Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 327 (Distribution Theory and Applications)

The Schwartz space consists of the smooth functions $\phi$ for whichfor all multi-indices $\alpha ,\beta$. Its topology is generated by these seminorms; ${\phi }_j\to \phi$ means convergence in every seminorm. A tempered distribution is a continuous linear functional on this space. Convergence in ${\mathcal{S}}^{\prime }$ here means weak convergence: $⟨u_j,\phi ⟩\to ⟨u,\phi ⟩$ for every Schwartz function.

Use the unnormalized angular-frequency Fourier transformThe transform is a continuous automorphism of the Schwartz space, so this duality defines a tempered distribution. Put $D_j=-i{\partial }_j$, as required by the stated identities. Integration by parts and differentiation under the integral giveThus, keeping the transpose factor of $D=-i\partial$ explicit,andIterating gives the Fourier derivative identities for distributions

The logarithm defines a tempered distribution because it is locally integrable and grows more slowly than a polynomial. Its derivative is $u^{\prime }(x)=x/(1+x^2)$. The standard transform pair $\widehat{(1+x^2{)}^{-1}}=\pi e^{-|\lambda |}$, obtainable by the residue theorem, givesAs $\widehat{u^{\prime }}=i\lambda \hat{u}$, we have $\lambda \hat{u}=-\pi \mathrm{sgn}(\lambda )e^{-|\lambda |}$. DefineThis integral is absolutely convergent: the numerator is $O(|\lambda |)$ near zero, and the exponential controls the tails. It defines a tempered distribution satisfying the same multiplication equation. The kernel of multiplication by a coordinate therefore gives $\hat{u}=T+c{\delta }_0$.

To determine $c$, use the supplied Gaussian Fourier transform pair with ${\phi }_n(0)=1/(2\sqrt{\pi })$. ThenAlso $⟨T,{\phi }_n⟩\to 0$ by the dominated convergence theorem: the relevant integrand is bounded by a constant times $|\lambda |e^{-|\lambda |}$ for $n\ge 1$. ConsequentlyThe cutoff limit agrees with the absolutely convergent integral; this particular subtraction convention fixes the otherwise ambiguous delta term. The reusable identity is Fourier transform of the logarithm of one plus x squared.

The space of test functions is $\mathcal{D}(\mathbb{R})=C_c^{\infty }(\mathbb{R})$. A sequence ${\phi }_j$ converges to $\phi$ when its supports eventually lie in a single compact set and every derivative converges uniformly. The distribution space ${\mathcal{D}}^{\prime }(\mathbb{R})$ is its continuous linear dual; weak convergence means $⟨u_j,\phi ⟩\to ⟨u,\phi ⟩$ for every test function. We use $⟨u^{\prime },\phi ⟩=-⟨u,{\phi }^{\prime }⟩$ and $⟨au,\phi ⟩=⟨u,a\phi ⟩$ for multiplication of a distribution by a smooth function.

If a test function $\phi$ has integral zero, thenis another compactly supported smooth test function with ${\psi }^{\prime }=\phi$. Hence $v^{\prime }=0$ implies $⟨v,\phi ⟩=-⟨v^{\prime },\psi ⟩=0$ for every such $\phi$.

Choose a test function $\eta$ with integral one. For arbitrary $\phi$, the function $\phi -\eta \int \phi$ has zero integral, soThus $v$ is the constant distribution $⟨v,\eta ⟩$. This proves a distribution with zero derivative is constant.

The principal-value reciprocal distribution isThe limit exists because the constant part of the numerator cancels symmetrically at zero. Multiplying by $x$ gives $x\mathrm{pv}(1/x)=1$.

If $v$ is any other solution, $w=v-\mathrm{pv}(1/x)$ obeys $xw=0$. To identify this kernel of multiplication by a coordinate, choose a cutoff function $\eta$ equal to one near zero. Every test function has the form $\phi =\phi (0)\eta +x\psi$ with $\psi \in \mathcal{D}$. Thus $⟨w,\phi ⟩=\phi (0)⟨w,\eta ⟩$, proving $w=c{\delta }_0$. ThereforeFor real distributions the constants are real.

The Leibniz rule rewrites the left side as $(xu{)}^{\prime }$. By a distribution with zero derivative is constant,The Dirac delta multiplication identity $x{\delta }_0^{\prime }=-{\delta }_0$ supplies a particular solution, and the preceding division result supplies all solutions of $xu=c$. HenceMultiplying by $x$ and then differentiating verifies the equation. The coordinate-multiplication kernel and the constant-derivative kernel show that no additional solutions are missing.

The distributional derivative of the Heaviside step function gives $H^{\prime }={\delta }_0$, while $((x-1)H(x-1){)}^{\prime \prime }={\delta }_1$. A particular solution is therefore $H(x)-2(x-1)H(x-1)$.

The difference from any other solution has second derivative zero. Applying a distribution with zero derivative is constant twice shows that difference is affine. ThusThe ramp function is continuous at $x=1$, and its first derivative jumps there, producing the indicated delta source in its second derivative.

Since $(x^2-1)(-{\delta }_0)={\delta }_0$, write $u^{\prime }=-{\delta }_0+w$, where $(x^2-1)w=0$. The polynomial has two simple zeros. The distribution annihilated by a function with simple zeros lemma givesIntegrating with the Heaviside step function yieldsThis includes every solution: its derivative has exactly the required delta terms, and any two primitives differ only by a constant distribution.

The Paley–Wiener–Schwartz theorem in its closed-ball form says that a tempered distribution $u$ is supported in ${\overline{B}}_R(0)$ if and only if its Fourier transform is the real restriction of an entire function $F$ on ${\mathbb{C}}^n$ satisfying, for some $C>0$ and integer $N\ge 0$,The distribution is unique by Fourier inversion. More generally a compact convex support set $K$ replaces $R|\mathrm{Im}\zeta |$ by its support function $H_K(\mathrm{Im}\zeta )$.

For the forward implication, a compactly supported distribution has finite order of a distribution, say $m$, on a fixed neighborhood of its support. Defineusing a cutoff equal to one near the support. Differentiation with respect to each complex coordinate is allowed in this pairing and gives ${\partial }_{{\zeta }_j}F=⟨u,-i{x}_j{e}^{-ix\cdot \zeta }⟩$, so $F$ is entire.

To obtain the exact radius $R$, rather than an enlarged radius, choose a cutoff function ${\chi }_{\epsilon }$ equal to one near ${\overline{B}}_R$ and supported in $B_{R+\epsilon }$, with derivatives through order $m$ bounded by $C_{\gamma }{\epsilon }^{-|\gamma |}$. Set $\epsilon =(1+|\zeta |{)}^{-1}$. The finite-order bound and the Leibniz rule giveThis proves the required growth estimate.

For the converse, $F$ has at most polynomial growth on real frequencies, so define its inverse tempered distribution byLet a compactly supported test function lie in a half-space $x\cdot \omega \ge R+\eta$, with $|\omega |=1$ and $\eta >0$. By contour shifting,Here is the decay needed to justify the shift and then let $t$ grow. Repeated integration by parts, applying $1-{\Delta }_x$ to $e^{-tx\cdot \omega }\phi (x)$, gives for every integer $M$,Choose $2M>N+n$. For each fixed $t$, this decay makes the vertical sides of a large rectangle vanish in the one complex coordinate parallel to $\omega$; Cauchy integral theorem then supplies the contour identity. Combining the two bounds yieldsThus $u$ vanishes on such test functions. Every point outside ${\overline{B}}_R$ has a neighborhood of this form. A partition of unity therefore proves $\mathrm{supp}u\subseteq {\overline{B}}_R$, completing the converse. The same argument with separating half-spaces proves the compact-convex-set form. This is the contour-shift proof of the Paley–Wiener–Schwartz theorem.

For a regular function $u$, the change of variables $y=tx$ givesThe map $\phi \mapsto \phi (\cdot /t)$ is continuous on the Schwartz space, so duality extends the dilation of a distribution toIts Fourier transform obeys $\widehat{{\delta }_{t}u}(\zeta )=t^{-n}F(\zeta /t)$. If the original support is in the unit ball, the theorem bounds $F$ by $C(1+|\zeta |{)}^N{e}^{|\mathrm{Im}\zeta |}$. HenceThe converse theorem gives the support law for distributional dilation