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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.

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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).

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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

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