6

Is there a way to reduce the following functional equation

$$ f(x+y)=f(x+1)f(y),\qquad x,y>0, $$

to the equation

$$ f(x+y)=f(x)f(y),\qquad x,y>0, $$

whose solutions are known?

Thanks in advance.

Martin Sleziak
  • 53,687
axl
  • 63

1 Answers1

13

Let $g(x)=f(x+1)$ then $g(x+y)=f(x+(y+1))=f(x+1)f(y+1)=g(x)g(y)$

Mark Bennet
  • 100,194
  • For $ f :R \to R$ ,if $ f$ (or $ g$) is assumed to be continuous then $g(x)=A^x$ for some constant $A > 0 $,( or $ g=0$ for all $ x)$. But there are many other discontinuous solutions for $g$. – DanielWainfleet Aug 31 '15 at 18:30
  • I think measurability of f (or g) suffices. Do you agree? – axl Sep 01 '15 at 13:59