Tempered Distributions and the Fourier Transform

Suppose $u$ is continuous. There is a basic zero-neighborhood $U=\{\phi :p_{{\alpha }_j{\beta }_j}(\phi )<{\epsilon }_j,1\le j\le r\}$ on which $\mathrm{\mid }u(\phi )\mathrm{\mid }<1$. If $r=0$, then $U=\mathcal{S}$ and linearity forces $u=0$, so take $C=0$. Otherwise put $Q(\phi )={\max }_j{p}_{{\alpha }_j{\beta }_j}(\phi )\mathrm{/}{\epsilon }_j$.

If $Q(\phi )>0$, then $\phi \mathrm{/}(2Q(\phi ))\in U$, whence $\mathrm{\mid }u(\phi )\mathrm{\mid }<2Q(\phi )$. If $Q(\phi )=0$, every scalar multiple of $\phi$ lies in $U$; boundedness of those scalar multiples of $u(\phi )$ forces $u(\phi )=0$. Choose $N$ and $M$ dominating the finitely many $\mathrm{\mid }{\alpha }_j\mathrm{\mid }$ and $\mathrm{\mid }{\beta }_j\mathrm{\mid }$. Then $Q(\phi )$ is at most ${\max }_j{\epsilon }_j^{-1}$ times the rectangular maximum in the statement, which proves the required estimate.

Conversely, assume the displayed estimate. For every $\epsilon >0$, the set on which its finite maximum is less than $\epsilon \mathrm{/}(C+1)$ is a zero-neighborhood and is carried by $u$ into the disk of radius $\epsilon$. Thus $u$ is continuous and hence tempered.

For each integer $N\ge 0$, expansion of $(1+\mathrm{\mid }{x}_1\mathrm{\mid }+\cdots +\mathrm{\mid }{x}_n\mathrm{\mid }{)}^N$ gives a finite constant $A_{n,N}$ for which the following estimate holds.

The estimate in step 1.1 makes the integral absolutely convergent for every Schwartz test and proves that $u_f$ is tempered. It also shows directly that changing $f$ on a null set changes no pairing.

The integer shells $m\le \mathrm{\mid }x\mathrm{\mid }<m+1$ have measure at most $(2m+2{)}^n$. Hence $\int (1+\mathrm{\mid }x\mathrm{\mid }{)}^{-s}dx<\mathrm{\infty }$ whenever $s>n+1$, by comparison with ${\sum }_{m\ge 1}{m}^{n-s}$. If $\mathrm{\mid }f(x)\mathrm{\mid }\le C(1+\mathrm{\mid }x\mathrm{\mid }{)}^d$ off a compact set, choose an integer $N>d+n+1$. Local integrability handles the compact part, and the shell estimate handles its complement, so step 2.1 applies.

If $f\in L^1$, take $N=0$. If $1<p<\mathrm{\infty }$ and $q$ is conjugate to $p$, choose $N$ with $Nq>n+1$; Hölder gives $\int \mathrm{\mid }f\mathrm{\mid }(1+\mathrm{\mid }x\mathrm{\mid }{)}^{-N}\le \parallel f{\parallel }_p\parallel (1+\mathrm{\mid }\cdot \mathrm{\mid }{)}^{-N}{\parallel }_q<\mathrm{\infty }$. If $p=\mathrm{\infty }$, choose $N>n+1$ and use any finite essential bound for $\mathrm{\mid }f\mathrm{\mid }$. The same estimates on bounded balls give local integrability of the chosen representatives.

For $n\ge 1$, as a boundary calculation, define $f_r(x)=\mathrm{\mid }x{\mathrm{\mid }}^r$ for $x\mathrm{\ne }0$ and assign any finite value, say $f_r(0)=0$, when $r<0$; set $f_0\equiv 1$, and for $r>0$ use the usual value $f_r(0)=0$. The value at this measure-zero point has no effect on local integrability or the induced distribution. The function $f_r$ is locally integrable at the origin exactly when $r>-n$: for $r<0$, the dyadic annuli $2^{-j-1}\le \mathrm{\mid }x\mathrm{\mid }<2^{-j}$ give a series comparable to ${\sum }_j{2}^{-j(n+r)}$; for $r\ge 0$ there is no singularity, while the reverse bound on a fixed cone gives divergence when $r\le -n$. At infinity it has polynomial growth. Therefore $f_r$ is among the tempered regular examples precisely for the locally meaningful range $r>-n$.

By [F2], $\mathcal{F}\phi$ is a Schwartz test and depends continuously and linearly on $\phi$. Thus $\phi \mapsto u(\mathcal{F}\phi )$ is a continuous complex-linear functional. This proves well-definedness, and linearity in $u$ follows directly from the pairing.

For a single test $\phi$, $p_{\phi }(\mathcal{F}u)=\mathrm{\mid }u(\mathcal{F}\phi )\mathrm{\mid }=p_{\mathcal{F}\phi }(u)$. Every target weak seminorm therefore pulls back to a source weak seminorm, so $\mathcal{F}$ is weakly continuous.

If $B\subseteq \mathcal{S}$ is bounded, continuity and linearity of the Schwartz transform imply that $\mathcal{F}(B)$ is bounded: each output seminorm is bounded by finitely many input seminorms.

Evaluate the second transform on an arbitrary Schwartz test $\phi$.

Reflection on the dual satisfies $R^2=I$. Since step 1.1 gives $R={\mathcal{F}}^2$ as operators on ${\mathcal{S}}^{\mathrm{\prime }}$, associativity gives the following two-sided inverse calculation.

The inverse $R\mathcal{F}={\mathcal{F}}^3$ is a composition of weakly continuous maps and also of strongly continuous maps. Therefore $\mathcal{F}$ is a topological automorphism for both topologies. Countable Choice is used only through [F1]–[F2].

Let $f\in L^1$ and $\phi \in \mathcal{S}$. Schwartz decay makes $\phi \in L^1$, so $\int \text{\hspace{0.17em}⁣}\int \mathrm{\mid }f(x)\phi (\xi )\mathrm{\mid }dxd\xi <\mathrm{\infty }$. Absolute Fubini is therefore applicable.

Let $f\in L^2$ and choose $f_j\in \mathcal{S}$ with $f_j\to f$ in $L^2$. For a fixed $\phi \in \mathcal{S}$, also $\mathcal{F}\phi \in L^2$, and Hölder gives the first convergence below.

Each $f_j$ belongs to $L^1\cap L^2$, and the two agreement results give the displayed identity.

Let $\phi \in \mathcal{S}({\mathbb{R}}^n)$. Apply the defining transpose and the comb pairing.

Poisson summation changes the last sum to ${\sum }_k\phi (k)=⟨{\mathrm{III}}_{{\mathbb{Z}}^n},\phi ⟩$. Equality on every Schwartz test proves the stated identity. Countable Choice is used only through the published Fourier and Poisson suppliers.

Fix compact $K\subseteq {\mathbb{R}}^n$ and $\phi \in {\mathcal{D}}_K$. Each Schwartz seminorm has the following fixed-support bound.

Step 1.1 makes every restricted inclusion ${\mathcal{D}}_K\to \mathcal{S}$ continuous. The LF universal property therefore makes $\iota$ continuous on all of $\mathcal{D}$.

The approximants in [F3] lie in $\mathcal{D}$ and converge in the Schwartz topology to the prescribed Schwartz function. Thus the image of $\iota$ is dense. This density argument is separate from continuity, and neither uses a choice axiom.

Restrict the $C^{\mathrm{\infty }}$ extension from [F1] to $\mathcal{S}$. Its continuity gives a compact $K$, an integer $m$, and $C\ge 0$ satisfying the following estimate.

For $\psi \in \mathcal{D}$, the extension property in [F1] gives $\tilde{v}(\psi )=v(\psi )$; this includes the zero distribution and empty support. Hence $\tilde{v}$ really extends $v$.

If $w\in {\mathcal{S}}^{\mathrm{\prime }}$ is another extension, then $w-\tilde{v}$ vanishes on $\mathcal{D}$. This difference is continuous on $\mathcal{S}$, and $\mathcal{D}$ is dense there, so it vanishes on all of $\mathcal{S}$. Therefore the extension is unique.

Define $\rho (u)(\psi )=u(j\psi )$. Since $j$ and $u$ are continuous complex-linear maps, $\rho (u)$ is a distribution, and $\rho$ is complex-linear. If $\rho (u)=0$, then $u$ vanishes on the dense image of $j$; continuity makes $u=0$ on $\mathcal{S}$. Thus $\rho$ is injective.

For each $\psi \in \mathcal{D}$, $p_{\psi }(\rho u)=p_{j\psi }(u)$. Hence every weak seminorm on the target pulls back to a weak seminorm on the source, proving weak continuity.

If $B\subseteq \mathcal{D}$ is bounded, continuity and linearity of $j$ imply that $j(B)$ is bounded in $\mathcal{S}$.

Fix $\alpha ,\beta$ and apply the multi-index Leibniz formula.

Step 1.1 proves simultaneously that $a\phi \in \mathcal{S}$ and that $M_a$ is continuous. A polynomial has only finitely many nonzero derivatives and each grows polynomially. If $a\in \mathcal{S}$, each derivative is bounded, so the hypothesis holds with exponent zero.

For $u\in {\mathcal{S}}^{\mathrm{\prime }}$, the composition $u\circ M_a$ is continuous and linear, hence tempered. For a single test, $p_{\phi }(au)=p_{a\phi }(u)$, proving weak continuity of the transpose.

If $B$ is bounded in $\mathcal{S}$, the finite estimates of step 1.1 show that $M_a(B)$ is bounded. Thus $p_B(au)=p_{M_a(B)}(u)$, proving strong continuity. The derivative hypothesis is essential: for example $a(x)=e^{\mathrm{\mid }x{\mathrm{\mid }}^2}$ is smooth but does not map every Schwartz function to a Schwartz function. No choice axiom is used.

Each displayed functional is the composition of $u$ with a continuous Schwartz endomorphism, followed in the derivative case by a scalar sign. It is therefore continuous and complex-linear on $\mathcal{S}$, hence belongs to ${\mathcal{S}}^{\mathrm{\prime }}$.

For a single test $\phi$, the absolute value after either operation is a source weak seminorm evaluated at the transformed test. Thus each operation is weakly continuous.

A continuous linear Schwartz endomorphism sends bounded sets to bounded sets. For a bounded $B$, the target strong seminorm is therefore the source strong seminorm on ${\mathrm{\partial }}^{\alpha }B$ or $PB$ (the derivative sign disappears under absolute values). This proves strong continuity.

If $\psi \in \mathcal{D}$, then ${\mathrm{\partial }}^{\alpha }\psi$ and $P\psi$ are again compactly supported tests. Evaluating the two displayed definitions on $\psi$ gives exactly the formulas in [F3]. Hence restriction to ${\mathcal{D}}^{\mathrm{\prime }}$ commutes with both operations. The cases $\alpha =0$, constant $P$, $P=0$, and $u=0$ follow from the same formulas. No choice axiom is used.

Fix a coordinate $j$ and a Schwartz test $\phi$; transposition gives the following calculation.

The first formula in [F3] gives $x_j\mathcal{F}\phi =(2\pi i{)}^{-1}\mathcal{F}({\mathrm{\partial }}_j\phi )$, so transposition yields the second calculation.

Coordinate derivatives and coordinate multipliers commute among themselves in the relevant families. Iterating steps 1.1 and 1.2 $\mathrm{\mid }\alpha \mathrm{\mid }$ times therefore gives the two multi-index identities. For $\alpha =0$ both reduce to $\mathcal{F}u=\mathcal{F}u$. Countable Choice is used only through the published Schwartz Fourier identity.

Evaluate the transform of ${\delta }_a$ on an arbitrary $\phi \in \mathcal{S}$.

Apply $\mathcal{F}$ to $\mathcal{F}{\delta }_0=1$. Since $R{\delta }_0={\delta }_0$, [F3] gives $\mathcal{F}1={\delta }_0$. Similarly step 1.1 with $a=-b$ gives $\mathcal{F}{\delta }_{-b}=e^{2\pi ib\cdot x}$, so applying $\mathcal{F}$ once more gives $\mathcal{F}(e^{2\pi ib\cdot x})=R{\delta }_{-b}={\delta }_b$.

Apply the derivative identity in [F4] to ${\delta }_0$ and use $\mathcal{F}{\delta }_0=1$ to obtain $\mathcal{F}({\mathrm{\partial }}^{\alpha }{\delta }_0)=(2\pi i\xi {)}^{\alpha }$. Apply the multiplication identity to the constant distribution and use $\mathcal{F}1={\delta }_0$ to obtain the formula for $x^{\alpha }$.

Every polynomial is a finite complex linear combination of monomials, so linearity completes the polynomial assertion. The zero multi-index recovers $\mathcal{F}{\delta }_0=1$ and $\mathcal{F}1={\delta }_0$. Countable Choice is used only through the published Fourier suppliers.

Fourier transformation, distributional differentiation, and finite addition are complex-linear, giving the following finite expansion.

Substitute [F1] into the finite sum from step 1.1 and factor the common distribution $\mathcal{F}u$ to obtain ${\sum }_{\alpha }{a}_{\alpha }(2\pi i\xi {)}^{\alpha }\mathcal{F}u=P(2\pi i\xi )\mathcal{F}u$. This includes constant and zero polynomials. The statement is only an algebraic equivalence: it asserts no division by $P(2\pi i\xi )$ and no existence or regularity theorem for a PDE. Countable Choice is used only through [F1].

Fix a compact set of parameters $K$. The inequality $1+\mathrm{\mid }y\mathrm{\mid }\le (1+{\sup }_{x\in K}\mathrm{\mid }x\mathrm{\mid })(1+\mathrm{\mid }x-y\mathrm{\mid })$ transfers every polynomial weight in $y$ to one in $x-y$, uniformly for $x\in K$. Apply the one-variable integral Taylor remainder along each coordinate.

For the integral clause, differentiate under the integral and apply the integral triangle inequality pointwise in $y$.

Let $Q_R=[-R,R{]}^r$. Subdivide it into the canonical equal mesh and form lower-corner finite sums $S_{R,m}$ for $H$. Uniform continuity in each seminorm and [F3] make these sums converge to $G_R(y)={\int }_{Q_R}H(t,y)dt$ in that seminorm. Continuity of $u$ may therefore be passed through this explicit limit.

The finite estimate [F2] involves only finitely many seminorms. Their $L^1$ majorants show both $G_R\to G$ in those seminorms and ${\int }_{Q_R}u(H(t))dt\to {\int }_{{\mathbb{R}}^r}u(H(t))dt$ as $R\to \mathrm{\infty }$. Passing to the limit in step 1.3 proves the interchange formula. Countable Choice is used exactly through [F3]'s Lebesgue interface; the finite-sum and continuity argument adds no stronger choice.

For every $x$ and coordinate $j$, Taylor's integral remainder and $1+\mathrm{\mid }y\mathrm{\mid }\le (1+\mathrm{\mid }x\mathrm{\mid })(1+\mathrm{\mid }x-y\mathrm{\mid })$ show that the $x_j$-difference quotients of $y\mapsto \phi (x-y)$ converge in every Schwartz seminorm to $y\mapsto {\mathrm{\partial }}_j\phi (x-y)$. Continuity of $u$ permits differentiation of the scalar pairing, and iteration gives every multi-index derivative.