A non-constant periodic solution to Cauchy's functional equation

Question

Is there a non-constant periodic solution to Cauchy's function equation i.e. $f(x+y) = f(x) + f(y)$?

My attempts:

Obviously $f(0)=0$, but if $k$ is non-zero: $ f(x+k) = f(x) \implies 0=f(k) = f(\frac{n-1}{n}k + \frac{1}{n}k) =...=nf(\frac{1}{n}k) \implies 0=f(\frac{1}{n}k)$

from which: $0*m = 0 = mf(\frac{1}{n}k) = f(\frac{m}{n}k)$ i.e $f$ is zero for all $x \in k\mathbb Q$

If $f$ is non-zero, then it is not bounded (goo.gl/d670wu). If $f(x_0)$ is a non-zero point, then by similar reasoning $f$ is non-zero for all $x \in x_0\mathbb Q_{>0}$, and since $0 = f(x + (-x)) = f(x) + f(-x) \implies f(x) = -f(-x)$, it follows that $f$ is non-zero for all $x \in x_0\mathbb Q \setminus \{0\}$

if $|f(x_1)|$ is infinite, then by the equation and similar reasoning $|f|$ is infinite for all $x \in x_1\mathbb Q\setminus\{0\}$

If we continue this type of conquer-divide stratergy, we will get a partition $\mathbb R = k\mathbb Q \cup \bigcup_{i}[ x_i\mathbb Q \setminus\{0\}]$, where $ \{k\} \cup \{x_i\}_i$ is some sort of set where all the numbers are not rational multiplies of each other, nor are expressible as a countable sum of each other.

I don't even know if this makes sense.

Answer 1

Assuming $f$ is supposed to be $\mathbb R\to\mathbb R$:

If you require $f$ to be continuous, then it has to be everywhere zero -- namely it is easy to see that it must be zero at every rational multiple of the period, and the continuity assumption then takes care of the rest.

If $f$ does not have to be continuous, then we can construct an example using the Axiom of Choice. Consider $\mathbb R$ to be a vector space over $\mathbb Q$, and choose a basis for it. Let $b_1,b_2$ be two different elements of the basis, and consider $$ f(x) = \text{the coefficient of $b_1$ when $x$ is a rational combination of basis elements} $$ Then $f$ is a linear transformation $\mathbb R\to\mathbb Q$, and therefore satisfies the Cauchy equation. It satisfies $f(0)=0$ and $f(b_1)=1$, so it is not constant. And because $f(b_2)=0$, it has $b_2$ as a period.

(It is difficult to visualize this $f$ intuitively; its graph is dense in $\mathbb R^2$).

Without some form of Choice, I think it is impossible to prove the existence of such a function. But I don't have an argument for this ready.

Answer 2

I think you can prove that the axiom of choice is necessary by the following sketch, I got this in mind, didn't really go into the details, but it should work. Consider the set {(x,f(x))|x is in [0,1]}, and use similar arguements to the proof that the Vitali set is not Lebesgue Measurable, to show such set cannot be Lebesgue Measurable if exists, since there's an axiomatic system called the axiom of determinacy which implies that every subset of R^2 is measurable, if ZF is consistent to AD (which is impossible to prove within ZF+AD) then f cannot be proven to exists in ZF alone.