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A real functional $\Phi$ is Hadamard differentiable at $s_0$ if there exists a continuous linear functional $L={\dot{\Phi }}_{s_0}:S\to \mathbb{R}$ such that, whenever $t_j\ne 0$, $t_j\to 0$ and $h_j\to h$ in the normed vector space,If $\Phi$ is defined only on a subset, require $s_0+t_j{h}_j$ to lie in its domain. The variation of the directions $h_j$ is essential: convergence along each fixed direction alone is a weaker differentiability condition. Tangential Hadamard differentiability restricts the limit directions to a specified subspace; here the derivative is defined on all of $S$.

We prove the functional delta method using compact sets, so do not need an unjustified application of the Skorokhod representation theorem on an arbitrary nonseparable normed vector space. As usual for function-space convergence in distribution, the limiting random element is tight, and the real statistics are measurable, or convergence is understood through outer expectations. These are automatic under the usual separable complete random-element formulation; the empirical-process formulation used below has a tight separably supported limit.

For $t\ne 0$ putHadamard differentiability says $R_{t_j}(h_j)\to 0$ for every convergent sequence of directions, since continuity gives $L(h_j)\to L(h)$. It implies a Hadamard remainder near a compact set: for every compact $K\subset S$ and $ϵ>0$ there are $a>0$ and $\delta >0$ such thatTo prove it, suppose it failed. Choose $t_j\to 0$ and $h_j$ at distance less than $1/j$ from $K$ with $|R_{t_j}(h_j)|\ge ϵ$. Choose $k_j\in K$ with $\Vert h_j-k_j\Vert <2/j$. A subsequence of $k_j$ converges by compactness, so the corresponding $h_j$ converge too. The defining derivative limit contradicts the lower bound on $R_{t_j}(h_j)$.

Now let $Z_n=r_n(X_n-s_0)\stackrel{d}{\to }X$ and $t_n=1/r_n$. Given $\eta >0$, tightness of a probability measure provides a compact $K$ with $\mathbb{P}(X\in K)>1-\eta$. For the open $\delta$-neighbourhood $K^{\delta }$, the Portmanteau theorem givesOn $K^{\delta }$ the preceding uniform remainder is less than $ϵ$ for all sufficiently large $n$. Consequently ${\mathrm{limsup}}_n\mathbb{P}(|R_{t_n}(Z_n)|\ge ϵ)\le \eta$. Since $\eta$ is arbitrary, this remainder tends to zero in probability. We have proved the actual expansionContinuity of $L$ and the continuous mapping theorem give $L(Z_n)\stackrel{d}{\to }L(X)$. The vanishing remainder and Slutsky theorem therefore yieldThis derivation uses the derivative definition, rather than assuming a uniform Fréchet remainder over every bounded set. If outer expectations are needed, the same compact-neighbourhood argument uses outer probabilities and gives the corresponding empirical-process functional delta method.

For the statistical application, use the sample's empirical distribution function and the plug-in estimatorHere the printed $L^{\infty }$ denotes bounded pointwise functions with the supremum norm, namely ${\ell }^{\infty }(\mathbb{R})$, rather than equivalence classes modulo Lebesgue null sets. Assume Hadamard differentiability at the true $F$ and the usual measurability of the scalar estimator. The Donsker theorem for empirical distribution functions and the proved functional delta method giveNo continuity of $F$ is needed for this application. Since the derivative is continuous and linear and the bridge is a centred Gaussian process, $Z_F$ is a centred Gaussian real random variable, possibly identically zero.

Finally, a finite real weak limit ensures the required boundedness in probability. Choose $M$ so large that $\mathbb{P}(|Z_F|\ge M)<ϵ/2$. Applying the closed-set part of the Portmanteau theorem to $\{z:|z|\ge M\}$ givesHence there is $n_0$ such that, for every $n\ge n_0$,In particular the upper-tail condition in the question holds. The plug-in estimator has estimation error $O_{\mathbb{P}}(n^{-1/2})$, even when its limiting variance is zero.

A map $\Phi :S_D\to \mathbb{R}$ is Hadamard differentiable at $s_0$ if there is a continuous linear functional ${\dot{\Phi }}_{s_0}:S\to \mathbb{R}$ such that, whenever $t_j\to 0$ with $t_j\ne 0$, $a_j\to a$ in $S$, and $s_0+t_j{a}_j\in S_D$,For tangential Hadamard differentiability, the limit directions $a$ are restricted to a specified tangent subspace and the derivative is continuous linear on that subspace. The domain condition is indispensable.

Assume $X_n\in S_D$ almost surely and use the usual measurable, separably supported setting for norm-valued convergence in distribution, with the limit in the tangent space. The functional delta method givesTo justify this by Skorokhod representation theorem, a sufficient precise version is: for weakly convergent random elements in a separable complete metric space, one can construct copies of the whole sequence and its limit on one probability space which converge almost surely in the metric. More generally the separable-limit-support version of the representation theorem can be used. Apply it to $Z_n=r_n(X_n-s_0)$, obtaining ${\tilde{Z}}_n\to \tilde{X}$ almost surely. The variables $s_0+r_n^{-1}{\tilde{Z}}_n$ have the same laws as $X_n$ and lie in $S_D$ on a common probability-one event. The definition of Hadamard differentiability, with $t_n=1/r_n$, gives convergence almost surely of the transformed quotients to ${\dot{\Phi }}_{s_0}(\tilde{X})$, and hence the claimed convergence in distribution. If $X$ is a Gaussian measure, the limit is Gaussian because the derivative is linear. Without membership of $X_n$ in $S_D$, the expression $\Phi (X_n)$ in the question is not defined.

The final function space is $C_b(\mathbb{R})$ with supremum norm; it is not separable as a whole. This does not affect the deterministic differentiability proof below. Any statistical application of the preceding representation argument must use its separability or tightness hypotheses, rather than assuming that every normed space automatically satisfies them.

Fix admissible $F,G$ and set $T(F,G)={\int }_0^{u}G(x)F^{\prime }(x)dx$. The candidate derivative, defined even for continuous directions $a,b$ without derivatives, isThis follows formally from the product rule and integration by parts. We now establish the uniform-direction limit required by Hadamard differentiability.

Let $F_j=F+t_j{a}_j$ and $G_j=G+t_j{b}_j$ belong to the stated domain, with $a_j\to a$, $b_j\to b$ in supremum norm. The domain gives $\int |F_j^{\prime }|\le 1$ and $\int |F^{\prime }|\le 1$, while $F_j\to F$ uniformly. We first claim that, for every continuous $b$ on $[0,u]$,For a continuously differentiable test function $v$, integration by parts gives convergence directly from uniform convergence of $F_j$: both boundary terms and $\int v^{\prime }{F}_j$ converge. By the Weierstrass approximation theorem, approximate $b$ uniformly on $[0,u]$ by a polynomial $v$. The difference between the two integrals introduced by this approximation is at most $2\Vert b-v{\Vert }_{\infty }$, by the uniform bounded variation bounds. Letting the approximation error decrease proves the claim. This is weak convergence of bounded-variation integrators.

Now decompose the quotient in the order that retains the bounded-variation bound:The first term differs from $\int b{F}_j^{\prime }$ by at most $\Vert b_j-b{\Vert }_{\infty }$, so it tends to $\int b{F}^{\prime }$. For the second term, integration by parts yieldswhich tends to the corresponding expression with $a$. Hence the quotient tends to ${\dot{T}}_{F,G}(a,b)$ for every admissible perturbation sequence. Finally,This proves that the derivative is a continuous linear functional on the product space. Therefore $T$ is Hadamard differentiable at every point of the stated domain. The bounded-variation restriction is what controls the apparently dangerous interaction between uniformly small perturbations and possibly large perturbation derivatives; no convergence of $a_j^{\prime }$ or $b_j^{\prime }$ has been assumed.