$f(a+b,\lambda) = f (a,\lambda) \cdot f(b,\lambda)$

Asked May 13 '13 at 14:39

Active May 13 '13 at 14:42

Viewed 106 times

Consider the equation:

$f(a+b,\lambda) = f (a,\lambda) \cdot f(b,\lambda)$, for $a \geq 0$ and $b \geq 0$.

Is my understanding that this simple functional equation is important in analysis. Can someone give me examples where this can be used in practice, its theoretical importance and maybe some literature where I can find more?

Also, if $f(\cdot,\lambda)$ is continuous, then one can show that $f(a,\lambda) =$exp$(ag(\lambda))$, for some function $g(\lambda)$. Is this the only type of function that will satisfy the equation above?

edited May 13 '13 at 14:42

MJD

65,394
39
298
580

asked May 13 '13 at 14:39

Cherufe

$f=0$ seem to be solution too ... – wece May 13 '13 at 14:43
2

Of relevance is this question (which provides a solution for fixed $\lambda$). – Lord_Farin May 13 '13 at 14:44
@Lord_Farin: Thank you! – Cherufe May 13 '13 at 14:50
@wece: It seams to me that is either zero for all $x$, or it is of the type exp$(ag(\lambda)$. – Cherufe May 13 '13 at 14:53
Since nothing happens to $\lambda$, set $g(x)\equiv f(x,\lambda)$ and you see that this is the exact same question as in Lord_Farin's comment. The solution there is $g(x)=\text e^{c x}$, hence you have the "pullback" $f(x,\lambda)=\text e^{c_\lambda x}$ – Nikolaj-K May 13 '13 at 14:54
yep really good link ... – wece May 13 '13 at 14:55
@Cherufe Is there anything else you'd like to know or can this question be closed as a duplicate of the referenced one? (The second answer there provides some perspective on the uses/theoretical importance in analysis.) – Lord_Farin May 13 '13 at 16:36
@Lord_Farin The link was explanatory enough :) – Cherufe May 14 '13 at 16:50
possible duplicate of Is there a name for function with the exponential property $f(x+y)=f(x) \times f(y)$? – Lord_Farin May 14 '13 at 16:52

$f(a+b,\lambda) = f (a,\lambda) \cdot f(b,\lambda)$

0 Answers0