[step:Pass the conditional bootstrap law through the derivative]Let $\mathrm{BL}_1(\mathbb R)$ denote the set of all functions $\varphi:\mathbb R\to\mathbb R$ such that $|\varphi(u)|\le1$ for all $u\in\mathbb R$ and
\begin{align*}
|\varphi(u)-\varphi(v)|\le |u-v|
\end{align*}
for all $u,v\in\mathbb R$. Let $\mathrm{BL}_1(\ell^\infty(\mathcal F))$ denote the set of all functions $\psi:\ell^\infty(\mathcal F)\to\mathbb R$ such that $|\psi(h)|\le1$ for all $h\in\ell^\infty(\mathcal F)$ and
\begin{align*}
|\psi(h)-\psi(k)|\le \|h-k\|_{\ell^\infty(\mathcal F)}
\end{align*}
for all $h,k\in\ell^\infty(\mathcal F)$. Let $(\Omega^*,\mathcal A^*,\mathbb P^*)$ denote the auxiliary bootstrap probability space used conditionally on the observed data $X_1,\dots,X_n$. Let $\mathbb E_n^*[\cdot]$ denote expectation with respect to the auxiliary bootstrap law conditional on the observed data $X_1,\dots,X_n$. Conditional weak convergence in probability of $Z_n^*$ to $G_{P_0}$ means that
\begin{align*}
\sup_{\psi\in\mathrm{BL}_1(\ell^\infty(\mathcal F))}
\left|
\mathbb E_n^*\!\left[\psi(Z_n^*)\right]
-
\mathbb E\!\left[\psi(G_{P_0})\right]
\right|
\to0
\end{align*}
in probability; if measurability is unavailable, $\mathbb E_n^*$ and the displayed supremum are interpreted as outer [conditional expectation](/page/Conditional%20Expectation) and outer bounded-Lipschitz distance.
Define the operator norm constant $L\in[0,\infty)$ by
\begin{align*}
L=\sup\left\{\left|\widetilde{\dot\theta}_{P_0}(h)\right|:h\in\ell^\infty(\mathcal F),\ \|h\|_{\ell^\infty(\mathcal F)}\le1\right\}.
\end{align*} For each $\varphi\in\mathrm{BL}_1(\mathbb R)$, define $\psi_\varphi:\ell^\infty(\mathcal F)\to\mathbb R$ by
\begin{align*}
\psi_\varphi(h)=\varphi\bigl(\widetilde{\dot\theta}_{P_0}(h)\bigr)
\end{align*}
for $h\in\ell^\infty(\mathcal F)$. Then $|\psi_\varphi|\le1$ and
\begin{align*}
|\psi_\varphi(h)-\psi_\varphi(k)|
\le
L\|h-k\|_{\ell^\infty(\mathcal F)}.
\end{align*}
Let $M:=\max\{1,L\}$. If $L>0$, then $M^{-1}\psi_\varphi$ belongs to $\mathrm{BL}_1(\ell^\infty(\mathcal F))$, because its sup-norm is at most $M^{-1}\le1$ and its Lipschitz constant is at most $L/M\le1$. If $L=0$, the map $\widetilde{\dot\theta}_{P_0}$ is zero and the convergence below is immediate because $\psi_\varphi$ is constant. Therefore the conditional bounded-Lipschitz convergence of $Z_n^*$ yields
\begin{align*}
\sup_{\varphi\in\mathrm{BL}_1(\mathbb R)}
\left|
\mathbb E_n^*\!\left[\varphi\bigl(\widetilde{\dot\theta}_{P_0}(Z_n^*)\bigr)\right]
-
\mathbb E\!\left[\varphi\bigl(\widetilde{\dot\theta}_{P_0}(G_{P_0})\bigr)\right]
\right|
\le
M
\sup_{\psi\in\mathrm{BL}_1(\ell^\infty(\mathcal F))}
\left|
\mathbb E_n^*\!\left[\psi(Z_n^*)\right]
-
\mathbb E\!\left[\psi(G_{P_0})\right]
\right|
\to0
\end{align*}
in probability. Since $G_{P_0}\in T$ almost surely, $\widetilde{\dot\theta}_{P_0}(G_{P_0})=\dot\theta_{P_0}(G_{P_0})$.
Since $r_n^*\to0$ in conditional probability, in probability, the conditional form of [Slutsky's Lemma](/theorems/1850) permits replacing $\widetilde{\dot\theta}_{P_0}(Z_n^*)$ by
\begin{align*}
\sqrt n\bigl(\theta(P_n^*)-\theta(P_n)\bigr)
\end{align*}
without changing the conditional weak limit. Therefore, conditionally on the observed data $X_1,\dots,X_n$, the law of
\begin{align*}
\sqrt n\bigl(\theta(P_n^*)-\theta(P_n)\bigr)
\end{align*}
converges in probability to the law of $\dot\theta_{P_0}(G_{P_0})$.[/step]