Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $S$ be a semigroup with no identity element and $m:S\to \Bbb C$ be given function($m\not\equiv 0$) satisfying the exponential functional equation $$ m(x+y)=m(x)m(y) $$ for all $x, y\in S$. Find all solutions $f:S\to \Bbb C$ satisfying the equation \begin{equation} f(x+y)=f(x)m(y)+f(y)m(x) \tag 1 \end{equation} for all $x, y\in S$.

Remark. If $S$ is a group, then using the fact that $m(x)\ne 0$ for all $x\in S$ and dividing $(1)$ by $m(x+y)$, we have $$ f(x)=m(x)A(x) $$ for all $x\in S$, where $A$ is an additive function.

share|improve this question

1 Answer 1

The subset $I=\{x\in S|m(x)=0\}$ is a prime ideal in $S$ (i.e. $xy\in I \Rightarrow x\in I \vee y\in I$), so $S\setminus I$ is a subsemigroup to which you can apply your reasonings concerning groups.

Addendum: Here is a counter-example showing that generally speaking $f$ has not the form $f(x)=m(x)A(x)$, where $A$ is additive on whole $S$.

Take an arbitrary set $J$ with a fixed element $z$. Let ${\bar J}$ be a set such that $|J|=|{\bar J}|, J\cap {\bar J}=z$, and $x\to {\bar x}$ a bijection from $J$ to ${\bar J}$ (${\bar z}=z$). Let $I=J\cup {\bar J}$ and $S=I\cup \{e\}$ where $e$ is an extra element.

Define the addition:

$$ I+I=z, \ \ e+e=e, \ \ e+x={\bar x} $$ so that $S$ will be a semigroup.

Set $m(a)=0$ for $a\in I$, $m(e)=1$. Then an arbitrary $f$ with $f(x)=f({\bar x})$ yields the equation (I).

share|improve this answer
    
Thank you very much for your concern, however, I have no idea how to apply it. Let $A(x)=f(x)/m(x)$ for $x\in S\setminus I$. Then $A$ is additive only on $S\setminus I$, which can not extends to $S$, in general. –  Chung. J May 17 '13 at 13:56
    
Evidently $f|_{I+I}=0$, it remains to find $f|_{I\setminus (I+I)}$. Have you some additional information about $S$? Is it commutative? –  Boris Novikov May 17 '13 at 14:34
    
Yes! $S$ is commutative semigroup with no identity. Sorry to omit the fact. I want to know that $f$ has the form $f(x)=m(x)A(x)$, where $A$ is additive on whole $S$. –  Chung. J May 17 '13 at 14:36
    
Is $S$ cancellative? –  Boris Novikov May 17 '13 at 14:48
    
No cancelative, if does, probably we can prove that $m$ vanishes at no point. –  Chung. J May 17 '13 at 14:56

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.