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)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)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)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:
- 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:
- 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:
- 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:
- And also:
- 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:
- 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. $\blacksquare$