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

Question

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.

Answer 1

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 $\frak c$?