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On a compact interval, positive quadrature rule weights whose sum is the interval length define uniformly bounded functionals on continuous functions. If the rules eventually integrate every fixed polynomial exactly, the Weierstrass approximation theorem proves convergence for every continuous function.

If each $(n+1)$-node quadrature rule is exact through degree $n$ and has positive weights, then it converges on every continuous function. The Weierstrass approximation theorem reduces the error to the uniform polynomial-approximation error, while exactness on constants controls the sum of the weights.

Two probability measures supported on one compact interval are equal if all their moments are equal. Indeed, equality on monomials gives equality on polynomials, the Weierstrass approximation theorem extends it to every continuous function on the interval, and compactly supported continuous functions determine a finite Borel measure.

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.

An explicit construction gives Bernstein's lethargy theorem in the requested inequality form. SetStrict decrease makes every coefficient positive. Telescoping and the limit assumption giveThusdefines a continuous function by the Weierstrass M-test and the uniform limit theorem. Apply the positive lacunary Chebyshev series calculation. At $n=0$,For $n\ge 1$, choose $K\ge 1$ so that $3^{K-1}\le n<3^K$. Its error isTherefore the function satisfiesThis explicit Chebyshev construction for Bernstein lethargy also has $E_n(f_{ϵ})\to 0$. It shows that continuity imposes no universal speed of convergence of best polynomial approximation, even though convergence itself follows from the Weierstrass approximation theorem.

To prove convergence of positive quadrature on continuous functions, for a continuous $f$, the Weierstrass approximation theorem supplies a polynomial $p$ with $\Vert f-p{\Vert }_{\infty }<\epsilon$. For every sufficiently large $n$, the Gaussian quadrature rule is exact for this fixed polynomial. Positivity of its weights and their sum $2$ then giveSince $\epsilon$ is arbitrary, $Q_n(f)\to {\int }_{-1}^{1}f$. Notice that no differentiability of $f$ is needed; uniform polynomial approximation and positivity provide the whole argument.