A finite-array normal limit from vanishing fourth moments

For each row put $a_{n,j}=\mathbb{E}{X}_{n,j}^2$ and $v_n={\sum }_j{a}_{n,j}$. Finite products in [F1] allow us to integrate replacement coordinates with independent centered Gaussian laws of variances $a_{n,j}$; a zero variance means the point mass at zero. These products need not be coupled across rows. The supplied scalar density already has integral one, so [F3] gives its scaled probability laws without a separate Gaussian-normalization theorem. For $\varphi (t)=(2\pi {)}^{-1\mathrm{/}2}{e}^{-t^2\mathrm{/}2}$, direct differentiation gives ${\varphi }^{\mathrm{\prime }}(t)=-t\varphi (t)$. On $[-R,R]$, apply [F5] with $(u,v)=(t,\varphi (t))$ and then $(u,v)=(t^3,\varphi (t))$ to obtain ${\int }_{-R}^R{t}^2\varphi (t)dt=[-t\varphi (t){]}_{-R}^R+{\int }_{-R}^R\varphi (t)dt,$ ${\int }_{-R}^R{t}^4\varphi (t)dt=[-t^3\varphi (t){]}_{-R}^R+3{\int }_{-R}^R{t}^2\varphi (t)dt\mathrm{.}$ The boundary terms tend to zero by [F6]; [F7] and the supplied normalization $\int \varphi =1$ then give $\mathbb{E}{G}^2=1$ and $\mathbb{E}{G}^4=3$. Odd moments vanish by symmetry. Consequently a replacement coordinate $G_{n,j}$ has mean zero, variance $a_{n,j}$ and fourth moment $3a_{n,j}^2$.

The sum of the replacement coordinates has law $N(0,v_n)$. One can verify this using only scalar integration: for positive variances $a,b$, Fubini and the substitution $y=t-x$ make the sum density the convolution of the two scalar densities. Completing the square in $x^2\mathrm{/}a+(t-x{)}^2\mathrm{/}b$ and using [F3] together with the given unit Gaussian integral makes that convolution $(2\pi (a+b){)}^{-1\mathrm{/}2}\exp (-t^2\mathrm{/}(2(a+b)))$. Induction gives the assertion for every finite row, and zero-variance coordinates have no effect. The absolute integrals here are finite because they are probability densities.

Let $f$ be bounded and $C^3$ on $\mathbb{R}$, with bounded first, second and third derivatives. In the finite product integral replace $X_{n,j}$ by $G_{n,j}$ one coordinate at a time. Conditional on the other coordinates, the constant, linear and quadratic terms of the Taylor expansion of $f$ at their sum have the same integrals, since both coordinates have mean zero and variance $a_{n,j}$. The two remainders are bounded by [F4]. Fubini is legitimate since $f$ is bounded and the displayed moments are finite. Summing the telescoping differences therefore gives $\mathrm{\mid }\mathbb{E}f(S_n)-\mathbb{E}f({\sum }_j{G}_{n,j})\mathrm{\mid }\le \mathrm{\parallel }{f}^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}{\sum }_j(\mathbb{E}\mathrm{\mid }{X}_{n,j}{\mathrm{\mid }}^3+\mathbb{E}\mathrm{\mid }{G}_{n,j}{\mathrm{\mid }}^3)\mathrm{/}6$.

Cauchy--Schwarz, first for each coordinate integral and then for the finite index sum, gives ${\sum }_j\mathbb{E}\mathrm{\mid }{X}_{n,j}{\mathrm{\mid }}^3\le ({\sum }_j\mathbb{E}{X}_{n,j}^2{)}^{1\mathrm{/}2}({\sum }_j\mathbb{E}{X}_{n,j}^4{)}^{1\mathrm{/}2}\to 0$. Also $a_{n,j}^2\le \mathbb{E}{X}_{n,j}^4$, by expanding the nonnegative integral of $(X_{n,j}^2-a_{n,j}{)}^2$. Hence ${\sum }_j{a}_{n,j}^2\to 0$, and the same calculation with the Gaussian fourth moments in step 1.1 gives ${\sum }_j\mathbb{E}\mathrm{\mid }{G}_{n,j}{\mathrm{\mid }}^3\le (3v_n{\sum }_j{a}_{n,j}^2{)}^{1\mathrm{/}2}\to 0$. Both Cauchy--Schwarz inequalities follow by expanding a nonnegative square and minimizing its scalar coefficient, for the coordinate integral and the finite index sum respectively. Thus the replacement error in step 2.2 tends to zero for each such test $f$.

For $\eta >0$, a concrete $C^3$ cutoff is obtained by integrating the nonnegative polynomial $s^3(1-s{)}^3$ on $[0,1]$, normalizing its integral to one, and extending the resulting cumulative function by zero and one outside that interval. Its first three derivatives join continuously at the endpoints. Translate, rescale and subtract this function from one to bound ${\mathbf{1}}_{(-\mathrm{\infty },t]}$ from below and above by bounded $C^3$ functions that differ only on $[t-\eta ,t+\eta ]$. Apply step 3.1 to these two fixed cutoffs. By step 2.1 the replacement sum has variance $v_n\to v>0$; its probability of this interval is at most $2\eta \mathrm{/}\sqrt{2\pi v_n}$, and its distribution at each fixed endpoint converges to $\mathrm{\Phi }$ scaled by $\sqrt{v}$. Taking $n\to \mathrm{\infty }$ and then $\eta \downarrow 0$ proves $\mathbb{P}(S_n\le t)\to \mathrm{\Phi }(t\mathrm{/}\sqrt{v})$ for every real $t$.

Finally fix $\delta >0$. Choose finitely many increasing real grid points so that the two tails of the limiting normal law each have mass below $\delta$ and its mass between consecutive grid points is below $\delta$. Such a finite grid exists because its density is bounded and its total integral is one. Step 4.1 gives convergence at every grid point simultaneously. Between consecutive points monotonicity of the two distribution functions bounds their difference by the largest grid-point difference plus $\delta$; the tail intervals have the same bound using zero and one. Thus the supremum in the Statement has limsup at most $\delta$. Letting $\delta \downarrow 0$ proves the asserted uniform convergence. All replacements were finite products of supplied probability laws, and the only additional choice hypothesis remains [A1].