Thursday, 25 August 2016

calculus - Solving functional equation f(x)f(y)=f(x+y)





I'm having some trouble solving the following equation for f:AB where AR and BC such as:



f(x)f(y)=f(x+y)x,yA



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.


No comments:

Post a Comment

real analysis - How to find limhrightarrow0fracsin(ha)h

How to find lim without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...