# Show that L^1(R) a Banach algebra Commutative

Show that ${L}^{1}\left(R\right)$ a Banach algebra Commutative
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Juan Spiller
We will define the multiplicative operation of Banach algebra in case of ${L}^{1}\left(R\right)$ and prove its commutativity.
Let: $A\in {L}^{1}\left(\mathbb{R}\right)$ be a Banch algebra
Let's define the multiplication: $A×A\to A$ as:
$f,g\in A:f×g={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f\left(\tau \right)g\left(t-\tau \right)d\tau$
{This is also called the convolution operation}
Noe observe that:
$g×f={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}g\left(\tau \right)f\left(t-\tau \right)d\tau$
Now substituting: $t-\tau \to x$ we get:
$g×f={\int }_{t+\mathrm{\infty }}^{t-\mathrm{\infty }}g\left(t-x\right)f\left(x\right)d\left(t-x\right)$
$⇒g×f={\int }_{\mathrm{\infty }}^{-\mathrm{\infty }}g\left(t-x\right)f\left(x\right)\left(-dx\right)$
$⇒g×f=-{\int }_{\mathrm{\infty }}^{-\mathrm{\infty }}f\left(x\right)g\left(t-x\right)dx$
$⇒g×f={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f\left(x\right)g\left(t-x\right)dx$
Again substituting: $x\to \tau$ we obtain:
$g×f={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f\left(\tau \right)g\left(t-\tau \right)d\tau$
$⇒g×f=f×g$
$\therefore$ Multiplication in A is Commutative
$⇒{L}^{1}\left(\mathbb{R}\right)$ is Commutative
censoratojk