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

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?

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 $$y \in S_\mathcal{F}^2$$.