Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $R([0,1])$ be the unital commutative $C^*$-algebra of complex valued Riemann integrable functions on $[0,1]$ with pointwise operations and the supremum norm.

In the 1980 paper The Gelfand space of the Banach algebra of Riemann integrable functions by Jörg Blatter (MR0602719) the author states, that no satisfactory representation is known for the space of maximal ideals of $R([0,1])$.

For the smaller space of regulated functions on $[0,1]$ the maximal ideal space consists of the functionals $\beta_x,\delta_x, \gamma_x$ given by $\beta_x(f)=f(x)$, $\delta_x(f)=\lim_{y\to x^+}f(y)$ and $\gamma_x(f)=\lim_{y\to x^-}f(y)$ where $x\in [0,1]$, see e.g. The character space of the algebra of regulated functions by S. K. Berberian (MR0487932).

I wonder whether some progress has been made concerning the maximal ideal space of $R([0,1])$ since Blatter's paper. In fact I would be very thankful if someone could provide me with an example of a character on $R([0,1])$ which is not of the form $\beta_x$, $\delta_x$, $\gamma_x$ as above or some intuition why the determination of the maximal ideal space of $R([0,1])$ is so much harder than the corresponding problem for the space of regulated functions.

share|cite|improve this question
lvb, just a remark: your last sentence seems to imply that the characters $\beta_x$, $\delta_x$, $\gamma_x$ extend to $R([0,1])$ which I don't think they do. By the way, what's the topology on $[0,1]\sqcup [0,1]\sqcup [0,1]$ regarded as the promitive ideal space of the regulated functions? – Rasmus Jun 26 '11 at 18:51
In fact every maximal ideal of R is an extension of one of $\beta_x$, $\delta_x$, $\gamma_x$. It's just that they do not extend in a unique way. You can construct the maximal ideals via ultrafilters. – George Lowther Jun 26 '11 at 22:36
Thank you. You are right. I should have seen that. Do you know a reference where this construction is carried out? – lvb Jun 27 '11 at 7:50
lvb, if you want to make sure that George Lowther notices your comment, use @George Lowther. – Rasmus Jun 27 '11 at 20:35
May be his can help you. Consider algebra of sets $\Sigma=\{E\subset[0,1]: \lambda(\operatorname{int}(E))=\lambda(\operatorname{cl}(E))\}$. Then $L_\infty([0,1],\Sigma,\lambda|_\Sigma)$ is isomorphic to the space Riemann integrable functions with $\sup$-norm. (see exercise 363Yi in Measure theory D. H. Fremlin Vol 3) – Norbert Dec 9 '13 at 16:48

Your Answer


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

Browse other questions tagged or ask your own question.