Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\mathcal A$ be commutative unital Banach algebra denote by $M(A)$ the space of all non-zero multiplicative functionals on $\mathcal A$.

I want to show that $M(A)$ is closed in the weak-* topology.

Now, for a general, non-separable $\mathcal A$, one has to use nets, i.e. $M(\mathcal A)$ is closed if and only if it contains the limit points of all its nets.

How can one technically deal with those nets, i.e. how can one show that the multiplicative property is still valid in the limit?

share|cite|improve this question
up vote 1 down vote accepted

Using nets does not generate any problem. Suppose that $f_j\to f$ pointwise in $M(A)$. Given $x,y\in A$, $$ f(xy)=\lim f_j(xy)=\lim f_j(x) \,f_j(y)=f(x)f(y). $$ The non-obvious equality is the last one. The only difference with the case of sequences is that a convergent net need not be bounded; but it is eventually bounded, and so the proof that the limit of the product is the product of the limit is the same as in the case of sequences.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.