The Group Algebra of R, L1(R)

# The Group Algebra of R, L1(R)

Let $L^1(\mathbb{R})$ be the set of all Lebesgue integrable functions on $\mathbb{R}$, that is:

(1)
\begin{align} \quad L^1(\mathbb{R}) = \left \{ f : \mathbb{R} \to \mathbb{C} \biggr | \int_{\mathbb{R}} |f(t)| \: dt < \infty \right \} \end{align}

We have already noted that $L^1(\mathbb{R})$ is a normed linear space with the operations of function addition and scalar multiplication. We have defined the norm on $L^1(\mathbb{R})$ for each $f \in L^1(\mathbb{R})$ to be:

(2)
\begin{align} \quad \| f \|_1 = \int_{\mathbb{R}} |f(t)| \: dt \end{align}

Moreover by The Riesz-Fischer Theorem, $(L^1(\mathbb{R}), \| \cdot \|)$ is a Banach space. We aim to define a product of $L^1(\mathbb{R})$ to make it into a Banach algebra.

For each $f, g \in L^1(\mathbb{R})$, define the product of $f$ and $g$ to be the convolution of $f$ and $g$. That is, $(f * g)$ is defined for each $x \in \mathbb{R}$ by:

(3)
\begin{align} \quad (f * g)(x) = \int_{\mathbb{R}} f(t)g(x - t) \: dt \end{align}
 Definition: The Group Algebra on $\mathbb{R}$ is the Banach algebra $L^1(\mathbb{R})$ with the operations of addition, scalar multiplication, and convolution product, and with the algebra norm $\| \cdot \|_1 : L^1(\mathbb{R}) \to [0, \infty)$ defined for each $f \in L^1(\mathbb{R})$ by $\displaystyle{\| f \|_1 = \int_{\mathbb{R}} |f(x)| \: dx}$.

The following proposition tells us that $L^1(\mathbb{R})$ with the notion of convolution product defined above is indeed a Banach algebra.

 Proposition 1: The space $(L^1(\mathbb{R}), \| \cdot \|)$ is a Banach algebra.
• Proof: We already know that $L^1(\mathbb{R})$ is a Banach space. All that remains to show is that (A) $L^1(\mathbb{R})$ with the convolution product is an algebra, and that (B) the norm $\| \cdot \|_1$ on $L^1(\mathbb{R})$ is an algebra norm.
• (A) Showing that $L^1(\mathbb{R})$ is an algebra:
• 1. Showing that $(f * g)*h = f * (g * h)$: Let $f, g, h \in L^1(\mathbb{R})$. Then for every $x \in \mathbb{R}$ we have that:
(4)
\begin{align} \quad [(f * g) * h](x) &= \int_{\mathbb{R}} (f * g)(t)h(x - t) \: dt \\ &= \int_{\mathbb{R}} \left [ \int_{\mathbb{R}} f(s)g(t - s) \: ds \right ] h(x - t) \: dt \\ &= \int_{\mathbb{R}} \int_{\mathbb{R}} f(s)g(t - s) h(x - t) \: ds \: dt \end{align}
• Let $A = \int_{\mathbb{R}} \int_{\mathbb{R}} |f(x)| \: dx$, $B = \int_{\mathbb{R}} |g(x)| \: dx$, and $C = \int_{\mathbb{R}} |h(x)| \: dx$.
• Then $A, B, C < \infty$ and so $\int_{\mathbb{R}} \int_{\mathbb{R}} |f(s)g(t - s) h(x - t)| \: ds \: dt \leq \int_{\mathbb{R}} |h(x - t)| \int_{\mathbb{R}} |f(s)||g(t - s)| \: ds \: dt \leq \int_{\mathbb{R}} |h(x - t)| AB \: dt \leq ABC < \infty$. So Fubini's theorem tells us that we can switch the order of integration, that is:
(5)
\begin{align} \quad [(f * g) * h](x) &= \int_{\mathbb{R}} \int_{\mathbb{R}} f(s)g(t - s)h(x - t) \: dt \: ds \\ &= \int_{\mathbb{R}} f(s) \left [ \int_{\mathbb{R}} g(t - s)h(x - t) \right ] \: dt \: ds \\ &= \int_{\mathbb{R}} f(s) \left [ \int_{\mathbb{R}} g((t+s) - s)h(x - (t + s)) \right ] \: dt \: ds \\ &= \int_{\mathbb{R}} f(s) \left [ \int_{\mathbb{R}} g(t)h((x - s) - t) \right ] \: dt \: ds \\ &= \int_{\mathbb{R}} f(s) \left [ (g * h)(x - s) \right ] \: ds \\ &= [f * (g * h)](x) \end{align}
• Therefore $(f * g) * h = f * (g * h)$ (since they both agree on all of $\mathbb{R}$.
• 2. Showing that $f * (g + h) = f * g + f * h$: Let $f, g, h \in L^1(\mathbb{R})$. Then for every $x \in \mathbb{R}$ we have by the linearity of the integral that:
(6)
\begin{align} \quad \quad [f * (g + h)](x) = \int_{\mathbb{R}} f(t)(g + h)(x - t) \: dt = \int_{\mathbb{R}} [f(t)g(x - t) + f(t)h(x - t)] \: dt = \int_{\mathbb{R}} f(t)g(x - t) \: dt + \int_{\mathbb{R}} f(t)h(x - t) \: dt = (f * g)(x) + (f * h)(x) \end{align}
• Therefore $f * (g + h) = f * g + f * h$.
• 3. Showing that $(\alpha f)* g = \alpha (f * g) = f * (\alpha g)$: Let $f, g \in L^1(\mathbb{R})$ and let $\alpha \in \mathbf{F}$. Then for every $x \in \mathbb{R}$ we have that:
(7)
\begin{align} \quad [(\alpha f) * g](x) \int_{\mathbb{R}} \alpha f(t) g(x - t) \: dt = \alpha \int_{\mathbb{R}} f(t)g(x - t) \: dt = [\alpha (f * g)](x) \end{align}
• And also:
(8)
\begin{align} \quad [\alpha (f * g)](x) = \alpha \int_{\mathbb{R}} f(t) g(x - t) \: dt = \int_{\mathbb{R}} f(t) \alpha g(x - t) \: dt = [f * (\alpha g)](x) \end{align}
• Therefore $(\alpha f) * g = \alpha (f * g) = f * (\alpha g)$. So $L^1(\mathbb{R})$ is an algebra.
• (B) Showing that $\| \cdot \|_1$ is an algebra norm on $L^1(\mathbb{R})$:
• We already know it is a norm on the linear space $L^1(\mathbb{R})$, so all that remains to show is that $\| f * g \| \leq \| f \| \| g \|$. Let $f, g \in L^1(\mathbb{R})$. Then:
(9)
\begin{align} \quad \| f * g \|_1 = \int_{\mathbb{R}} |(f * g)(t)| \: dt = \int_{\mathbb{R}} \left | \int_{\mathbb{R}} f(s)g(t - s) \: ds \right | \: dt \leq \int_{\mathbb{R}} \int_{\mathbb{R}} |f(s)||g(t - s)| \: ds \: dt &=\int_{\mathbb{R}} \int_{\mathbb{R}} |f(s)||g(t - s)| \: dt \: ds \\ &= \int_{\mathbb{R}} |f(s)| \left [ \int_{\mathbb{R}} |g(t - s)|\: dt \right ] \: ds \\ &= \int_{\mathbb{R}} |f(s)| \| g \|_1 \: ds \\ &= \| f \|_1 \| g \|_1 \end{align}
• Thus, $\| \cdot \|_1$ is an algebra norm on $L^1(\mathbb{R})$.
• We conclude that $(L^1(\mathbb{R}), \| \cdot \|_1)$ is a Banach algebra. $\blacksquare$