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As a tempered distribution on ${\mathbb{R}}^3$,for a nonzero convention-dependent constant $C$. One proof writesand applies the Fourier transform of a Gaussian before substituting $u=|\xi {|}^2/(4t)$.

The Fourier transform of a Gaussian and the convolution theorem give this multiplier with the angular-frequency convention $e^{-ix\cdot \xi }$. In cycles-per-length frequencies $k$, it is $e^{-2{\pi }^2{\sigma }^2|k{|}^2}$. Each coordinate has variance ${\sigma }^2$, and higher frequencies receive stronger attenuation.

Take the Fourier transform in $x$. The transform of the source is $\delta (t-t^{\prime })e^{-ik{x}^{\prime }}$, so the Green function equation becomesChoose the causal Green function, zero for $t<t^{\prime }$. Away from $t^{\prime }$ the solution is a decaying exponential, and integrating across $t^{\prime }$ gives the jump $\tilde{G}(t^{\prime }+)-\tilde{G}(t^{\prime }-)=e^{-ik{x}^{\prime }}$. ThereforeThe phase shifts the kernel's centre to $x^{\prime }$. Applying the inverse Fourier transform and the Fourier transform of a Gaussian with $a=D(t-t^{\prime })$ givesFor positive elapsed time this heat kernel solves the homogeneous diffusion equation and has unit integral; as elapsed time tends to zero it approaches the spatial delta distribution, supplying the required impulse.

There is a literal domain error in the printed formula. A semivariogram indexed by a signed displacement must satisfy $\gamma (h)=\gamma (-h)$, since reversing the two observations leaves the squared increment unchanged. For positive ${\sigma }^2$ or ${\tau }^2$, the displayed positive-lag value is nonzero whereas the displayed negative-lag value is zero. Thus the claim on all of $\mathbb{R}$ is false as written.

Interpret the argument as a nonnegative distance, or use the even extension of the Gaussian semivariogram. For ${\sigma }^2,{\tau }^2\ge 0$ and $\varphi >0$, the intended model isIt admits the covariogramThe Gaussian kernel is a positive-definite kernel, as can be seen from the Fourier transform of a Gaussian: in one dimensionand the nonnegative spectral density makes every finite covariance quadratic form nonnegative. In higher dimensions use the product Gaussian density. Adding independent Gaussian white noise of variance ${\tau }^2$ adds the diagonal nugget covariance. Thus a zero-mean stationary Gaussian random field exists with this covariance, and $C(0)-C(h)$ gives the corrected semivariogram.

For nonzero structured variance, the nugget effect is the right limit ${\tau }^2$ at zero, the sill of a semivariogram is ${\tau }^2+{\sigma }^2$, and the range of a semivariogram is infinite if defined as the distance at which the sill is exactly reached. A 95-percent practical range solves $1-e^{-{\varphi }^2{h}^2}=0.95$:The scale $1/\varphi$ is sometimes called the range parameter; it is neither the exact range nor the 95-percent practical range. If ${\sigma }^2=0$, there is no nontrivial structured range.

The fact that a semivariogram does not determine stationarity is a further distinction: a semivariogram specifies increment variation, not a unique process or absolute covariance. Adding an independent random constant adds a constant to $C$ without changing $\gamma$. More strongly, subtracting $W(0)$ from a stationary field $W(s)$ preserves its semivariogram but generally makes its variance depend on $s$. The corrected statement is that this semivariogram admits a second-order stationary model, with the displayed covariance using the usual convention $C(h)\to 0$ at large distance; it does not force every compatible process to be stationary.

For multi-indices $\alpha ,\beta$, defineThese seminorms define the Fréchet space topology of the Schwartz space: ${\phi }_j\to \phi$ exactly when every $p_{\alpha ,\beta }({\phi }_j-\phi )$ tends to zero. An equivalent increasing family isA tempered distribution is a continuous linear functional on this space. Equivalently, $|⟨T,\phi ⟩|\le C{q}_m(\phi )$ for some $C,m$. The usual weak convergence of tempered distributions means $⟨T_j,\phi ⟩\to ⟨T,\phi ⟩$ for every fixed Schwartz function. The strong dual topology instead requires uniform convergence on every subset of the Schwartz space that is a bounded set in a topological vector space; the Fourier maps below are continuous in both topologies.

Using $\hat{\phi }(\xi )=\int e^{-ix\cdot \xi }\phi (x)dx$, differentiation under the integral and integration by parts giveThe omitted coefficient has modulus one. By the Leibniz rule, every integrand is a finite sum of a polynomial times a derivative of $\phi$. Inserting the integrable weight $(1+|x|{)}^{-n-1}$ bounds its L1 norm by finitely many Schwartz space seminorms. In particular $q_m(\hat{\phi })\le C_m{q}_{m+n+1}(\phi )$. Thus the Fourier transform maps $\mathcal{S}$ continuously into itself.

For completeness, the Fourier inversion theorem follows here by Gaussian regularization. The inverse transform of $e^{-t|\xi {|}^2}$ is $g_t(x)=(4\pi t{)}^{-n/2}{e}^{-|x{|}^2/(4t)}$. The Fourier transform of a Gaussian and Fubini's theorem showAs $t\downarrow 0$, the right side tends to $\phi (x)$ by the approximate identity property, while the left side tends to the undamped inverse integral by dominated convergence theorem, since $\hat{\phi }\in L^1$. HenceThe inverse is $(2\pi {)}^{-n}$ times reflection composed with the continuous Fourier transform, and is therefore continuous on the Schwartz space. This proves a continuous linear isomorphism with continuous inverse.

Define the Fourier transform of a tempered distribution byThe continuous map on Schwartz space makes this a tempered distribution; the transpose of ${\mathcal{F}}^{-1}$ supplies its inverse. Pointwise convergence of pairings proves weak continuity. For the strong dual topology, the transform of a bounded set of Schwartz functions is bounded, so uniform convergence of pairings on bounded sets proves continuity of both Fourier maps there too.

Now let $R\in SO(n)$ be a rotation matrix in the special orthogonal group and write ${\rho }_{R}T=T\circ R$. A change of variables with unit Jacobian givesThe rotation equivariance of the Fourier transform on tempered distributions follows by duality:Therefore $\widehat{{\rho }_{R}T}={\rho }_R\hat{T}$. Applying this identity and the inverse Fourier transform gives the two directions:This is precisely preservation of radial tempered distributions. For $n>1$, $SO(n)$ acts transitively on spheres, so an invariant smooth function is an ordinary radial function.

For the Fourier transform convention of equation 3, insert the given power spectrum as $S_F$. With $t=\nu L/2$,The Fourier transform of a Gaussian, or the order-zero Hankel transform, also givesThus the coherent attenuation in a white-noise random medium isFor a unit incident plane wave, the Fresnel propagator leaves the initial envelope unchanged, so the answer reduces to $\boxed{m=e^{-k_0^2{\mu }^{2}Lx/(2\pi )}}$. If instead the given numerical function is interpreted as the self-reciprocal Hankel transform spectrum suggested by the printed equations 4 and 5, then $Q(r)=2{\mu }^{2}L{e}^{-r^2/L^2}$ and the alternative isThe factor $2\pi$ ambiguity is present in the original PDF. No unique numerical attenuation coefficient follows until the power spectrum convention is fixed. The covariance strength $Q(0)$ has units of length in this longitudinal white noise model, making both attenuation exponents dimensionless.

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.

Although $x\mapsto |x{|}^{-1}$ is not integrable on all of ${\mathbb{R}}^3$, it defines a tempered distribution. For $x\ne 0$, the Gamma function integral givesThe three-dimensional form of the Fourier transform of a Gaussian isTherefore, up to a nonzero constant depending on the Fourier-transform convention,For $\xi \ne 0$, the substitution $u=|\xi {|}^2/(4t)$ turns the last integral intoThus, as asserted by the Fourier transform of inverse distance in three dimensions,as a tempered distribution, where $C^{\prime }\ne 0$.

Choose a nonzero $f\in L^p({\mathbb{R}}^n)\cap L^1({\mathbb{R}}^n)$ whose transform belongs to $L^q$, for example a Gaussian by the Fourier transform of a Gaussian. For $\lambda >0$, set $f_{\lambda }(x)=f(\lambda x)$. The scaling property of the Fourier transform givesThe assumed estimate, applied for every $\lambda >0$, becomesSince both norms are nonzero, this can hold as $\lambda$ tends both to zero and to infinity only if the powers agree. Thus the scaling necessity for an Lp to Lq Fourier bound yieldsHence $q$ is uniquely the conjugate exponent of $p$.

The Cauchy integral theorem states that if $f$ is holomorphic on a simply connected domain, then its integral around every closed piecewise smooth contour in that domain is zero.

Put $q=b-k$. Completing the square givesAfter the change of variable $z=x-iq/(2a)$, the integral runs along the horizontal line $\mathrm{Im}z=-q/(2a)$. The integrand $e^{-a{z}^2}$ is entire. Apply Cauchy's theorem to a rectangle joining this line to the real axis; the two vertical integrals tend to zero as their real parts tend to $\pm \infty$, because $a>0$. The contour may therefore be shifted to the real line. Hence the Fourier transform of a Gaussian gives