@Srivatsan I don't see a direct relation, but yes. It is related in that it's a reverse problem. JDH's answer reminded me of some constructions of topologies on (nice) measure spaces, but there the existence of a measure is quite essential. I think what he considers is called the canonical density topology, but I need to check. (I'm somewhat dubious that two people have already checked the details of the answer, though)
@tb Well, it reminded me of that question because this was one more thing I was wondering about, but I ended up not posting somehow. I think it turned out well, because Borel somehow strays into regions I understand even less.
At some point last week, I was thinking of giving a small bounty just to attract attention. I was thinking of these two posts for bounties (Mike's and your answers). =) [Mike's post got 6 more votes since the bounty, which is encouraging.]
@tb You got your money now: that thread just got three upvotes. =)
@Srivatsan That's very kind of you, but I think Mike's question is much more worthy of a bounty. By the way: would you be interested in seeing the generalization I spoke of, i.e., localizable measure spaces?
(not that your question wouldn't deserve a bounty, but there's more interesting stuff to be expected from a bounty on Mike's question)
@Srivatsan Nowhere. I guess I misunderstood you. I interpreted your "push this" as you didn't know how to ask for further explanations after having received an (apparently) satisfactory answer... :)
@Srivatsan It's a bit specialized. Localizable measure spaces are quite useful. Let me think of something that might be interesting and somewhat on-topic.
@JonasTeuwen assume no or little prior knowledge, I will be interested =)
So, let's think of the hypothetical situation where you would want to consider the space $L^2$ with the Gaussian measure $\mathrm{d}\gamma(x) = \pi^{-d/2} e^{-|x|^2} \, \mathrm{d}x$.
So, you know that for $L^2(\mathbf R^d)$ that the Laplacian $\Delta$ is a symmetric operator. So you think: Great! I can use that operator and do all my usual harmonic analysis arguments.
$\mathbf R^d$, yes.
Alas! The Laplacian is not symmetric in $L^2(\gamma)$!
So we could try to find something that looks like the Laplacian and that is symmetric in $L^2(\gamma)$. It seems obvious to try something like $\sum \partial_i^* \partial i$ where $\partial_i^* = -\partial_i + 2 x_i$ is the formal adjoint in $L^2(\gamma)$ of $\partial_i$.
@Srivatsan We are not interested in the Laplacian since it is not symmetric.
Well, this operator is obviously symmetric. We tack on $-\frac12$ to get $L := -\frac12 \sum_{i = 1}^d \partial_i^* \partial_i$.
We call this: The Ornstein-Uhlenbeck operator (on $\mathbf R^d$).
Why?
I believe this comes from the Ornstein-Uhlenbeck process from SDEs.
Wait. You're defining it on $C_{c}^{\infty}(\mathbf{R}^d)$ and using integration by parts we see that $\langle Lf,g \rangle_{L^2(\gamma)} = \langle f,Lg \rangle_{L^2(\gamma)}$ for all $f,g \in C_{c}^\infty$. (this would also answer @Srivatsan's question, as non-symmetry of the Laplacian can be seen by the failure of this equation).
I usually add a tiny piece of ice during the summer. You see, the whisky gets to a temperature of nearly 30 C, which is about 12 degrees higher than its intended drinking temperature.
I guessing that when x=y, we would want it to look like a gaussian, hence the weird normalisation. [What I said seems coincidental, but it's not an explanation of anything, I guess.]
@Jonas: Okay. I think I'm with you again: we now pass to the Hardy space and the semigroup turns out to be bounded there. Now Gaussian measure is not doubling, which is of course annoying.
I haven't done covering arguments yet, we have adapted the definition of the cones in harmonic analysis to "admissible cones", that is we cut of the top with $m(x) = \min(1, |x|^{-1})$ where $x$ is our base point of the cone.
@AsafKaragila "if we have maths question we can ask here.. but even for this simple question you are not ready to answer means then what kind of mind you're having to do the research on maths.. nothing in this world started as big from the beginning itself. Little drops of water makes the ocean.. try understand the little drops first.. thanks."
@JonasTeuwen Okay, got that. I'm still a bit lost with the admissibility condition. So if $x$ is large you don't want the ball to come too close to where the Gaussian measure is concentrated, right?
We now have the following two operators $T_a u(x) = \sup_{(y, t) \in \Gamma_x^a(\gamma)} |e^{t^2 L} u(y)|^2$, the maximal operator and the conical square function operator $$S_a u(x) = \left ( \int_{\Gamma_x^a} \frac1{\gamma(B(y, t))} |t \nabla e^{t^2 L} u(y)|^2 \, \textrm{d}\gamma(y) \, \frac{\textrm{d} t}{t}\right )^{\frac12}.$$
@JonasTeuwen Okay. The maximal operator seems fine, but the square function operator needs a bit more of staring at $$S_a u(x) = \left ( \int_{\Gamma_x^a} \frac1{\gamma(B(y, t))} |t \nabla e^{t^2 L} u(y)|^2 \, \textrm{d}\gamma(y) \, \frac{\textrm{d} t}{t}\right )^{\frac12}.$$
@JonasTeuwen Can you say a few words about it? you have $\gamma \times (\text{Haar measure on } \mathbf{R}_+)$ you take a weighted $L^2$-norm for that and you average over the cone $\Gamma_{x}^a$, but I don't quite see how the weight $\gamma(B(y,t))$ comes in.
So now we can define our Hardy spaces with norm $\|u\|_{H^1} = \|F u\|_1 + \|u\|_1$ where $F$ is either $S_a$ or $M_a$.
These are analogues for the normal Hardy space where we have the Laplacian instead of the Ornstein-Uhlenbeck operator. We would now like to prove that $\|T_a u\|_1 \simeq \|S_a u\|_1$.
(Actually we define those norms on $C_c^\infty(\mathbf R^d)$ and take the completion.)
@Srivatsan It's similar to the definition of the product topology. You look at the $\sigma$-algebra generated by cylinder sets. These are of the form $\prod_{a \in A} C_a$ where $C_a \neq \{0,1\}$ only for countably many $a \in A$. Now you take the $\sigma$-algebra generated by that.
I clicked on Improve, and made more edits. Now I see Paul's edits, followed by mine. Is it because someone else approved it? This is the question I am talking about: math.stackexchange.com/questions/89397/…
One feature request would be view the latex source of the old and edited post. Sometimes it's impossible to figure out what edit was made. Has this ever happened to you?
Most of Shelah's results are difficult to understand. He's on a whole other level as a person too.
I discussed Shelah's papers once with a former postdoc in our department. He suggested that refereeing a paper by Shelah is equivalent to co-authoring it.
