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Transform pair. In the paper's symmetric normalization, the inverse Fourier transform isSplit the transform of $e^{-c|x|}$ at zero, with $c>0$:The inverse formula applied to this result givesBecause the kernel is even, replacing the transform variable by $x$ and the argument by $k$ yieldsThe factors of $\sqrt{2\pi }$ depend on the normalization and will also enter the convolution theorem.

Convolution theorem. For integrable functions define the convolutionSubstitute this into the Fourier transform, interchange the two absolutely integrable factors using Fubini's theorem, and set $z=x-y$:Therefore

Integral equation. Put $f_a(x)=1/(x^2+a^2)$. Since this is even, the left side is $f_a\ast g$. The transformed equation issoThe assumption $b>a$ makes the new scale $d=b-a$ positive. Inverting with the previously established transform pair givesThe Cauchy convolution semigroup gives a quick verification. With $K_c(x)=c/[\pi (x^2+c^2)]$, the formula is $g=(a/b)K_{b-a}$ and $f_a=(\pi /a)K_a$. Hence $f_a\ast g=(\pi /b)K_b=1/(x^2+b^2)$, as required. The solution is unique among integrable functions because the multiplier ${\tilde{f}}_a(k)$ never vanishes.

Use the Fourier transform convention $\hat{f}(k)={\int }_{\mathbb{R}}f(t)e^{-ikt}dt$, with inverse factor $1/(2\pi )$. For functions in L1 space, their convolution is $h(t)={\int }_{\mathbb{R}}f(s)g(t-s)ds$. Fubini's theorem applies because the absolute double integral is $\Vert f{\Vert }_1\Vert g{\Vert }_1$. Substituting $r=t-s$ therefore proves the convolution theorem:For the specified rectangular pulse, the support condition is $|t|<1$. Its Fourier transform isThe convolution $f\ast f$ is the length of the overlap of $(-1,1)$ and $(t-1,t+1)$. Thus it is the triangular pulseThe transform's value at zero is four, given by its continuous extension. In this convention, the Parseval identity for square-integrable functions isFor the rectangular pulse the left side is two, giving the sinc-squared integralFor the triangular pulse, $\int |f\ast f{|}^{2}dt=2{\int }_0^2(2-t{)}^{2}dt=16/3$. Apply the same Parseval identity with transform $4{\sin }^{2}k/k^2$ to obtain the sinc fourth-power integralBoth integrands have removable singularities at zero and integrable tails.

For causal functions the convolution is $(f\ast g)(t)={\int }_0^{t}f(\tau )g(t-\tau )d\tau$. In a half-plane of absolute convergence, changing variables $u=t-\tau$ in its Laplace transform and using Fubini's theorem givesThis proves the convolution theorem for Laplace transforms. Transforming the differential equation gives $(s^2+1)Y(s)=sy(0)+y^{\prime }(0)+F(s)$. The direct transform pairs $\mathcal{L}(\cos t)=s/(s^2+1)$ and $\mathcal{L}(\sin t)=1/(s^2+1)$ then giveFor a continuous forcing, call the last term $w(t)$. The Leibniz integral rule gives $w^{\prime }={\int }_0^{t}f(\tau )\cos (t-\tau )d\tau$ and $w^{\prime \prime }=f(t)-w(t)$. Also $w(0)=w^{\prime }(0)=0$. Thus both the original equation and the initial values hold.

For an impulse at positive time $T$, the Dirac delta evaluates the convolution, yieldingHere $H$ is the Heaviside step function. The displacement remains continuous at $T$, while its derivative jumps upward by one, exactly as required by integrating the differential equation across the unit impulse.

With the stated angular-frequency convention, the inverse Fourier transform is $f(x)=(2\pi {)}^{-1}{\int }_{-\infty }^{\infty }\tilde{f}(k)e^{ikx}dk$. Direct integration gives ${\tilde{f}}_0(k)={\int }_0^{\infty }{e}^{-(\alpha +ik)x}dx=1/(\alpha +ik)$. For $n\ge 1$, integration by parts has zero boundary terms and gives ${\tilde{f}}_n=n{\tilde{f}}_{n-1}/(\alpha +ik)$. ThereforeThis also follows from the Fourier transform of a polynomial multiple, since multiplication by $x$ corresponds to $i{\partial }_k$.

To prove Parseval's identity, substitute the inverse transform of $g$ into $\int f^{\ast }g$. Under the assumed integrability, Fubini's theorem permits interchange, and the inner integral is $\int f(x{)}^{\ast }{e}^{ikx}dx=\tilde{f}(k{)}^{\ast }$. HenceApply this with $f=g=f_n$. The spatial Gamma integral is ${\int }_0^{\infty }{x}^{2n}{e}^{-2\alpha x}dx=(2n)!/(2\alpha {)}^{2n+1}$. The transformed squared modulus is $(n!{)}^2/({\alpha }^2+k^2{)}^{n+1}$, giving the scaled even rational Parseval integral

The one-sided Laplace transform is $\bar{f}(p)={\int }_0^{\infty }{e}^{-pt}f(t)dt$. Its Bromwich inversion formula is $f(t)=(2\pi i{)}^{-1}{\int }_{\gamma -i\infty }^{\gamma +i\infty }{e}^{pt}\bar{f}(p)dp$, with an upward vertical contour to the right of all singularities and within the appropriate convergence half-plane.

To derive the convolution theorem for Laplace transforms, transform $x(t)={\int }_0^{t}f(\tau )g(t-\tau )d\tau$. With $u=t-\tau \ge 0$, the triangular integration region becomes a quadrant. Fubini's theorem gives $\bar{x}(p)={\int }_0^{\infty }{\int }_0^{\infty }{e}^{-p(u+\tau )}f(\tau )g(u)dud\tau =\bar{f}(p)\bar{g}(p)$, under the usual absolute convergence hypotheses. Inversion gives the required convolution formula.

For the differential equation, the transformed initial conditions give $(p^2-1)\bar{y}=\bar{g}+p+1$, hence $\bar{y}=\bar{g}/(p^2-1)+1/(p-1)$. Since $\mathcal{L}[\sinh t]=1/(p^2-1)$ for $\mathrm{Re}p>1$, the convolution yieldsThe integral and its first derivative vanish at zero, so the homogeneous term supplies both initial values. Differentiating twice gives the forcing through the unit derivative of the hyperbolic-sine kernel at zero.

Choose the Fourier transform convention $\hat{f}(k)={\int }_{\mathbb{R}}{e}^{-ikx}f(x)dx$, with inverse $(2\pi {)}^{-1}\int e^{ikx}\hat{f}(k)dk$. For absolutely integrable $f_1,f_2$, Fubini's theorem permits the substitution $s=x-y$ in the convolution transform:This proves the convolution theorem in the chosen convention.

Take $q(x)=\frac{1}{2}{e}^{-|x|}$. Splitting its transform at zero gives $\hat{q}(k)=\frac{1}{2}[(1+ik{)}^{-1}+(1-ik{)}^{-1}]=(1+k^2{)}^{-1}$. Thus the desired inverse is $q\ast q$. For $x\ge 0$, split the convolution integral at $y=0$ and $y=x$:The result is even, soDifferent normalization conventions alter the transform prefactors, which is why the convention was stated explicitly.

First prove the convolution theorem with the stated negative-exponential Fourier transform. Assuming absolute integrability sufficient for Fubini's theorem, change variables $u=t-s$:

For the differential equation, integration by parts givesThe boundary terms vanish by the assumed decay of $y,y^{\prime }$. As with the convolution step, the transforms can first be justified for integrable functions and derivatives; the resulting kernel gives the desired decaying response. ThusThe convolution theorem and Fourier inversion theorem imply $y=f\ast G$.

For the inverse transform of $\tilde{G}$, useThe poles are at $ib,ia$. For $t>0$, close the inverse-transform contour above; the residues of $e^{ikt}\tilde{G}(k)$ there are $-i{e}^{-bt}/(a-b)$ and $i{e}^{-at}/(a-b)$. The inverse-transform factor $1/(2\pi )$ makes the answer $i$ times their sum. For $t<0$, close below, enclosing no poles. The large arcs vanish: the complete rational factor is $O(|k{|}^{-2})$ and the exponential is bounded on the appropriate semicircle. Therefore the causal Green function of an overdamped oscillator isIt is continuous, with $G(0)=0$ and $G^{\prime }(0^{+})=1$, so $(D_t+a)(D_t+b)G=\delta (t)$. The response can also be written $y(t)={\int }_{-\infty }^{t}G(t-s)f(s)ds$.

For the bilateral exponential forcing, split the real integral at zero:ConsequentlyFor negative time the lower half-plane contains only $k=-i$. The residue of $e^{ikt}\tilde{y}$ there is $i{e}^t/[(a+1)(b+1)]$. Clockwise closure makes the inverse transform $-i$ times that residue. For positive time the three upper residues, at $i,ia,ib$ respectively, areMultiplication of their sum by $i$ gives the bilateral exponential forcing of an overdamped oscillator:Both expressions have $y(0)=y^{\prime }(0)=1/[(a+1)(b+1)]$. This verifies the matching that prevents an extra Dirac delta function at zero. Direct substitution gives the forcing on either side, and all terms decay at their corresponding infinities. A causal Green function does not imply zero response before $t=0$ when the forcing has already been acting at negative times. The assumptions $a,b\ne 1$ avoid repeated poles; their limiting cases would contain polynomial times exponential terms.

For integrable $f,g$, define their convolution by $(f\ast g)(x)={\int }_{\mathbb{R}}f(y)g(x-y)dy$. Absolute integrability of the double integral permits Fubini's theorem. With $s=x-y$,The convolution theorem, together with uniqueness of the Fourier transform, therefore identifies the inverse transform of that product with $f\ast g$. With the stated convention the inverse carries $1/(2\pi )$, so there is no extra factor in the convolution itself.

Now take $f=e^{-|x|}$. For $x\ge 0$, split the convolution at $y=0$ and $y=x$:The convolution of two even functions is even, so the self-convolution of a bilateral exponential givesIn particular the centre value is one; both pieces extend continuously there.

Splitting the Fourier integral at zero givesHence $\tilde{h}(k)=4/(1+k^2{)}^2$. The Fourier inversion theorem yieldsThe integral is absolutely convergent. At $x=0$ it gives ${\int }_{\mathbb{R}}(1+k^2{)}^{-2}dk=\pi /2$, providing a normalization check.

First take $f,g\in L^1(\mathbb{R})$, so Fubini's theorem applies to the cross-correlation double integral. With $s=x+y$, the Fourier transform isThe complex conjugation and the sign in $x+y$ are both essential to this convention.

Suppose initially that $f,g$ are absolutely integrable, so Fubini's theorem justifies exchanging the integrations. The transform of their convolution isSet $s=z-t$, so $z=s+t$ and $dz=ds$. The exponential separates:Therefore the convolution theorem isNo extra $2\pi$ occurs in this direction with the specified normalization.

Assume initially that $f,g$ are integrable, so Fubini's theorem justifies interchanging the integrals. Transform the convolution and put $u=x-z$:This is the convolution theorem. No factor $2\pi$ appears in this direction because the forward transform in the question is unnormalized.

For integrable functions $f,g$, Fubini's theorem justifies exchanging the two integrals in the Fourier transform of the convolution. Set $s=x-t$ to obtainThis proves the convolution theorem with the stated normalization:

Use the Fourier transform conventionThe convolution is $(f\ast g)(t)={\int }_{-\infty }^{\infty }f(\tau )g(t-\tau )d\tau$. For integrable $f,g$, Fubini's theorem and the substitution $s=t-\tau$ giveHence the convolution theorem is

Assume first that $f$ and $g$ are integrable functions, so that Fubini's theorem justifies changing the order of the two integrals. In the defining Fourier transform of the convolution, put $y=x-x^{\prime }$:This proves the convolution theorem with the stated normalization:

Read the differential order $dxdy$ literally: take quadrature nodes $X=(c,(c+d)/2,d)$ in $x$ and $Y=(a,(a+b)/2,b)$ in $y$. Applying Simpson's rule first in $x$ and then in $y$ multiplies the one-dimensional quadrature weights. With $s=(1,4,1)$, this tensor-product quadrature isEquivalently, the nine sampled function values have the quadrature weight arraywith rows indexed by the $x$ quadrature nodes and columns by the $y$ quadrature nodes. The quadrature weights sum to the rectangle's area. If the intended coordinate assignment is instead $x\in [a,b]$, $y\in [c,d]$, exchange $X$ and $Y$ in the formula. Both conventions give the same calculation on the unit square used below.

For the separable integrand, tensor-product quadrature factors into two one-dimensional applications of Simpson's rule:soThe exact double integral, evaluated by Fubini's theorem and the power rule for integration, isThe $x$ factor is always integrated exactly. All quadrature error therefore comes from applying Simpson's rule to the $y$ factor.

Write $P=\omega {\omega }^T$. Since $P^T=P$ and $P^2=P$, the orthogonal reflection $R_{\omega }=I-2P$ satisfies $R_{\omega }^T{R}_{\omega }=I$ and $R_{\omega }^{-1}=R_{\omega }$. Thus, for every Borel set $E$, invariance under the orthogonal group gives

The spherical average must use normalized surface measure without an additional area factor. Let $d\sigma$ be the normalized surface measure on a sphere, and setThe correct radial disintegration of an orthogonally invariant measure isIndeed, take normalized Haar measure $H$ on the compact orthogonal group $O(d)$. Invariance and Fubini's theorem giveRight invariance of Haar measure and transitivity of the orthogonal group on each sphere show that the inner average depends only on $r=|x|$. Averaging it once more over $x=r\theta$, using invariance of surface measure on a sphere and Fubini's theorem, identifies its value as $A\eta (r)$. At $x=0$ the same identity reads $\eta (0)=\eta (0)$. The definition of pushforward measure now proves the boxed formula. This argument also works for invariant probability measures with an atom of a measure at the origin.

For the printed normalization, choose a smooth cutoff function $\eta$ equal to one on a ball containing the support of a measure $\mu$. Its left integral is one, whereas the printed right integral is ${\sigma }_d$. Thus the extra ${\sigma }_d$ cannot be retained. Equivalently, with unnormalized area measure $dS$, the average is ${\sigma }_d^{-1}\int \eta (r\theta )dS(\theta )$.

The reflection-averaging hint also fails in general dimension. For $x=e_1$,The first equality uses $\int {\omega }_i{\omega }_{j}d\sigma ={\delta }_{ij}/d$, which follows from coordinate reflections, coordinate permutations, and ${\sum }_i{\omega }_i^2=1$. Taking a test function equal to the first coordinate near the unit sphere disproves the hint when $d\ne 2$. Averaging over the full orthogonal group, as above, proves the required corrected result without that hint.

Write $d\mu =f(x)dx$, with $f\ge 0$. Applying polar coordinates and Fubini's theorem gives, for bounded nonnegative measurable $h$,Also $\tilde{\mu }(\{0\})=\mu (\{0\})=0$ by absolute continuity of measures. Extending $g$ by zero on the negative half-line proves

Both radial pushforward measures have compact support of a measure, so Brenier's theorem on the line supplies a unique optimal transport plan and a transport map $\tau$, unique $\tilde{\mu }$-almost everywhere, withIn one dimension, the gradient of a convex function is a nondecreasing function. Nonnegativity holds $\tilde{\mu }$-almost everywhere because ${\tau }_{\#}\tilde{\mu }=\tilde{\nu }$ is carried by $[0,\infty )$. To choose a nonnegative, nondecreasing representative everywhere, let $F,G$ be the radial cumulative distribution functions and use the monotone rearrangementAt $F(r)=0$ set $\tau (r)=0$, and at $F(r)=1$ use the largest point of the support of a measure $\tilde{\nu }$. These choices preserve monotonicity and boundedness. Since $F$ is continuous, $F(R)$ has the uniform distribution on $(0,1)$ when $R$ has law $\tilde{\mu }$, and the quantile function $G^{-1}$ sends this uniform distribution to $\tilde{\nu }$. Thus this representative has the required pushforward measure, even if the target has atoms.

The indicated measurable function is nonnegative. Tonelli's theorem applies without first assuming integrability and givesThus $h$ is integrable. Fubini's theorem now allows the reverse order. For $x>1$,ConsequentlyAt $x=1$ the quotient has the removable limit $b-a$; that endpoint does not create an additional divergence.

For sigma-finite measures, Tonelli's theorem states that the double integral of a nonnegative measurable function equals either iterated integral, allowing $+\infty$. Fubini's theorem states that if a measurable function is absolutely integrable, its sections are integrable almost everywhere, both iterated integrals exist, and they equal the double integral.

Assign any value at $(0,0)$, which does not affect Lebesgue integrability. Near that corner set $s=x+y$ and $t=(x-y)/(x+y)$. Then $x=s(1+t)/2$, $y=s(1-t)/2$, and the absolute Jacobian determinant is $s/2$. On $0<s<1$, the full range is $-1<t<1$, soThe remaining part of the unit square has $x+y\ge 1$ and is harmless. The improper power integral is finite precisely when $\alpha <3$. Under the given restriction,For those values, the change $(x,y)\mapsto (y,x)$ preserves the square and changes the sign of $F$. Absolute integrability justifies this change of variables, givingFor $\alpha \ge 3$ the positive and negative parts each diverge, so symmetry does not define a Lebesgue integral.

Fubini's theorem states that an absolutely integrable measurable function on a product of two $\sigma$-finite measure spaces has integrable sections almost everywhere, the corresponding iterated integrals exist, and both orders equal its product integral. In particular, for $f\in L^1({\mathbb{R}}^2)$,To establish integrability of the nonnegative function here, first use Tonelli's theorem: for a nonnegative measurable function the product integral equals either iterated integral with infinity allowed. This avoids assuming the conclusion needed for Fubini's theorem.

If $\alpha \ge -1$, ${\int }_0^{\infty }(1+x+y{)}^{\alpha }dx=\infty$ for every fixed $y$. If $\alpha <-1$, it isThe outer integral is finite exactly when $\alpha +1<-1$, that is, $\alpha <-2$. In that caseFor $-2\le \alpha <-1$ the outer integral diverges, including logarithmic divergence at $\alpha =-2$. Nonnegativity means finiteness of this integral is exactly absolute integrability; after proving it, Fubini's theorem allows either order.