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For the angular-frequency Fourier transform and Schwartz seminorms $p_{m,n}(\varphi )={\sup }_x|x^m{\varphi }^{(n)}(x)|$, every output seminorm is bounded by finitely many input seminorms. To prove this, differentiation under the integral sign and integration by parts giveThe generalized Leibniz rule, $\Vert \hat{g}{\Vert }_{\infty }\le \Vert g{\Vert }_1$, and $\Vert x^a{\varphi }^{(b)}{\Vert }_1\le \pi (p_{a,b}+p_{a+2,b})$ implyThis proves continuity of the Fourier transform on the Schwartz space and rapid decay of every transformed derivative.

For a $2\pi$ periodic distribution $u\in {\mathcal{S}}^{\prime }(\mathbb{R})$, choose a smooth periodic partition of unity $\Psi$, and define $c_n=(2\pi {)}^{-1}⟨u,\Psi e^{-inx}⟩$. A finite Schwartz seminorm bound on the fixed compact support gives $|c_n|\le C(1+|n|{)}^N$. For a Schwartz function $\varphi$, the Poisson summation formula and Fourier inversion imply $P\hat{\varphi }(x)={\sum }_n\varphi (-n)e^{inx}$. Localizing the pairing for $u$ givesproving the displayed Fourier transform formula. The series converges in tempered distributions because its coefficients have polynomial growth while $\varphi (n)$ decreases rapidly. Applying the inverse Fourier transform gives $u={\sum }_n{c}_n{e}^{inx}$ in tempered distributions. These are the smooth-window Fourier coefficients of a periodic distribution, and the transform formula shows independence of $\Psi$.

Use the Fourier transform convention $\hat{f}(\xi )={\int }_{\mathbb{R}}f(x)e^{-ix\xi }dx$, with complex-linear distribution pairings throughout. The Riemann-Lebesgue lemma says that, for an integrable function $f\in L^1(\mathbb{R})$, its Fourier transform is continuous and tends to zero as $|\xi |\to \infty$. It also satisfies $\Vert \hat{f}{\Vert }_{\infty }\le \Vert f{\Vert }_1$. The product rule, or convolution theorem, isHere $(f\ast g)(x)=\int f(y)g(x-y)dy$ belongs to $L^1(\mathbb{R})$ by Tonelli's theorem, and Fubini's theorem proves the displayed identity. The companion Fourier product formula, for Schwartz functions, is $\widehat{fg}=(2\pi {)}^{-1}\hat{f}\ast \hat{g}$.

A Schwartz function is a smooth function $\varphi :\mathbb{R}\to \mathbb{C}$ for which every Schwartz seminormis finite. These seminorms define the topology of the Schwartz space $\mathcal{S}(\mathbb{R})$. In particular,Thus every Schwartz function is an integrable function.

The Fourier bounds for Schwartz functions assert that each Schwartz seminorm of $\hat{\varphi }$ is bounded by finitely many Schwartz seminorms of $\varphi$. Explicitly, differentiation under the integral sign and integration by parts giveso the generalized Leibniz rule and the preceding integrability estimate giveIn particular the Fourier transform maps the Schwartz space continuously into itself.

Write $U=\hat{u}$, $V=\hat{v}$, and $⟨\xi ⟩=(1+{\xi }^2{)}^{1/2}$. By the Sobolev space hypotheses, $A=⟨\xi {⟩}^{s}U$ and $B=⟨\xi {⟩}^{t}V$ belong to the L2 space. Their product is an integrable function by Cauchy-Schwarz inequality. Thus $UV=⟨\xi {⟩}^{-s-t}AB$ is a locally integrable function defining a tempered distribution, with estimateThe supremum is controlled by finitely many Schwartz seminorms. Define the convolution of Sobolev distributions byIt is independent of the representatives chosen for $U,V$, and the convolution theorem shows that it agrees with ordinary convolution for Schwartz functions. More generally, approximate $u,v$ by Schwartz functions in their respective Sobolev norms; the displayed estimate shows convergence of the products in tempered distributions, so their ordinary convolutions converge to this definition. For $u,v\in L^1$ that also satisfy the stated Sobolev space hypotheses, the ordinary convolution theorem gives the same result.

There need not be any real $r$ with $u\ast v\in H^r(\mathbb{R})$. ChooseHere $1_{(0,1)}$ is an indicator function. Since ${\int }_0^1{\xi }^{-2/3}d\xi =3$ and every weight $(1+{\xi }^2{)}^q$ is bounded above and below by positive constants on $[0,1]$, both $u$ and $v$ belong to every Sobolev space. But their convolution has Fourier transform $UV={\xi }^{-2/3}{1}_{(0,1)}$, and for every real $r$,Thus $\boxed{u,v\in \bigcap _{q\in \mathbb{R}}{H}^q(\mathbb{R})\text{but}u\ast v\notin \bigcup _{r\in \mathbb{R}}{H}^r(\mathbb{R})}$. The obstruction is failure of local square integrability at zero; a weight controlling large frequencies cannot fix it.

A distribution $u$ is a $2\pi$ periodic distribution whenIteration gives invariance under all integer multiples of $2\pi$. For a tempered distribution the identity also holds for every Schwartz function, since test functions are dense in the Schwartz space: multiplying by cutoff functions $\chi (x/R)$ equal to one near zero gives such an approximation in every Schwartz seminorm.

For $\varphi \in \mathcal{S}(\mathbb{R})$ define $a_k(x)=\Psi (x+2\pi k)\varphi (x)$. The smooth periodic partition of unity gives $\varphi ={\sum }_k{a}_k$ with convergence in the Schwartz space. Indeed, each $a_k$ is supported where $|x+2\pi k|\le 2\pi$, and the generalized Leibniz rule and arbitrarily rapid decay of the derivatives of $\varphi$ give, for every $m,n,M$,Consequently continuity and periodicity of the tempered distribution giveFor the last interchange, all summands have the fixed compact support of $\Psi$ and their derivatives decrease faster than any power of $|k|$. Hence the series converges in the space of test functions, as well as in the Schwartz space. Notice that $P\varphi$ itself need not be a Schwartz function; the factor $\Psi$ makes the pairing legitimate.

The $2\pi$ periodization of a Schwartz function isFor every nonnegative integer $j$, the Schwartz seminorms give $|{\varphi }^{(j)}(x+2\pi k)|\le C_j(1+|k|{)}^{-2}$ uniformly for $x\in [0,2\pi ]$. The Weierstrass M-test gives absolute and uniform convergence there of the defining series and all its differentiated series. Termwise differentiation proves that $P\varphi$ is a smooth function, with $(P\varphi {)}^{(j)}=P({\varphi }^{(j)})$. Reindexing the absolutely convergent series proves $P\varphi (x+2\pi )=P\varphi (x)$, so it is a periodic function.

For an explicit smooth periodic partition of unity, choose the bump functionIt is a test function, positive on $[-\pi ,\pi ]$. The smooth function $S(x)=P\eta (x)$ is $2\pi$ periodic and strictly positive, since every $x$ has a translate in $[-\pi ,\pi ]$. SetDivision preserves smoothness because $S>0$, and the support remains contained in $[-2\pi ,2\pi ]$.

Use the smooth periodic partition of unity $\Psi$ above and define the Fourier coefficients of the periodic distribution byThese are the smooth-window Fourier coefficients of a periodic distribution for period $2\pi$. The finite Schwartz seminorm bound for $u$, restricted to the fixed compact support of $\Psi$, gives $|c_n|\le C(1+|n|{)}^N$ for some $C,N$. Indeed, only finitely many derivatives of $\Psi e^{-inx}$ occur, and each contributes at most a fixed power of $|n|$.

For a Schwartz function $\varphi$, the preceding Fourier series formula and $\hat{\hat{\varphi }}(n)=2\pi \varphi (-n)$ yieldAll derivatives of this series converge uniformly. The localization formula in the previous answer therefore givesConsequently the Fourier transform of a periodic distribution isThe polynomial bound on $c_n$ and rapid decay of $\varphi (n)$ make every displayed pairing absolutely convergent. In particular there are no derivatives of Dirac delta distributions in this formula.

The inverse Fourier transform of ${\delta }_n$ is $(2\pi {)}^{-1}{e}^{inx}$. Applying continuous Fourier inversion on tempered distributions to the symmetric partial sums givesExplicitly, for every Schwartz function $\theta$, the pairing of the partial sum with $\theta$ is ${\sum }_{|n|\le M}{c}_n\hat{\theta }(-n)$, which converges absolutely to $⟨u,\theta ⟩$. The Fourier transform formula also proves that $c_n$ is independent of the chosen smooth periodic partition of unity. For a locally integrable $2\pi$ periodic function it reduces, by splitting the integral into periods, to $c_n=(2\pi {)}^{-1}{\int }_0^{2\pi }u(x)e^{-inx}dx$.

Use the unitary normalization of the Fourier transform,The Schwartz space $\mathcal{S}({\mathbb{R}}^n)$ consists of smooth functions for which every Schwartz seminorm ${\sup }_x|x^{\alpha }{\partial }^{\beta }f(x)|$ is finite. Such functions and all their derivatives are integrable. Differentiating under the integral and integrating by parts giveConsequentlyExpand the derivative by the product rule and insert the integrable weight $(1+|x|{)}^{-n-1}$; the right side is bounded by finitely many Schwartz seminorms of $f$. This proves continuity $\mathcal{F}:\mathcal{S}\to \mathcal{S}$. The identical argument with the opposite sign proves continuity of $\mathcal{G}$.

For surjectivity and injectivity an inversion proof is needed. If $f\in \mathcal{S}$, multiply $\mathcal{F}f(\xi )$ by $e^{-\epsilon |\xi {|}^2/2}$. Fubini's theorem and the elementary Gaussian integral giveThe Gaussian has total integral one and its mass outside any fixed ball tends to zero. The right side therefore tends to $f(x)$, by uniform continuity of $f$. On the left, $\mathcal{F}f\in \mathcal{S}\subset L^1$, so dominated convergence gives $\mathcal{GF}f(x)$. This proves $\mathcal{GF}=I$. Reversing the signs in the same calculation proves $\mathcal{FG}=I$. ThusIts inverse is continuous by the seminorm estimate already proved. This establishes the requested bijection without assuming Fourier inversion as an unexplained theorem.

Inversion gives ${\mathcal{F}}^{2}f(x)=f(-x)$ and the Parseval identity. Indeed, inserting the integral for $\mathcal{F}f$ into $\int \mathcal{F}f\overline{\mathcal{F}g}$ and using $f,\mathcal{F}g\in L^1$ gives $\int f\overline{g}$. Density of Schwartz functions in $L^2$ extends $\mathcal{F}$ to a unitary operator on $L^2$, the Plancherel theorem. Translation of $f$ becomes multiplication by a phase, modulation becomes frequency translation, andThese formulas explain why the Fourier transform diagonalizes constant-coefficient differential operators. By transpose it also defines a bijection of the tempered distributions: with the complex-linear distributional pairing, $⟨\mathcal{F}T,\varphi ⟩=⟨T,\mathcal{F}\varphi ⟩$, since the Fourier kernel is symmetric in its variables.

The connection with creation and annihilation operators is especially transparent for the harmonic oscillator. On the common invariant dense domain $\mathcal{S}$, setIntegration by parts shows that these are formal Hilbert-space adjoints. They are closable unbounded operators; commutators below are computed on $\mathcal{S}$, avoiding undefined products on arbitrary $L^2$ vectors. The identity $[{\partial }_j,x_k]={\delta }_{jk}I$ gives the canonical commutation relationsIt also givesThe normalized vacuum $h_0(x)={\pi }^{-n/4}{e}^{-|x{|}^2/2}$ satisfies $a_j{h}_0=0$. Define the Hermite functionsMoving annihilation operators past creation operators with the commutation relations provesThese same relations show that the Hermite functions are orthonormal: unequal occupations are orthogonal eigenvectors of one of the commuting symmetric number operators $a_j^{†}{a}_j$, and the ladder relations recursively give norm one.

For completeness, let $g\in L^2$ be orthogonal to every $h_{\alpha }$. The Hermite polynomials have nonzero leading monomials and triangular lower-degree terms, so these orthogonality relations say that all polynomial moments of $h(x)=g(x)e^{-|x{|}^2/2}\in L^1$ vanish. Its Fourier integral extends to an entire function of $\zeta \in {\mathbb{C}}^n$:The Cauchy-Schwarz inequality, with the Gaussian absorbing every exponential linear weight and every polynomial, justifies differentiation uniformly on compact sets of $\zeta$. Every Taylor coefficient at zero is a vanishing moment, so the entire function is zero. Fourier uniqueness on $L^1$ follows here without circularity: if an $L^1$ function has zero Fourier transform, convolving it with ${\rho }_{\epsilon }$ gives zero by the Gaussian integral and Fubini, and the approximate-identity limit in $L^1$ gives the original function zero. Thus $h=0$ and $g=0$. The Hermite functions form an orthonormal basis of $L^2$.

Finally, the elementary Gaussian integral gives $\mathcal{F}{h}_0=h_0$. The derivative and multiplication identities above implywhere the operators on the right act on the frequency variable. HenceThus creation and annihilation organize both the oscillator's spectral decomposition and the four possible Fourier eigenvalues. On Schwartz functions the oscillator powers control weighted derivatives, and Hermite coefficients decay faster than every power of $1+|\alpha |$; conversely this rapid coefficient decay implies all Schwartz seminorms are finite. The Fourier phases preserve this decay, providing another explanation of the Fourier automorphism of Schwartz space.

Use $e(t)=e^{2\pi it}$ and the Fourier transform convention $\hat{f}(\xi )={\int }_{\mathbb{R}}f(x)e(-x\xi )dx$, initially for an integrable function. For the interval indicator function, direct integration givesThe integral bound is one, and the displayed numerator has modulus at most two. By the convolution theorem, the Fourier transform of the $m$-fold convolution is its $m$th power. Thus the Fourier decay of repeated interval convolutions givesAt zero, interpret the second term of the minimum as infinity.

The Schwartz space $\mathcal{S}(\mathbb{R})$ consists of smooth functions for which every Schwartz seminorm ${\sup }_x|x^a{f}^{(b)}(x)|$ is finite, for nonnegative integers $a,b$. Differentiation under the integral sign and integration by parts giveAll differentiated integrands are integrable, and their boundary terms vanish because of rapid decay. The last transform is bounded by its integrand's $L^1$ norm. The same argument applies at every derivative order, so $\hat{f}\in \mathcal{S}(\mathbb{R})$.

For $f\in \mathcal{S}(\mathbb{R})$, the Poisson summation formula isTo prove it, form the periodization of a Schwartz function $F(x)={\sum }_{n}f(x+n)$. Rapid decay gives uniform convergence of this series and of every differentiated series on $[0,1]$, so $F$ is a smooth periodic function. Its Fourier coefficients areThe series ${\sum }_k\hat{f}(k)e(kx)$ converges absolutely and uniformly. Its Fourier coefficients equal those of $F$. The uniqueness theorem for Fourier series says that two integrable functions with the same coefficients agree almost everywhere; as both functions here are continuous, they agree everywhere. Evaluating at zero proves Poisson summation.

The Gaussian Fourier transform of $e^{-\pi t{x}^2}$, $t>0$, is $t^{-1/2}{e}^{-\pi {\xi }^2/t}$. Apply Poisson summation to obtain the transformation of the Jacobi theta functionFor $\Re s>1$, termwise integration in the Mellin transform gives the Mellin representation of the completed Riemann zeta functionSplit at one and substitute $t=1/u$ in the lower integral. The theta transformation givesThe integral is an entire function of $s$, since $\vartheta (t)-1$ decays exponentially. This pole-subtracted theta integral for the completed zeta function supplies a meromorphic continuation with simple poles at zero and one. Both the rational term and the integral are unchanged by $s\mapsto 1-s$, proving the functional equation of the Riemann zeta functionEquivalently the Riemann xi function $\xi (s)=\frac{1}{2}s(s-1)\Lambda (s)$ is entire and satisfies $\xi (s)=\xi (1-s)$. The reflection and duplication identities for the Gamma function also give $\zeta (s)=2^s{\pi }^{s-1}\sin (\pi s/2)\Gamma (1-s)\zeta (1-s)$ by meromorphic continuation.

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.

A distribution has period $T>0$ if the displayed equality holds for every test function $\varphi$. This agrees with $f(x+T)=f(x)$ almost everywhere for a regular distribution. If $u$ is a tempered distribution, density of test functions in the Schwartz space extends the identity to all Schwartz functions.

For a test function $\Psi$ with ${\sum }_k\Psi (x+kT)=1$, one hasIndeed, expand $\varphi ={\sum }_k\Psi (x+kT)\varphi (x)$ in the Schwartz space and move each translation to $\varphi$ by periodicity. All Schwartz seminorms of the localized terms decay faster than any power of $|k|$, justifying the sum of pairings. This localization permits rigorous Fourier coefficients without multiplying a distribution by a discontinuous window.