Distance between SDE solutions with different initial values 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!
 A: 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 $$ E[\sup_{t\le T} X_t^2] \le C_T (1+x^2).$$ 
