Distribution Theory

Existence of smooth partition of unity: Let $Ω$ be an open subset of $ℝ^{n}, K$ a compact subset of $Ω$ and suppose $Ω=Ω_{1} ∪ … ∪ Ω_{J}$, where $Ω_{j}$ are open nonempty subsets of $Ω$. Then there exist $ϕ_{j} ∈ 𝒟(Ω)$, $j ∈\{1, …, J\}$, satisfying for each $j, \supp(ϕ_{j}) ⊂ Ω_{j}, 0 ≤ ϕ_{j} ≤ 1$ and such that $\sum_{j=1}^{J} ϕ_{j} ≤ 1$ on $Ω$ and $\sum_{j=1}^{J} ϕ_{j}=1$ on $K$.
The restriction of $u$ to $ω$ is the distribution $u|_{ω} ∈ 𝒟'(ω)$ defined by $⟨u|_{ω}, ϕ⟩=⟨u, ϕ⟩$, where $ϕ ∈ 𝒟(ω)$ on the right-hand side is extended by 0 on $Ω ∖ ω$. This extension is clearly a test function on $Ω$ making the right-hand side well-defined. It is then also clear that $u|_{ω}: 𝒟(ω) → ℂ$ is linear and $𝒟$ continuous.
In order to prove that $v=0$ we fix $ϕ ∈ 𝒟(Ω)$. Now the set $K=\supp(ϕ)$ is compact and contained in $Ω$ and the family $\{ω_{x}: x ∈ K\}$ is an open cover of $K$, so admits a finite subcover. Say already $K ⊆ ω_{1} ∪ … ∪ ω_{J}$, where we abbreviated $ω_{j}=ω_{x_{j}}$. According to (a)(i) we find a smooth partition of unity $\{ϕ_{j}\}_{j=1}^{J}$ for $K$ subordinated to the covering $\{ω_{j}\}_{j=1}^{J}$. Now $ϕ ϕ_{j}$ is a test function supported in $ω_{j}$, so by assumption $⟨v, ϕ ϕ_{j}⟩=⟨v|_{ω_{j}}, ϕ ϕ_{j}⟩=0$, and consequently $$ ⟨v, ϕ⟩=\sum_{j=1}^{J}⟨v, ϕ ϕ_{j}⟩=0 $$
The support of $u$ is $$ \supp(u)=\{x ∈ Ω: ∀ r>0 ∃ ϕ ∈ 𝒟(Ω ∩ B_{r}(x)) \text{ such that }⟨u, ϕ⟩ ≠ 0\} $$ Thus its complement in $Ω$ equals $$ Ω ∖ \supp(u)=\{x ∈ Ω: ∃ r>0 \text{ such that }u|_{Ω ∩ B_{r}(x)}=0\} $$ and consequently if $x ∈ Ω ∖ \supp(u)$, then for some $r>0$ also $B_{r}(x) ⊂Ω ∖ \supp(u)$. The complement is therefore an open subset of $Ω$ and so the set itself must be relatively closed in $Ω$. It follows from (ii) that $u|_{Ω ∖ \supp(u)}=0$. The singular support of $u$ is $$ \singsupp(u)=\{x ∈ Ω: ∀ r>0 \text{ such that }u|_{Ω ∩ B_{r}(x)} ∉ \mathrm{C}^{∞}(Ω ∩ B_{r}(x))\} $$ Thus its complement in $Ω$ equals $$ Ω ∖ \singsupp(u)=\{x ∈ Ω: ∃ r>0 \text{ such that }u|_{B_{r}(x)} ∈ \mathrm{C}^{∞}(B_{r}(x))\} $$ and consequently if $x ∈ Ω ∖\singsupp(u)$, then for some $r>0$ also $B_{r}(x) ⊂Ω ∖\singsupp(u)$. The complement is therefore an open subset of $Ω$ and so the set itself must be relatively closed in $Ω$. In order to check that $u$ is $\mathrm{C}^{∞}$ on the complement $Ω ∖ \singsupp(u)$ we note that for each $x ∈ Ω ∖ \singsupp(u)$ there exist $r_{x}>0$ with $B_{x}=B_{r_{x}}(x) ⊂ Ω$ and $f_{x} ∈ \mathrm{C}^{∞}(B_{x})$ such that $u|_{B_{x}}=f_{x}$. Now for $x, y ∈ Ω ∖\singsupp(u)$ with $B_{x} ∩ B_{y} ≠ ∅$, then for all $ϕ ∈ 𝒟(B_{x} ∩ B_{y})$ we have $∫_{B_{x} ∩ B_{y}} f_{x} ϕ \mathrm{d} t=⟨u, ϕ⟩=∫_{B_{x} ∩ B_{y}} f_{y} ϕ \mathrm{d} t$ and therefore $f_{x}=f_{y}$ in $B_{x} ∩ B_{y}$ because the functions in particular are continuous. Now observe that $\bigcup B_{x}=Ω ∖\singsupp(u)$ and define $f: Ω ∖\singsupp(u) → ℂ$ by $f=f_{x}$ on $B_{x}$. Then $f$ is a well-defined $\mathrm{C}^{∞}$ function on $Ω ∖ \singsupp(u)$ and by (ii) applied to $u-f$ we see that $u|_{Ω ∖ \singsupp(u)}=f$.
If $u=𝟏_{ω}$, then clearly $u=0$ on the open set $Ω ∖ \bar{ω}$ and it is clearly the largest open subset of $Ω$ with this property, so $\supp(u)=Ω ∩ \bar{ω}$. Next, note that $u$ is $\mathrm{C}^{∞}$ on the open set $Ω∖\bar{ω}$ because it is 0 there and on the open set $ω$ because it is 1 there. It is clearly not continuous on any larger open subset of $Ω$, hence $\singsupp(u)=Ω ∩ ∂ ω$.
The convolution rule: If $u, v ∈ 𝒟'(ℝ^{n})$ and one of them has compact support, then $\supp(u * v) ⊆ \supp(u)+\supp(v)$.
The singular support rule: If $u, v ∈ 𝒟'(ℝ^{n})$ and one of them has compact support, then $\singsupp(u * v) ⊆\singsupp(u)+\singsupp(v)$.

We use $Δ=4 \frac{∂}{∂ \bar{z}} \frac{∂}{∂ z}$ and note that for $z ≠ 0$ the function $\frac{1}{4 π} \log |z|^{2}$ is $\mathrm{C}^{∞}$ and $\frac{∂}{∂ z} \log (z \bar{z})=\frac{1}{z}$. Here $\frac{1}{z} ∈ \mathrm{L}_\text{loc}^{1}(ℝ^{2})$ and we claim the above differentiation is also valid in $𝒟'(ℝ^{2})$. Put $ξ=ρ * 𝟏_{(-∞, \frac{1}{2}]}$ and for each $ε>0, ψ_{ε}=ξ(\frac{|z|^{2}}{ε^{2}})$. Hereby $ψ_{ε} ∈ 𝒟(ℝ^{2})$ and $𝟏_{B_{ε}(0)} ≤ ψ_{ε} ≤ 𝟏_{B_{\sqrt{2} ε}(0)}$, and so $2 \log |z|(1-ψ_{ε}) ∈ \mathrm{C}^{∞}(ℝ^{2})$. Furthermore we check that since $\log |z| ∈\mathrm{L}_\text{loc}^{1}(ℝ^{2})$ we have $2 \log |z|(1-ψ_{ε}) → 2 \log |z|$ in $𝒟'(ℝ^{2})$ as $ε ↘ 0$ and so by $𝒟'$ continuity of differentiation that $\frac{∂}{∂ z}(2 \log |z|(1-ψ_{ε})) → \frac{∂}{∂ z} 2 \log |z|$ in $𝒟'(ℝ^{2})$ as $ε ↘ 0$. Now $\frac{∂}{∂ z}(2 \log |z|(1-ψ_{ε}))=\frac{1-ψ_{ε}}{z}-2 \log |z| ξ'(\frac{|z|^{2}}{ε^{2}}) \frac{\bar{z}}{ε^{2}}$. We have $\frac{1-ψ_{ε}}{z} → \frac{1}{z}$ in $𝒟'(ℝ^{2})$ as $ε ↘ 0$ because $\frac{1}{z} ∈ \mathrm{L}_\text{loc}^{1}(ℝ^{2})$. For the second term we estimate: \begin{aligned} ∫_{ℝ^{2}}|2 \log | z|ξ'(\frac{|z|^{2}}{ε^{2}}) \frac{\bar{z}}{ε^{2}}| \mathrm{d} x \mathrm{~d} y &≤ 4 π ∫_{0}^{\sqrt{2} ε} \log r \max |ξ'| \frac{r^{2}}{ε^{2}} \mathrm{~d} r \\ &≤ 8 π \max |ξ'| ∫_{0}^{\sqrt{2} ε} \log r \mathrm{~d} r → 0 \end{aligned} as $ε ↘ 0$, and consequently $\frac{∂}{∂ z}(2 \log |z|)=\frac{1}{z}$ in $𝒟'(ℝ^{2})$. But then we get $δ_{0}=\frac{∂}{∂ \bar{z}} E$ as required.
Since $\singsupp(E)=\{0\}$ the operator $\frac{∂}{∂ \bar{z}}$ is elliptic and thus the elliptic regularity theorem implies that all solutions to the homogeneous equation $\frac{∂ h}{∂ \bar{z}}=0$ in $𝒟'(ℝ^{2})$ are $\mathrm{C}^{∞}$, hence are holomorphic by the Cauchy-Riemann equations. But then the fundamental solutions are exactly $E+h$, where $h: ℂ → ℂ$ is holomorphic.
Since $f$ has compact support, $E * f$ is well-defined in $𝒟'(ℝ^{2})$ and by the singular support rule $\singsupp(E * f) ⊆\singsupp(f)$. Since $\frac{∂}{∂ \bar{z}}(E * f)=f$ we also get the opposite inclusion so $\singsupp(E * f)=\singsupp(f)$ holds. The general solution in $𝒟'(ℝ^{2})$ is $E * f+h$, where $h: ℂ → ℂ$ is any holomorphic function. When $\singsupp(f) ≠ ∅$ the function $E * f$ is then also not a $\mathrm{C}^{∞}$ function and so there can be no $\mathrm{C}^{∞}$ solutions.
By symmetry of $z$ and $\bar{z}$ in $Δ=4 \frac{∂}{∂ \bar{z}} \frac{∂}{∂ z}$ we see that $\bar{E}+\bar{h}=\frac{1}{π \bar{z}}+\bar{h}$ are the fundamental solutions in this case, where again $h: ℂ → ℂ$ is any holomorphic function.