14
$\begingroup$

A function is additive if $f(x+y) = f(x) + f(y)$. Intuitively, it might seem that an additive function from R to R must be linear, specifically of the form $f(x) = kx$. But assuming the axiom of choice, that is wrong, and the proof is rather simple: you just take a Hamel basis of $\mathbb{R}$ as a vector space over $\mathbb{Q}$, and then you define your function f to be different in at least two distinct elements of the basis.

But my question is this: if there is no Hamel basis of $\mathbb{R}$, then must $f$ be linear? To put it another way, does ZF + the existence of a nonlinear additive function imply the existence of Hamel basis of $\mathbb{R}$?

I checked the Consequences of the Axiom of Choice Project, a database of choice axioms and their relationships here, and it said that it didn't know.

$\endgroup$
8
  • $\begingroup$ I believe this is equivalent, but not sure where to find a proof or how to construct one. $\endgroup$
    – Bombyx mori
    Commented Apr 19, 2013 at 2:38
  • $\begingroup$ What reason do you have for thinking they're equivalent? It certainly seems intuitive that they'd be equivalent. But then again it may be intuitive that you could never have a nonlinear additive function under any circumstances. $\endgroup$
    – Keshav Srinivasan
    Commented Apr 19, 2013 at 3:53
  • $\begingroup$ No, it is not by intuition only. This is an old problem in Rudin's book, and I think if you do a proper search (for example, on the book "counterexamples in analysis") they consider the two being equivalent. I do not have time to think deeper, however. $\endgroup$
    – Bombyx mori
    Commented Apr 19, 2013 at 4:39
  • 1
    $\begingroup$ @user32240: I have doubts that people like Herrlich and Paul Howard who dedicated much of their career to AC related problems wouldn't know about that problem (unless Rudin didn't know it was open, and underestimated its difficulty). $\endgroup$
    – Asaf Karagila ♦
    Commented Apr 19, 2013 at 8:25
  • 1
    $\begingroup$ A copy of this question at MO: mathoverflow.net/questions/128187/… $\endgroup$
    – Martin Sleziak
    Commented Jun 9, 2014 at 18:12

1 Answer 1

Reset to default
10
$\begingroup$

To my knowledge this is an open problem.

If one looks at Herrlich The Axiom of Choice, there is a diagram (7.23, p. 156) of implications related to non-measurable sets (which include discontinuous solution to the Cauchy functional equation problem), one can see that this is pretty far down below the existence of a Hamel basis.

Had it been known to be equivalent, an arrow back would be there -- and the book is not that old.

Related threads

  1. Strength of the statement "$\mathbb R$ has a Hamel basis over $\mathbb Q$"
  2. Does the existence of a $\mathbb{Q}$-basis for $\mathbb{R}$ imply that choice holds up to $\frak c$?
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .