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

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.