varepsilon > 0, f$ is bounded by $\varepsilon$ on an open interval around zero". Thus, you are neither proving or disproving Conjecture 4. Is that correct? – SRobertJames Nov 29 '22 at 22:18
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    If $|f(x)|\le M$ on $(-a,a)$ then $|f(x)|\le M/n$ on $(-a/n,a/n)$ for any $n.$ That gives continuity at $x=0.$ By additivity continuity at $0$ implies continuity at any point. Concerning finite sums: for any infinite countable set $X$ of the real numbers we may get $0=\sum_Xb_xx$ with rational $b_x\neq 0$ for every $x.$ Thus infinite rationally independent set in your sense does not exist. – Ryszard Szwarc Nov 30 '22 at 01:01

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