Consider the semimartingale $Z$, which by Doob decomposition can be written as $Z=M+A$, where M is a martingale and A is the process of total bounded variation. I am trying to make sense of the following definition of $Q$, but I do not understand the intuition behind the author defining $Q$ in the following way. I know that something approximate would be derived considering $\langle Z,Z\rangle=\langle M+A,M+A\rangle$, by using inner product properties, in which this inner product would be the quadratic variation. But in that sense the author does not really define the terminology being employed for example || stands for quadratic variation, but why not write $\langle M^{\alpha},M^{\alpha}\rangle_t$ as $|M^{\alpha}|_t$ instead?
Let $Z$ be the canonical decomposition of semimartingale $Z$ into a local martingale and a process of locally bounded variation. Define $$Q_t=\sum \langle M^{\alpha},M^{\alpha}\rangle_t+\sum |A^{\alpha}|^3_t+|A^{\alpha}|_t+t$$
where $|A^{\alpha}|_t$ stands for the total variation of $A^{\alpha}$ on [0,t].Adding $t$ in the definiton makes $Q$ strictly increasing.
Then he proves the following inequality and he uses $Q$ in order to create upper bounds with respect to the martingale and bounded variation integral. I have no idea where this is coming from.
Lemma: There is a constant $C$ depending only on $m$ and $l$ such that for any $F*$ adapted, $M(m,l)$ valued continous process $F$ and an $F*$ stopping time $\tau$:
$$E \max_{0\leqslant t\leqslant\tau}|\int_{0}^{t}F_s dZ_s|^2\leqslant C E \int_0^{\tau}|F_s|^2dQ_s$$
In the proof the Ito isometry holds $E \left(\int_0^{\tau}|F_s|^2dQ_s\right)^2=\int_0^{\tau}|F_s|^2 dt$
I do not even know how $dt$ can be inferred from d$Q$
Question:
What is the intuition behind this?
Thanks in advance.
The quadratic variation and the total $2$-variation are two different things. The former is a property over a partition that becomes smaller and smaller, whereas the latter takes the supremum over all partitions not relying on the partition's interval's size. That is, $$ \lim_{\delta \to 0^+} \sum_{[u,v] \in \mathcal{P}_\delta([0,1])} |X_v-X_u|^2 \neq \sup_{\mathcal{P}([0,1])} \sum_{[u,v]\in \mathcal{P}([0,1])} |X_v-X_u|^2$$ where I denoted with $\mathcal{P}_\delta$ any partition of $[0,1]$ such that each adjacent point of the partition has distance less than $\delta$ and without the $\delta$, means that it is just a partition of the given interval.
From what I gathered from the book you reference, the definition of $Q$ is pretty much defined in order for the proof of the lemma below to work. Indeed, because it gets the integral $$ \int_0^\tau |F_s|^2 (|A_s|^2 + 1) \; dA_s \le \frac{1}{3} \int_0^\tau |F_s|^2 d(|A_s|^3 + |A_s|)$$ and wants to collect this with respect to the same driving path of the integral part dependent on the martingale. So, instead of integrating wrt $\langle M, M \rangle$, we integrate with respect to a ''larger'' path, namely $Q$ (note that every term of $Q$ is an increasing process by definition so that the inequality $\int_0^s f(u) \, d (|A_s|^3 - |A_s|) \le \int_0^s f(u) \; dQ_s$ is trivial).
And, because the other part of the estimate is given by Ito integral over the martingale, i.e. $$ \mathbb{E} \left\{ \max_{t\le \tau} \left| \int_0^t F(s)\; dM_s\right|^2 \right\} \le 4 \mathbb{E} \left\{\int_0^\tau |F(s)|^2 \; d\langle M, M\rangle_S\right\}$$ using Doob's inequality and Ito's isometry. Finally, because we defined $Q$ in such a way that ''contains'' the path generated by $\langle M, M \rangle$ we can bound it with respect to $Q$. Therefore, in the end the author collects both estimates as integrals with respect to $Q$ and obtain the claim.
Hope this helps.