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](http://consequences.emich.edu/CONSEQ.HTM), and it said that it didn't know.



Does a nonlinear additive function on R imply a Hamel basis of R?
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.


I doubt Herrlich would have omitted a reference to such fact, had it been known, and the last time I looked around, I couldn't find anything newer which proved anything related to that (and I did look around several times when answering a couple of questions).


