Theorem. Suppose $T$ is continuously $\rho_r$-Frechet differentiable at $F$ with the influence function satisfying $0<E[\phi_F(X_1)]^2<\infty$ and that $\int F(x)[1-F(x)]^{r/2}dx<\infty$. Define $$\hat G_n(t)=P_{F_n}\Bigg{(}\sqrt{n}\Big{[}T(F_n^*)-T(F_n)\Big{]}\le t\Bigg{)}\\G_n(t)=P_{F_n}\Bigg{(}\sqrt{n}\Big{[}T(F_n)-T(F)\Big{]}\le t\Bigg{)}$$ where, $F_n$ is the emperical distribution corresponding to $X_1,X_2,...X_n\overset{\text{iid}}{\sim}F$ and $F_n^*$ is the emperical distrubtion corresponding to $X_1^*,...X_n^*\overset{\text{iid}}{\sim}F_n$. Then, $$\rho_r(\hat G_n,G)\overset{as}{\rightarrow}0$$
If specifically, $T$ is continuously $\rho_\infty$-Frechet differentiable at $F$. Then, $$\rho_\infty(\hat G_n,G)\overset{P}{\rightarrow}0$$
I am only interested in the proof for the special case: $T$ is continuously $\rho_\infty$-Frechet differentiable at $F$. Then, (known result) $$T(F_n)-T(F)=\frac{1}{n}\sum_{i=1}^n\phi(X_i)+o_p(n^{-1/2})$$ Let $r_n^*=T(F_n^*)-T(F_n)-\left(\frac{1}{n}\sum_1^n\phi(X_i^*)-\frac{1}{n}\sum_i^n\phi(X_i)\right)$. We first show that $\forall \eta>0$,$$P_*\left\{\sqrt{n}|r_n^*|>\eta\right\}\overset{as}{\rightarrow}0$$ I do not understand how we arrive at the following claim.
From the differentiability of $T$, for any $\epsilon>0$, there is a $\delta_\epsilon>0$ such that $$P_*\left\{\sqrt{n}|r_n^*|>\eta\right\}\le P_*\left\{\sqrt{n}\rho_\infty(F_n^*,F_n)>\eta/\epsilon\right\}+P_*\left\{\rho_\infty(F_n^*,F_n)>\delta_\epsilon\right\}.$$
Definition. $T$ is continuously $\rho$-Frechet differentiable at $G$ if $$\frac{T(H_k)-T(G_k)-L_T(H_k-G_k;G)}{\rho(H_k,G_k)}\rightarrow0$$ for any $G_k,H_k$ satisfying $\rho(H_k,G_k)\rightarrow0,\rho(G_k,G)\rightarrow0$.
Reference. Theorem 3.6 The Jackknife & Bootstrap, Jun Shao
