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Assume the usual positive initial stock price and strike, so the logarithm is defined, and interpret $U$ as a classical solution of the displayed partial differential equation. For a fixed complex $z$, put $F_t(z)=S_t^{z}U(t,{\sigma }_t,z)$, defining $S_t^z=\exp (z\log S_t)$. The Itô formula for the two Brownian motions with correlation coefficient $\rho$ givesThe partial differential equation cancels the entire drift. By the permission to treat the resulting local martingales as true martingales, and the terminal condition, $F_t(z)=\mathbb{E}[S_T^z\mid {\mathcal{F}}_t]$.

The bounded spot volatility ensures $\mathbb{E}{S}_T^{x_0}<\infty$: the Itô formula for a real power and localization bound its moment by $S_0^{x_0}\exp (x_0(x_0-1)\Vert \sigma {\Vert }_{\infty }^{2}T/2)$. Rewrite the proposed integrand asNow $|F_t(x_0+iy)|\le \mathbb{E}[S_T^{x_0}\mid {\mathcal{F}}_t]$, an integrable bound independent of $y$. Conditional Fubini's theorem and part (b) therefore implyThis route justifies the contour exchange without assuming bounds on $U$ uniform in all complex $z$. Cash is constant, $S$ is a true martingale under the stated allowance, and the displayed $C$ is a true martingale. The original measure is thus an equivalent martingale measure for all three assets. The fundamental theorem of asset pricing gives the market has no arbitrage under the usual admissible trading convention.

A multiset $\mathcal{C}$ of subsets of $[n]$ is a uniform cover of multiplicity $k$ when each coordinate occurs in exactly $k$ members. The uniform covers theorem states that every Euclidean body $S\subseteq {\mathbb{R}}^n$ satisfieswhere $S_A$ is the coordinate projection of a Euclidean body onto the coordinates in $A$.

We prove it by induction on $n$. Split the cover into ${\mathcal{C}}^{-}$, whose members omit $n$, and ${\mathcal{C}}^{+}$, whose members contain $n$. Exactly $k$ members lie in ${\mathcal{C}}^{+}$. For a last-coordinate value $x$, let $S(x)\subseteq {\mathbb{R}}^{n-1}$ be the corresponding slice. Removing $n$ from the members of ${\mathcal{C}}^{+}$ and retaining the members of ${\mathcal{C}}^{-}$ gives a $k$-uniform cover of $[n-1]$. The inductive hypothesis and Fubini's theorem giveApply Hölder's inequality to the $k$ factors in the integral. Sincewe obtain $|S|\le {\prod }_{A\in \mathcal{C}}|S_A{|}^{1/k}$, which is the result after taking the $k$th power. The one-dimensional base case is immediate.

The Bollobas--Thomason box theorem states that for every body $S\subseteq {\mathbb{R}}^n$ there is an axis-parallel box $B$ such thatAn irreducible uniform cover cannot be decomposed into two smaller uniform covers. There are only finitely many such covers of $[n]$: encode a cover by its multiplicity vector in ${\mathbb{N}}^{2^n}$ and apply the Dickson lemma.

Choose a componentwise minimal positive array $(x_A{)}_{A\subseteq [n]}$ satisfyingfor every irreducible $k$-uniform cover $\mathcal{C}$, together withThe actual projection volumes are feasible by the uniform covers theorem, and finiteness gives a minimal array. Every uniform cover is a disjoint union of irreducible ones, so its cover inequality also holds for this array.

Minimality implies that, for each coordinate $i$, some tight uniform-cover inequality can be chosen whose cover contains the singleton $\{i\}$. Indeed, either such an inequality already blocks decreasing $x_{\{i\}}$, or a tight product inequality $x_A={\prod }_{j\in A}{x}_{\{j\}}$ does; in the latter case take a tight cover containing $A$ and replace that occurrence of $A$ by its singleton coordinates. Let these tight covers be ${\mathcal{C}}_i$, of multiplicities $k_i$, and let $K={\sum }_i{k}_i$. Their multiset union is a $K$-uniform cover. Removing one copy of every singleton leaves a $(K-1)$-uniform cover, so comparison of its cover inequality with the product of all the tight equalities yieldsThe singleton cover gives the reverse inequality, henceFor any $A\subseteq [n]$, the one-uniform cover consisting of $A$ and the singletons $\{i\}$ for $i\notin A$ now gives $x_A\ge {\prod }_{i\in A}{x}_{\{i\}}$. The defining product inequality gives the reverse bound. Thus all these quantities are equal. Taking the side lengths of $B$ to be $x_{\{1\}},\dots ,x_{\{n\}}$ proves the theorem.

Finally suppose that the proper body $S\subseteq {\mathbb{R}}^3$ satisfiesThis is equality in the three-dimensional Loomis--Whitney inequality. In its proof, equality must hold in both applications of Cauchy-Schwarz inequality. Their equality conditions force the three projection indicators to factor through one-dimensional measurable sets $E_1,E_2,E_3$, and forceup to a set of Lebesgue measure zero; this is Equality in the three-dimensional Loomis--Whitney inequality.

Because $S$ is connected, each one-coordinate projection is connected and hence is an interval. Because $S$ is a finite union of positive-volume axis-parallel boxes, a proper difference between $S$ and the product of those three intervals would contain a positive-volume rectangular cell in a common finite subdivision. That would contradict equality up to measure zero. Consequently the equality is exact and

Condition on $S$ and use the characteristic function of a standard normal distribution:The integrand has modulus one, so Fubini's theorem is immediate. Taking expectation over $S$ yields

By the bound in part (a), Fubini's theorem applies. Conditional on $S$,where the Characteristic function of the Cauchy distribution was used. Since $\mathbb{E}S=1$,The formula expresses a European call option value through complex moments of $S$. In an affine stochastic-volatility model such as the Heston model, those moments are available from an explicit transform, so call prices reduce to a one-dimensional Fourier expectation or integral.

The notation $P_0=\mathcal{N}(0,I)$ denotes generalized Gaussian white noise over $H=L^2({\mathbb{R}}^2)$, not a Gaussian measure supported on $H$: the identity is not a trace-class operator there. For example, realize the observation on a suitable space of tempered distributions. Its Cameron-Martin space of a Gaussian measure is $H$. For deterministic $h\in H$, the stochastic pairing $W_m(h)$ has normal distribution $\mathcal{N}(0,\Vert h{\Vert }_H^2)$ under $P_0$ and is the isonormal Gaussian process indexed by $H$.

Interpret the stated maps between Sobolev spaces as bounded linear operators, as usual. In particular $K:H\to H^2({\mathbb{R}}^2)\subset H$ is bounded, so its adjoint operator $K^{\ast }$ is bounded on $H$. Hence $A=K^{\ast }K$ maps $H$ into $H$, which is the property needed here. The stronger smoothing follows by Sobolev duality: boundedness of $K:H^{-4}\to H^{-2}$ gives $K^{\ast }:H^2\to H^4$, and therefore $A:H\to H^4\subset H$. As $\Pi (H)=1$, the shift $Au$ belongs to the noise Cameron-Martin space of a Gaussian measure for $\Pi$-almost every $u$.

The white-noise likelihood for a square-integrable shift, obtained from the Cameron-Martin theorem for a Gaussian measure in its white-noise form, isHere $P_u$ is the law of $Au+\eta$. In an orthonormal basis $(e_j)$ of $H$, with white-noise coordinates $m_j$, the pairing under $P_0$ isFor fixed $u$ this has mean-square convergence and converges almost surely, since ${\sum }_j|⟨Au,e_j⟩{|}^2=\Vert Au{\Vert }_H^2$. The stipulated joint measurability allows the likelihood to be used under ${\nu }_0(du,dm)=\Pi (du)P_0(dm)$. One must not replace this expression by a finite $\Vert m-Au{\Vert }_H^2$, because white noise is not $H$-valued. On the unbounded domain ${\mathbb{R}}^2$, membership of white noise in a global unweighted negative Sobolev space must not be assumed either; the stochastic pairing avoids that issue.

Let $\nu (du,dm)=\Pi (du)P_u(dm)$ be the actual joint law. The shift formula gives $\nu \ll {\nu }_0$ with Radon-Nikodym derivative $L$. DefineFor each fixed admissible $u$, the moment-generating function of a normal distribution givesBy Tonelli theorem, $\int Z(m)P_0(dm)=1$, so $Z(m)<\infty$ for $P_0$-almost every $m$. The likelihood is finite and strictly positive for ${\nu }_0$-almost every $(u,m)$; Fubini's theorem then gives $Z(m)>0$ for $P_0$-almost every $m$. The marginal observation law is $Z(m)P_0(dm)$ and is an equivalent probability measure to $P_0$.

The Bayes formula for a dominated observation model therefore definesTo verify that it is the conditional distribution, for measurable sets $B$ of unknowns and $C$ of data,Thus the posterior distribution is well defined for almost every observation under the actual data law. This is the appropriate almost-everywhere assertion supplied by Bayes theorem; the stated assumptions do not prescribe a canonical posterior at every exceptional generalized datum.

The integrands are absolutely integrable because $f$ has compact support, $g\in L^1(\mathbb{R})$, and $|{\phi }_Z|\le 1$. Fubini's theorem therefore permits us to insert ${\phi }_Z(s)=\mathbb{E}[e^{isZ}]$ and interchange the expectation and integrals:With $r=\epsilon s$, the inner integral in $s$ isNow set $t=(Z-x)/\epsilon$. The change of variables $x=Z-\epsilon t$ giveswhich is the required identity.

Choose any $g$ with $g,\hat{g}\in L^1$, $g(0)\ne 0$, and for which Fourier inversion theorem applies, for example a Gaussian function. The left side factors through the Fourier transform of $f$ asOn the right side, $f(Z-\epsilon t)\to f(Z)$ andThe dominated convergence theorem and Fourier inversion at zero giveConsequently the Fourier-cutoff recovery of expectations from a characteristic function isThe right side depends only on ${\phi }_Z$. Equal characteristic functions therefore give equal expectations for every compactly supported continuous $f$, and part a yields the uniqueness theorem for characteristic functions.

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 Poincare 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 $Z\sim N(0,1/3)$.

For positive $u$, integrate $t-f$ from zero to $u$; for negative $u$, reverse the integral from $u$ to zero. This proves the identity without a positivity restriction on the data. For $u,f\in L^2$, the absolute integral is bounded by $u^2/2+|fu|$, so Fubini's theorem applies. The zero-function baseline removes the divergence from the negative-height tail.