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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.

Take the Fourier transform in $x$. Since $\widetilde{{\theta }_{xx}}=-k^2\tilde{\theta }$, the transformed heat equation isThereforeThe given transform pair and the convolution theorem yield the heat-kernel solution

The convolution isWe claim that the $n$-fold convolution isThis is true for $n=1$. If it holds for $n-1$ and $x\ge 0$, thenand the convolution vanishes for $x<0$. This is the gamma density from repeated exponential convolution.

The Fourier transform is

The convolution theorem statesIndeed, Fubini and $u=x-y$ giveSince $\hat{F}(k)=(1+ik{)}^{-1}$, induction immediately verifies

For Parseval identity, take $g(x)=\overline{f(-x)}$. ThenUsing Fourier inversion at zero, which follows from the supplied delta identity,and the convolution theorem gives

Apply this to $F_{n+1}$. Sinceone obtains the Even rational Parseval integral

Insert the transform definitions and use Fubini:For the indicator of $(-1,1)$,Putting $g=f$ and $x=0$ in the convolution theorem givesand hence

Use normalized convolution and Fourier coefficients, and define the large spectrumBy the Parseval identity and $0\le f\le 1$,so $|\Gamma |\le {ϵ}^{-2}{\delta }^{-1}$.

The convolution theorem givesIf $y\in B(\Gamma ,ϵ)$, the part over $\Gamma$ is at mostbecause $|\hat{f}(\gamma )|\le \delta$. On the complementary spectrum, $|\hat{f}(\gamma )|<ϵ\delta$ and $|\gamma (y)-1|\le 2$, so the contribution is less than $2ϵ{\delta }^2$. The required difference is therefore less than $3ϵ{\delta }^2$.

With the convention in convolution theorem, absolute integrability permits Fubini:

If $g\in L^2$ is orthogonal to every translate of $f$, then the correlation $g\ast \tilde{f}$ vanishes. By convolution theorem,
$\hat{g}\overline{\hat{f}}=0$ almost everywhere. Since $\hat{f}$ vanishes only at one point, a null set, $\hat{g}=0$ almost everywhere. By Plancherel theorem, $g=0$. Orthogonal-complement characterization proves density in $L^2$.

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.

The convolution is $(u\ast v)(x)={\int }_{-\infty }^{\infty }u(t)v(x-t)dt$. In this transform convention, the convolution theorem says $\widehat{u\ast v}=\hat{u}\hat{v}$, and Parseval's identity saysTo match the integral's scale exactly, set $u=f/2$ and $h(x)=\frac{1}{2}{1}_{[-1,1]}(x)=g(2\pi x)/2$. Then $\hat{u}=1/(1+k^2)$ and $\hat{h}=\sin k/k$, using the value one at zero. Their convolution is the box smoothing of a two-sided exponential with both scales equal to one:Its transform is precisely $\sin k/[k(1+k^2)]$. Since $w$ is even, Parseval's identity reduces the requested integral toUse ${\int }_0^1\cosh xdx=\sinh 1$, ${\int }_0^1{\cosh }^{2}xdx=1/2+\sinh 2/4$ and ${\int }_1^{\infty }{e}^{-2x}dx=e^{-2}/2$. Simplifying givesThe apparent singularity at $k=0$ is removable, and the integrand decays as $O(|k{|}^{-6})$, so the integral converges.

Use the conventionThese are the Fourier transform and inverse Fourier transform for suitably integrable data. Assuming the required ordinary transforms exist and the lower derivatives vanish at both ends, integration by parts gives $\widehat{g^{\prime }}=[g{e}^{-ikx}{]}_{-\infty }^{\infty }+ik\hat{g}=ik\hat{g}$. Repeating this argument yieldsThe decay assumptions are used to discard the boundary terms at each step.

For positive diffusivity $\alpha$, transforming the heat equation gives ${\partial }_t\hat{f}=-\alpha k^2\hat{f}$. Thus $\hat{f}(k,t)=e^{-\alpha k^{2}t}{\hat{f}}_0(k)$, where $f_0(x)=f(x,0)$. The convolution theorem and the Gaussian heat kernel giveFor $f_0(x)=e^{-x^2}$, the Fourier transform of a Gaussian is $\sqrt{\pi }{e}^{-k^2/4}$. Multiplying by the heat factor and inverting gives the Gaussian initial data for the heat equation solutionIt recovers the initial profile as $t\downarrow 0$ and preserves its spatial integral $\sqrt{\pi }$.

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.

Split the Fourier transform integral at zero:By Fourier inversion applied to this pair, or by the residue theorem, the other transform isFor example, inversion with $\lambda =\mu$ gives $\int e^{ikx}(k^2+{\mu }^2{)}^{-1}dk=(\pi /\mu )e^{-\mu |x|}$; evenness supplies the desired sign convention.

Take the spatial Fourier transform of the Laplace equation. For $y>0$ it gives ${\hat{\psi }}_{yy}-k^2\hat{\psi }=0$. Removing the exponentially growing solution and matching the boundary givesAt $k=0$ this formula is interpreted by continuity; a single transform value does not alter the inverse integral. The previous transform pair identifiesthe Poisson kernel for the upper half-plane. The convolution theorem therefore gives the Poisson integralThis is the bounded harmonic solution for bounded decaying data, or the corresponding solution in an appropriate Fourier class. Such a growth condition matters if vertical decay is interpreted only pointwise: the nonzero harmonic function $e^{x^2-y^2}\sin (2xy)=\mathrm{Im}{e}^{(x+iy{)}^2}$ vanishes on the real axis and tends to zero for fixed $x$ as $y\to \infty$, but grows rapidly in $x$ and has no ordinary spatial Fourier transform. It can be excluded by requiring boundedness; the stated pointwise limits alone would not establish uniqueness.

For the interval data, direct integration gives the Poisson extension of interval boundary dataAs $y\downarrow 0$ it approaches $c$ for $|x|<a$, zero for $|x|>a$, and $c/2$ at $x=\pm a$. At those jump points the prescribed isolated values cannot be attained by a continuous extension; they do not affect the integral. The solution decays vertically, with $\psi \sim 2ac/(\pi y)$ for fixed $x$ and large $y$.

The convolution theorem turns the product $e^{-k^{2}t}\tilde{\theta }(k,0)$ into a spatial convolution. Its kernel isComplete the square:After rescaling the variable, the supplied translated Gaussian integral givesThis is the heat-kernel solution for diffusivity one. Its kernel integrates to one, preserving the total temperature integral, and approaches a Dirac delta function as $t\downarrow 0$. Its variance is $2t$; the factor is appropriate to ${\theta }_t={\theta }_{xx}$, rather than the alternative convention with diffusivity $1/2$.

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.

The filter is a real even triangular function. Its Fourier transform iswith $\tilde{f}(0)=1$. Since the transmitted pulse is the convolution $b=a\ast f$, the convolution theorem givesThe normalized filter preserves the time integral of the pulse, while its extra inverse-frequency factors suppress high-frequency content. It is a smoothing filter rather than an exact compact-bandwidth cutoff.

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.

For an integrable applied voltage, the convolution theorem givesThe integrated electric current is the zero-frequency transform, $Q=\tilde{I}(0)$. Since $\tilde{V}(0)$ can be arbitrary, zero total electric charge for every such input is equivalent toFor nonzero $p_1,p_2$, this can be writtenBecause both amplitudes are positive, nonzero $p_1,p_2$ must have opposite signs. If exactly one $p_i$ is zero there is no cancellation with positive amplitudes; if both vanish the kernel is identically zero and any amplitudes work. In particular, if both frequency parameters were additionally required to be positive, no such nontrivial system would exist. The PDF requires positivity of $a_i,{\gamma }_i$, but allows real fixed $p_i$.

This is zero-frequency cancellation of a causal response kernel. The phrase arbitrary voltage is understood within a class for which the net-charge integrals exist; a permanent DC input need not have finite total charge.

Let $\beta =\sqrt{{\Omega }^2-{\kappa }^2}$, interpreted as $i\sqrt{{\kappa }^2-{\Omega }^2}$ in the overdamped case. Factoring the denominator givesFourier inversion and the convolution theorem givewhereThus the requested numerator isFor all positive $\kappa ,\Omega$, both poles lie in the upper half-plane. In the overdamped case their imaginary parts are $\kappa \pm \sqrt{{\kappa }^2-{\Omega }^2}$, both positive. At equality they merge into a double pole.