Marla Payton
2022-01-07
Answered

Show that ${L}^{1}\left(R\right)$ a Banach algebra Commutative

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Juan Spiller

Answered 2022-01-08
Author has **38** answers

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:$A\times A\to A$ as:

$f,g\in A:f\times g={\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}f\left(\tau \right)g(t-\tau )d\tau$

{This is also called the convolution operation}

Noe observe that:

$g\times f={\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}g\left(\tau \right)f(t-\tau )d\tau$

Now substituting:$t-\tau \to x$ we get:

$g\times f={\int}_{t+\mathrm{\infty}}^{t-\mathrm{\infty}}g(t-x)f\left(x\right)d(t-x)$

$\Rightarrow g\times f={\int}_{\mathrm{\infty}}^{-\mathrm{\infty}}g(t-x)f\left(x\right)(-dx)$

$\Rightarrow g\times f=-{\int}_{\mathrm{\infty}}^{-\mathrm{\infty}}f\left(x\right)g(t-x)dx$

$\Rightarrow g\times f={\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}f\left(x\right)g(t-x)dx$

Again substituting:$x\to \tau$ we obtain:

$g\times f={\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}f\left(\tau \right)g(t-\tau )d\tau$

$\Rightarrow g\times f=f\times g$

$\therefore$ Multiplication in A is Commutative

$\Rightarrow {L}^{1}\left(\mathbb{R}\right)$ is Commutative

Let:

Let's define the multiplication:

{This is also called the convolution operation}

Noe observe that:

Now substituting:

Again substituting:

censoratojk

Answered 2022-01-09
Author has **46** answers

Good explanation, thanks a lot!

asked 2022-04-25

How to solve a cyclic quintic in radicals?

Galois theory tells us that

$\frac{{z}^{11}-1}{z-1}={z}^{10}+{z}^{9}+{z}^{8}+{z}^{7}+{z}^{6}+{z}^{5}+{z}^{4}+{z}^{3}+{z}^{2}+z+1$ can be solved in radicals because its group is solvable. Actually performing the calculation is beyond me, though - here what I have got so far:

Let the roots be $\zeta}^{1},{\zeta}^{2},\dots ,{\zeta}^{10$, following Gauss we can split the problem into solving quintics and quadratics by looking at subgroups of the roots. Since 2 is a generator of the group [2,4,8,5,10,9,7,3,6,1] we can partition into the five subgroups of conjugate pairs [2,9],[4,7],[8,3],[5,6],[10,1].

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Once one has $A}_{0},\dots ,{A}_{4$ one easily gets $x}_{1},\dots ,{x}_{5$. It's easy to find $A}_{0$. The point is that $\tau$ takes $A}_{j$ to $\zeta}^{-j}{A}_{j$ and so takes $A}_{j}^{5$ to $A}_{j}^{5$. Thus $A}_{j}^{5$ can be written down in terms of rationals (if that's your starting field) and powers of $\zeta$. Alas, here is where the algebra becomes difficult. The coefficients of powers of $\zeta$ in $A}_{1}^{5$ are complicated. They can be expressed in terms of a root of a "resolvent polynomial" which will have a rational root as the equation is cyclic. Once one has done this, you have $A}_{1$ as a fifth root of a certain explicit complex number. Then one can express the other $A}_{j$ in terms of $A}_{1$. The details are not very pleasant, but Dummit skilfully navigates through the complexities, and produces formulas which are not as complicated as they might be. Alas, I don't have the time nor the energy to provide more details.

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