[proofplan] We prove the theorem by deriving the uniform asymptotic linear expansion of the empirical quantile map on the compact set $I$. First we use the positive lower bound on the density to obtain local invertibility and uniform control of $F^{-1}$ near $I$. Then we combine the quantile inversion inequalities with the empirical-process representation to show that $F_n^{-1}(t)$ stays uniformly close to $F^{-1}(t)$. Finally, Taylor expansion of $F$ at $F^{-1}(t)$ and stochastic equicontinuity of the empirical process yield the linear approximation, and the Donsker theorem for empirical distribution functions gives the Brownian-bridge limit. [/proofplan]

[step:Localize the quantile map on an interval where $F$ is strictly increasing]Define the deterministic generalized inverse $F^{-1}:(0,1)\to\mathbb{R}$ by \begin{align*} F^{-1}(t)=\inf\{x\in\mathbb{R}:F(x)\ge t\},\qquad t\in(0,1). \end{align*} Define the deterministic quantile function $q:I\to\mathbb{R}$ by $q(t)=F^{-1}(t)$ for $t\in I$. Let $V\subset\mathbb{R}$ be an open neighbourhood of $q(I)$ on which $F$ is continuously differentiable with derivative $f_X$. For $t\in I$, the generalized inverse relation gives $F(q(t)-)\le t\le F(q(t))$, where $F(q(t)-)$ denotes the left limit of $F$ at $q(t)$. Since $q(t)\in V$ and $F$ is continuous at $q(t)$, we have $F(q(t)-)=F(q(t))$. Therefore $F(q(t))=t$ for every $t\in I$. We next prove that $q$ is continuous on $I$. Fix $t_0\in I$ and set $x_0:=q(t_0)$. Since $f_X(x_0)\ge c$ and $f_X$ is continuous at $x_0$, there exists an open interval $J_{t_0}\subset V$ containing $x_0$ such that $f_X(x)\ge c/2$ for every $x\in J_{t_0}$. If $x,y\in J_{t_0}$ with $x\ge y$, the [Mean Value Theorem](/theorems/186) applied to $F|_{J_{t_0}}$ gives \begin{align*} F(x)-F(y)\ge \frac{c}{2}(x-y). \end{align*} Thus $F|_{J_{t_0}}$ is strictly increasing. Choose $\rho_{t_0}>0$ such that $[x_0-\rho_{t_0},x_0+\rho_{t_0}]\subset J_{t_0}$, and define \begin{align*} \delta_{t_0}:=\min\{F(x_0+\rho_{t_0})-F(x_0),F(x_0)-F(x_0-\rho_{t_0})\}>0. \end{align*} If $s\in I$ and $|s-t_0|<\delta_{t_0}$, then $F(x_0-\rho_{t_0})<s<F(x_0+\rho_{t_0})$. Since $F(q(s))=s$, global monotonicity of the distribution function $F$ rules out $q(s)<x_0-\rho_{t_0}$, for then $F(q(s))\le F(x_0-\rho_{t_0})<s$. It also rules out $q(s)>x_0+\rho_{t_0}$, for then $F(q(s))\ge F(x_0+\rho_{t_0})>s$. Hence $q(s)\in[x_0-\rho_{t_0},x_0+\rho_{t_0}]$. Letting $\rho_{t_0}\downarrow0$ within such intervals proves continuity of $q$ at $t_0$. Since $I$ is compact and $q$ is continuous, $K:=q(I)\subset\mathbb{R}$ is compact. Because $f_X$ is continuous and $f_X\ge c$ on $K$, there exists an open neighbourhood $W\subset V$ of $K$ such that $f_X(x)\ge c/2$ for every $x\in W$. By compactness of $K$, choose $\varepsilon_0>0$ such that \begin{align*} \{x\in\mathbb{R}:\operatorname{dist}(x,K)<\varepsilon_0\}\subset W. \end{align*} For every $t\in I$ and every $u\in[-\varepsilon_0,\varepsilon_0]$, the interval joining $q(t)$ to $q(t)+u$ is contained in $W$. Hence the [Mean Value Theorem](/theorems/186) gives, for $0\le h\le\varepsilon_0$, \begin{align*} F(q(t)+h)-F(q(t))\ge \frac{c}{2}h. \end{align*} It also gives \begin{align*} F(q(t))-F(q(t)-h)\ge \frac{c}{2}h. \end{align*}[/step]

[guided]We first make the inverse convention explicit. The deterministic generalized inverse is the map $F^{-1}:(0,1)\to\mathbb{R}$ defined by \begin{align*} F^{-1}(t)=\inf\{x\in\mathbb{R}:F(x)\ge t\},\qquad t\in(0,1), \end{align*} and $q:I\to\mathbb{R}$ is the restriction $q(t)=F^{-1}(t)$. Let $V\subset\mathbb{R}$ be an open neighbourhood of $q(I)$ on which $F$ is continuously differentiable with derivative $f_X$. For $t\in I$, the generalized inverse relation gives $F(q(t)-)\le t\le F(q(t))$. Since $q(t)\in V$ and $F$ is continuous at $q(t)$, the left limit equals the value, so $F(q(t))=t$. We next prove that $q$ is continuous. Fix $t_0\in I$ and set $x_0=q(t_0)$. Since $f_X(x_0)\ge c$ and $f_X$ is continuous at $x_0$, there is an open interval $J_{t_0}\subset V$ containing $x_0$ such that $f_X(x)\ge c/2$ for every $x\in J_{t_0}$. If $x,y\in J_{t_0}$ and $x\ge y$, the [Mean Value Theorem](/theorems/186) applies to $F|_{J_{t_0}}$ because this restriction is continuously differentiable, and gives \begin{align*} F(x)-F(y)\ge \frac{c}{2}(x-y). \end{align*} Thus $F$ is strictly increasing on $J_{t_0}$. Choose $\rho_{t_0}>0$ with $[x_0-\rho_{t_0},x_0+\rho_{t_0}]\subset J_{t_0}$ and define \begin{align*} \delta_{t_0}=\min\{F(x_0+\rho_{t_0})-F(x_0),F(x_0)-F(x_0-\rho_{t_0})\}>0. \end{align*} If $s\in I$ and $|s-t_0|<\delta_{t_0}$, then $F(x_0-\rho_{t_0})<s<F(x_0+\rho_{t_0})$. Since $F(q(s))=s$ and $F$ is monotone, neither $q(s)<x_0-\rho_{t_0}$ nor $q(s)>x_0+\rho_{t_0}$ can occur. Hence $q(s)\in[x_0-\rho_{t_0},x_0+\rho_{t_0}]$. Letting the radius shrink within admissible intervals proves continuity of $q$ at $t_0$. Because $I$ is compact and $q$ is continuous, $K=q(I)$ is compact. The function $f_X$ is continuous and bounded below by $c$ on $K$, so there is an open neighbourhood $W\subset V$ of $K$ such that $f_X(x)\ge c/2$ for every $x\in W$. By compactness of $K$, choose $\varepsilon_0>0$ such that \begin{align*} \{x\in\mathbb{R}:\operatorname{dist}(x,K)<\varepsilon_0\}\subset W. \end{align*} For $t\in I$ and $0\le h\le\varepsilon_0$, the segments from $q(t)$ to $q(t)+h$ and from $q(t)-h$ to $q(t)$ are contained in $W$. Applying the [Mean Value Theorem](/theorems/186) on those segments gives \begin{align*} F(q(t)+h)-F(q(t))\ge \frac{c}{2}h \end{align*} and \begin{align*} F(q(t))-F(q(t)-h)\ge \frac{c}{2}h. \end{align*} These are the uniform one-sided lower bounds needed for quantile inversion.[/guided]

[step:Prove uniform consistency of the empirical quantiles on $I$]Let $(\Omega,\mathcal F,\mathbb{P})$ be the probability space on which the sample is defined, and let $\mathcal B(\mathbb{R})$ denote the Borel $\sigma$-algebra on $\mathbb{R}$. Let $X_1,X_2,\dots:(\Omega,\mathcal F)\to(\mathbb{R},\mathcal B(\mathbb{R}))$ denote the i.i.d. sample with distribution function $F$. For $n\in\mathbb{N}$, define the empirical distribution function $F_n:\mathbb{R}\to[0,1]$ by \begin{align*} F_n(x)=\frac{1}{n}\sum_{j=1}^{n}\mathbb{1}_{(-\infty,x]}(X_j),\qquad x\in\mathbb{R}, \end{align*} and define the empirical quantile function $F_n^{-1}:(0,1)\to\mathbb{R}$ by \begin{align*} F_n^{-1}(t)=\inf\{x\in\mathbb{R}:F_n(x)\ge t\},\qquad t\in(0,1). \end{align*} Let $Q_n:I\to\mathbb{R}$ be the empirical quantile process restricted to $I$, defined by $Q_n(t)=F_n^{-1}(t)$ for $t\in I$. We show that \begin{align*} \sup_{t\in I}|Q_n(t)-q(t)|\xrightarrow{\mathbb{P}}0, \end{align*} meaning that for every $\eta>0$, $\mathbb{P}(\sup_{t\in I}|Q_n(t)-q(t)|>\eta)\to0$. Fix $\varepsilon>0$ with $\varepsilon\le\varepsilon_0$, where $\varepsilon_0$ is the tubular-neighbourhood radius constructed in the localization step. The lower bound from the previous step gives, uniformly in $t\in I$, \begin{align*} F(q(t)+\varepsilon)-t &=F(q(t)+\varepsilon)-F(q(t)) \ge \frac{c}{2}\varepsilon. \end{align*} It also gives \begin{align*} t-F(q(t)-\varepsilon) &=F(q(t))-F(q(t)-\varepsilon) \ge \frac{c}{2}\varepsilon. \end{align*} On the event \begin{align*} \sup_{x\in\mathbb{R}}|F_n(x)-F(x)|<\frac{c\varepsilon}{4}, \end{align*} we have \begin{align*} F_n(q(t)+\varepsilon)>t \end{align*} and \begin{align*} F_n(q(t)-\varepsilon)<t \end{align*} for every $t\in I$. By the definition of the generalized inverse, these two inequalities imply \begin{align*} q(t)-\varepsilon\le Q_n(t)\le q(t)+\varepsilon,\qquad t\in I. \end{align*} The [Glivenko--Cantelli theorem](/theorems/2004) for empirical distribution functions gives \begin{align*} \sup_{x\in\mathbb{R}}|F_n(x)-F(x)|\xrightarrow{\mathbb{P}}0. \end{align*} Equivalently, this also follows from the empirical-process functional [central limit theorem](/theorems/521) invoked below, because tight convergence of $\sqrt n(F_n-F)$ in $\ell^\infty(\mathbb{R})$ implies $\|F_n-F\|_{\ell^\infty(\mathbb{R})}=O_{\mathbb{P}}(n^{-1/2})$. Therefore the event above has probability tending to $1$, and the displayed uniform consistency follows.[/step]

[guided]The point of this step is to turn uniform closeness of distribution functions into uniform closeness of their inverse functions. We first recall the objects used in the argument. The deterministic quantile map is $q:I\to\mathbb{R}$, $q(t)=F^{-1}(t)$, and the empirical quantile process is $Q_n:I\to\mathbb{R}$, $Q_n(t)=F_n^{-1}(t)$. From the localization step, there are an open neighbourhood $W\subset\mathbb{R}$ of $q(I)$ and a number $\varepsilon_0>0$ such that every segment joining $q(t)$ to $q(t)+u$, with $t\in I$ and $|u|\le\varepsilon_0$, lies in $W$, and $f_X(x)\ge c/2$ for every $x\in W$. Also $F(q(t))=t$ for every $t\in I$. We prove $\sup_{t\in I}|Q_n(t)-q(t)|\xrightarrow{\mathbb{P}}0$. Choose $\varepsilon>0$ with $\varepsilon\le\varepsilon_0$. Since $F$ is continuously differentiable on $W$ and $f_X\ge c/2$ on $W$, the [Mean Value Theorem](/theorems/186) applied on the segment from $q(t)$ to $q(t)+\varepsilon$ gives, uniformly in $t\in I$, \begin{align*} F(q(t)+\varepsilon)-t=F(q(t)+\varepsilon)-F(q(t))\ge \frac{c}{2}\varepsilon. \end{align*} Applying the same theorem on the segment from $q(t)-\varepsilon$ to $q(t)$ gives \begin{align*} t-F(q(t)-\varepsilon)=F(q(t))-F(q(t)-\varepsilon)\ge \frac{c}{2}\varepsilon. \end{align*} Now suppose that the empirical distribution function is uniformly within $c\varepsilon/4$ of $F$: \begin{align*} \sup_{x\in\mathbb{R}}|F_n(x)-F(x)|<\frac{c\varepsilon}{4}. \end{align*} Then \begin{align*} F_n(q(t)+\varepsilon) \ge F(q(t)+\varepsilon)-\frac{c\varepsilon}{4} > t. \end{align*} Also, \begin{align*} F_n(q(t)-\varepsilon) \le F(q(t)-\varepsilon)+\frac{c\varepsilon}{4} < t. \end{align*} The generalized inverse is the first point where the empirical distribution reaches level $t$. Since $F_n$ is below $t$ at $q(t)-\varepsilon$ and above $t$ at $q(t)+\varepsilon$, monotonicity of $F_n$ and the definition of $Q_n(t)=F_n^{-1}(t)$ give \begin{align*} q(t)-\varepsilon\le Q_n(t)\le q(t)+\varepsilon. \end{align*} This holds for every $t\in I$ on the same event, so \begin{align*} \sup_{t\in I}|Q_n(t)-q(t)|\le \varepsilon. \end{align*} The [Glivenko--Cantelli theorem](/theorems/2004) for empirical distribution functions gives \begin{align*} \sup_{x\in\mathbb{R}}|F_n(x)-F(x)|\xrightarrow{\mathbb{P}}0. \end{align*} Thus the probability of the event above tends to $1$, proving uniform consistency.[/guided]

[step:Linearize $F$ uniformly along the empirical quantiles]Define the deterministic reciprocal density map $a:I\to(0,\infty)$ by \begin{align*} a(t)=\frac{1}{f_X(q(t))},\qquad t\in I. \end{align*} Since $f_X$ is continuous and bounded below by $c$ on $K=q(I)$, the function $a$ is bounded and continuous on $I$. By uniform consistency, with probability tending to $1$ all points $Q_n(t)$ lie in $W$. On that event, the one-dimensional mean-value secant formula for the continuously differentiable map $F|_W:W\to\mathbb{R}$ gives, for every $t\in I$, \begin{align*} F(Q_n(t))-F(q(t)) = f_X(q(t))(Q_n(t)-q(t)) + r_n(t)(Q_n(t)-q(t)), \end{align*} where $r_n:I\to\mathbb{R}$ is the random remainder function defined as follows: for $t\in I$ with $Q_n(t)\ne q(t)$, \begin{align*} r_n(t)=\frac{F(Q_n(t))-F(q(t))}{Q_n(t)-q(t)}-f_X(q(t)), \end{align*} and for $t\in I$ with $Q_n(t)=q(t)$, set $r_n(t)=0$. If $Q_n(t)\ne q(t)$, the secant quotient equals $f_X(\xi_n(t))$ for some point $\xi_n(t)$ in the closed interval with endpoints $q(t)$ and $Q_n(t)$, by the [Mean Value Theorem](/theorems/186). This function satisfies \begin{align*} \sup_{t\in I}|r_n(t)|\xrightarrow{\mathbb{P}}0. \end{align*} The convergence of $r_n$ follows from the [uniform continuity](/page/Uniform%20Continuity) of $f_X$ on a compact neighbourhood of $K$ and the [uniform convergence](/page/Uniform%20Convergence) $Q_n-q\to 0$ in probability.[/step]

[guided]The goal of this step is to replace the nonlinear increment $F(Q_n(t))-F(q(t))$ by its first-order part. Define $a:I\to(0,\infty)$ by \begin{align*} a(t)=\frac{1}{f_X(q(t))},\qquad t\in I. \end{align*} Because $q(I)=K$ is compact, $f_X$ is continuous on $K$, and $f_X\ge c$ on $K$, the reciprocal density map $a$ is bounded and continuous on $I$. Uniform consistency gives $\sup_{t\in I}|Q_n(t)-q(t)|\xrightarrow{\mathbb{P}}0$. Hence, with probability tending to $1$, all points $Q_n(t)$ lie in the neighbourhood $W$ where $F$ is continuously differentiable. On that event, fix $t\in I$. If $Q_n(t)\ne q(t)$, the [Mean Value Theorem](/theorems/186) applies to $F$ on the closed interval with endpoints $q(t)$ and $Q_n(t)$, because that interval is contained in $W$ on the event under consideration. Thus the secant quotient equals $f_X(\xi_n(t))$ for some point $\xi_n(t)$ between $q(t)$ and $Q_n(t)$. Define $r_n:I\to\mathbb{R}$ by \begin{align*} r_n(t)=\frac{F(Q_n(t))-F(q(t))}{Q_n(t)-q(t)}-f_X(q(t)) \end{align*} when $Q_n(t)\ne q(t)$, and set $r_n(t)=0$ when $Q_n(t)=q(t)$. Then, for every $t\in I$, \begin{align*} F(Q_n(t))-F(q(t)) = f_X(q(t))(Q_n(t)-q(t))+r_n(t)(Q_n(t)-q(t)). \end{align*} It remains to see that the remainder is uniformly small. Since $K$ is compact and $Q_n$ is uniformly close to $q$ in probability, the points between $q(t)$ and $Q_n(t)$ eventually lie in a fixed compact neighbourhood of $K$. The density $f_X$ is uniformly continuous on that compact neighbourhood. Therefore \begin{align*} \sup_{t\in I}|r_n(t)|\xrightarrow{\mathbb{P}}0. \end{align*} This is the uniform $C^1$ linearization needed for the delta-method argument.[/guided]

[step:Use the quantile inversion inequalities to identify the empirical-process term]Choose $\varepsilon_1\in(0,\varepsilon_0)$ such that the compact set \begin{align*} K_1:=\{x\in\mathbb{R}:\operatorname{dist}(x,K)\le \varepsilon_1\} \end{align*} is contained in $W$. Define the local empirical distribution function process $G_n:K_1\to\mathbb{R}$ by $G_n(x)=\sqrt n\{F_n(x)-F(x)\}$ for $x\in K_1$. Define the semimetric $d_F:K_1\times K_1\to[0,\infty)$ by $d_F(x,y)=|F(x)-F(y)|$. The indexed class \begin{align*} \mathcal{C}_{K_1}:=\{(-\infty,x]\cap K_1:x\in K_1\} \end{align*} is a subclass of the half-line class and is therefore a Donsker class for the i.i.d. sample $(X_j)_{j\ge1}$. By the Donsker theorem for empirical distribution functions, applied to the measurable indicators $\mathbb{1}_{(-\infty,x]}$ with $x\in K_1$, and using that $F$ is continuous on $K_1$, the process $G_n$ is asymptotically uniformly equicontinuous with respect to $d_F$ on the fixed compact set $K_1$. The same theorem gives convergence of $G_n\circ q$ in distribution in $\ell^\infty(I)$ to the Brownian bridge process $B$ restricted to $I$. We use the following stochastic-equicontinuity consequence of the same Donsker theorem; this is the standard random-index form of asymptotic equicontinuity in empirical-process theory, and the verification below records the reduction to deterministic $d_F$-balls. If $A_n:I\to W$ and $C_n:I\to W$ are random maps, defined on events whose probabilities tend to one, and \begin{align*} \sup_{t\in I}d_F(A_n(t),C_n(t))\xrightarrow{\mathbb{P}}0, \end{align*} then \begin{align*} \sup_{t\in I}|G_n(A_n(t))-G_n(C_n(t))|\xrightarrow{\mathbb{P}}0. \end{align*} To justify the random indexing, view $A_n$ and $C_n$ as maps into the totally bounded semimetric space $(K_1,d_F)$ on the high-probability event where their images lie in $K_1$; the displayed convergence says that the random index pairs are $d_F$-asymptotically diagonal uniformly over $I$. If measurability of a supremum is not automatic, the convergence statements are understood in outer probability, which is the standard formulation of asymptotic equicontinuity in $\ell^\infty$ empirical-process theory. For every $\eta>0$ and $\delta>0$, split the probability according to whether $\sup_{t\in I}d_F(A_n(t),C_n(t))\le\delta$. On that event the displayed supremum is bounded by $\sup\{|G_n(x)-G_n(y)|:x,y\in K_1,\ d_F(x,y)\le\delta\}$. Taking first $n\to\infty$ and then $\delta\downarrow0$ gives convergence to zero by asymptotic uniform equicontinuity. For $x\in\mathbb{R}$, define $F_n(x-)$ to be the left limit $\lim_{y\uparrow x}F_n(y)$. The generalized inverse inequalities give $F_n(Q_n(t)-)\le t\le F_n(Q_n(t))$ for $t\in I$. On the event $Q_n(I)\subset K_1$, the distribution of each observation has no atoms in $K_1$ because $F$ is continuous there. For any distinct indices $j,k\in\{1,\dots,n\}$, the event $\{X_j=X_k\in W\}$ has probability zero; taking the finite union over pairs shows that, with probability one, no empirical atom in $W$ has multiplicity greater than one. Hence each jump of $F_n$ inside $W$ has size at most the displayed quantity \begin{align*} \frac{1}{n}. \end{align*} Therefore $\sup_{t\in I}|F_n(Q_n(t))-t|\le n^{-1}$ on this event, and $\sup_{t\in I}|\sqrt n\{F_n(Q_n(t))-t\}|\xrightarrow{\mathbb{P}}0$. For every $t\in I$ on the event $Q_n(I)\subset K_1$, the identity $F_n(Q_n(t))-F(Q_n(t))=n^{-1/2}G_n(Q_n(t))$ gives \begin{align*} \sup_{t\in I}\left|\sqrt n\{F(Q_n(t))-t\}+G_n(Q_n(t))\right|\xrightarrow{\mathbb{P}}0. \end{align*} Since $Q_n-q\to0$ uniformly in probability and $F$ is uniformly continuous near $K$, we have \begin{align*} \sup_{t\in I}d_F(Q_n(t),q(t)) = \sup_{t\in I}|F(Q_n(t))-F(q(t))| \xrightarrow{\mathbb{P}}0. \end{align*} The stochastic-equicontinuity consequence just stated therefore applies to the two random maps $Q_n:I\to K_1$ and $q:I\to K_1$ on events whose probabilities tend to one, and yields \begin{align*} \sup_{t\in I}|G_n(Q_n(t))-G_n(q(t))|\xrightarrow{\mathbb{P}}0. \end{align*} Consequently, if $\alpha_n:I\to\mathbb{R}$ is defined by $\alpha_n(t)=G_n(q(t))=\sqrt n\{F_n(q(t))-t\}$, then $\sup_{t\in I}|\sqrt n\{F(Q_n(t))-t\}+\alpha_n(t)|\xrightarrow{\mathbb{P}}0$.[/step]

[guided]The purpose of this step is to replace the empirical process evaluated at the random point $Q_n(t)$ by the same empirical process evaluated at the deterministic point $q(t)$. We first fix the index set carefully. Choose $\varepsilon_1\in(0,\varepsilon_0)$ such that \begin{align*} K_1=\{x\in\mathbb{R}:\operatorname{dist}(x,K)\le\varepsilon_1\}\subset W. \end{align*} This set is compact, contains $q(I)$, and contains $Q_n(I)$ with probability tending to $1$ by uniform consistency. Define $G_n:K_1\to\mathbb{R}$ by \begin{align*} G_n(x)=\sqrt n\{F_n(x)-F(x)\},\qquad x\in K_1, \end{align*} and define $d_F:K_1\times K_1\to[0,\infty)$ by \begin{align*} d_F(x,y)=|F(x)-F(y)|. \end{align*} The Donsker theorem for empirical distribution functions applies to the half-line indicators $\mathbb{1}_{(-\infty,x]}$ indexed by $x\in K_1$. Since $F$ is continuous on $K_1$, it gives asymptotic uniform equicontinuity of $G_n$ with respect to $d_F$ on $K_1$, and also gives $G_n\circ q\xrightarrow{d}B$ in $\ell^\infty(I)$. We now record the random-index consequence. If $A_n:I\to K_1$ and $C_n:I\to K_1$ satisfy \begin{align*} \sup_{t\in I}d_F(A_n(t),C_n(t))\xrightarrow{\mathbb{P}}0, \end{align*} then \begin{align*} \sup_{t\in I}|G_n(A_n(t))-G_n(C_n(t))|\xrightarrow{\mathbb{P}}0. \end{align*} Indeed, on the event $\sup_{t\in I}d_F(A_n(t),C_n(t))\le\delta$, the last supremum is bounded by \begin{align*} \sup\{|G_n(x)-G_n(y)|:x,y\in K_1,\ d_F(x,y)\le\delta\}. \end{align*} Taking $n\to\infty$ and then $\delta\downarrow0$ gives convergence to zero by asymptotic uniform equicontinuity. Supremum measurability is understood in outer probability when necessary, which is the standard empirical-process formulation. For the quantile term, the generalized inverse inequalities give \begin{align*} F_n(Q_n(t)-)\le t\le F_n(Q_n(t)),\qquad t\in I. \end{align*} On the event $Q_n(I)\subset K_1$, the distribution has no atoms in $K_1$ because $F$ is continuous there. Hence, with probability one, no empirical jump in $K_1$ has multiplicity greater than one, so every such jump has size at most $1/n$. Therefore \begin{align*} \sup_{t\in I}|F_n(Q_n(t))-t|\le n^{-1}, \end{align*} and consequently \begin{align*} \sup_{t\in I}|\sqrt n\{F_n(Q_n(t))-t\}|\xrightarrow{\mathbb{P}}0. \end{align*} Using $F_n(Q_n(t))-F(Q_n(t))=n^{-1/2}G_n(Q_n(t))$, we obtain \begin{align*} \sup_{t\in I}\left|\sqrt n\{F(Q_n(t))-t\}+G_n(Q_n(t))\right|\xrightarrow{\mathbb{P}}0. \end{align*} Since $Q_n\to q$ uniformly in probability and $F$ is uniformly continuous on $K_1$, \begin{align*} \sup_{t\in I}d_F(Q_n(t),q(t))\xrightarrow{\mathbb{P}}0. \end{align*} The random-index equicontinuity result with $A_n=Q_n$ and $C_n=q$ yields \begin{align*} \sup_{t\in I}|G_n(Q_n(t))-G_n(q(t))|\xrightarrow{\mathbb{P}}0. \end{align*} Thus, for $\alpha_n:I\to\mathbb{R}$ defined by \begin{align*} \alpha_n(t)=G_n(q(t))=\sqrt n\{F_n(q(t))-t\}, \end{align*} we have \begin{align*} \sup_{t\in I}|\sqrt n\{F(Q_n(t))-t\}+\alpha_n(t)|\xrightarrow{\mathbb{P}}0. \end{align*}[/guided]

[step:Combine the linearization with Donsker convergence]Since $F(q(t))=t$, the Taylor expansion from the linearization step gives \begin{align*} \sqrt n\{F(Q_n(t))-t\} = f_X(q(t))\sqrt n\{Q_n(t)-q(t)\} + r_n(t)\sqrt n\{Q_n(t)-q(t)\}. \end{align*} The previous step shows that $\sqrt n\{F(Q_n(t))-t\}$ is uniformly equal to $-\alpha_n(t)+o_{\mathbb{P}}(1)$ on $I$. Since $(\alpha_n)$ is tight in $\ell^\infty(I)$, the family $\sqrt n\{F(Q_n)-t\}$ is tight in $\ell^\infty(I)$. On the event $\sup_{t\in I}|r_n(t)|<c/2$, the Taylor identity and the lower bound $f_X(q(t))\ge c$ give \begin{align*} \frac{c}{2}\|\sqrt n(Q_n-q)\|_{\ell^\infty(I)}\le \|\sqrt n\{F(Q_n)-t\}\|_{\ell^\infty(I)}. \end{align*} Because $\sup_{t\in I}|r_n(t)|\xrightarrow{\mathbb{P}}0$, this absorption estimate proves that $\sqrt n(Q_n-q)$ is tight in $\ell^\infty(I)$ without using the desired conclusion. Combining this tightness with $\sup_{t\in I}|r_n(t)|\to0$ in probability gives \begin{align*} \sup_{t\in I} \left| \sqrt n\{Q_n(t)-q(t)\} + \frac{\alpha_n(t)}{f_X(q(t))} \right| \xrightarrow{\mathbb{P}}0. \end{align*} The multiplier map $M:\ell^\infty(I)\to\ell^\infty(I)$ defined by \begin{align*} M(h)(t)=-\frac{h(t)}{f_X(q(t))},\qquad h\in\ell^\infty(I),\quad t\in I, \end{align*} is continuous and linear, because $a(t)=1/f_X(q(t))$ is bounded on $I$. By the empirical distribution function Donsker theorem and the [Continuous Mapping Theorem](/theorems/1847), \begin{align*} M(\alpha_n)\xrightarrow{d}M(B) \end{align*} in $\ell^\infty(I)$. The asymptotic linear expansion and Slutsky's theorem therefore imply \begin{align*} \sqrt n\{F_n^{-1}(t)-F^{-1}(t)\}_{t\in I} \xrightarrow{d} \left\{-\frac{B(t)}{f_X(F^{-1}(t))}\right\}_{t\in I}. \end{align*} This is the desired convergence in $\ell^\infty(I)$.[/step]

[guided]We now assemble the preceding estimates in a way that keeps the limiting map visible. Since $F(q(t))=t$ for every $t\in I$, the $C^1$ secant expansion obtained above gives \begin{align*} \sqrt n\{F(Q_n(t))-t\} = f_X(q(t))\sqrt n\{Q_n(t)-q(t)\} + r_n(t)\sqrt n\{Q_n(t)-q(t)\}. \end{align*} The stochastic-equicontinuity step proved \begin{align*} \sup_{t\in I}\left|\sqrt n\{F(Q_n(t))-t\}+\alpha_n(t)\right|\xrightarrow{\mathbb{P}}0, \end{align*} where $\alpha_n:I\to\mathbb{R}$ is the empirical process along the deterministic quantile curve, \begin{align*} \alpha_n(t)=G_n(q(t))=\sqrt n\{F_n(q(t))-F(q(t))\}=\sqrt n\{F_n(q(t))-t\}. \end{align*} The Donsker theorem for empirical distribution functions gives $\alpha_n\xrightarrow{d}B$ in $\ell^\infty(I)$, where $B$ is the Brownian bridge restricted to $I$. This invocation uses the i.i.d. sample $(X_j)_{j\ge1}$ and the half-line indicator class indexed by $q(I)\subset W$. In particular, $(\alpha_n)$ is tight in $\ell^\infty(I)$, and the preceding display implies that $\sqrt n\{F(Q_n)-t\}$ is tight in $\ell^\infty(I)$. We still need to make sure the Taylor remainder does not hide the unknown quantity $\sqrt n(Q_n-q)$. On the event $\sup_{t\in I}|r_n(t)|<c/2$, the lower bound $f_X(q(t))\ge c$ gives, for every $t\in I$, \begin{align*} |\sqrt n\{F(Q_n(t))-t\}| \ge \left(f_X(q(t))-|r_n(t)|\right)|\sqrt n\{Q_n(t)-q(t)\}| \ge \frac{c}{2}|\sqrt n\{Q_n(t)-q(t)\}|. \end{align*} Taking suprema over $t\in I$ yields \begin{align*} \frac{c}{2}\|\sqrt n(Q_n-q)\|_{\ell^\infty(I)} \le \|\sqrt n\{F(Q_n)-t\}\|_{\ell^\infty(I)}. \end{align*} Since $\sup_{t\in I}|r_n(t)|\xrightarrow{\mathbb{P}}0$, this proves that $\sqrt n(Q_n-q)$ is tight. Multiplying the secant identity by the bounded reciprocal density $a(t)=1/f_X(q(t))$ and using tightness gives \begin{align*} \sup_{t\in I} \left| \sqrt n\{Q_n(t)-q(t)\} + \frac{\alpha_n(t)}{f_X(q(t))} \right| \xrightarrow{\mathbb{P}}0. \end{align*} Define the multiplier map $M:\ell^\infty(I)\to\ell^\infty(I)$ by \begin{align*} M(h)(t)=-\frac{h(t)}{f_X(q(t))},\qquad h\in\ell^\infty(I),\quad t\in I. \end{align*} This map is linear and continuous because $1/f_X(q(t))$ is bounded on the compact set $I$. The [Continuous Mapping Theorem](/theorems/1847) applied to $M$ gives $M(\alpha_n)\xrightarrow{d}M(B)$ in $\ell^\infty(I)$. The preceding supremum-norm approximation tends to zero in probability, so Slutsky's theorem implies \begin{align*} \sqrt n\{F_n^{-1}(t)-F^{-1}(t)\}_{t\in I} \xrightarrow{d} \left\{-\frac{B(t)}{f_X(F^{-1}(t))}\right\}_{t\in I}. \end{align*} This is exactly the claimed convergence in $\ell^\infty(I)$.[/guided]