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For a complex symmetric three-dimensional $A=G+iB$ with real symmetric $B$ and positive-definite $G$, the reciprocal is a regular tempered distribution since $|x^{T}Ax|\ge x^{T}Gx$. Holomorphic continuation of the real quadratic transform proves the displayed expression. The determinant branch is the analytic determinant square root for accretive symmetric matrices. The other root has positive real part because $\mathrm{Re}({\xi }^T{A}^{-1}\xi )={\xi }^T{G}^{-1/2}(1+C^2{)}^{-1}{G}^{-1/2}\xi >0$ for nonzero real $\xi$. Both the original and transformed singularities are locally integrable, so the continued identity holds as distributions at the origin too.

For the forward implication, finite order of a distribution makes $F(\zeta )=⟨u,e^{-ix\cdot \zeta }⟩$ entire. Use a smooth cutoff in an $\epsilon$-neighborhood of the support set, with derivatives of order $j$ bounded by $C_j{\epsilon }^{-j}$. Taking $\epsilon =(1+|\zeta |{)}^{-1}$ in the finite-order estimate gives the polynomial factor and adds at most $e$ to the desired exponential bound. This shrinking cutoff recovers the exact support indicator, instead of an arbitrarily enlarged one.

For the converse, the real restriction of $F$ has polynomial growth and defines an inverse tempered distribution. If a test function $\phi$ is supported in $x\cdot \omega \ge R+\eta$ for a unit vector $\omega$ and $\eta >0$, contour shifting givesRepeated integration by parts gives, for every integer $M$,Choosing $2M>N+n$ justifies the shift by Cauchy integral theorem and bounds the pairing by $C^{\prime }(1+t{)}^{N+2M}{e}^{-\eta t}$, which tends to zero. Half-spaces of this form cover the complement of the ball, and a partition of unity proves the support inclusion. For general compact convex $K$, replace $R$ by $H_K(\omega )$ in each separating direction. Fourier inversion gives uniqueness. A related proof outline appears in Richard Melrose's distribution-theory problem set.

For $T\in {\mathcal{S}}^{\prime }$ and $\psi \in \mathcal{S}$, define $(T\ast \psi )(x)=⟨T_y,\psi (x-y)⟩$. Translations are smooth in the Schwartz space, so every derivative is ${\partial }^{\alpha }(T\ast \psi )(x)=⟨T,{\partial }^{\alpha }\psi (x-\cdot )⟩$. The finite-seminorm continuity estimate for $T$ implies $|{\partial }^{\alpha }(T\ast \psi )(x)|\le C_{\alpha }(1+|x|{)}^m$ for one $m$. Thus the result is a smooth function defining a tempered distribution, but it need not be a Schwartz function: $1\ast \psi =1$ when $\int \psi =1$. If both factors are radial, testing the rotation action on the translated $\psi$ shows that the convolution is radial.

The Fourier transform with negative exponential and inverse factor $(2\pi {)}^{-n}$ is a continuous isomorphism of the Schwartz space. Weighted frequency derivatives are transforms of derivatives of weighted input functions, so their suprema are controlled by finitely many Schwartz seminorms. Fourier inversion gives ${\mathcal{F}}^2=(2\pi {)}^n\mathcal{R}$, where $\mathcal{R}f(x)=f(-x)$. Transposition gives a continuous isomorphism on tempered distributions as well.

For $p>-1$ this formula is interpreted away from $k=0$ as a tempered distribution or by exponential regularization. It also supplies local high-frequency cusp terms after multiplying by a smooth cutoff. Even nonnegative integer $p$ give smooth polynomials and only contact terms at $k=0$, so their localized forms have no algebraic tail. For example, the cusp expansion of $(1+|x{|}^3{)}^{-1}$ gives $-12|k{|}^{-4}+29!|k{|}^{-10}+O(|k{|}^{-16})$.

For a real symmetric positive-definite three-dimensional matrix, the reciprocal quadratic form is locally integrable and defines a regular tempered distribution. With the Fourier kernel $e^{-ix\cdot \xi }$, radial Abel regularization gives $\mathcal{F}(|x{|}^{-2})=2{\pi }^2/|\xi |$. The linear change of variables $y=G^{1/2}x$ gives the displayed formula. No principal value or origin-supported correction is required. The result is homogeneous of degree $-1$ on the frequency side.

With $\hat{\phi }(\xi )=\int e^{-ix\cdot \xi }\phi (x)dx$, the Fourier transform of a tempered distribution is defined byContinuity of the Fourier transform on the Schwartz space makes this well defined. For regular integrable functions, Fubini's theorem shows that it agrees with the ordinary transform. The inverse convention has factor $(2\pi {)}^{-n}$ and the positive exponential.

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 $u=|\xi {|}^2/(4t)$.

For $u(x)=\frac{1}{2}\log (1+x^2)$, the angular-frequency transform is the tempered distributionThe numerator cancels the singularity at zero and makes this integral absolutely convergent. Differentiating $u$ and using $\widehat{(1+x^2{)}^{-1}}=\pi e^{-|\xi |}$ determines this expression up to a Dirac delta distribution. That delta coefficient is zero: testing with the expanding Gaussian ${\phi }_n(x)=(2\sqrt{\pi }{)}^{-1}{e}^{-x^2/(4n)}$ makes both the proposed expression and $⟨u,{\hat{\phi }}_n⟩$ tend to zero, while ${\phi }_n(0)$ stays fixed. Changing the subtraction convention would change the delta coefficient.

A Schwartz function is a smooth function for which every seminormis finite. These seminorms define the Fréchet topology of the Schwartz space $\mathcal{S}({\mathbb{R}}^n)$. The tempered distributions form its continuous linear dual ${\mathcal{S}}^{\prime }$. We use bilinear distribution pairing, with no complex conjugation on the test function.

Differentiation under the Fourier integral and integration by parts giveTherefore each Schwartz seminorm of $\hat{\phi }$ is bounded by a finite sum of $L^1$ norms of polynomially weighted derivatives of $\phi$. Insert the integrable weight $⟨x{⟩}^{-n-1}$ to bound those norms by finitely many Schwartz seminorms. This proves that $\mathcal{F}:\mathcal{S}\to \mathcal{S}$ is continuous.

To justify inversion without a merely formal exchange of integrals, insert $e^{-\epsilon |\xi {|}^2/2}$ in the inverse integral. The Gaussian Fourier integral givesThe normalized Gaussian is an approximate identity, so the right side tends to $\phi (x)$; the left side converges by dominated convergence theorem since $\hat{\phi }$ is integrable. It follows thatReflection is continuous in Schwartz seminorms, so ${\mathcal{F}}^{-1}=(2\pi {)}^{-n}\mathcal{R}\mathcal{F}$ is continuous as well. This proves the Fourier transform isomorphism of the Schwartz space.

Define the distributional Fourier transform byIt agrees with the ordinary integral transform whenever Fubini applies. Transposing the Schwartz inverse gives an inverse on ${\mathcal{S}}^{\prime }$, and the same squared-transform identity holds. For the usual strong dual topology, a seminorm is $p_B(u)={\sup }_{\phi \in B}|⟨u,\phi ⟩|$ for a bounded set $B\subset \mathcal{S}$. Since a continuous linear Schwartz map takes bounded sets to bounded sets, $p_B(\hat{u})=p_{\mathcal{F}B}(u)$ proves continuity, and similarly for the inverse. Continuity also holds in the weak dual topology. Thus the transform is a continuous isomorphism on tempered distributions, with the stated normalization.

For a real symmetric positive-definite matrix $G$, there is $c>0$ with $g(x)=x^{T}Gx\ge c|x{|}^2$. The reciprocal is locally integrable in dimension three: the radial factor near zero is $r^2{r}^{-2}dr$. At infinity, rapid decay of a Schwartz test function makes it integrable. More quantitatively,This single seminorm bound proves a tempered distribution. No principal-value extension is needed at the origin.

First calculate the isotropic transform. For $f_{\epsilon }(x)=e^{-\epsilon |x|}/|x{|}^2$ and $k=|\xi |>0$, spherical integration and the supplied sine-integral identity giveThe original functions converge in ${\mathcal{S}}^{\prime }$ to $|x{|}^{-2}$ by dominated convergence theorem against tests. The transforms are bounded by $2{\pi }^2/|\xi |$, which is locally integrable in three dimensions and integrable against Schwartz functions at infinity. HenceLet $y=G^{1/2}x$. The Jacobian is $(\det G{)}^{-1/2}$ and the dual vector is $G^{-1/2}\xi$. The Fourier transform of a reciprocal positive quadratic form is consequentlyThe frequency-origin value is understood distributionally; there is no additional delta term.

For the complex extension use a symmetric matrix $A=G+iB$ with real symmetric $B$ and positive real part $G$. Symmetry is natural for a quadratic form; a skew-symmetric part contributes nothing. The bound $|x^{T}Ax|\ge x^{T}Gx$ again gives a regular tempered reciprocal. Both sides of the prospective formula depend holomorphically on the matrix while its real part is positive: compact parameter sets give a common $|x{|}^{-2}$ bound for pairing and differentiated integrands. Continue from the real positive matrices along $A_z=G+izB$. In a complex neighborhood of $z=0$, imaginary $z$ gives real positive matrices, so the one-variable identity theorem supplies the equality; connected continuation along $0\le z\le 1$ reaches $A$.

The analytic determinant square root for accretive symmetric matrices must follow that continuation, rather than an arbitrary scalar principal root of $\det A$. Explicitly, put $C=G^{-1/2}B{G}^{-1/2}$ and let its real eigenvalues be $b_j$. Thenwhere each factor has positive real part. This is the determinant square root normalized positively on real positive matrices. AlsoThus the quadratic-form square root has the unambiguous branch with positive real part. The accretive complex quadratic reciprocal Fourier transform isUniform local integrability of the right side justifies its analytic continuation as a tempered distribution, not merely pointwise away from the origin.

For the particular form,Substitution gives the required normalized expressionFor every nonzero real $\xi$, the radicand has positive real part, so the specified square root exists uniquely. Both sides are regular tempered distributions, despite their locally integrable singularities at zero.