Past exam of the distribution theory course of the University of Oxford / 2023 / 3

Let $\Omega \subset {\mathbb{R}}^n$ be an open set and $(U_{\alpha }{)}_{\alpha \in A}$ an open cover of $\Omega$. There exist nonnegative smooth functions $({\chi }_j{)}_{j\in J}$ and choices $\alpha (j)$ such thatand the family of supports is locally finite: every point has a neighborhood meeting only finitely many of them. This is a smooth partition of unity subordinate to the cover, allowing a refinement of its indexing. In Euclidean open sets, one can choose $J$ countable and the individual supports compact in their assigned cover members. Local finiteness ensures that the sum is smooth and that only finitely many terms meet any given compact set.

If $\varphi \in \mathcal{D}(\omega )$, let $\tilde{\varphi }$ be its extension by zero to $\Omega$. Its compact support is contained in $\omega$, so this extension remains a test function. The restriction of a distribution isExtension by zero preserves convergence in the space of test functions, so this defines a distribution on $\omega$.

Choose a partition of unity $({\chi }_j)$ subordinate to the neighborhoods ${\omega }_x$ from the hypothesis. For an arbitrary $\varphi \in \mathcal{D}(\Omega )$, local finiteness and compactness of $\mathrm{supp}\varphi$ give a finite decomposition $\varphi ={\sum }_j{\chi }_j\varphi$. Each term is a test function supported in one ${\omega }_x$. ThusSince this holds for every test function, $v=0$. This is the locality of distributions.

Define two open sets byThe support of a distribution and its singular support areBoth are relatively closed because the removed sets are open. The locality of distributions gives $u{|}_{Z(u)}=0$. On an overlap of two open smoothness regions, the two smooth functions represent the same distribution, so their difference is zero almost everywhere and hence everywhere by continuity. They therefore join to one smooth function on $G(u)$. The same partition of unity argument shows that it represents $u$ there.

For $u=1_{\omega }$, openness of $\omega$ impliesIndeed every neighborhood of a point in this relative closure meets $\omega$ in an open set containing a nonnegative test function with positive integral; outside the closure the indicator function is identically zero near the point.

The singular support of an indicator function needs a measure-sensitive boundary. For a Lebesgue measurable set $A\subseteq \Omega$, define its essential closure byThen the general answer isOutside this intersection, some ball has either no points of $\omega$ or a complement of Lebesgue measure zero. The regular distribution is respectively the constant zero or one on that ball and is smooth. Conversely, at a point of the intersection, every sufficiently small ball has positive Lebesgue measure in both sets. If a smooth function represented the indicator function there, its values would be in $\{0,1\}$ everywhere: any other value, by continuity, would give a ball of positive Lebesgue measure on which equality almost everywhere fails. On a connected ball such a continuous two-valued function is constant, contradicting the two positive measures.

In particular, the answer need not be the full topological boundary. For $\Omega ={\mathbb{R}}^n$ and $\omega ={\mathbb{R}}^n\setminus \{0\}$, the indicator function is the constant one almost everywhere, so its singular support is empty although the boundary is $\{0\}$. The full relative boundary is the answer when every neighborhood of each boundary point contains a positive Lebesgue measure portion of the complement, as for an open set with smooth boundary.

Let $u,v\in {\mathcal{D}}^{\prime }({\mathbb{R}}^n)$, with at least one a compactly supported distribution, so their convolution is defined. The support and singular support rules for convolution areHere $A+B=\{a+b:a\in A,b\in B\}$ is the Minkowski sum. The compact-factor hypothesis also ensures that these sums of closed sets are closed. More generally, the same rules hold when addition is a proper map on $\mathrm{supp}u\times \mathrm{supp}v$, which is another sufficient condition for defining the convolution. The second inclusion explains why a compactly supported smooth function smooths every distribution under convolution.

Write $z=x+iy$, $r=|z|$, and $L=(4\pi {)}^{-1}\log (x^2+y^2)$. The Wirtinger derivatives satisfy $4{\partial }_{\bar{z}}{\partial }_z=\Delta$. Away from zero, ordinary differentiation givesThis equality also holds for distributional derivatives. To check that no Dirac delta distribution was lost at zero, integrate by parts outside the disk $r\le \epsilon$ against a test function. The new boundary term has absolute value at most $C\epsilon |\log \epsilon |\Vert \varphi {\Vert }_{\infty }\to 0$. Both $L$ and its displayed first derivatives, of size $O(r^{-1})$, are locally integrable functions in two dimensions, so the remaining integrals converge to their full-plane values.

Using the supplied fundamental solution of the Laplace equation identity $\Delta L={\delta }_0$, and commutation of distributional derivatives, now givesThus the Cauchy kernel fundamental solution has exactly the required normalization.

If $F$ is any other fundamental solution, then $H=F-E$ satisfies ${\partial }_{\bar{z}}H=0$ as a distribution. Applying $4{\partial }_z$ gives $\Delta H=0$. Elliptic regularity for the Laplacian implies that the initially arbitrary distribution $H$ is a smooth function. Its equation is then the ordinary Cauchy-Riemann equation, so $H$ is an entire function.

For a direct justification of the needed regularity, choose a radial mollifier ${\eta }_r$. Locally, $H\ast {\eta }_{\epsilon }$ is a smooth harmonic function. On a ball with sufficient room for both radii, the mean value property gives $(H\ast {\eta }_{\epsilon })\ast {\eta }_r=H\ast {\eta }_{\epsilon }$. Passing to the weak convergence of distributions limit $\epsilon \to 0$ yields $H\ast {\eta }_r=H$ on a smaller ball. The left side is smooth, proving the claim locally. This establishes the distributional Cauchy-Riemann lemma without assuming in advance that $H$ is a locally integrable function.

Conversely every entire function $h$ satisfies ${\partial }_{\bar{z}}h=0$ and can be added to $E$. Hence all fundamental solutions areNo growth condition at infinity was imposed, so $h$ need not be a polynomial.

Because $f$ is a compactly supported distribution, the convolution of distributions with a compactly supported factor defines $E\ast f$. The convolution derivative identity and part (i) giveThe difference between any two solutions has zero ${\partial }_{\bar{z}}$ distributional derivative. Part (ii) therefore proves that the general solution of the partial differential equation isThis formula is an identity of distributions; a pointwise integral is not required for arbitrary $f$.

If $u$ were a smooth function, its Wirtinger derivative ${\partial }_{\bar{z}}u=f$ would also be smooth, contradicting $\mathrm{singsupp}f\ne \varnothing$. In fact the singular support is exactly preserved here:One inclusion follows because a differential operator cannot create a singularity where $u$ is smooth. For the reverse inclusion, $E$ is smooth off zero, and its singular support is exactly $\{0\}$: if it were smooth near zero, ${\partial }_{\bar{z}}E={\delta }_0$ would be smooth there. The support and singular support rules for convolution give $\mathrm{singsupp}(E\ast f)\subseteq \{0\}+\mathrm{singsupp}f$, and adding the smooth entire function $h$ leaves this singular support unchanged. Thus there is no globally smooth solution when the given singular support is nonempty.

Interchange the two Wirtinger derivatives in part (i). The same real fundamental solution of the Laplace equation $L$ givesThe calculation is valid for distributions by the same vanishing boundary term estimate. Any other fundamental solution differs by a distribution $H$ with ${\partial }_{z}H=0$. Then $\Delta H=0$, so the regularity argument of part (ii) makes $H$ smooth, and its Cauchy-Riemann equations make it an antiholomorphic function. On the whole plane it can be written $H(z)=h(\bar{z})$ with $h$ an entire function. Conversely every such $H$ has ${\partial }_{z}H=0$. Therefore