Tempered Distributions and the Fourier Transform — Examples

For $\phi \in \mathcal{S}$, $⟨\mathcal{F}{\delta }_0,\phi ⟩=\widehat{\phi }(0)=\int \phi =⟨1,\phi ⟩$. Thus $\mathcal{F}{\delta }_0=1$.

The constant-distribution formula in [F1] gives $\mathcal{F}1={\delta }_0$ directly. Together with step 1.1, this checks that the reciprocal pair carries no factor of $(2\pi {)}^n$ in the repository normalization. Countable Choice is used only through [F1].

Set $a=-b$ in [F1]. Then $\mathcal{F}{\delta }_{-b}=e^{2\pi ib\cdot \xi }$; the positive exponential sign therefore corresponds to a delta initially placed at $-b$.

The plane-wave clause of [F1] directly gives $\mathcal{F}(e^{2\pi ib\cdot x})={\delta }_b$. Step 1.1 checks the sign by locating the pre-transform delta at $-b$. For $b=0$, the formula reduces to $\mathcal{F}1={\delta }_0$. Countable Choice is used only through [F1].

Evaluate the transform of ${\delta }_0^{\mathrm{\prime }}$ on an arbitrary $\phi \in \mathcal{S}(\mathbb{R})$.

The monomial clause of [F1], specialized to the one-dimensional multi-index $1$, directly gives $\mathcal{F}(x)=-(2\pi i{)}^{-1}{\delta }_0^{\mathrm{\prime }}$, which is the second formula. Together with step 1.1 this records both directions of the delta-derivative/monomial pair. Countable Choice is used only through [F1].

Symmetry cancels the constant term near zero, reducing the defining limit to two absolutely convergent integrals.

Test multiplication by the smooth coordinate function $x$.

Put $U=\mathcal{F}(\mathrm{pv}(1\mathrm{/}x))$. Transform step 2.1 and use the exact Fourier multiplication law and constant transform.

Integration by parts on the two half-lines gives $(\mathrm{sgn}{)}^{\mathrm{\prime }}=2{\delta }_0$: indeed $-\int \mathrm{sgn}(x){\psi }^{\mathrm{\prime }}(x)dx=2\psi (0)$ for every compact test $\psi$. Thus $S=-i\pi \mathrm{sgn}$ satisfies $S^{\mathrm{\prime }}=-2\pi i{\delta }_0$, and $W=U-S$ has zero derivative. The bounded function $\mathrm{sgn}$ is itself regular tempered by the elementary estimate $\mathrm{\mid }\int \mathrm{sgn}\phi \mathrm{\mid }\le \int \mathrm{\mid }\phi \mathrm{\mid }\le C{p}_{2,0}(\phi )$.

By [F4], the restriction of $W$ is a constant distribution $c$. The principal value is odd under reflection, Fourier transformation commutes with reflection by its defining integral, and $S$ is odd; hence $W$ is odd. A constant distribution is even, so $c=-c$ and $c=0$. Injectivity in [F4] then makes $W=0$ already in ${\mathcal{S}}^{\mathrm{\prime }}$. This proves $U=S$. Countable Choice is used only through the cited Fourier, integration, and zero-derivative interfaces.

Evaluate comb invariance on an arbitrary $\phi \in \mathcal{S}$.

Conversely, suppose the displayed lattice-sum identity holds for every $\phi \in \mathcal{S}$.

Apply step 1.1 to $g_t$ and substitute [F2]. The left and right lattice sums become exactly the two sides of the theta transformation. At $t=1$ the Gaussian is itself Fourier invariant; as $t$ varies, the formula exchanges $t$ and $1\mathrm{/}t$ with the dimension factor $t^{-n\mathrm{/}2}$. Countable Choice is used only through [F1]–[F3].

Since $E\in L^1(\mathbb{R})$, [F1] applies. Split at zero and use the elementary decaying exponential antiderivative.

For $P(z)=1-z^2$, apply the Fourier-symbol identity to step 1.1.

The denominator is strictly positive on the real frequency axis, so this particular division produces a smooth bounded multiplier. The calculation does not assert that an arbitrary polynomial symbol can be divided in ${\mathcal{S}}^{\mathrm{\prime }}$, nor any general PDE existence or regularity theorem. Countable Choice is used only through [F1]–[F3].

Assume for contradiction that $(A,J,\mathrm{\partial })$ satisfies the three stated requirements. Let $H={\mathbf{1}}_{(0,\mathrm{\infty })}$. It is locally integrable and piecewise smooth; evaluate its derivative on an arbitrary $\phi \in \mathcal{D}(\mathbb{R})$.

Put $h=J(u_H)$ and $d=J({\delta }_0)$. Since $H^2=H$ and $H^3=H$ pointwise, property 3 gives $h^2=h$ and $h^3=h$. Property 2 and step 1.1 give $\mathrm{\partial }h=d$.

Apply the derivation to $h^2=h$ and $h^3=h$.

Yet ${\delta }_0\mathrm{\ne }0$: choose a test function with $\phi (0)=1$ and use [F3]. Injectivity of $J$ therefore implies $d=J({\delta }_0)\mathrm{\ne }0$, contradicting step 3.1.

The contradiction concerns only the simultaneous requirements above. Special products, including [F4], and separately chosen nonlinear regularizations are not ruled out. Countable Choice is used only through the published regular-distribution and integration interfaces.

The constant function $1$ has polynomial growth of order zero, so each factor defines a tempered distribution by [F1]. Neither factor is compactly supported, so [F2] does not itself define their convolution.

Choose a nonnegative $\psi \in \mathcal{D}({\mathbb{R}}^n)$ with $c:={\int }_{{\mathbb{R}}^n}\psi (z)dz>0$. The formal distributional convolution pairing would require the following addition-pullback integral.

Hence the ordinary integral construction does not produce a finite pairing even on this one nonnegative test function, and $1\ast 1$ is undefined as an ordinary distributional convolution. This does not say that no separately chosen regularization can assign an object to the pair; such an assignment is additional structure, not the ordinary convolution supplied by [F2].