MFRH015

The Cauchy Exponential Equation involves the equation f(x+y) = f(x)f(y). There are actually discontinuous functions that satisfy this equation, but we will show that continuous functions satisfying the above equation are f(x) = 0 (or f(x) identically equal to 0) and f(x) = b^x for some positive constant b.

Clearly, f(x) = 0 is a continuous function that satisfies the equation. Moreover, if there exists a point a such that f(a) = 0, then f(x) = 0 for all x. To see this, suppose f(a) = 0, then

f(x) = f(x-a)f(a) = f(x-a) * 0 = 0.

Hence, if f(x) is 0 at even one point, then f(x) is 0 everywhere.

Now we need to find a continuous function such that f is not identically equal to 0. Note that

f(x) = f(x/2)f(x/2) = [f(x/2)]^2, which implies that f(x) > 0 for all x.

Since f is positive, we can take (natural) logarithms to get

log f(x+y) = log f(x) + log f(y).

Let g(x) = log f(x), then our equality above tells us that g(x + y) = g(x) + g(y). This is in fact known as Cauchy’s Equation, which has f(x) = ax for some constant a, as a solution (easily checked). Hence, log f(x) = g(x) = ax for some constant a, which implies that f(x) = e^(ax). But since a is arbitrary, we might as well let b be some other constant so that f(x) = b^x, for some positive constant b. This concludes our discussion on the Cauchy Exponential Equation, for now.

Type: art.discourse.mathematics
Produced by: The Numbers