The Dirichlet space of martingales

The Hardy space of martingales can be defined in terms of martingale differences. I'll stick to the simplest case of dyadic martingales.

Notation The underlying probability space can be taken to be $$[0,1]$$, the filtration $$\{F_n\}$$ has $$F_n$$ generated by dyadic intervals $$F_n\ni I=I_+\cup I_-$$ ($$I_\pm\in F_{n+1}$$) with $$|I|=2^{-n}$$ and the probability measure is Lebesgue measure. A martingale $$f=\{F_n\ni f_n=\sum_0^n d_j\}$$ satisfies $$d_n(I_+)+d_n(I_-)=0$$ if $$I\in F_{n-1}$$, where $$d(J)=d_n(J)$$ is the value of the martingale difference $$d_n=f_n-f_{n-1}$$ on $$J\in F_n$$.

The martingale $$f$$ is in the Hardy space $$M^2$$ if $$\infty>\|f\|_2^2=\lim_{n\to\infty}{\mathbb E}f_n^2=\sum_{j=0}^\infty\sum_{|J|=2^{-j}}d_j^2(J)2^{-j}=\sum_Jd(J)^2|J|.$$ There is a vast literature on this and related martingale spaces ($$M^p$$'s, $$BMO$$, etcetera). The starting point is that $$f=\lim_{n\to\infty}f_n$$ exists a.e. on $$[0,1]$$ and $$f_n={\mathbb E}[f|F_n]$$.

Consider instead the Dirichlet martingale space $$D\subset M^2$$ defined by the condition $$\infty>\|f\|_D^2=\sum_Jd(J)^2.$$ It is related to $$M^2$$ the same way the holomorphic Dirichlet space is related to the holomorphic Hardy space, or the Sobolev space $$W^{1/2,2}({\mathbb R})$$ ($$1/2$$ derivative in $$L^2({\mathbb R})$$) is related to $$L^2({\mathbb R})$$.It is easy to show that $$f=\lim_{n\to\infty}f_n$$ exists outside a set of null logarithmic capacity in $$[0,1]$$. In a sense, it is the "right" space if one wants to do calculus of variations for martingales. Also, it can be defined via martingale transforms of Hardy martingales.

The questions are the following. (1) Is there some literature on this space? (2) The fact that the boundary values are defined outside a set of zero capacity has some easy to understand probabilistic meaning?

[I have often used $$D$$ as a toy model for testing facts and proofs I wanted to verify or use dealing with holomorphic functions].

• Probabilists usually don't care about how big a set is once it's measure is zero... – Kostya_I Jun 13 at 19:04
• That's why I asked. As a matter of fact, if the norm is changed to $\sum_Jd(J)^2/|J|$, the martingale has limits everywhere, which is perhaps more poignant. Notwithstanding that a.e. is the right notion of negligibility, the fact that Dirichlet martingales have limits remains, and might hide some phenomenon. – Nicola Arcozzi Jun 13 at 21:02