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

I am currently studying for a first course in operator theory, and I was wondering about the following that occurred to me whilst doing my reading:

It seems to me that, by talking about maps in general we have a process that allows us to change from very "basic" mathematical objects, such as a point, to more and more complex objects.

For example:

Suppose $n$ is a natural number. We can think of the set $\mathbb{R}$ as the vector space of all the possible (continuous) functions $n \mapsto x_n \in \mathbb{R}$. In the next step we have the vector space $C(\mathbb{R})$ of all possible continuous functions $\mathbb{R} \ni x \mapsto f(x)$. And then there is the vector space $\mathcal{C}(C(\mathbb{R}))$ of all possible continuous operators $C(\mathbb{R}) \ni f \mapsto Af$.

So in each step we take the previous object and think about all possible continuous functions on it. The thing I was wondering about is - what are the objects that we use in the next step ? That is, what objects act on the space $\mathcal{C}(C(\mathbb{R}))$ ?

\begin{equation} n \to \mathbb{R} \to C(\mathbb{R}) \to \mathcal{C}(C(\mathbb{R})) \to \, ? \end{equation}

I realize this question might be very stupid, I hope it makes sense at least .. appologies in case not, I am just starting to learn about operators. Thanks for your help !

EDIT: the notation $\mathcal{C}(C(\mathbb{R}))$ might not be the one that is used commonly, so far I have only seen linear operators and their common notation, in case somebody could hint towards the proper notation that would be great ! I shall also have a look at books to find out, but until that I appologize if the notation looks extravagant

share|cite|improve this question
This does not look convincing to me at all. The passage from $C(\mathbb{R})$ to $\mathcal{B}(C(\mathbb{R}))$ is not that much analogous than the one from $\mathbb{R}$ to $C(\mathbb{R})$, because one case is concerned with linearity and the other one isn't. And I don't understand what you mean by $n\to \mathbb{R}$ or $n\mapsto \mathbb{R}$. – Florian Apr 12 '12 at 11:49
@Florian: For any singleton, e.g. $\{n\}$, there is an obvious correspondence between $\Bbb R$ and the set of maps from the singleton to $\Bbb R$; that’s what harlekin is talking about for his first step. – Brian M. Scott Apr 12 '12 at 12:03
@Florian: thanks for your comment! The map you wondered about looked strange indeed, what Brian M. Scott suggests is what I wanted to write. I edited the post accordingly. With regards to the second step - you are right the process looses generality in that it restricts to linear maps (operators) on the preceeding space. I could probably reduce the post to the question "what are the maps that act on bounded linear operators ?" .. – harlekin Apr 12 '12 at 12:20
Well, once you have a vector space, you can always take the dual. But this may not be what you are looking for. – M Turgeon Apr 12 '12 at 12:43
@M Turgeon: that's a very good point - though the dual would then "collapse" everything back to $\mathbb{R}$ (roughly speaking). I was rather looking for a space of maps $\mathcal{B}(C(\mathbb{R})) \to \mathcal{B}(C(\mathbb{R})) $. But in any case, what you say partly answers the question, thanks for pointing this out! – harlekin Apr 12 '12 at 13:15

When you say "continuous functions on $C(\mathbb{R})$", you don't say what topology you are taking. The most natural choice is probably uniform convergence (because it makes it complete); but then you get a topological space which is not particularly nice, as it is not locally compact. And I don't think much can be said about the set of continuous functions over a non-LC space.

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.