Using BDG inequality to show the solution to a BSDE belongs to $S^2_{\mathscr{F}}$

Question

For a BSDE:

$$y_t = \xi + \int_{t}^{T}g_0(s)ds - \int_{t}^{T}z_sdB_s$$

Which has a fixed $\xi \in L^2(\mathscr{F}_T)$ and $g_0(\cdot)$ satisfying $E(\int_{0}^{T}|g_0(t)|dt)^2 < \infty$. There exists a unique pair of process $(y., z.)\in L_{\mathscr{F}}^2(0,T;R^{1+d})$ satisfies the BSDE shown before.

In addition, if $g_0(\cdot)\in L_{\mathscr{F}}^2(0,T)$, then $(y., z.)\in S_{\mathscr{F}}^2(0,T) \times L_{\mathscr{F}}^2(0,T;R^d)$.

Here comes the question: Why $y.\in S_{\mathscr{F}}^2(0,T)$?

We know that:

$L_{\mathscr{F}}^2(0,\tau;R^m)$={$R^m$-valued and $\mathscr{F}_t$-adapted and stochastic processes such that $E[\int_0 ^\tau|\phi_t|^2dt]<\infty$}
$D_{\mathscr{F}}^2(0,\tau;R^m)$={all RCLL(right continuous with left limit) processes in $L_{\mathscr{F}}^2(0,\tau;R^m)$ such that $E[\sup_{0 \leq t \leq \tau}|\phi_t|^2]< \infty$}
$S_{\mathscr{F}}^2(0,\tau;R^m)$ = {all continuous processes in $D_{\mathscr{F}}^2(0,\tau;R^m)$}

Therefore, in order to prove the question, we need to show that $E[\sup_{0\leq t \leq T}|y_t|^2] < \infty$.

The paper suggests to use B-D-G inequality to prove that, so I go to look for it and find something different with the question.

The B-D-G inequality is that: for $g\in L^2(R)$:

$$E[\sup_{0\leq t \leq T}|\int _0 ^t g(s)dB_s|^2]\leq 4E[\int _0 ^T |g(s)|^2 ds]$$

So, how to use B-D-G inequality to prove the question? Do I need to use Jensen’s inequality to transform it first?

Answer 1

By an application of the BDG inequality or Doob's maximal inequality, one can show that $$\mathbb{E}\left[ \sup_{0 \leq t \leq T} |y_t|^2 \right] < \infty$$ This was shown in your other recent question.

What remains for you to show that $y_t \in S_{\mathcal{F}}^2$ is showing that $y_t$ has continuous paths. To see this, simply observe that the function $$t \mapsto \int_t^T g(s)ds$$ is continuous by virtue of being an integral. Finally, recall that stochastic integrals against Brownian motion admit a continuous version, so that for (almost surely) $\omega \in \Omega$ the term $$t \mapsto \int_t^T z_sdB_s$$ is continuous. Altogether, this implies that $y \in S_\mathcal{F}^2$.