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Ideals in Banach algebras

I came across the following statement while reading Murphy’s book (Theorem 1.3.1): “If I is a proper ideal in a Banach algebra, then \overline{I} is also proper.”

Quick recap: a Banach algebra is a Banach space with a multiplication operation, such that the norm is submultiplicative (\|ab\| \leq \|a\|\|b\|). A prototypical (non-abelian) finite dimensional Banach space is that of all n\times n matrices with the spectral norm, and a prototypical (abelian) infinite dimensional Banach space is that of all continuous functions on \R with the sup norm. An ideal is a vector subspace that ‘absorbs’ under multiplication: if a is in the space and b is in the ideal, both ab and ba 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 \R 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?

This entry was posted on Monday, June 30th, 2008 at 12:38 am and is filed under Mathematics. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

2 Responses to “Ideals in Banach algebras”

  1. George Says:
    June 30th, 2008 at 2:17 pm

    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.

  2. Alex Says:
    June 30th, 2008 at 4:44 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.

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