Norm in the space of square integrable martingales

Question

I was learning stochastic integration and encountered two different norms used in the space of square integrable martingales They are as follows:

1.Let M be a square integrable martingale, then $|M|_t$ : = $\sqrt{ \mathbb{E}[M_t^2]}$

$$|M|:=\sum_1^\infty \frac{{1\wedge |M|_n}}{2^n}$$ ($\wedge$ is minimum)

And,

2. Let M be a square integrable martingale, then $|M|$ : =$\sqrt{\mathbb{E}[\sup_{t < \infty}M_t^2]}$.

Are these two norms equivalent

Answer 1

For clarity I will denote the first norm as $\left|\cdot\right|_a$ and the second as $\left|\cdot\right|_b$, i.e. for a square-integrable martingale, $M$ $$\left|M\right|_a = \sum_{n = 1}^\infty \frac{1\wedge \sqrt{\mathbb{E}\left[M_n^2\right]}}{2^n} \quad\quad \left|M\right|_b = \sqrt{\mathbb{E}\left[\sup_{t \geq 0}M_t^2\right]}.$$ The second norm is strictly stronger than the first meaning that there exists a $C>0$ such that for any square-integrable martingale $M$, $\left|M\right|_a \leq C\left|M\right|_b$, but the reverse, $\left|M\right|_b \leq C' \left|M\right|_b$, is not true.

Since for both norms $|M| = \left|\left(M - M_0\right)\right| + \left\|M_0\right\|_{L^2}$, we can assume that $M_0 = 0$. Let $\left<M, M\right>$ denote the quadratic variation of $M$. We will apply two facts about quadratic variation 1) $\mathbb{E}\left[M^2_t\right] = \mathbb{E}\left<M,M\right>_t$ and 2) $\left(\left<M, M\right>_t\right)_{t \geq 0}$ is an increasing process.

$$\left|M\right|_a = \sum_{n = 1}^\infty \frac{1\wedge \sqrt{\mathbb{E}\left[M_n^2\right]}}{2^n} = \sum_{n = 1}^\infty \frac{1\wedge \sqrt{\mathbb{E}\left<M,M\right>_n}}{2^n} \leq \sum_{n = 1}^\infty \frac{\sqrt{\mathbb{E}\left<M,M\right>_n}}{2^n} \leq \sqrt{\mathbb{E}\left<M, M\right>_\infty}\leq \frac{1}{\sqrt{c_2}}\sqrt{\mathbb{E}\left[\sup_{t \geq 0}M_t^2\right]} = \frac{1}{\sqrt{c_2}}\left|M\right|_b$$ The last inequality follows from the Burkholder–Davis–Gundy inequality for $p = 2$ and $c_2$ is the corresponding universal constant (see for example IV.4 in Revuz and Yor or 5.3.3 in Le Gall).

The other direction necessary for equivalence fails for Brownian motion, $B$, as $|B|_a = 1$ but $|B|_b = \infty$.