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The Fourier inversion theorem makes decomposition into frequencies reversible: a complete Fourier transform, including its wave phases, determines the original function up to equality almost everywhere. With the angular-frequency conventionif $f$ and $\hat{f}$ belong to the L1 space, thenalmost everywhere. The right side is a continuous representative of $f$. If $f$ itself is continuous, equality holds everywhere, as explained by the continuous version of Fourier inversion. The integrability hypotheses ensure that both displayed integrals are ordinary absolutely convergent integrals. For arbitrary elements of the L2 space, the Plancherel theorem instead supplies inversion in the L2 space norm.

Each factor $e^{ix\cdot \xi }$ is a Fourier mode; the inverse integral adds them with the complex weights $\hat{f}(\xi )/(2\pi {)}^n$. The normalization is part of the recipe. In one dimension, writing ordinary frequency as $\nu =\xi /(2\pi )$ gives the equivalent pairwhere $F(\nu )=\hat{f}(2\pi \nu )$. Reversing both exponential signs gives another equivalent convention, but the forward and inverse formulas must use opposite signs consistently.

For example, the Fourier transform of a Gaussian givesFourier inversion says that a weighted continuum of oscillations reconstructs this smooth Gaussian function. Narrowing the original Gaussian function by increasing $a$ broadens its frequency spectrum. This is a concrete demonstration that a sharply localized shape needs a wider range of frequencies.

Inversion is what makes a frequency calculation useful: transform a function, solve the simpler problem there, then return to the original variables. A brief proof outline is to multiply $\hat{f}(\xi )$ by $e^{-ϵ|\xi {|}^2}$, $ϵ>0$. The inverse integral becomes convolution with a normalized Gaussian function, an approximate identity. As $ϵ$ tends to zero, dominated convergence gives the unregularized inverse integral, while the approximate identity recovers $f$ almost everywhere. Joseph Fourier's Théorie analytique de la chaleur (1822), Chapter IX, article 417, contains a cosine double-integral representation antecedent to this modern formula; the hypotheses above are the modern statement.^[ref]

For $\alpha >0$, put $a=\alpha /2$. The regular distribution of $(1+x^2{)}^{-a}$ is a tempered distribution, since the function is bounded. Its Fourier transform is the integrable functionwith arbitrary finite value at zero when the displayed integral diverges there. To prove the identity, the gamma function gives $(1+x^2{)}^{-a}=\Gamma (a{)}^{-1}{\int }_0^{\infty }{t}^{a-1}{e}^{-t}{e}^{-t{x}^2}dt$. Pair this with a Schwartz function, use the Fourier transform of a Gaussian, and interchange integrals by Fubini's theorem. The absolute bound for the transformed pairing follows from Tonelli's theorem:The positive kernel and dominated convergence give positivity and continuity away from zero. It has a continuous representative at zero exactly when $\alpha >1$, in which case $h_{\alpha }(0)=\sqrt{\pi }\Gamma ((\alpha -1)/2)/\Gamma (\alpha /2)$. For $\alpha \le 1$, Fatou's lemma gives $h_{\alpha }(\xi )\to \infty$ as $\xi \to 0$, so no change on a set of zero Lebesgue measure can give continuity.

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 Gaussian function $f_a(x)=e^{-\pi a{x}^2}$ with $a>0$. Every derivative is a polynomial times the same Gaussian function, so $f_a$ is a Schwartz function. The Fourier transform of a Gaussian gives, in the cycles convention,Thus the Poisson summation formula gives the exact identityThe Gaussian Fourier transform is given in scaled form in Exercise 41 of Terence Tao's 245C, Notes 2: The Fourier transform.^[ref] The identity is also the imaginary-axis transformation of the Jacobi theta function. At $a=1$ the Gaussian function is its own Fourier transform; at reciprocal values of $a$ its spatial and spectral widths exchange.

For a short derivation of the Gaussian Fourier transform, write $H(\xi )={\hat{f}}_a(\xi )$. Differentiation under the integral and integration by parts, using $f_a^{\prime }(x)=-2\pi ax{f}_a(x)$, giveThe Gaussian integral gives $H(0)=a^{-1/2}$, so this ordinary differential equation yields the claimed formula. The decay of the Gaussian function justifies these steps and removes the boundary term.^[ref]

For $a=0.04$, the physical Gaussian function has width parameter $1/\sqrt{a}=5$, while its Fourier transform has width parameter $\sqrt{a}=0.2$. Many physical samples matter, but the dual zero sample already givesFor $a=25$, use the physical side instead:These are bounds on the full positive tails, obtained below, rather than estimates based only on the first omitted term. Ordinary floating-point arithmetic rounds these sums to $5$ and $1$; the exact real numbers are slightly larger. A Gaussian sum truncation bound lets us choose the cheaper side and certify its truncation error before evaluating it.

Let $\mu$ be a finite positive surface measure and $|f|\le 1$. Expand $|\widehat{f\mu }(x){|}^{2k}$ and integrate against $e^{-\pi |x{|}^2/R^2}$. The Fourier transform of a Gaussian turns the oscillatory kernel into $R^d{e}^{-\pi R^2|{\sum }_{j\le k}{\omega }_j-{\sum }_{j>k}{\omega }_j{|}^2}$, which is nonnegative. Taking absolute values of the bounded amplitudes gives an upper bound by the same moment with $f=1$. This weighted positivity is valid even when $f$ is merely measurable; it transfers a scalar surface-decay estimate to a local bounded-density moment estimate.

Complete the square inside the Fourier transform of the Gaussian heat kernel:After $z=x/(2\sqrt{t})$,The supplied shifted Gaussian integral equals $\sqrt{\pi }$ for real $\sqrt{t}\omega$. HenceAt zero this says $\int q_t=1$, so convolution with the heat kernel averages rather than rescales the initial mass. The formula is an instance of the Fourier transform of a Gaussian.

For $\epsilon >0$, insert a Gaussian function into the proposed inverse Fourier transform:The Fourier bounds for Schwartz functions show that $\hat{\varphi }$ is an integrable function, so dominated convergence givesOn the other hand, Fubini's theorem applies because $\Vert \varphi {\Vert }_1\int e^{-\epsilon {\xi }^2/2}d\xi <\infty$. The given Fourier transform of a Gaussian, rescaled by $\sqrt{\epsilon }$, yieldsThe last integrand is bounded by $\Vert \varphi {\Vert }_{\infty }{e}^{-z^2/2}/\sqrt{2\pi }$. Since $\varphi$ is continuous, another application of dominated convergence makes this tend to $\varphi (x)$. Equating the limits proves the Fourier inversion theorem:The Gaussian functions above form an approximate identity; no Fourier inversion was used in evaluating their integrals.

Since $0<f_{\alpha }\le 1$, its regular distribution satisfiesso $f_{\alpha }$ is a tempered distribution for every $\alpha >0$.

Put $a=\alpha /2$. The defining integral of the gamma function, with $r=(1+x^2)t$, givesThe Fourier transform of a Gaussian suggests the following representative for the Fourier transform of an inverse power of one plus x squared:All factors are positive. Tonelli's theorem evaluates its total integral, allowing an infinite value at the single point zero:Assign any finite value at zero if necessary. This gives an integrable function with $\Vert h_{\alpha }{\Vert }_1=2\pi$.

To verify equality as tempered distributions, take any Schwartz function $\varphi$. Substituting the mixture into $⟨f_{\alpha },\hat{\varphi }⟩$ is justified byFor each $t>0$, Fubini's theorem and the Fourier transform of a Gaussian identify the inner pairing with $\int \sqrt{\pi /t}{e}^{-{\xi }^2/(4t)}\varphi (\xi )d\xi$. Interchanging the resulting integrals is valid because their absolute value is bounded by $2\pi \Vert \varphi {\Vert }_{\infty }$ after integrating the positive kernel. ThereforeFinally, on a compact set with $|\xi |\ge \delta >0$, the mixture integrand is bounded by $t^{a-3/2}{e}^{-t-{\delta }^2/(4t)}$, which is integrable both at zero and at infinity. Dominated convergence proves continuity there. Hence $\boxed{\widehat{f_{\alpha }}=h_{\alpha }\in L^1(\mathbb{R}),h_{\alpha }(\xi )>0\text{ and continuous for }\xi \ne 0}$.

Mode locking fixes the relative phases of a laser's longitudinal cavity modes, so they interfere constructively in a regularly repeating short pulse rather than producing an uncontrolled temporal waveform. Active mode locking uses an externally driven modulation of loss, gain or phase at the cavity repetition frequency; an electro-optic modulator acting as a periodic transmission gate is one example. Passive mode locking uses the pulse itself to select a high-intensity waveform: a saturable absorber has lower loss at the pulse peak and can lock the phases without an external clock. A semiconductor saturable-absorber mirror is a practical example. Gain recovery, dispersion and saturation must support a stable circulating pulse.

Using the complex-field convention in the question, add the fields of all the oscillating modes. Set $m=p-q_0$, so that ${\omega }_p={\omega }_{q_0}+m\Delta \omega$. ThenThere are $N+1$ terms with the printed inclusive indexing. The sum is periodic with periodwhich is the separation of adjacent pulses when many modes contribute. The carrier factor need not repeat over this interval.

To specify the carrier-envelope phase sign, write the real field near pulse $j$, with $\tau =t-jT$, as $A(\tau )\cos ({\omega }_{q_0}\tau +{\phi }_j)$. Its cosine phase is ${\phi }_j={\omega }_{q_0}jT-{\varphi }_{q_0}$. Since $q_0\Delta \omega T=2\pi q_0$,Equivalently, the phase ${\varphi }_j$ in the complex convention $E_j=A(\tau )e^{-i{\omega }_{q_0}\tau +i{\varphi }_j}$ advances by $-{\omega }_{\mathrm{ce}}T$. This is the negative of the cosine phase, not a different physical prediction. The carrier-envelope offset frequency fixes this pulse-to-pulse slip. It matters for a few-cycle field's peak electric waveform and for locating an optical frequency comb's lines; a stable intensity envelope alone does not fix the slip.

If $\delta /\Delta \omega \gg 1$ and $N\Delta \omega /2\gg \delta$, many closely spaced modes lie under the spectral Gaussian and the cutoffs have negligible weight. Near one pulse, replace the sum by its continuum integral:The Fourier transform of a Gaussian therefore givesThis is a local single-pulse approximation. The discrete spectrum produces its periodic replicas; extending one Gaussian over all time would lose them. The temporal light intensity is proportional to $e^{-{\delta }^2{t}^2/2}$, while the spectral intensity envelope is proportional to $e^{-2{\Omega }^2/{\delta }^2}$. Their full widths at half maximum areIn ordinary frequency this time-bandwidth product is $\Delta {\nu }_L\tau =2\ln 2/\pi \simeq 0.441$. The spectral phases are flat, so this is a transform-limited pulse; chirp would generally increase its temporal width.

A chirp is variation of the instantaneous frequency through a signal. Here the group delay relative to the pulse centre at the optical lens isConsequently $\eta >0$ orders lower frequencies before higher frequencies; $\eta <0$ reverses that ordering. A representative strongly chirped optical pulse iswith a smooth localized envelope $A(t)$. The carrier oscillations become closer together towards later times for positive $\eta$ and farther apart for negative $\eta$.

The relation between the spectral phase and a locally assigned instantaneous frequency is a stationary phase method, not an exact reconstruction from spectral phase alone: the spectral wave amplitude also matters. For example, if the spectral envelope is $e^{-(\omega -{\omega }_0{)}^2/(2{\sigma }^2)}$, a Fourier transform of a Gaussian gives the exact real-field phase ${\omega }_{0}t+\eta t^2/(4|a{|}^2)$ up to a constant, where $a=1/(2{\sigma }^2)-i\eta$. The slope retains the sign of $\eta$ and approaches $1/(2\eta )$ for $|\eta |{\sigma }^2\gg 1$.

Put $b=\Delta {\omega }_b>0$ and $s=t-t_m$. At the peak time, $e^{-ip\Delta \omega t_m}=1$, so the exact comb sum becomesFor many occupied normal modes, $b\gg \Delta \omega$, replace the sum by an integral with density $1/\Delta \omega$. Setting $x=\omega -{\omega }_0$ and evaluating the Fourier transform of a Gaussian givesThus the isolated-optical pulse approximation hasThe optical pulse has a Gaussian function envelope and no chirp. Its carrier angular frequency is ${\omega }_0$, and its light intensity full width at half maximum is $2\sqrt{2\ln 2}/b$. Successive envelope maxima have different carrier-envelope phases, consistent with b.

Approximation in the printed assertion. A discrete optical frequency comb gives a train, not one exact isolated Gaussian pulse. Applying the Poisson summation formula to the ideal sum over all integers instead givesNear $t_m$, the $j=m$ term dominates when $bT\gg 1$. The other terms are exponentially small but not identically zero. A single Gaussian function used globally would decay between $0$ and $T$, whereas the exact light intensity has equal maxima there; this supplies a concrete reason the isolated form cannot be an exact global identity. The extension to negative frequencies is the usual narrow-positive-band analytic-field approximation when ${\omega }_0\gg b$.

Use $\widehat{f\sigma }(x)={\int }_{S^1}f(\omega )e^{-2\pi ix\cdot \omega }d\sigma (\omega )$, with arclength measure. We first bound the transform for constant density. By rotational symmetry, $\hat{\sigma }(x)={\int }_0^{2\pi }{e}^{-2\pi ir\cos \theta }d\theta$, $r=|x|$. For $r\ge 1$, remove neighborhoods of width $h=r^{-1/2}$ about the stationary points $0,\pi$. Their contribution is $O(h)$. On each remaining interval integrate by parts using the phase derivative $2\pi r\sin \theta$. The boundary terms are $O((rh{)}^{-1})$, and the integral of the derivative of its reciprocal is bounded bySplitting at the stationary points gives the same bound on the other half of the circle. Together with the trivial estimate for bounded $r$, this proves $|\hat{\sigma }(x)|\lesssim (1+|x|{)}^{-1/2}$.

For merely measurable $f$ there is no reason that its amplitude permits this integration by parts. Instead use Gaussian positivity for even extension moments. SetExpand the fourth power and integrate first in $x$. The Fourier transform of a Gaussian givesAll exchanges of integrals are justified by bounded amplitudes, finite surface measure and the integrable Gaussian. The kernel is nonnegative and $|f|\le 1$ almost everywhere. Taking the absolute value of the integral therefore gives $0\le I_R(f)\le I_R(1)$. The preceding scalar decay estimate and polar coordinates yieldFor the last bound split at $1$ and $R$; the tail becomes an integrable $e^{-\pi s^2}/s$ integral after $r=Rs$. Since the Gaussian is at least $e^{-\pi }$ on the radius-$R$ ball, the local fourth-moment restriction estimate for the circle follows:Every constant here is independent of $f$ and $R$.

For the requested connection to Besicovitch sets, let $\delta$ be small and dilate the unit segments of a bounded planar Kakeya set by ${\delta }^{-2}$. Their $\delta$-neighborhoods become direction-separated rectangles of length ${\delta }^{-2}$ and width ${\delta }^{-1}$. A smooth circle cap of angular width $c\delta$ has a Fourier extension operator of magnitude at least $c^{\prime }\delta$ on such a rectangle: after removing the constant phase, the tangential phase variation is $O(c)$ and the normal variation is $O(c^2)$. A frequency modulation translates the rectangle to any desired location. Choose separated caps for the rectangle directions, and multiply their modulated densities by independent Rademacher random variables. Because the caps are disjoint, the input remains bounded by one.

Apply the local fourth-moment estimate in a ball of radius $C_E{\delta }^{-2}$ containing the rectangles, then take expectation over the Rademacher random variables. The fourth moment dominates the square of the sum of squared packet magnitudes, givingThere are $M\asymp {\delta }^{-1}$ rectangles, each of area $\asymp {\delta }^{-3}$, so the integral of their sum is $\asymp {\delta }^{-4}$. Cauchy-Schwarz forces their union area to be at least $c_E{\delta }^{-4}/\log (1/\delta )$. Scaling back gives $|E_{C\delta }|{\gtrsim }_{E}1/\log (1/\delta )$, the planar Kakeya neighborhood lower bound. The covering-number argument in Question 1 then gives both Minkowski dimensions equal to two.

Use the Gaussian pole transition for a large gamma parameter. With $t=1+s/\sqrt{n}$ and $\lambda =n+i\mu \sqrt{n}+\nu +o(1)$, Taylor expansion givesThe upper semicircle is traversed from left to right, so $\int ds/s=-i\pi$ on the indentation. The minus sign in $dt/(1-t)$ therefore makes its contribution $+i\pi$. The remaining leading integral isAt $\mu =0$ its Cauchy principal value vanishes by oddness. Differentiating with respect to $\mu$ removes the pole, leaving the Fourier transform of a Gaussian:Integrating from zero givesThis is uniform for bounded real $\mu ,\nu$ (and extends analytically over bounded complex values where the local contour is continued consistently). Contributions away from the saddle are exponentially small after localization. The bounded shift $\nu$ enters at the next order, not the leading error function. For example, retaining the displayed cubic correction gives $\sqrt{2\pi /n}{e}^{-{\mu }^2/2}[\nu +(2+{\mu }^2)/3]$ as the next term. This also checks the constant and contour sign.

The connected generating functional is $W_0[J]=\log Z_0[J]$. Its second functional source derivative gives the connected correlation function; the source-independent determinant drops out. Translational invariance therefore givesThis covariance is independent of the source in a purely Gaussian theory; at zero source the mean field vanishes.

Fourier inversion can be evaluated with the Schwinger parameterization. Since $(p^2+{\mu }^2{)}^{-1}={\int }_0^{\infty }ds{e}^{-s(p^2+{\mu }^2)}$, the elementary Fourier transform of a Gaussian giveswhere $K$ is the Modified Bessel function of the second kind. This integral formula also gives both asymptotic regimes directly.

For $\mu r\ll 1$, set $u=r^2/(4s)$. The integral tends to the gamma function integral ${\int }_0^{\infty }{u}^{D/2-2}{e}^{-u}du=\Gamma (D/2-1)$; this is finite for $D>2$. Hence, at distances larger than the microscopic cutoff,For $\mu r\gg 1$, the exponent ${\mu }^{2}s+r^2/(4s)$ has its minimum at $s=r/(2\mu )$, with value $\mu r$. Gaussian expansion about that saddle, or the large-argument asymptotic expansion of a modified Bessel function, gives $K_{\rho }(z)\sim \sqrt{\pi /(2z)}{e}^{-z}$. ThereforeThis is the massive Gaussian field correlation tail with its normalization displayed. For example $D=3$ gives exactly $G_0(r)=\alpha e^{-r/\xi }/(4\pi r)$, verifying both regimes and the exponential length scale.

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.