User wolfgang loehr - MathOverflow

reference for "X compact <=> C_b(X) separable" (X metric space)

2013-01-09T11:51:38Z

I know (and am able to prove via Stone-Čech compactification) that the following is correct:

Theorem: A metric space is compact if and only if its space of bounded, continuous, real-valued functions is separable in the uniform topology.

I use it in a paper for readers who are presumably not familiar with this kind of topology, so I cannot call it "obvious" or "well-known". I would be thankful for a name and/or good reference to cite this theorem!

Answer by Wolfgang Loehr for Proposed Counterexample to a Theorem of Differential Geometry (on Banach manifolds)

2012-09-10T15:46:16Z

Sub-question 2 is easy: $X$ is connected. Assume $A\subseteq X$ is open and closed. $Y\setminus \ker\lambda$ carries the topology inherited from $\ell^2$, hence is connected, hence we may assume w.l.o.g. that it is contained in $A$ (otherwise we take the complement of $A$). Now fix $y$ with $\lambda(y)=1$. For any $x\in\ker\lambda$, $x_n:=x+\frac1n y\in A$ and $x_n\to x$ , hence $x\in A$ and $A=X$.

Does the law of a Feller Process on a non-locally-compact Polish space depend continuously on the initial condition (in Skorohod path-space)?

2012-08-16T14:03:57Z

I am sure this is written down somewhere but cannot find it. Consider a Polish space $E$ and a strong Markov process $(X_t)_{t\ge 0}$ with values in $E$ and cadlag paths. More precisely, we have a family $X^x=(X^x_t)_{t\ge 0}$ , $x\in E$, of Markov processes with a common transition kernel and $X^x_0=x$ almost surely. Further assume the Feller property, i.e. the corresponding semi-group maps bounded continuous functions to bounded continuous functions. Equivalently, the law of $X_t^x$ depends continuously on $x$ (for fixed $t$), hence the same is true for finite-dimensional distributions.

I know that in the case of locally compact $E$ this is enough to deduce that the law of $X^x$ depends continuously on $x$ w.r.t. weak convergence on the path-space with Skorohod topology (Edit: it is not, see below).

Now my question is if the assumption of local compactness is really necessary:

Does the law (in weak topology on Skorohod space) of a Feller Process on a general Polish space depend continuously on the initial condition?

If not, are there reasonable conditions to ensure this continuity in paths-pace?

Edit: As George pointed out, contrary to my claim the above does not even hold for locally compact spaces if we do not assume that the functions vanish at $\infty$, which makes no sense in non-locally-compact spaces. Still it would be nice to have a reference about processes with the continuity property defined above.

I would be very thankful for any good reference for Markov processes with the Feller property on non-locally-compact spaces!

Most sources I found already include local compactness in the definition of "Feller".

Is the set of real-valued lower semi-continuous functions measurable in epigraph topology (= topology of Gamma convergence)?

2012-08-12T22:48:31Z

Let LSC = LSC([0,1]) be the set of non-negative, lower semi-continuous functions on the unit interval which take values in $\mathbb{R}_+ \cup \{\infty\}$ . We use epigraph topology on LSC, i.e. a sequence of functions converges iff the sequence of epigraphs converges in Fell (or equivalently Kuratowski) topology. This convergence coincides with $\Gamma$-convergence (i.e. $f_n\to f$ iff for all $x_n \to x$ we have $\liminf f_n(x_n)\ge f(x)$ and for all $x$ there exists $x_n\to x$ with $\lim f_n(x_n) = f(x)$ ). LSC with this topology is a compact, metrisable space.

Now consider the subspace of bounded functions in LSC. This is clearly a countable union of closed subsets, hence Borel-measurable in LSC. Now my question is:

Is the subset of $\mathbb{R}_+$ -valued functions in LSC Borel-measurable?

Edit: I am asking this, because I naturally came to the space of semi-continuous functions that are not allowed to take on the value $\infty$ , and need a "nice" topology on it. Epigraph topology makes the things converge which I want to converge, but being separable, metrisable is just not "nice" enough, while being a measurable subset of a compact metric space (= Lusin space) would do. Maybe continuous image of a Polish space (= Souslin space) would also be enough.

So if anyone had, alternatively, an idea of a "similar" topology with better properties, this would also be great!

P.S. I found out that on the space of $\mathbb{R}_+$ -valued functions in LSC, epigraph topology is also equivalent to convergence of the epigraphs in Hausdorff metric (which is not the case on the whole of LSC). Unfortunately, Hausdorff distance of the epigraphs is also an incomplete metric.

Answer by Wolfgang Loehr for Convolution of a continuous function and a finitely additive measure

2012-08-14T10:13:33Z

Edit: As Mateusz pointed out, it becomes more interesting if $f$ does not need to vanish at $\infty$ . For uniformly continuous $f$ , the above still works and for $\sigma$ -additive $\mu$ we can use dominated convergence as suggested by Davide. For arbitrary, bounded continuous $f$ and non- $\sigma$ -additive $\mu$ , $f*\mu$ need not be continuous:

Let $\mu$ be defined by $\mu(f) = \lim_{n\to\infty, n\in\mathbb{N}} f(n)$ if the limit exists and extend $\mu$ by Hahn Banach to a positive linear functional (a Banach-Mazur limit). Let $f$ be a continuous function which is zero on $[n-\frac1{|n|}, n+\frac1{|n|}]$ for every $n\in\mathbb{Z}$ and 1 for points which are more than $\frac2{|n|}$ away from every $n\in \mathbb{Z}$ . Then $f*\mu(0) = 0 $ but $f*\mu(x) = 1$ for $x$ close to but unequal zero.

Answer by Wolfgang Loehr for Infinite linear span vs closed linear span

2012-08-14T09:33:18Z

In the case of linear $A$ , which seems to be your case of interest, you can simply do it as follows. For $x$ in the closure of $A$ take a sequence $(x_n)$ in $A$ converging to $x$ such that $\|x_n-x_{n+1}\|<2^{-n}$ . Then set $a_i=x_i-x_{i-1}$ , $\beta_i=1$ and the sum will converge absolutely, hence also unconditionally to $x$ . Or did I miss something?

Answer by Wolfgang Loehr for Standard notation/symbol for an embedding function

2012-08-13T12:26:29Z

I usually use $\iota$ (\iota) for all kinds of embeddings.

Are affine continuous functions on Bauer sub-simplices of the probability measures given by integration over continuous functions?

2011-05-24T15:52:22Z

Let $X$ be a compact (non-metrizable) Hausdorff space and $\mathcal{P}(X)$ the set of Radon probability measures with weak- $*$ topology (weak topology induced by the continuous functions). Consider a compact subset $S\subseteq \mathcal{P}(X)$ that is itself a Bauer simplex, i.e. it is convex, the set $\mathrm{ex}(S)$ of extreme points is compact, and the barycenter map (called resultant by Choquet), $r\colon \mathcal{P}\bigl(\mathrm{ex}(S)\bigr) \to S$ , $r(\mu)(A) = \int \nu(A)\, \mu(\mathrm{d}\nu)$ is injective (thus a homeomorphism).

In this situation, every continuous real valued function on $\mathrm{ex}(S)$ can be extended to a continuous affine function on $S$ .

Now my question is the following:

Can every continuous affine function $F\colon S\to \mathbb{R}$ on a Bauer simplex $S\subseteq\mathcal{P}(X)$ be extended to a continuous affine function on $\mathcal{P}(X)$ (and therefore $F(\mu) = \int f\,\mathrm{d}\mu$ for some continuous $f\colon X\to \mathbb{R}$ )?

I think the following question is equivalent: Can $F$ be extended to a continuous linear function on the vector space spanned by $S$ in the space $\mathcal{M}(X)$ of signed measures of bounded variation? (Then we can use Hahn Banach).

I am also interested in the following more general formulation. Every Bauer simplex $S$ is affinely homeomorphic to a probability simplex, namely $\mathcal{P}\bigl(\mathrm{ex}(S)\bigr)$ . If $S$ is given as a subset of a closed hyperplane (that does not contain $0$ ) of a locally convex topological vector space, can the affine homeomorphism be extended to a linear homeomorphism of the vector space spanned by $S$ into $\mathcal{M}\bigl(\mathrm{ex}(S)\bigr)$ ?

I guess this is extension cannot be done in general, but I do not know.

Any references, partial solutions, counter examples, and ideas are welcome.

Comment by Wolfgang Loehr

2013-01-09T16:57:00Z

Thank you very much! I think one can cite exercises in Dunford & Schwartz ;-)

Comment by Wolfgang Loehr

2013-01-09T15:11:49Z

@Gerald: Thanks, I'll check Dunford & Schwartz @Theo: This proof is more elementary than mine, thank you.

Comment by Wolfgang Loehr

2012-09-14T09:54:19Z

For me as a probabilist this terminology sounds strange (and I did not know it). In probability a kernel would be more something like a (not necessarily normalized) Markov kernel and we would say ``finite measure''.

Comment by Wolfgang Loehr

2012-09-13T09:14:04Z

The assumption that the generator is a compact operator appears quite restrictive to me. What example do you have in mind?

Comment by Wolfgang Loehr

2012-09-12T23:00:34Z

At least something seems to be wrong in [Abraham, Marsden, and Ratiu], for after the disputed theorem they claim in Corollary 6.5.13 (3rd edition, 2007 pdf-version): ``Paracompact (or second countable) Hausdorff manifolds modeled on separable real Hilbert spaces admit Riemannian metrics.'' This would make our $X$ a counter example to their theorem. However, they prove the corollary only in the paracompact case and $X$ is second countable but of course not paracompact.

Comment by Wolfgang Loehr

2012-09-12T08:52:07Z

How could $(X_t)$ be stationary when $(B^H_t)$ obviously isn't?

Comment by Wolfgang Loehr

2012-09-11T09:45:54Z

Have you put backticks around formulas with underscores? Sorry, I do not have enough reputation to edit.

Comment by Wolfgang Loehr

2012-09-11T09:05:59Z

You should edit your post to make the math formulas readable. See the Box "how to write math" in the lower right corner of the page.

Comment by Wolfgang Loehr

2012-09-10T16:31:20Z

Not quite. Everyone agrees that the boundary of the Mandelbrot set is a fractal, but it has Hausdorff dimension 2. Of course its topological dimension is 1, so a better definition would be the one of Mandelbrot, namely Hausdorff dimension strictly bigger than topological dimension. But even this definition is not universally agreed upon...

Comment by Wolfgang Loehr

2012-09-10T15:19:08Z

If $u$ is fixed, why do you specify that $a_t$ is a continuous process?

Comment by Wolfgang Loehr

2012-09-10T14:17:48Z

I just want to remark that every metrisable space is regular, so you don't need separability/second-countability to conclude non-metrisability (but of course you also use it for the Riemannian metric).

Comment by Wolfgang Loehr

2012-09-10T12:46:24Z

If I understand 1. correctly, the answer is yes (assuming of course that $a_u$ is integrable), because $a_u$ is just a fixed random variable an you can use Jensen for conditional probabilities.

Comment by Wolfgang Loehr

2012-08-24T14:28:38Z

Consider $f(x):=\sin(\frac1x)$ for $x\ne0$, $f(0):=-1$. Then clearly $f$ is lower semi-continuous but has neither a left nor a right limit at $0$.

Comment by Wolfgang Loehr

2012-08-24T09:49:38Z

A semi-continuous function needs to have neither left nor right limits (though it is continuous on a dense $\mathcal{G}_\delta$ -set).

Comment by Wolfgang Loehr

2012-08-22T13:14:44Z

Sure it's OK to repost here, because you didn't get an answer at MSE, and made a link to the other post. You might also want to link this post at MSE.