## The functional equation f(x+y) = f(x)f(y)

Suppose satisfies . What can we say about ?

Putting gives

which can happen if either or . Note that the function which is identically zero satisfies the functional equation. If is not this function, i.e., if for at least one value of , then plugging that value of (say ) into the equation gives . Also, for any , the equation forces as well. Further, so for all .

Next, putting gives , and by induction . Putting in place of in this gives which means (note we’re using here). And again, . So , which completely defines the function at rational points.

[As , it can be written as for some constant , which gives for rational .]

To extend this function to irrational numbers, we need some further assumptions on , such as continuity. It turns out that being continuous at any point is enough (and implies the function is everywhere): note that . Even being Lebesgue-integrable/measurable will do.

Else, there are discontinuous functions satisfying the functional equation. (Basically, we can define the value of the function separately on each “independent” part. That is, define the equivalence class where and are related if for rationals and , pick a representative for each class using the axiom of choice (this is something like picking a basis for , which corresponds to the equivalence class defined by the relation ), define the value of the function independently for each representative, and this fixes the value of on . See this article for more details.)

To step back a bit: what the functional equation says is that is a homorphism from , the additive group of real numbers, to , the multiplicative monoid of real numbers. If is not the trivial identically-zero function, then (as we saw above) is in fact a homomorphism from , the additive group of real numbers, to , the multiplicative group of positive real numbers. What we proved is that the exponential functions are precisely all such functions that are nice (nice here meaning either measurable or continuous at least one point). (Note that this set includes the trivial homomorphism corresponding to : the function identically everywhere. If is not this trivial map, then it is in fact an isomorphism.)

Edit [2013-10-11]: See also Overview of basic facts about Cauchy functional equation.

Very nice!

KVMWed, 2013-04-10 at 19:28:36

[…] writing a series of posts leading up to an understanding of the exponential function (here, here, here), but it seems to have got abandoned. Consider this one a contribution to that […]

The idea of logarithms, and the first appearance of e | The Lumber RoomWed, 2013-11-27 at 10:53:02

Really lovely proof. Thanks for sharing!

Jess WenFri, 2016-12-09 at 13:00:08