Ideals in Banach algebras
I came across the following statement while reading Murphy’s book (Theorem 1.3.1): “If 
is a proper ideal in a Banach algebra, then 
is also proper.”
Quick recap: a Banach algebra is a Banach space with a multiplication operation, such that the norm is submultiplicative (
). A prototypical (non-abelian) finite dimensional Banach space is that of all 
matrices with the spectral norm, and a prototypical (abelian) infinite dimensional Banach space is that of all continuous functions on 
with the sup norm. An ideal is a vector subspace that ‘absorbs’ under multiplication: if 
is in the space and 
is in the ideal, both 
and 
are in the ideal. A proper ideal is one strictly contained in the ambient space.
This statement caught my eye because it gives a way in which ideals topologically differ from vector subspaces. It is easy to come up with examples of proper subspaces whose closures are the entire space; e.g. take the set of differentiable functions in the above infinite dimensional Banach space: every continuous function on is the uniform limit of differentiable functions, so the closure of this subspace is the space itself. It’s harder for me to come up with ideals nontrivially exemplifying the above statement.
The question is: exactly how does one use the multiplicative closure property of ideals to prove the above?
Possibly relevant posts:
- The Gelfand representation theorem (7/4/2008)
- the insanity that is CDS202 (1/26/2008)
- Kvetches about Janich’s book (7/2/2007)
In the space of continuous functions, a nontrivial example of an ideal is $\{ f | f(x) = 0 \}$, for a fixed $x$ in the domain.
The result follows from the standard result that any proper ideal is contained in a maximal ideal. For this, apply Zorn’s Lemma to the set of proper ideals containing $I$, ordered by inclusion. Then, show that any maximal ideal is closed.
Comment by George — 6/30/2008 @ 2:17 pm
Funny, I was thinking of using this result to show that maximal ideals are closed. I didn’t think of going the other way; I’ll give it a try.
I consider that example of an ideal trivial, because clearly its closure is not the entire set of continuous functions. Ideally
we could find an example where that isn’t so obvious.
Comment by Alex — 6/30/2008 @ 4:44 pm