### Pri la Funkcia Ekvacio *f(x + y) = f(x) + f(y)*

#### Posted by Tom Leinster

Jam de longe, Cauchy pruvis ke *kontinua* funkcio $f(x)$ kiu verigas la
funkcian ekvacion
$f(x + y) = f(x) + f(y)$
kiuj ajn estu la nombroj $x, y$, necese estas homogena, unuagrada funkcio $f(x)
\equiv A x$.

So begins Maurice Fréchet’s 1913 paper in *L’Enseignement
Mathématique*.
I came across it when I was trying to find the right reference for the
solution of this functional equation. Apparently Cauchy was the first to prove
that when $f$ is *continuous*, the only solutions are $f(x) = A x$ for
some constant $A$. Mark Meckes had told me that *Lebesgue measurability*
of $f$ was sufficient, and found a nice explanation at the
Tricki.

But I needed to find the original reference. And when I tracked it down, I was surprised and intrigued to find that it was in Esperanto.

It turns out that Fréchet, apart from making numerous contributions to analysis, was a keen Esperantist, publishing many papers in the subject and (as Mark pointed out to me) serving as president of the Internacia Scienca Asocio Esperantista. I’m looking forward to citing it.

Incidentally, Fréchet’s result can be improved further. Mark Kormes, in 1926, showed that one only needs to assume $f$ to be bounded on some set of positive measure. (This is a weaker condition than measurability.) But, boringly, his paper is in English.

## Re: Pri la Funkcia Ekvacio f(x + y) = f(x) + f(y)

Wow! Cool! But you should have posted this blog entry using rot13.

A cute fact, sort of obvious from what you said, is that starting with Zermelo-Fraenkel set theory, it’s perfectly consistent to assume either that $f(x + y) = f(x) + f(y)$

doeshave solutions other than the obvious $f(x) = A x$ ones, or that itdoesn’t. The Axiom of Choice implies itdoeshave other solutions… which unfortunately we are unable to write down. The Axiom of Determinancy implies itdoesn’t.Guess which axiom most mathematicians prefer.

And another, slightly harder puzzle: guess which famous mathematician wrote sentences like this: