I'm having some trouble solving the following equation for f:A→B where A⊆R and B⊆C such as:
f(x)f(y)=f(x+y)∀x,y∈A
The solution is supposed to be the form f(x)=ekx where k can be complex but I don't see how to show it.
Note: I'm trying to solve this equation to find irreductible representation of U(1) but it doesn't really matter I think
Answer
The general solution of the given functional equation depends on conditions imposed to f.
We have:
The general continuous non vanishing complex solution of f(x+y)=f(x)f(y) is exp(ax+bˉx), where ˉx is the complex conjugate of x. Note that this function is not holomorphic.
If we request that f be differentiable then the general solution is f(x)=exp(ax), as proved in the @doetoe answer.
Added:
here I give a similar but a bit simpler proof:
Note that f(x+y)=f(x)f(y)⇒f(x)=f(x+0)=f(x)f(0)⇒f(0)=1 ( if f is not the null function). We have also: f′(x+y)=f′(x)f(y) (where f′ is the derivative with respect to x).
Now, setting x=0 and dividing by f(y) yields:
f′(y)=f′(0)f(y)⇒f(y)=kexp(f′(0)y)
and, given f(0)=1 we must have k=1.
The continuity request can be weakened but, if f is not measurable then, with the the aid of an Hamel basis, we can find more general ''wild'' solutions.
For a proof see: J.Aczél : lectures on functional equations and their applications, pag. 216.
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