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Fix $p\in [1,2]$, a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_t,\mathbb{P})$, and let $\mathcal{H}_{\mathscr{S}}^p$ denote the space of semimartingales $X$ such that the norm $$ \|X\|_{\mathcal{H}_{\mathscr{S}}^p}:= \inf\big\{ \mathbb{E}\big[\big( [M]_{\infty}^{1/2} + \int_0^{\infty} |dA|_s \big)^p\big]^{1/p} \big\} $$ is finite, where the infimum is taken over all decompositions of $X$ into a local martingale with $M_0=0$ and a process $A$ of finite variation with $A_0=X_0=\Delta A_0$. It is known (Lemma 16.2.25 of this book) that every semi-martingale in ${\mathcal{H}_{\mathscr{S}}^p}$ is special and the above norm can be written as $$ \|X\|_{\mathcal{H}_{\mathscr{S}}^p} = \mathbb{E}[([\tilde{M}]^{1/2})^p]^{1/p} + \mathbb{E}[(\int_0^{\infty} |d\tilde{A}_s|)^p]^{1/p}, $$ where $X=\tilde{M}+\tilde{A}$ is the canonical decomposition of $X$.

We call $\mathcal{H}_{\mathscr{S}}^p$ the semi-martingale-hardy space.


Clearly, $\mathcal{H}_{\mathscr{S}}^p$ is a separable Banach space and it is a Hilbert space when $p=2$.


Question: What is an explicit Schauder basis of this space?

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There can be many possibilities. Let me focus on the martingale part. For each different Martingale one can construct a basis using the difference functions eg see Theorem 4.2.6. in Schauder Bases and the Factorization Property where they also reference chapter 7 in "Introduction to Banach spaces: analysis and probability".

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So there can be many possibilities. In terms of standard construction, you can push the Schauder basis from here Proof of existence of Schauder basis for $L^p(\Omega)$? by mapping it to the probability space $\Omega$.

For simplicity I focused only the martingale part since it is a subspace of the above. By Burkholder–Davis–Gundy inequality, the norm in terms of quadratic variation is comparable to the Lp of the sup of the martingale.

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  • $\begingroup$ But is this even for the same space? It seems like a martingale (difference) basis of $L^p$. $\endgroup$
    – ABIM
    Commented Jan 12, 2023 at 19:44
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    $\begingroup$ For simplicity I focused only the martingale part. By Burkholder–Davis–Gundy inequality, the norm in terms of quadratic variation is comparable to the Lp of the sup of the martingale. So the above answer is basically showing that there are many possible unconditional basis. $\endgroup$ Commented Jan 12, 2023 at 20:54
  • $\begingroup$ I see but I feel to get the points this should be written out in the answe $\endgroup$
    – ABIM
    Commented Jan 13, 2023 at 11:35

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