Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $H^p$ be the Hardy space of analytic functions on the open unit disk $\mathbb{D}$: $f \in H^p$ if $f$ is analytic on $\mathbb{D}$ and $\sup_{r < 1} \int_0^{2\pi} |f(re^{i\theta})|^p d\theta < \infty$.

Consider a filtration generated by a 2-d (complex) Brownian Motion $B$. The martingale hardy space $\mathcal{H}^p$ defined on some time interval $[0,T]$, say, is the set of martingales $M$ such that $M^* = \sup_{t \in [0,T]} |M_t| \in L^p$. This definition is mostly interesting for $p=1$, as for $p>1$, $\mathcal{H}^p$ can be associated with a regular $L^p$ space of martingales.

If $B$ starts at zero, let $\tau$ be the hitting time of the boundary of $\mathbb{D}$. Then a connection between these two spaces is the following: for $f$ analytic on the unit disk, $f(B_{t \wedge \tau}) \in \mathcal{H}^p$ if and only if $f \in H^p$, and this mapping is continuous.

This allows you to associate $H^p$ to a subspace of $\mathcal{H}^p$. For studying $\mathcal{H}^p$, it would be useful to have a more complete representation of part of $\mathcal{H}^p$ in terms of functions evaluated on $B$. Specifically, for martingales that run on the whole time interval. Can this be obtained by using another hardy space, such as the Hardy space $h^p$ on $\mathbb{R}^2$? Can anything else be said relating hardy spaces of martingales and hardy spaces of functions?

share|cite|improve this question
maybe helpful – john mangual Jun 3 at 17:28
The standard text on this subject is "Isomorphisms of $H^1$ spaces" by P.F.X. Müller. – priel Aug 20 at 12:40

1 Answer 1

Charles Fefferman proved that the space of functions of bounded mean oscillation $BMO$ is the dual of the Hardy space $H^1$. The key was the observation that the classical Hardy space $H^1$ is a natural substitution for $L^1$, and $BMO$ is a natural substitution for $L^{\infty}$.

One result of comparing classical Hardy and martingale Hardy spaces is that we can establish that the dual of the martingale Hardy space $\mathcal{H}^1$ is martingale $BMO$.

Check out "Martingale Inequalities: Seminar Notes on Recent Progress (Mathematics Lecture Note Series)" by Adriano Garsia

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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