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Yes. Fubini's theorem and the triangle inequality giveThus $T$ is Lipschitz continuous with respect to $d_1$ as well.

Since $X_1-X_s$ is centered Gaussian with variance $1-s$,Fubini's theorem therefore shows that the defining integral for $A_1$ is absolutely finite almost surely.

Because $0<\alpha <n$, $|x{|}^{-\alpha }$ is locally integrable at the origin and has only polynomial growth at infinity, so it defines a tempered distribution. It is a homogeneous distribution of degree $-\alpha$ and is radial. Its Fourier transform is therefore radial and homogeneous of degree $\alpha -n$, so it must have the form $c_{\alpha }|\lambda {|}^{\alpha -n}$. In particular, $\boxed{\beta =n-\alpha }$.

To determine the constant, use the stated Gamma integral representation, Fubini's theorem, and the Fourier transform of a Gaussian:The change of variables formula $s=|\lambda {|}^2/(4\tau )$ then givesThis is precisely the Fourier transform of the Riesz kernel.

Let $X\sim \mathrm{Bin}(n,t)$ and $Y\sim \mathrm{Bin}(n,s)$ be independent. The displayed Bernstein polynomial isGiven $\epsilon >0$, uniform continuity of $f$ supplies $\delta >0$ such that changing both coordinates by less than $\delta$ changes $f$ by less than $\epsilon$. If $M=\Vert f{\Vert }_{\infty }$, then Chebyshev inequality and
$\mathrm{Var}(X/n),\mathrm{Var}(Y/n)\le 1/(4n)$ give, uniformly in $(t,s)$,Thus $B_{n}f\to f$ uniformly.

Each $B_{n}f$ is a polynomial, whose iterated integrals may be interchanged term by term. Uniform convergence permits passage to the limit in both iterated integrals, proving the asserted continuous-function form of Fubini's theorem.

Now suppose every monomial moment of $f$ vanishes. By linearity,
$\int \int pf=0$ for every two-variable polynomial $p$. The just-proved Bernstein approximation gives polynomials $p_n\to f$ uniformly. Thereforeso $\boxed{f=0}$. Statement (i) is true.

Because the centered normal distribution density $g_t$ is nonnegative and has integral one, Fubini's theorem and translation invariance giveThus $f\ast g_t\in L^1$.

Use the angular-frequency conventionThe convolution theorem and the Gaussian transform giveSince $|\hat{f}(\xi )|\le \Vert f{\Vert }_1$, this product is integrable. Moreover, absolute integrability justifies exchanging the integrals:This proves the Fourier inversion theorem for the convolution directly.

Now additionally suppose $\hat{f}\in L^1$. For every $x$, the dominated convergence theorem givesFourier inversion identifies the right-hand side with $f(x)$ for Lebesgue almost everywhere $x$. Therefore the Gaussian approximate identity satisfies

By the Fourier inversion theorem and Fubini's theorem,Thus multiplication of Fourier transforms corresponds to the convolution $h=f\ast g$, which is the convolution theorem in the convention used here.

Compact support and the fundamental theorem of calculus give, for every $(x,y)$,whereApplying the same argument in the other coordinate gives

Multiplication and Fubini's theorem now yieldFinally, the Cauchy-Schwarz inequality givesThus the required estimate holds with $C=4$; it is the Ladyzhenskaya inequality in two dimensions.

Use the conventionThe Fourier inversion theorem says that, when $f,\hat{f}\in L^1({\mathbb{R}}^d)$,for almost every $x$.

Sincethe Fubini's theorem and Fourier inversion giveThus the Plancherel theorem identity in this convention is

The inverse integral is a continuous function of $x$, since $\hat{f}\in L^1$ permits dominated convergence. If $f$ is continuous, it and the inverse integral are continuous and agree almost everywhere. They must agree everywhere: otherwise their continuous difference would be nonzero on an open set of positive measure. This is the continuous version of Fourier inversion.

Define on $H$ the bilinear formand the linear functionalThe Hardy inequality on an interval, Cauchy-Schwarz inequality, and the one-sided Poincaré inequality show that $a$ is a bounded bilinear form and that

For $v\in H$, $v(x)={\int }_0^x{v}^{\prime }(t)dt$, so Cauchy--Schwarz and Fubini's theorem giveHenceThus $a$ is a coercive bilinear form. The Lax-Milgram theorem supplies a unique $u\in H$ satisfying $a(u,v)=\ell (v)$ for every $v\in H$. This is precisely the unique weak solution described by the weak boundary value problem with an inverse-square potential.

LetThe integral is absolutely convergent. For a Schwartz function $\phi$, Fubini's theorem and integration by parts in $\lambda$ giveEquivalently, $v^{\prime }=-\hat{u}$ as a distributional derivative, and hence

With densities $p,q$,By Cauchy-Schwarz inequality,Also $(\sqrt{p}-\sqrt{q}{)}^2\le |p-q|$, so $H^2\le 2\mathrm{TV}$.

The Hellinger affinity is $\rho (P,Q)=\int \sqrt{pq}d\mu =1-H^2/2$. Product densities and Fubini's theorem give $\rho (P^n,Q^n)=\rho (P,Q{)}^n$, hence

Le Cam two-point lemma states, for squared-error estimation at parameter points ${\theta }_0,{\theta }_1$, thatTake ${\theta }_0=0$, ${\theta }_1=\delta =1/(4n)$. The one-observation uniform densities overlap on length $1-\delta$, so $H^2(P_0,P_1)=2\delta$. ThereforeThe first distance inequality gives $\mathrm{TV}(P_0^n,P_1^n)\le 1/\sqrt{2}$. Le Cam's lemma now yieldsThis proves the claim with the displayed universal positive constant.

The kernel $k(x,y)=e^{-|x-y|}$ is real and symmetric. For $f,g\in L^2[0,1]$, Fubini's theorem givesso $K$ is self-adjoint. It is linear, and because $k\in L^2([0,1{]}^2)$ it is a Hilbert-Schmidt operator, hence bounded, with

The variable $Z={\int }_0^1{B}_{t}dt$ is the $L^2$ limit of Riemann sums of the Gaussian process $B$, so it is a Gaussian random variable. Its mean is zero, and Fubini's theorem with $\mathbb{E}[B_s{B}_t]=\min (s,t)$ givesTherefore $Z\sim N(0,1/3)$.

The original double integral covers the triangle $0\le y\le x\le 1$. By Fubini's theorem, after reversing the order, $y$ runs from $0$ to $1$ and $x$ from $y$ to $1$. Since a continuous function is integrable on this bounded region,For the square pyramid, a horizontal section at height $z$ is a square of side $2(1-z)$: each vertex moves linearly from the base towards $(1,1,1)$. Hence its cross-sectional area is $4(1-z{)}^2$, and its volume is $4{\int }_0^1(1-z{)}^{2}dz$. To use the identity just proved, choose $f(y)=1-y$:

The positive integrand factors into three one-variable functions on a rectangular domain. Fubini's theorem therefore gives

Differentiate the rewritten expression. Its new kernel vanishes at the moving upper endpoint, and its partial derivative is an integrable inverse square root. The derivative of an Abel integral isFor a rigorous route, rewrite $2{\int }_0^x{h}^{\prime }(t)\sqrt{x-t}dt$ as ${\int }_0^x{\int }_0^s{h}^{\prime }(t)/\sqrt{s-t}dtds$ using Fubini's theorem, and apply the fundamental theorem of calculus to the continuous inner integral. The assumption $h(0)\ne 0$ implies $g(x)=2h(0)\sqrt{x}+O(x^{3/2})$, so the extension $g(0)=0$ has no finite right derivative at zero.

The region is $0\le y\le 1$, $2y\le x\le 2$. Reversing the order describes the same triangle as $0\le x\le 2$, $0\le y\le x/2$. Since the integrand is continuous on this compact region, Fubini's theorem permitsThe new inner integral introduces the factor $x$ needed for the change of variables $u=x^2$.

The integrand is continuous on the compact rectangle, so Fubini's theorem allows integrating in $y$ first. For $x\ne 0$, ${\int }_0^{1}x{e}^{xy}dy=e^x-1$; the same formula holds by continuity at $x=0$. Hence

Translate the variables by $u=x-x_0$, $v=y-y_0$, $w=z-z_0$. The numerator becomes $u+x_0$. The part odd in $u$ integrates to zero, while the constant part factors into three Gaussian integrals. ThusThe absolute integrability of a polynomial times a Gaussian justifies this separation by Fubini's theorem. Neither $y_0$ nor $z_0$ affects the result; when $x_0=0$, odd symmetry makes the integral zero.

For a vertical slice through the triangle, $0\le x\le 2$ and $0\le y\le x/2$. Reversing the order by Fubini's theorem givesThe agreement checks that both orders describe the same region.