The Discrete Semigroup Algebra of a Semigroup S, ℓ1(S, α)

# The Discrete Semigroup Algebra of a Semigroup S, ℓ1(S, α)

Let $S$ be a semigroup. That is, $S$ is a set with an associative binary operation defined on $S$. Let $\alpha$ be a real-valued function on $S$ such that $\alpha(x) > 0$ for every $x \in S$, and such that for every $x, y \in S$:

(1)
\begin{align} \quad \alpha(x \cdot y) \leq \alpha (x) \alpha (y) \end{align}

Let:

(2)
\begin{align} \quad \ell^1(S, \alpha) = \left \{ f : S \to \mathbb{C} \biggr | \sum_{x \in S} |f(s)| \alpha(s) < \infty \right \} \end{align}

For each $f, g \in \ell^1(S, \alpha)$ and all $a \in \mathbf{F}$ we define addition $f + g$, scalar multiplication $a f$, and convolution multiplication $f * g$ for each $x \in S$ by:

(3)
\begin{align} \quad (f + g)(x) &= f(x) + g(x) \\ \quad (af)(x) &= af(x) \\ \quad (f * g)(x) &= \sum_{tu = x} f(t)g(u) \quad \mathrm{where} \quad (f * g)(x) = 0 \: \mathrm{if} \: tu = x \: \mathrm{has \: no \: solutions} \end{align}

Define $\| \cdot \|_1 : \ell^1(S, \alpha) \to [0, \infty)$ for each $f \in \ell^1(S, \alpha)$ by:

(4)
\begin{align} \quad \| f \|_1 = \sum_{x \in S} |f(x)| \alpha(x) \end{align}
 Definition: Let $S$ be a semigroup and let $\alpha$ be a positive real-valued function on $S$. The Discrete Semigroup Algebra of $S$ Relative to $\alpha$ is the Banach algebra of $\ell^1(S, \alpha)$ with the operations of pointwise function addition, scalar multiplication, and product defined above, and with the norm $\| \cdot \|_1 : \ell^1(S, \alpha) \to [0, \infty)$ defined for each $f \in \ell^1(S, \alpha)$ by $\displaystyle{\| f \|_1 = \sum_{x \in S} |f(x)| \alpha(x)}$. The Discrete Semigroup Algebra of $S$ is simply $\ell^1(S, 1)$ (that is, $\alpha = 1$).

Note that when $S$ is a group we have that the discrete semigroup algebra of $S$ coincides with the discrete group algebra of $S$.

The following proposition tells us that $\ell^1(S, \alpha)$ is indeed a Banach algebra.

 Proposition 1: Let $S$ be a semigroup and let $\alpha$ be a positive real-valued function on $S$. Then $\ell^1(S, \alpha)$ is a Banach algebra.
• Proof: It is easy to verify that $\ell^1(S, \alpha)$ with the operations of addition and scalar multiplication, as well as the norm $\| \cdot \|_1$ is a normed linear space. We will omit proving that it is a Banach space (albiet, the proof is similar to showing that the discrete group algebra of a group is a Banach space).
• With that said, all that remains to show is that (A) $\ell^1(S, \alpha)$ with the operation of product defined above forms an algebra, and that (B) $\| \cdot \|_1$ is an algebra norm on $\ell^1(S, \alpha)$.
• (A) Showing that $\ell^1(G)$ is an algebra:
• 1. Showing that $(f * g) * h = f * (g * h)$: Let $f, g, h \in \ell^1(S, \alpha)$. Then for every $x \in S$:
(5)
\begin{align} \quad [(f* g) * h](x) &= \sum_{tu = x} (f * g)(t)h(u) \\ &= \sum_{tu = x} \left [ \sum_{vw = t} f(v)g(w) \right ] h(u) \\ &= \sum_{tu = x} \sum_{vw = t} f(v)g(w)h(u) \\ &= \sum_{vwu = x} f(v)g(w)h(u) \quad (*) \end{align}
• And also:
(6)
\begin{align} \quad [f * (g * h)](x) &= \sum_{ut = x} f(u)(g * h)(t) \\ &= \sum_{ut = x} f(u) \left [\sum_{vw = t} g(v)h(w) \right ] \\ &= \sum_{ut = x} \sum_{vw = t} f(u)g(v)h(w) \\ &= \sum_{uvw = x} f(u)g(v)h(w) \quad (**) \end{align}
• These two sums are same by change of variables, and so we see that $(f * g) * h = f * (g * h)$.
• 2. Showing that $f * (g + h) = f * g + f * h$: Let $f, g, h \in \ell^1(S, \alpha)$. Then for every $x \in S$:
(7)
\begin{align} \quad [f * (g + h)](x) = \sum_{tu = x} f(t)(g + h)(u) = \sum_{tu = x} [f(t)g(u) + f(t)h(u)] = \sum_{tu = x} f(t)g(u) + \sum_{tu = x} f(t)h(u) = (f * g)(x) + (f * h)(x) \end{align}
• Therefore $f * (g + h) = f * g + f * h$.
• 3. Showing that $(a f) * g = a (f * g) = f * (ag)$: Let $f, g \in \ell^1(S, \alpha)$ and let $a \in \mathbf{F}$. Then for every $x \in S$ we have that:
(8)
\begin{align} \quad [(af) * g](x) = \sum_{tu = x} af(t)g(u) = a \sum_{tu = x} f(t)g(u) = [a (f * g)](x) \end{align}
• And also:
(9)
\begin{align} \quad [a(f * g)](x) = a \sum_{tu = x} f(t)g(u) = \sum_{tu = x} f(t) a g(u) = [f * (ag)](x) \end{align}
• Therefore $(af) * g = a (f * g) = f * (ag)$.
• Hence $\ell^1(S, \alpha)$ is an algebra over $\mathbf{F}$.
• (B) Showing that $\| \cdot \|_1$ is an algebra norm on $\ell^1(S, \alpha)$:
• 1. Showing that $\| f \|_1 = 0$ if and only if $f = 0$: Suppose that $\| f \|_1 = 0$. Then $\displaystyle{\sum_{x \in S} |f(x)|\alpha(x) = 0}$. Since $\alpha(x) > 0$ for all $x \in S$, this implies that $|f(x)| = 0$ for all $x \in S$ and so $f = 0$. On the other hand, if $f = 0$ then $\displaystyle{\sum_{x \in S} |f(x)|\alpha(x) = 0}$, i.e., $\| f \|_1 = 0$.
• 2. Showing that $\| a f \|_1 = |a| \| f \|_1$ for all $f \in \ell^1(S, \alpha)$ and all $a \in \mathbf{F}$: Let $f \in \ell^1(S, \alpha)$ and let $a \in \mathbf{F}$. Then:
(10)
\begin{align} \quad \| a f \|_1 = \sum_{x \in S} |af(x)| \alpha(x) = \sum_{x \in S}|a||f(x)| \alpha(x) = |a| \sum_{x \in S} |f(x)| \alpha(x) = |a| \| f \|_1 \end{align}
• 3. Showing that $\| f + g \|_1 \leq \| f \|_1 + \| g \|_1$: Let $f, g \in \ell^1(S, \alpha)$. Then:
(11)
\begin{align} \quad \| f + g \|_1 = \sum_{x \in S}|f(x) + g(x)|\alpha(x) \leq \sum_{x \in S} [|f(x)| + |g(x)|]\alpha(x) = \sum_{x \in S} [|f(x)| \alpha(x) + |g(x)| \alpha(x)] = \sum_{x \in S} |f(x)| \alpha(x) + \sum_{x \in S} |g(x)| \alpha(x) = \| f \|_1 + \| g \|_1 \end{align}
• 4. Showing that $\| f * g \|_1 \leq \| f \|_1 \| g \|_1$: Let $f, g \in \ell^1(S, \alpha)$. Then using the fact that $\alpha(tu) \leq \alpha(t) \alpha(u)$, we have that::
(12)
\begin{align} \quad \| f * g \|_1 = \sum_{x \in S} |(f * g)(x)| \alpha(x) = \sum_{x \in S} \left | \sum_{tu = x} f(t)g(u) \right | \alpha(x) \leq \sum_{x \in S} \left [ \sum_{tu = x} |f(t)g(u)| \right ] \alpha(x) = \sum_{tu \in S} |f(t)||g(u)| \alpha(tu) &\leq \sum_{tu \in S} |f(t)||g(u)|\alpha(t) \alpha(u) \\ &\leq \left [ \sum_{t \in S} |f(t)|\alpha(t) \right ] \left [ \sum_{u \in S} |f(u)| \alpha(u) \right ] \\ &\leq \| f \|_1 \| g \|_1 \end{align}
• Therefore $\| \cdot \|_1$ is an algebra norm on $\ell^1(S, \alpha)$.
• We conclude that $(\ell^1(S, \alpha), \| \cdot \|_1)$ is a Banach Algebra. $\blacksquare$