Distance between SDE solutions with different initial values

Question

I am reading the book of El Karoui & Mazliak "Backward stochastic differential equations". In section 3.1, we have the SDE $$ X^{t,x}_s = x + \int_t^s b(u,X^{t,x}_u) \mathrm du + \int_t^s \sigma(u,X^{t,x}_u) \mathrm dW_u. $$ with $b$ and $\sigma$ being uniformly Lipschitz in the second argument and $$ |b(s,x)|+|\sigma(s,x)| \le C (1 + |x|). $$ In the proof of proposition 3.1, it is stated that "the classical martingale inequalities [Karatzas and Shreve]" (without mentioning specific ones) implied $$ \mathbb E\bigg( \sup_{0\le s\le T} |X^{t,x}_s |^2 \bigg) \le C\big(1 + |x|^2\big) \\ \mathbb E\bigg( \sup_{0\le s\le T} |X^{t,x}_s - X^{t',x'}_s|^2 \bigg) \le C\big(1 + |x|^2\big) \big(|x-x'|^2 + |t-t'|\big). $$ I would like to know which inequalities are used to obtain that. Thank you for your help!

Answer 1

This is not an answer, but is too long for comment.

I will elaborate on my comment. We have for $t\le T$ $$X_t = x + \int_0^t b(X_s)\,d s + \int_0^t \sigma(X_s) \, d W_s $$ in $\mathbb{R}$. In the following $C$ changes from line to line. For each $t \le T$ then, \begin{align} E (X_t^2) &\le C \left[x^2 + E \left\{\left( \int_0^t b(X_s)\,d s\right)^2\right\} +E \left\{\left( \int_0^t \sigma(X_s)\,d W_s\right)^2\right\}\right]&\\ &\le C \left[x^2 + E \left\{t\int_0^t b(X_s)^2\,d s\right\} +E \left\{ \int_0^t \sigma(X_s)^2\,d s\right\}\right] &\\ & = C \left[x^2 + t\int_0^t E(b(X_s)^2)\,d s + \int_0^t E(\sigma(X_s)^2)\,d s\right]&\\ &\le C \left[x^2 + (t+1)\int_0^t \left(1+E(X_s^2)\right)\,d s\right] &\\ &\le C \left[x^2 + t(t+1) + (t+1)\int_0^t E(X_s^2)\,d s\right] & \end{align} where I have used Jensen's inequality and Ito isometry in in the second line and the linear growth conditions in the fourth line. Now apply Gronwall to the above to obtain \begin{align} E(X_t^2) \le C(x^2 + t(t+1))\exp(Ct(t+1)). \end{align} for all $t \le T$. Now \begin{align} E[\sup_{t\le T} X_t^2] &\le C \left[x^2 + E \left\{\sup_{t\le T}\left( \int_0^t b(X_s)\,d s\right)^2\right\} +E \left\{\sup_{t\le T}\left( \int_0^t \sigma(X_s)\,d W_s\right)^2\right\}\right]&\\ & \le C \left[x^2 + E \left\{\sup_{t\le T}\left( t\int_0^t b(X_s)^2\,d s\right)\right\} +E \left\{ \int_0^T \sigma(X_s)^2\,d s\right\}\right]&\\ & \le C \left[x^2 + E \left\{ T\int_0^T b(X_s)^2\,d s \right\} +E \left\{ \int_0^T \sigma(X_s)^2\,d s\right\}\right]&\\ &= C \left[x^2 + T\int_0^T E(b(X_s)^2)\,d s + \int_0^T E(\sigma(X_s)^2)\,d s\right] & \\ & \le C \left[x^2 + 2T+ (T+1)\int_0^T E(X_s^2)\,d s \right] \end{align} where the second line uses BDG and Jensen and the last line uses the linear growth conditions. Now plug in estimate on $E(X_t^2)$ obtained above to this and you have a bound of the form $$ E[\sup_{t\le T} X_t^2] \le C_T (1+x^2).$$