Continuous functions that vanish at infinity

[Fredrik]

I'm trying to understand the set $C_0(X)$, defined here as the set of continuous functions $f:X\rightarrow\mathbb C$ such that for each $\varepsilon>0$, $\{x\in X|\,|f(x)|\geq\varepsilon\}$ is compact. (If you're having trouble viewing page 65, try replacing the .se in the URL with your country domain). The book also defines $C_b(X)$ as the set of all bounded continuous functions from X into $\mathbb C$. The book makes additional assumptions about X, but clearly the definitions work even when X is just an arbitrary topological space.

It's always hard work to fill in the details that Conway leaves out. I think I have verified that $C_0(X)\subset C_b(X)$, and that if X is Hausdorff, $C_0(X)$ is closed under linear combinations. I still need to show that $C_0(X)$ is a closed set, and I would like to understand what any of this have to do with local compactness. [strike]Is $C_0(X)$ a subalgebra or just a vector subspace?[/strike]. I would appreciate any help with any of these details.

I have already LaTeXed the proof of the first part ($C_0(X)$ is a linear subspace, if X is Hausdorff) for my notes. I'll post it here if someone requests it. I have also LaTeXed the proof that $C_b(X)$ is a Banach algebra with identity, and wouldn't mind posting that too.

Edit: I think I proved that $C_0(X)$ is closed under multiplication (if X is Hausdorff), so we can scratch that item off the list.

 

[micromass]

The space $$C_0(X)$$ is a space that I find immensly interesting. So after we dealt with your problem, I'll say something more about it and it's relation to physics. But first I have some questions for you:

I'm trying to understand the set $C_0(X)$, defined here as the set of continuous functions $f:X\rightarrow\mathbb C$ such that for each $\varepsilon>0$, $\{x\in X|\,|f(x)|\geq\varepsilon\}$ is compact. (If you're having trouble viewing page 65, try replacing the .se in the URL with your country domain). The book also defines $C_b(X)$ as the set of all bounded continuous functions from X into $\mathbb C$. The book makes additional assumptions about X, but clearly the definitions work even when X is just an arbitrary topological space.

Well, of course the construction works for arbitrary topological spaces. However, $$C_0(X)$$ has the following nice characterization:

f:X-->R belongs to $$C_0(X)$$, if and only if we can extend f to $$f:\alpha X\rightarrow R$$ (where $$\alpha X$$ is the Alexandroff (one-point) compactification) and where $$f(\infty)=0$$. Thus we can extend the space X by one-point (= infinity) and f becomes 0 in this point.
Of course, the Alexandroff compactification is only interesting when X is a locally compact Hausdorff space. So this is why we often want X to be locally compact Hausdorff.

It's always hard work to fill in the details that Conway leaves out. I think I have verified that $C_0(X)\subset C_b(X)$, and that if X is Hausdorff, $C_0(X)$ is closed under linear combinations.

Question 1: Conway doesn't assume Hausdorff in proposition 1.7, does he? Or does he assume Hausdorff as one of the defining properties of locally compact?
Question 2: Can you post this proof? I actually find it weird that you need Hausdorff for that...

I still need to show that $C_0(X)$ is a closed set, and I would like to understand what any of this have to do with local compactness. [strike]Is $C_0(X)$ a subalgebra or just a vector subspace?[/strike]. I would appreciate any help with any of these details.

But the proof is given? Or do you want to know why you need local compactness for that? Well, I don't really think you need local compactness for it. I'll study the proof some more to see if I'm missing something, but I really don't think that you need local compactness for this part.

I have already LaTeXed the proof of the first part ($C_0(X)$ is a linear subspace, if X is Hausdorff) for my notes. I'll post it here if someone requests it. I have also LaTeXed the proof that $C_b(X)$ is a Banach algebra with identity, and wouldn't mind posting that too.

Edit: I think I proved that $C_0(X)$ is closed under multiplication (if X is Hausdorff), so we can scratch that item off the list.

Yes, if you can post the proof that $$C_0(X)$$ is a subalgebra of $$C_b(X)$$, that would be most helpful!

 

[micromass]

Ah yes, I've tried the proof myself and I think that I see why Hausdorff is necessairy. You basically use it because you need a compact set to be closed, right?

 

[Fredrik]

f:X-->R belongs to $$C_0(X)$$, if and only if we can extend f to $$f:\alpha X\rightarrow R$$ (where $$\alpha X$$ is the Alexandroff (one-point) compactification) and where $$f(\infty)=0$$. Thus we can extend the space X by one-point (= infinity) and f becomes 0 in this point.
Of course, the Alexandroff compactification is only interesting when X is a locally compact Hausdorff space. So this is why we often want X to be locally compact Hausdorff.

I haven't heard of the Alexandroff compactification. I will need to look that up.

Question 1: Conway doesn't assume Hausdorff in proposition 1.7, does he? Or does he assume Hausdorff as one of the defining properties of locally compact?
Question 2: Can you post this proof? I actually find it weird that you need Hausdorff for that...

Yes, if you can post the proof that $$C_0(X)$$ is a subalgebra of $$C_b(X)$$, that would be most helpful!

I will post the proofs about $C_0(X)$ that I've done so far in a separate post. (It will take me a few minutes to change the formatting and make a few minor edits).

Conway assumes that all his topological spaces are Hausdorff. There's a reminder of that in example 1.6, but he doesn't mention it in proposition 1.7.

Ah yes, I've tried the proof myself and I think that I see why Hausdorff is necessairy. You basically use it because you need a compact set to be closed, right?

That's close. I need closed subsets of compact subsets of X to be compact.

Edit: Maybe I don't need it to be Hausdorff for that. D'oh, now I have to think again. I may have confused the result I actually need with the one you guessed. OK, now I've thought about it. We don't need the space to be Hausdorff for that. Closed subsets of compact sets are always compact.

But the proof is given?

That was embarrassing. I read through that page rather quickly, and then didn't look at it again while I was doing my proofs, except to get a hint about a detail in the proof that $C_b(X)$ is a Banach algebra. When it was time to prove that $C_0(X)$ is closed, I somehow thought that the proof wasn't included on that page, and didn't even look at the page again. I will read through Conway's proof a few hours from now and then post an update about the status of my understanding.

Or do you want to know why you need local compactness for that? Well, I don't really think you need local compactness for it. I'll study the proof some more to see if I'm missing something, but I really don't think that you need local compactness for this part.

I'm wondering why he mentions it at all. Since $C_0(X)$ is defined in the statement of that theorem, he doesn't even define $C_0(X)$ for an arbitrary topological space. The other book I'm reading, "Functional analysis: Spectral theory" by V.S. Sunder, is adding to the confusion, by also defining $C_0(X)$ only for X that are locally compact and Hausdorff. This is on page 5. Item (4) on page 62 makes me even more confused, mainly by requiring X to be locally compact again. He also defines $C_c(X)$ as the algebra of continuous functions that vanish outside a compact set, and says that $C_0(X)$ is the completion of $C_c(X)$. I don't see how $C_c(X)$ can be different from $C_0(X)$. His comment at the end of item (4) also gives me a reason to think that he's sloppy with his assumptions, because he says that if X is compact and Hausdorff, C(X) is a Banach algebra. I have verified that you don't need the Hausdorff assumption here (with some help from others in this forum, in particular Jarle). You just need the compactness assumption to ensure that all continuous functions are bounded.

 

[mathwonk]

forgive me its a little late and it was a long time ago, but i liked this topic in my analysis course. there is a relation between compactifications of spaces with a few assumptions (completely regular T1?), and subalgebras of their algebra of continuous functions. basically the map one way takes a compactification to the subalgebra of functions that extend to the compactification. then in the other direction one starts from a subalgebra and tries to define a compactification to which all elements of the subalgebra do extend. the subalgebra of functions with limits at infinity correspond to the one point compactification. then in the compactification, there is a 1-1 correspondence between maximal ideals and points. so the extra point at infinity in the one point compactification corresponds to the ideal C0 of functions vanishing at infinity.

 

[Fredrik]

Here's the proof that $C_0(X)$ is a subalgebra of $C_b(X)$:

Suppose that $f\in C_0(X)$, and let $\epsilon>0$ be arbitrary. First we show that f is bounded. Define $S=\{x\in X|\, |f(x)|\geq\varepsilon\}$. By assumption, f is continuous and S is compact. Continuous functions take compact sets to compact sets, so $f(S)\subset\mathbb C$ is compact. Compact subsets of metric spaces are bounded, and every bounded subset of $\mathbb C$ is a subset of some open ball around 0. So there exists an $r\in\mathbb R$ such that $f(S)\subset B(0,r)$. This means that for all $x\in S$, $|f(x)|\leq r$. By definition of S, for all $x\in S^c$, $|f(x)|<\varepsilon$. So for all $x\in X$, $|f(x)|\leq\max\{r,\varepsilon\}$. This implies $C_0(X)\subset C_b(X)$.

We need to show that $C_0(X)$ is closed under linear combinations. Let $f,g\in C_0(X)$ and $a,b\in\mathbb C$ be arbitrary. Let $h=af+bg$, and define the sets

$$\begin{align*} A &=\Big\{x\in X\Big|\,|f(x)|\geq\frac{\varepsilon}{2|a|}\Big\}\\ B &=\Big\{x\in X\Big|\,|g(x)|\geq\frac{\varepsilon}{2|b|}\Big\}\\ S &=\big\{x\in X\big|\,|h(x)|\geq\varepsilon\big\} \end{align*} For all x\in A^c\cap B^c, \begin{align*} &|h(x)|=|af(x)+bg(x)|\leq|a||f(x)|+|b||g(x)|<\varepsilon\\ & x \notin S \end{align*}$$

So we have $A^c\cap B^c\subset S^c$, which is equivalent to $S\subset (A^c\cap B^c)^c=A\cup B$. By assumption, A and B are compact, and that implies that $A\cup B$ is compact. We want to show that S is compact, and since every closed subset of a compact set is compact, it's sufficient to show that S is closed. Let $\langle x_i\rangle_{i\in I}$ be an arbitrary convergent net in S, and define $x=\lim_i x_i$. The continuity of f, g, the vector space operations, and the absolute value function implies that

$$\begin{align*} |h(x)|=|h(\lim_i x_i)|=|\lim_i h(x_i)|=\lim_i \underbrace{|h(x_i)|}_{\mathclap{\displaystyle\geq\varepsilon\text{ because }x_i\in S}}\geq\varepsilon \end{align*}$$

The last inequality holds because $\langle|h(x_i)|\rangle_{i\in I}$ is a convergent net in $\mathbb C$ with every term $\geq\varepsilon$. Its limit |h(x)| can't be $<\varepsilon$, since that would imply that for every $t\in\mathbb R$ such that $|h(x)|<t<\varepsilon$, there's an $i_t\in I$ such that $i\geq i_t\Rightarrow |h(x_i)|-|h(x)|<t-|h(x)|$, and this inequality implies $|h(x_i)|<t<\varepsilon$, which contradicts $|h(x_i)|\geq\varepsilon$.

We also need to show that $C_0(X)$ is closed under multiplication. Let $f,g\in C_0(X)$ be arbitrary, and define the sets

$$\begin{align*} A &=\big\{x\in X\big|\,|f(x)|\geq\sqrt\varepsilon\big\}\\ B &=\big\{x\in X\big|\,|g(x)|\geq\sqrt\varepsilon\big\}\\ S &=\big\{x\in X\big|\,|(fg)(x)|\geq\varepsilon\big\} \end{align*}$$

For all $x\in A^c\cap B^c$,

$$\begin{align*} &|(fg)(x)|=|f(x)g(x)|=|f(x)||g(x)|<\sqrt\varepsilon\sqrt\varepsilon=\varepsilon\\ & x \notin S \end{align*}$$

So we have $A^c\cap B^c\subset S^c$, which is equivalent to $S\subset (A^c\cap B^c)^c=A\cup B$. This implies that S is compact. (The reasons are the same as in the proof that $C_0(X)$ is closed under linear combinations, except that now we have to use the continuity of the product instead of the continuity of the vector space operations).

 

[Fredrik]

I've read the proof that $C_0(X)$ is closed in the topology of $C_b(X)$ now. It gave me a strong sense of deja vu. I think I studied the proof before I typed up the rest for my notes, and when I got to this part of it, I simply forgot that I had already understood the proof, and then convinced myself that it would be difficult. Weird.

I think we might still need the Hausdorff condition, in spite of what I said in my edit above. My proof used that f is continuous if and only if for each x, $x_i\rightarrow x$ implies $f(x_i)\rightarrow f(x)$. But if X isn't Hausdorff, the net $\langle x_i\rangle_{i\in I}$ may have several distinct limits, while the net $\langle f(x_i)\rangle$ can't (because it's a net in $\mathbb C$, which is Hausdorff. I need to think about that some more.

I still don't see a use for local compactness.

I looked up the Alexandroff compactification, so now I know what it is.

 

[micromass]

I'm wondering why he mentions it at all. Since $C_0(X)$ is defined in the statement of that theorem, he doesn't even define $C_0(X)$ for an arbitrary topological space. The other book I'm reading, "Functional analysis: Spectral theory" by V.S. Sunder, is adding to the confusion, by also defining $C_0(X)$ only for X that are locally compact and Hausdorff. This is on page 5. Item (4) on page 62 makes me even more confused, mainly by requiring X to be locally compact again. He also defines $C_c(X)$ as the algebra of continuous functions that vanish outside a compact set, and says that $C_0(X)$ is the completion of $C_c(X)$. I don't see how $C_c(X)$ can be different from $C_0(X)$. His comment at the end of item (4) also gives me a reason to think that he's sloppy with his assumptions, because he says that if X is compact and Hausdorff, C(X) is a Banach algebra. I have verified that you don't need the Hausdorff assumption here (with some help from others in this forum, in particular Jarle). You just need the compactness assumption to ensure that all continuous functions are bounded.

Hmm, this is confusing me to. I don't think local compactness and Hausdorff is necessary anywhere. I've looked into my old course text, and they assume local compactness to, but there it is for a different kind of reason.

Anyway, why people often assume local compactness for $$C_0(X)$$ is probably simply because all these spaces are only interesting for local compact spaces. So it's a reasonable assumption. However, in this case, I feel that it is an unnecessary assumption. I will check more texts right now, to see if they do the same thing.

I've checked your proof of $$C_0(X)$$ being a subalgebra. I don't think you need local compactness or Hausdorff there anywhere...


Oh yes, for the difference between $$C_c(X)$$ and $$C_0(X)$$. Take the function $$f:\mathbb{R}\rightarrow \mathbb{R}:x\rightarrow e^{-x^2}$$. This belongs to $$C_0(X)$$, since $$\lim_{x\rightarrow \pm\infty}{f(x)}=0$$.
But this is not an element of $$C_c(X)$$, since it doesn't vanish anywhere.
Of course, if X is compact, then the two notions coincide...

I think we might still need the Hausdorff condition, in spite of what I said in my edit above. My proof used that f is continuous if and only if for each x, $x_i\rightarrow x$ implies $f(x_i)\rightarrow f(x)$. But if X isn't Hausdorff, the net $\langle x_i\rangle_{i\in I}$ may have several distinct limits, while the net $\langle f(x_i)\rangle$ can't (because it's a net in $\mathbb C$, which is Hausdorff. I need to think about that some more.

Well, your use of nets is certainly correct, but you can do without them. For notational issues, put $$H(x)=|h(x)|$$. Then $$S=H^{-1}([\epsilon,+\infty[)$$ is closed as inverse image of a closed set.

 

[micromass]

Maybe this is a good time to explain the usefulness of $$C_0(X)$$, as it will also become apparent why the local compactness assumption is made.

When quantum mechanics was developed, the physicists needed a new mathematical framework to describe their observations. So mathematicians developed the theory of C*-algebra's, in which every element represent some kind of observation.

However, a C*-algebra is a highly abstract notion, and mathematicians wanted to give representation for them in terms of things that were more concrete. If the C*-algebra was commutative and has an idenity, then the answer to this question was given by Gelfand-Naimark. Every C*-algebra can be seen as a ring of continuous functions on a compact Hausdorff space. Thus the C*-algebra is isomorph to $$C(X)$$. If the C*-algebra didn't have a unit, then the C*-algebra can be seen as $$C_0(X)$$ with X locally compact and Hausdorff. Thus, to study commutative C*-algebra's, is exactly thesame as studying $$C_0(X)$$ for locally compact spaces. This is why we often ask for our spaces to be locally compact: it is easier, and if the goal is to study C*-algebra's then the spaces are locally compact. Thus nonlocally compact spaces aren't really interesting.

Furthermore, a lot of good properties of the space X, can be translated into properties of $$C_0(X)$$. For example, compactness of X is exactly thesame as asking that $$C_0(X)$$ has a unit. And compactifications correspond nicely to unitizations of $$C_0(X)$$. Connectedness of X correspond the the existence of idempotent elements in $$C_0(X)$$. (a full list can be seen at planetmath.org/encyclopedia/NoncommutativeTopology.html ).
This means that we can actually do topology without needing topological spaces. The only thing we need is a C*-algebra (which represents the ring of vanishing continuous functions). This C*-algebra is called compact if it has a unit, etc. This gives rise to the field of noncommutative topology.

 

[Fredrik]

Oh yes, for the difference between $$C_c(X)$$ and $$C_0(X)$$. Take the function $$f:\mathbb{R}\rightarrow \mathbb{R}:x\rightarrow e^{-x^2}$$. This belongs to $$C_0(X)$$, since $$\lim_{x\rightarrow \pm\infty}{f(x)}=0$$.
But this is not an element of $$C_c(X)$$, since it doesn't vanish anywhere.

Of course. I don't know how I could fail to see that.

Well, your use of nets is certainly correct, but you can do without them. For notational issues, put $$H(x)=|h(x)|$$. Then $$S=H^{-1}([\epsilon,+\infty[)$$ is closed as inverse image of a closed set.

Thanks, that's a significant improvement of what I did. I think this makes it clear that we don't need X to be Hausdorff.

Regarding what you said in #9, I'm somewhat familiar with the Gelfand-Naimark stuff. I'm reading Sunder precisely because I want to learn the details of that. His book gets to those things much faster than other texts, and at the same time is much easier to read than Conway. I checked the book today (a few pages after the last ones I studied carefully), and as I suspected, he seems to be proving that the "spectrum" (i.e. the set of non-trivial algebra homomorphisms into ℂ) of a commutative Banach algebra is locally compact (and also compact if and only if the Banach algebra has an identity). So he's probably assuming that "X" has those properties too right at the start, just because the space we're really interested in (the spectrum) does. But I wish he could have been more clear about these things. This book has a lot of flaws too.

I had never heard of non-commutative topology, but it sounds pretty cool.

 

[Landau]

Have all questions been answered?

A reason for the local compactness Hausdorff assumption, is that you want to apply Urysohn's lemma in proving that C_0(X) is the completion of C_c(X) w.r.t. the sup-norm. (So you need X to be at least normal, and in functional analysis you always want Hausdorffness because non-unique limits are stupid. In most cases like these, l.c. Hausdorff is the most reasonable assumption.)

 

[micromass]

Yes, that should be true. Except that a local compact space is not normal in general. Luckily, a local compact space is completely regular, and thus satisfies a weaker form of the Urysohn lemma...

 

[Landau]

Yeah, I meant "Urysohn's weaker lemma", applicable to a l.c. Hausdorff X: if V is open in X and K a compact subset of V, then there exists a $f\in C_c(X)$ with $1_K\leq f\leq 1_V$ (with 1_A of course the characteristic function of A).

[Frankly, I detest many of the words used for all these separation axioms and properties, because everyone seems to be using different definitions. I wish we could make universal conventions for them (and please ban the terms T_1,T_2,T_3,... T_{3+1/2}, etc.! How could anyone ever remember what they mean?)]

 

[micromass]

[Frankly, I detest many of the words used for all these separation axioms and properties, because everyone seems to be using different definitions. I wish we could make universal conventions for them (and please ban the terms T_1,T_2,T_3,... T_{3+1/2}, etc.! How could anyone ever remember what they mean?)]

Haha, couldn't agree more

 

[Fredrik]

You guys are awesome. I think the questions I asked have been answered. I'm still confused about a lot of things, but I think it's time to start studying the Gelfand-Naimark stuff again (I studied some of it in 2010 but put it aside in December), and maybe it will all clear up when I do. If it doesn't, I know where to find you.

I should probably start with the completeness proof that Landau mentioned in #11. Do you know where I can find a proof? Even if I can figure it out on my own, I'd rather save some time right now. Will I find that weaker Uhrysohn lemma in Munkres? (I studied Uhrysohn's lemma in the appendix of Sunder. It's been my favorite theorem since then, because I think the proof is absolutely beautiful).

 

[micromass]

You guys are awesome. I think the questions I asked have been answered. I'm still confused about a lot of things, but I think it's time to start studying the Gelfand-Naimark stuff again (I studied some of it in 2010 but put it aside in December), and maybe it will all clear up when I do. If it doesn't, I know where to find you.

I should probably start with the completeness proof that Landau mentioned in #11. Do you know where I can find a proof? Even if I can figure it out on my own, I'd rather save some time right now. Will I find that weaker Uhrysohn lemma in Munkres? (I studied Uhrysohn's lemma in the appendix of Sunder. It's been my favorite theorem since then, because I think the proof is absolutely beautiful).

The completeness thing that Landau mentioned is basically saying that $$C_c(X)$$ is dense in $$C_0(X)$$. This indeed uses local compactness. I can't believe I didn't see this before...

Anyway, I checked the book of Sunder, and it seems that there is enough material in there to proof this. In particular, check A.6.6, this is the Urysohn thing that Landau referred to. (or closely resembles it).

Let me prove it for you. Take a function $$f\in C_0(X)$$. We need to find a function $$g\in C_c(X)$$ such that $$\|f-g\|_\infty<\varepsilon$$. We know that $$\{x~\vert~|f(x)|\geq \varepsilon\}$$ is compact. Call this space K. Also, the space $$\{x~\vert~|f(x)|\leq \varepsilon/2\}$$ is closed. Thus there exists a Urysohn function u (see lemma 1.6.6.(a)). Define g(x)=u(x)f(x). Then g(x)=0 outside a compact set, and the distance between g and f is smaller then epsilon.

 

[Fredrik]

Thanks. I had actually only read the part of the appendix that was before the measure theory section, but I have started reading the A.6 section now.

Unfortunately, I haven't been able to understand your proof. Let's denote the closed set $\{x~\vert~|f(x)|\leq \varepsilon/2\}$ by A. The function u takes the value 0 on A and 1 on K. f(x) can be >0 for all x. So g(x)=u(x)f(x) is only guaranteed to be =0 where u(x)=0, and this is just the set A. You say that g is zero outside a compact set. That compact set must be closed, and the smallest closed set outside which g is zero, is $\operatorname{cl}(A^c)$. Your claim can only be true if this set is compact, and I don't see why it must be, unless of course we assume that X is compact.

Edit: I think I get it, after reading about the one-point compactification stuff in section A.6. All closed subsets of a locally compact Hausdorff space are compact. One way of looking at it is that every locally compact Hausdorff space is a compact Hausdorff space with one point removed, so every closed subset of a locally compact Hausdorff space is a closed subset of a compact Hausdorff space as well.

 

[Landau]

Edit: I think I get it, after reading about the one-point compactification stuff in section A.6. All closed subsets of a locally compact Hausdorff space are compact.

That can't be true. Take X itself: it is closed in X, but of course not (necessarily) compact. The one-point compactification makes the whole space compact, so it changes the topology. What is true, is that all closed (and also all open, for that matter) subsets of a such a space are again l.c.Hausdorff with the subspace topology.

Anyway, it is quite easy. You don't even need the set A. Just apply Urysohn to the whole space X (instead of A) and K. You get $u\in C_c(X)$ (!) with $0\leq u\leq 1$ and u=1 on K. As u vanishes outside a compact set, so does h=fu.

 

[Landau]

So this proves that C_c is dense in C_0. You also need that C_0 is in fact complete. For that, you take a Cauchy sequence (f_n)_n in C_0(X). Then it converges uniformly to some continuous f (because the larger space C(X) is complete); we need to show f is in C_0. Given e>0, pick n with $\|f_n-f\|_\infty<e$, and a compact K such that f_n is <e outside K. Then f is <2e outside K. So f is indeed in C_0.

 

[Fredrik]

That can't be true. Take X itself: it is closed in X, but of course not (necessarily) compact. The one-point compactification makes the whole space compact, so it changes the topology.

That's a good point. I missed the fact that closed sets aren't necessarily closed in the new topology.

Anyway, it is quite easy. You don't even need the set A. Just apply Urysohn to the whole space X (instead of A) and K.

But Urysohn can only be used when the sets are disjoint. These links may be useful. Proposition A.6.7 looks like what we're trying to prove. I haven't read the proof yet, but I'll do that now.

The first two pages of section A.6 at Google Books. (I can't preview the pages after that. I don't know if you will be able to).
The whole book in pdf format. (From the Author's web page. The page numbers in the pdf don't match the actual book).

 

[Landau]

But Urysohn can only be used when the sets are disjoint.

No, we are talking about some other form of urysohn:

Yeah, I meant "Urysohn's weaker lemma", applicable to a l.c. Hausdorff X: if V is open in X and K a compact subset of V, then there exists a $f\in C_c(X)$ with $1_K\leq f\leq 1_V$ (with 1_A of course the characteristic function of A).

 

[Landau]

Equivalently, apply Prop.A.6.6. from Sunders to K=K and A=emptyset.

 

[micromass]

Well, Landau basically explained all of it. But let me explain my original proof

Thanks. I had actually only read the part of the appendix that was before the measure theory section, but I have started reading the A.6 section now.

Unfortunately, I haven't been able to understand your proof. Let's denote the closed set $\{x~\vert~|f(x)|\leq \varepsilon/2\}$ by A. The function u takes the value 0 on A and 1 on K. f(x) can be >0 for all x. So g(x)=u(x)f(x) is only guaranteed to be =0 where u(x)=0, and this is just the set A. You say that g is zero outside a compact set. That compact set must be closed, and the smallest closed set outside which g is zero, is $\operatorname{cl}(A^c)$. Your claim can only be true if this set is compact, and I don't see why it must be, unless of course we assume that X is compact.

The set A is

$$A=\{x~\vert~|f(x)|\leq\varepsilon/2\}$$

Of course $$A^c$$ is not compact, but we do have

$$A^c=\{x~\vert~|f(x)|> \varepsilon/2\}\subseteq \{x~\vert~|f(x)|\geq\varepsilon/3\}$$

And the latter set is compact from the condition. Furthermore, g vanishes outside that set. I probably should have mentioned that.

But then again, Landau's proof is probably better

 

[Landau]

But then again, Landau's proof is probably better

Well, I think it is easier, because you avoid the set A altogether. It is just the following simple observation:

If g=uf, then supp(g) is contained in supp(u). As u has compact support, so does g.

Differently put: if u vanishes then g vanishes; so if u vanishes outside a compactum, then so does g.

The great thing about (the weak) Urysohn is that u is already known to have compact support.

 

[Fredrik]

Of course $$A^c$$ is not compact, but we do have

$$A^c=\{x~\vert~|f(x)|> \varepsilon/2\}\subseteq \{x~\vert~|f(x)|\geq\varepsilon/3\}$$

And the latter set is compact from the condition. Furthermore, g vanishes outside that set. I probably should have mentioned that.

Of course. I feel like such a moron for not seeing this.

Equivalently, apply Prop.A.6.6. from Sunders to K=K and A=emptyset.

It didn't even occur to me that I could take one of the closed sets in Urysohn's lemma to be the empty set. But I have given it some thought now. I think that if A is empty, then the lemma guarantees the existence of a continuous function that's =1 on K, but it doesn't say that this function isn't =1 on all of $$\hat X$$.

 

[Landau]

Ah, I now see that Sunders' formulation of A.6.6. is too weak. Let's improve it!

[A.6.6a improved] Suppose X is a locally compact Hausdorff space. If A and K are disjoint subsets of X such that A is closed in X and K is compact, then there exists a continuous function $f :X\to [0,1]$ such that $f(A)\subseteq\{0\}$ and $f(K)\subseteq\{1\}$, with compact support.

Proof: Let X' be the 1-point compactification of X, and $B=A\cup\{\infty\}$. As in Sunders, we see that B and K are disjoint closed subsets in the compact Hausdorff X'. By Urysohn we obtain a continuous $g:X'\to [0,1]$ with $g(B)\subseteq\{0\}$ and $g(K)\subseteq\{1\}$. Note that $g(\infty)=0$, hence $\text{supp}(g)$ is a subset of X; as X' is compact, this is a closed subset of a compactum, so g has compact support. Now $f=g|_{X}$ does the job.

Remark: I said $f(A)\subseteq\{0\}$ and $f(K)\subseteq\{1\}$ instead of $f(A)=\{0\}$ and $f(K)=\{1\}$ to allow K or A to be empty. This is allowed, because the function g you get from Urysohn (A.4.23) already has this property: it says "g(x)=0 if x is in A", so if A or B is empty, this is vacuously true.

 

[Fredrik]

Crystal clear. Thanks.

I see that your improvement of Sunder's A.6.6. consists of replacing = with $\subseteq$ (and to explicitly mention the compact support). You are of course right that this is necessary if we want to consider empty sets. I didn't even notice that before. The mistake I made was to not realize that even if we start with an empty A, neither of the closed sets that we apply the "actual" Urysohn's lemma to are empty. If we treat the empty A like we would treat any other closed set, we would take the union with {∞} before we apply Urysohn. (When A isn't empty we have to do this because A isn't closed in the topology of X', but A⋃{∞} is). For non-empty A, we don't have to do this, but we should do it anyway, just to make sure that the function we find isn't constant =1.

 

[Fredrik]

However, a C*-algebra is a highly abstract notion, and mathematicians wanted to give representation for them in terms of things that were more concrete. If the C*-algebra was commutative and has an idenity, then the answer to this question was given by Gelfand-Naimark. Every C*-algebra can be seen as a ring of continuous functions on a compact Hausdorff space. Thus the C*-algebra is isomorph to $$C(X)$$. If the C*-algebra didn't have a unit, then the C*-algebra can be seen as $$C_0(X)$$ with X locally compact and Hausdorff. Thus, to study commutative C*-algebra's, is exactly thesame as studying $$C_0(X)$$ for locally compact spaces.

I have a question about these things. Unless I have misunderstood something, every Banach algebra is an ideal in a Banach algebra with identity, and given a Banach algebra with identity, we can always redefine the norm so that $\|1\|=1$ without changing the topology. Why doesn't this mean that we will only need to study those C*-algebras that have an identity and are such that $\|1\|=1$? It seems to me that the the whole theory could be considerably simplified if we do all that Gelfand-Naimark stuff only for those "nice" algebras. Is there a reason not to?

I'm sure there's some reason. What I'm wondering is if the less nice algebras are useful only in exotic applications like non-commutative topology, or are they needed for standard stuff like proving the spectral theorems for bounded normal operators?

(My knowledge about these things extend roughly to page 76 in Sunder, so I haven't yet reached the proof of the spectral theorem. It doesn't look like I will have time to continue studying this for at least another month).

 

[micromass]

First a small technicality:

I have a question about these things. Unless I have misunderstood something, every Banach algebra is an ideal in a Banach algebra with identity, and given a Banach algebra with identity, we can always redefine the norm so that $\|1\|=1$ without changing the topology.

It is true that you can extend any Banach algebra to a Banach algebra with identity. But you cannot be sure that this extended Banach algebra will be a C*-algebra. Indeed, the extended Banach algebra that you've likely seen, is NOT a C*-algebra. This can be solved however, by considering a quite different, and more complicated, norm. So what you say is true, but I just wanted to point out the subtleties.

Why doesn't this mean that we will only need to study those C*-algebras that have an identity and are such that $\|1\|=1$? It seems to me that the the whole theory could be considerably simplified if we do all that Gelfand-Naimark stuff only for those "nice" algebras. Is there a reason not to?

Yes, you basically want to add an identity to the C*-algebra. This is possible, and it is the analog of "compactifications" of the underlying space. A minimal unitization (= adding an identity) will correspond to an Alexandroff compactification, and a maximal unitization will be the Cech-Stone compactification.

But to answer your question: yes, it is enough to prove all the nice results for C*-algebra's with an identity, and we indeed do this. The Gelfand-Naimark will usually only be stated for C*-algebra's with an identity. So for almost all applications, you could probably assume that your algebra has an identity. However, I just wanted to point out that the theory could be extended to algebra's without a unit.