$n$ goes to infinity. Similarly, using Burkholder-Davis-Gundy Inequality, we have $$B^n\leq  \E\left[\left(\int_0^T \left\|(Z^n_s - Z_s)\circ \tau_h\right\|^2 ds\right)^\frac12\right]= \E\left[\left(\int_0^T \|Z^n_s - Z_s\|^2 ds\right)^\frac12 \circ \tau_h\right].$$ Therefore, we can use Proposition \[prop.cam\] and Cauchy-Schwarz Inequality, to also deduce that $B^n\underset{n\to +\infty}{\to} 0$. Finally, we have $$\begin{aligned}
C^n=&\ \E\left[\mathcal E\left(\int_0^T\dot{h}(s) \cdot dW_s\right){\left|\int_0^T (Z_s^n-Z_s) \cdot \dot{h}(s)ds\right|}\right]\\
\leq&\ \E\left[\mathcal E\left(\int_0^T\dot{h}(s)\cdot dW_s\right)^2\right]^{1/2}\E\left[\left(\int_0^T\|Z_s^n-Z_s\| \|\dot{h}(s)\| ds\right)^2\right]^{1/2}\\
\leq &\ \E\left[\mathcal E\left(\int_0^T\dot{h}(s) \cdot dW_s\right)^2\right]^{1/2}\E\left[\int_0^T\|Z_s^n-Z_s\|^2ds\right]^{1/2}\left(\int_0^T\|\dot{h}(s)\|^2ds\right)^{1/2},\end{aligned}$$ which also goes to $0$ as $n$ goes to infinity. Therefore the proof is complete.

This result entails the following useful consequence. Let $t$ in $(0,T]$ and $h$ in $H$ such that $\dot{h}_s=(0,\ldots,0)$ for $s\geq t$. Then for any $Z$ in $\H_{ d}^2$, it holds that: $$\label{eq:intshiftprev}
\int_t^T Z_s \cdot dW_s {\circ \tau_{h}} = \int_t^T Z_s {\circ \tau_{h}}\cdot  dW_s, \quad \P_0-a.s.,$$ since $\int_t^T Z_s {\circ \tau_{h}} \cdot \dot{h}(s) ds=0$.

A characterization of Malliavin differentiability {#section:Malliavin}
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Before going further, we would like to recall the main finding of [@Sugita]. Any Malliavin-Sobolev type space $\D^{1,p}$ as defined in Section \[sec:preli\] (originally defined by Malliavin [@Malliavin] and Shigekawa [@Shige]) agrees with the Sobolev space (due to Stroock [@stroock] and Kusuoka [@Kusuoka]) $\tilde{\D}^{1,p}$ consisting in the set of *Ray Absolutely Continuous* (RAC) and *Stochastically Gâteaux Differentiable* (SGD) r.v. $F$ 