Editing Openai/6922876a-7988-8007-9c62-5f71772af6aa (section)

===== \begin{lemma}[Displacement W2W_2W2 control]\label{lem:dyn-W2-correct} =====
Let P,QP,QP,Q be probability laws on R\RR with finite second moments, and TTT the increasing transport map pushing PPP to QQQ. For t∈[0,1]t\in[0,1]t∈[0,1], define Tt(x)=(1−t)x+tT(x)T_t(x)=(1-t)x+tT(x)Tt(x)=(1−t)x+tT(x) and Xt=Tt(X)X_t=T_t(X)Xt=Tt(X) with X∼PX\sim PX∼P; write PtP_tPt for the law of XtX_tXt.
For any C1C^1C1 function φ\varphiφ such that ∫01\EPt[φ′(Xt)2] dt<∞\int_0^1 \E_{P_t}[\varphi'(X_t)^2]\,dt<\infty∫01\EPt[φ′(Xt)2]dt<∞,

∣\EQ[φ]−\EP[φ]∣  ≤  (∫01\EPt[φ′(Xt)2] dt)1/2 W2(P,Q).\big|\E_Q[\varphi]-\E_P[\varphi]\big|
\;\le\;
\Big(\int_0^1 \E_{P_t}[\varphi'(X_t)^2]\,dt\Big)^{1/2}\, W_2(P,Q).\EQ[φ]−\EP[φ]≤(∫01\EPt[φ′(Xt)2]dt)1/2W2(P,Q).
\end{lemma}

\begin{proof}
By the chain rule and the fact that X˙t=T(X)−X\dot X_t=T(X)-XX˙t=T(X)−X,

ddt\E[φ(Xt)]=\E[φ′(Xt) (T(X)−X)].\frac{d}{dt}\E[\varphi(X_t)]=\E[\varphi'(X_t)\,(T(X)-X)].dtd\E[φ(Xt)]=\E[φ′(Xt)(T(X)−X)].
Integrating over t∈[0,1]t\in[0,1]t∈[0,1] and using Cauchy–Schwarz in (t,Ω)(t,\Omega)(t,Ω) gives

∣\EQ[φ]−\EP[φ]∣≤(∫01\E[φ′(Xt)2] dt)1/2(∫01\E[(T(X)−X)2] dt)1/2.\big|\E_Q[\varphi]-\E_P[\varphi]\big|
\le
\Big(\int_0^1 \E[\varphi'(X_t)^2]\,dt\Big)^{1/2}
\Big(\int_0^1 \E[(T(X)-X)^2]\,dt\Big)^{1/2}.\EQ[φ]−\EP[φ]≤(∫01\E[φ′(Xt)2]dt)1/2(∫01\E[(T(X)−X)2]dt)1/2.
The second factor equals W2(P,Q)W_2(P,Q)W2(P,Q), and \E[φ′(Xt)2]=\EPt[φ′(Z)2]\E[\varphi'(X_t)^2]=\E_{P_t}[\varphi'(Z)^2]\E[φ′(Xt)2]=\EPt[φ′(Z)2].
\end{proof}