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Let $S$ be a semigroup such that $S\ne S+S$ and let $f:S\to \Bbb C$ be an unbounded function satisfying $$ |f(s_1)f(s_2)-f(t_1)f(t_2)|\le 1 $$ for all $s_1, s_2, t_1, t_2 \in S$ such that $s_1+s_2=t_1+t_2$. Then, does it hold $$ f(s_1)f(s_2)-f(t_1)f(t_2)=0 $$ for all $s_1, s_2, t_1, t_2 \in S$ such that $s_1+s_2=t_1+t_2$ ?

Remark. If $S=S+S$, the above statement is true !

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1 Answer 1

Here are a few preliminary thoughts:

Lemma 1. If $f(z)=0$ then $f|_{z+S}=0$.

Proof: If $f(z)=0$, then for $s,x\in S$ we have $|f(x)f(s+z)|=|f(x)f(s+z)-f(x+s)f(z)|\le 1$. Since $f(x)$ becomes arbitrarily large, we conclude $f(z+s)=0$. $_\square$

Lemma 2. For $a\in S$, the restriction $f|_{a+S}$ is either zero or unbounded.

Proof: Assume $f(a+s)\ne 0$. Then$f(s)\ne0$ by lemma 1. From $|f(x)f(a+s)-f(a+x)f(s)|\le 1$ we see $|f(a+x)|\ge\left\vert\frac{f(a+s)}{f(s)}\right\vert \cdot |f(x)|-\frac 1{|f(s)|}$. The right hand side is unbounded as $x$ varies, hence so is the left hand side. $_\square$

Lemma 3. If $f(a+b)=0$ then $f|_{a+S}=0$ or $f|_{b+S}=0$.

Proof: Assume $f(a+b)=0$ and $f|_{a+S}\ne 0$. By lemma 2, $f|_{a+S}$ is unbounded. Then $|f(a+x)f(b)|=|f(a+b)f(x)-f(a+x)f(b)|\le 1$ implies $f(b)=0$ and hence by lemma 1, $f|_{b+S}=0$. $_\square$

Let $e\in S$. Then we can make $T=(e+S)\cup \{e\}$ a semingroup by letting $(a+e)\oplus(b+e)=a+b+e$ (NB: If $a'+e=a+e$ and $b'+e=b+e$, then $a'+b'+e=a'+b+e=a'+e+b=a+e+b=a+b+e$, so $\oplus$ is well-defined). Thus $T\oplus T=T$. If $s_1,s_2,t_1,t_2\in T$ with $s_1\oplus s_2=t_1\oplus t_2$, then also $s_1+s_2=(s_1\oplus s_2)+e=(t_1\oplus t_2)+e=t_1+t_2$, hence $|f(s_1)f(s_2)-f(t_1)f(t_2)|\le 1$. If $f|_{e+S}$ is nonzero, your remark applies to $T$ and $f$, i.e. $f(s_1)f(s_2)=f(t_1)f(t_2)$ for all $s_1,s_2,t_1,t_2\in T$ with $s_1+ s_2=t_1+ t_2$. Especially, $f(ne)=q^{n-1}f(e)$ with $q=\frac{f(2e)}{f(e)}$. As a matter of fact, this also holds (with $q=0$) if $f|_{e+S}=0$.

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