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

Marla Payton 2022-01-07 Answered
Show that L1(R) a Banach algebra Commutative
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

Juan Spiller
Answered 2022-01-08 Author has 38 answers
We will define the multiplicative operation of Banach algebra in case of L1(R) and prove its commutativity.
Let: AL1(R) be a Banch algebra
Let's define the multiplication: A×AA as:
f,gA:f×g=f(τ)g(tτ)dτ
{This is also called the convolution operation}
Noe observe that:
g×f=g(τ)f(tτ)dτ
Now substituting: tτx we get:
g×f=t+tg(tx)f(x)d(tx)
g×f=g(tx)f(x)(dx)
g×f=f(x)g(tx)dx
g×f=f(x)g(tx)dx
Again substituting: xτ we obtain:
g×f=f(τ)g(tτ)dτ
g×f=f×g
Multiplication in A is Commutative
L1(R) is Commutative
Not exactly what you’re looking for?
Ask My Question
censoratojk
Answered 2022-01-09 Author has 46 answers

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

You might be interested in

asked 2022-04-25

How to solve a cyclic quintic in radicals?
Galois theory tells us that
z111z1=z10+z9+z8+z7+z6+z5+z4+z3+z2+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 ζ1,ζ2,,ζ10, 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].
A0=x1+x2+x3+x4+x5A1=x1+ζx2+ζ2x3+ζ3x4+ζ4x5A2=x1+ζ2x2+ζ4x3+ζx4+ζ3x5A3=x1+ζ3x2+ζx3+ζ4x4+ζ2x5A4=x1+ζ4x2+ζ3x3+ζ2x4+ζx5
Once one has A0,,A4 one easily gets x1,,x5. It's easy to find A0. The point is that τ takes Aj to ζjAj and so takes Aj5 to Aj5. Thus Aj5 can be written down in terms of rationals (if that's your starting field) and powers of ζ. Alas, here is where the algebra becomes difficult. The coefficients of powers of ζ in A15 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 A1 as a fifth root of a certain explicit complex number. Then one can express the other Aj in terms of A1. 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.

asked 2021-09-28

Find the volume of the parallelepiped with adjacent edges PQ, PR, and PS. P(3,0,1),Q(1,2,5),R(5,1,1),S(0,4,2)

asked 2020-12-30

In the froup Z12, find |a|,|b|, and |a+b|
a=3,b=8

asked 2021-09-05
The average height of a 2 year old boy is 38 inches. an 8 year old averages 56 inches. Use this information to write a linear equation that models the height (in inches), y, in terms of the age (in years), x. Use the linear equation to predict the average height of a 5 year-old boy.
asked 2022-06-02
I just started my first upper level undergrad course, and as we were being taught vector spaces over fields we quickly went over fields. What confused me that the the set {0, 1, 2} was a field. However, to my understanding, that set doesn't satisfy the axiom, "For every element a in F, there is an element b such that a+b=0", among others. Can someone help clarify where my understanding is off.
Also one my friend states "for every prime power p^n, there exists a field with p n elements" and then doesn't expand on it. If someone could give an example or proof i would be very grateful.
asked 2021-01-15
In the froup Z12, find |a|, |b|, and |a+b|
a=6, b=2
asked 2020-11-27
Let R be a commutative ring. If I and P are idelas of R with P prime such that I!P, prove that the ideal P:I=P

New questions

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question