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Homework Statement
The problem is to prove the following:
If [itex]\mathcal A[/itex] is a Banach algebra, and [itex]\mathcal I[/itex] is a closed ideal in [itex]\mathcal A[/itex], then [itex]\mathcal A/\mathcal I[/itex] is a Banach algebra.
This is problem 3.1.3 (4)(b) in "Functional analysis: spectral theory", by V.S. Sunder. Link.
Homework Equations
A normed algebra is a vector space with a bilinear (i.e. distributive) multiplication operation, and a norm that satisfies the usual conditions on a norm and also [itex]\|xy\|\leq \|x\|\|y\|[/itex]. A Banach algebra is a normed algebra in which every Cauchy sequence is convergent.
An ideal I of a normed algebra A is a subspace (in the vector space sense) such that xi=ix is in I for all x in A and all i in I. A closed ideal is an ideal that's also a closed set. In particular, it means that if a sequence in I is convergent, its limit is in I. I'm familiar with a theorem that says that for any x in A, there's a unique i in I such that d(x,I)=d(x,i). This i is such that x-i is orthogonal to I. I suspect that maybe I should use this theorem, because it only holds when I is closed, and nothing I've come up with uses that assumption.
The members of the quotient algebra A/I are subsets of A, which we write as x+I={x+i|i is in I}. We define addition, multiplication by a scalar, multiplication, and the norm by
(x+I)+(y+I)=(x+y)+I
a(x+I)=ax+I
(x+I)(y+I)=xy+I
[itex]\|x+I\|=d(x,I)=\inf\{d(x,i)|i\in I\}[/itex]
In part (a), I proved that A/I is a normed algebra.
The Attempt at a Solution
The only strategy I've come up with is the following: Consider a Cauchy sequence [itex]x_n+I[/itex] in A/I and show that [itex]x_n[/itex] must be a Cauchy sequence too. Since A is complete, [itex]x_n\rightarrow x[/itex] for some x. The next step would be to show that [itex]x_n+\mathcal I\rightarrow x+I[/itex]
Unfortunately,
[tex]\|x_n-x_m\|=\|x_n-x_m-0\|\geq d(x_n-x_m,I)=\|(x_n-x_m)+I\|=\|(x_n+I)-(x_m+I)\|[/tex]
so I don't seem to be able to use the assumption that I can make that last thing arbitrarily small. However, if I can show that [itex]x_n\rightarrow x[/itex] for some x, I can handle the rest:
[tex]\|(x_n+I)-(x+I)\|=\|(x_n-x)+I\|=d(x_n-x,I)\leq d(x_n-x,0)=\|x_n-x\|<\varepsilon[/tex]