4
$\begingroup$

I have been studying the construction of the Martin boundary on a discrete set $X$ admitting an irreducible transient random walk $(X,P)$ from Wolfgang Woess' book titled "Random Walks on Infinte Graphs and Groups", where I came across the Poisson-Martin representation theorem which has been stated below:

For every positive harmonic function $h$ on a $X$, there is a positive Borel measure $\nu^h$ on the Martin boundary $\mathcal{M}$ of $X$ such that $$h(x)=\int_{\mathcal{M}} K(x,.) \mathrm{d} \nu^h$$ where $K(.,.)$ denotes the Martin kernel corresponding to $(X, P)$.

I also read about the Martin boundary on the Encyclopedia of Mathematics, where it is stated that for certain choices of $X$ (say $X$ is a bounded domain in $\mathbb{R}^n$ or a Green space), the Borel measure $\nu^h$ is actually a Radon measure. One question that intrigues me is whether we can always expect the Borel measure $\nu^h$ occurring in the above theorem to be a Radon measure? If not, can we expect that $\nu^h$ is a finite measure on $\mathcal{M}$? Is it finite on any compact subset of $\mathcal{M}$? Any help or reference is deeply appreciated. Thanks in advance!

$\endgroup$

1 Answer 1

5
$\begingroup$

There is one more ingredient in the definition of the representing measure $\nu^h$ missing in your description. This is a reference point $o$ from the state space (more generally, a probability measure on the state space), so that it would be more appropriate to use the notation $\nu_o^h$. The functions $\phi$ representing the points of the Martin boundary are normalized by requiring that $\phi(o)=1$. [In a more invariant language, in order to use the Choquet theory for the convex cone of positive superharmonic functions one has to fix a convex compact base of that cone.] It implies that the representing measure $\nu^h_o$ is actually finite, and its mass is $h(o)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.