- What does it mean to say that $u$ is positive? What does it mean to say that $u$ has order 0 ?
- Prove that a positive distribution on $ℝ^n$ has order 0.
- Assume $u$ has order 0 and denote by $\mathrm{C}_c(ℝ^n)$ the space of all continuous and compactly supported functions $ϕ: ℝ^n → ℝ$. Let $(ρ_{ε})_{ε>0}$ be the standard mollifier on $ℝ^n$. Prove that \[ ⟨U, ϕ⟩=\lim_{ε ↘ 0}⟨u, ρ_{ε} * ϕ⟩, ϕ ∈ \mathrm{C}_c(ℝ^n), \] defines a linear functional $U: \mathrm{C}_c(ℝ^n) → ℝ$ satisfying $⟨U, ϕ⟩=⟨u, ϕ⟩$ for $ϕ ∈ D(ℝ^n)$ and with the boundedness property that for each compact subset $K$ of $ℝ^n$ there exists a constant $c_K ⩾ 0$ such that \[ |⟨U, ϕ⟩| ⩽ c_K \max_{x ∈ ℝ^n}|ϕ(x)| \] holds for all $ϕ ∈ \mathrm{C}_c(ℝ^n)$ with $\supp(ϕ) ⊆ K$. Explain also why $U$ is the unique such extension.

- $u$ is positive if $⟨u, ϕ⟩ ≥ 0$ holds for all $ϕ ∈ 𝒟(ℝ^n)$ with $ϕ(x) ≥ 0$ for all $x ∈ ℝ^n$. $u$ has order 0 if for any compact set $K$ in $ℝ^n$ we can find a constant $c_K ≥ 0$ such that $|⟨u, ϕ⟩| ≤ c_K \max |ϕ|$ holds for all $ϕ ∈ 𝒟(K)$.
- Proof. Fix a compact subset $K$ of $ℝ^n$. Define $χ=ρ * 𝟏_{B_1(K)}$. Then $χ ∈ 𝒟(ℝ^n)$ satisfies $𝟏_K ≤ χ ≤ 𝟏_{B_2(K)}$. If $ϕ ∈ 𝒟(K)$, then we have that $Φ^±=χ(\max |ϕ| ± ϕ)$ are both non-negative and in $𝒟(ℝ^n)$, hence by assumption \[ 0 ≤⟨u, Φ^±⟩=⟨u, χ⟩ \max |ϕ| ±⟨u, χ ϕ⟩. \] But $χ ϕ=ϕ$ so this rearranges to $|⟨u, ϕ⟩| ≤ c \max |ϕ|$, where $c=⟨u, χ⟩$.
- Proof. First note that if $ϕ ∈ \mathrm{C}_c(ℝ^n)$, then $ρ_{ε} * ϕ ∈ 𝒟(ℝ^n)$, $\supp(ρ_{ε} *ϕ) ⊆ \overline{B_{ε}(\supp(ϕ))}$ and $ρ_{ε} * ϕ → ϕ$ uniformly on $ℝ^n$ as $ε ↘ 0$. Fix a compact set $K$ such that $ϕ$ is supported in $K$. Corresponding to the compact neighbourhood $\overline{B_1(K)}$ we find since $u$ has order 0 a constant $c ≥ 0$ such that \[ |⟨u, ψ⟩| ≤ c \max |ψ| \] holds for all $ψ ∈ \mathrm{C}_c(ℝ^n)$ supported in $\overline{B_1(K)}$. We apply this to the function $ψ=ρ_{ε'} * ϕ-ρ_{ε''} * ϕ$ for $ε', ε'' ∈(0,1)$: \begin{aligned} ⟨u, ρ_{ε'} * ϕ⟩-⟨u, ρ_{ε''} * ϕ⟩| &≤ c \max |ρ_{ε'} * ϕ-ρ_{ε''} * ϕ| \\ &≤ c \max |ρ_{ε'} * ϕ-ϕ|+c \max |ϕ-ρ_{ε''} * ϕ| \end{aligned} and here the right-hand side tends to 0 as $ε', ε'' ↘ 0$. Thus $⟨u, ρ_{ε} * ϕ⟩$ is Cauchy in $ℝ$ as $ε ↘ 0$ and so by completeness has a limit in $ℝ:⟨U, ϕ⟩:=\lim_{ε↘0}⟨u, ρ_{ε} * ϕ⟩$ exists in $ℝ$. Thus $U: \mathrm{C}_c(ℝ^n) → ℝ$ is a well-defined linear functional. Another application of the above boundedness property yields for $ϕ ∈ 𝒟(ℝ^n)$ supported in $K$ that $⟨U, ϕ⟩=⟨u, ϕ⟩$. Finally if $ϕ ∈ \mathrm{C}_c(ℝ^n)$ is supported in $K$, then applying the above bound to $ψ=ρ_{ε} * ϕ$ for $ε ∈(0,1)$ we get \[ |⟨U, ϕ⟩|=\lim_{ε ↘ 0}|⟨u, ρ_{ε} * ϕ⟩| ≤ \limsup_{ε ↘ 0} c \max |ρ_{ε} * ϕ|=c \max |ϕ| \] as required. Finally the uniqueness: Assume $V$ is another linear functional that extends $u$ to $\mathrm{C}_c(ℝ^n)$ and that has the boundedness property. Let $ϕ ∈ \mathrm{C}_c(ℝ^n)$ and take a compact neighbourhood $K$ of $\supp(ϕ)$. From the boundedness property and since $ρ_{ε} * ϕ ∈ 𝒟(ℝ^n)$ we infer that $⟨V, ϕ⟩=\lim_{ε↘0}⟨V, ρ_{ε} * ϕ⟩=\lim_{ε↘0} ⟨u, ρ_{ε} * ϕ⟩=⟨U, ϕ⟩$

- Define for $t>0$ the function $T_t(y)=\sqrt{y^2+t}, y ∈ ℝ$. Let $f ∈ \mathrm{C}^{∞}(ℝ^n)$ and consider for a fixed $t>0$ the composite function $T_t(f)(x)=T_t(f(x)), x ∈ ℝ^n$. Show that for each $j ∈\{1, …, n\}$, \[ (∂_j T_t(f))^2+T_t(f) ∂_j^2 T_t(f)=(∂_j f)^2+f ∂_j^2 f \] and deduce that \[ Δ T_t(f) ⩾ \frac{f}{T_t(f)} Δ f \]
- Prove, for instance by use of mollification, (a)(iii) and (b)(i), that \[ Δ T_t(u)-\frac{u}{T_t(u)} Δ u \] is a well-defined and positive distribution.
- Prove that $Δ|u|$, that is, the distributional Laplacian of the absolute value of $u$, is a distribution of order 0.

- For $f ∈ \mathrm{C}^{∞}(ℝ^n)$ we calculate using the chain and quotient rules \[ ∂_j T_t(f)=\frac{f ∂_j f}{T_t(f)} \text{and } ∂_j^2 T_t(f)=\frac{(∂_j f)^2+f ∂_j^2 f-(∂_j T_t(f))^2}{T_t(f)} \] and rearranging the last identity we arrive at \[ (∂_j T_t(f))^2+T_t(f) ∂_j^2 T_t(f)=(∂_j f)^2+f ∂_j^2 f, \] as required. Note that $(∂_j T_t(f))^2=\frac{f^2}{f^2+t^2}(∂_j f)^2≤(∂_j f)^2$, so necessarily, $T_t(f) ∂_j^2 T_t(f) ≥f ∂_j^2 f$, and hence summing over $j$ we get $T_t(f) Δ T_t(f) ≥ f Δ f$ or equivalently, \[ Δ T_t(f) ≥ \frac{f}{T_t(f)} Δ f. \]
- Proof. First we check that the distribution is well-defined. Clearly $T_t(u)$ being a continuous function is a regular distribution, so that the distributional Laplacian $Δ T_t(u)$ is well-defined. For the second term the issue is that $u / T_t(u)$ merely is a continuous function. That the product $\frac{u}{T_t(u)} Δ u$ is well-defined follows because $Δ u$ is a distribution of order 0. By (a)(iii) it then has a unique extension, again denoted $Δ u$, to $\mathrm{C}_c(ℝ^n)$ as a linear functional with the zero-order boundedness property. Now if $ϕ ∈ 𝒟(ℝ^n)$, then $\frac{u}{T_t(u)} ϕ ∈ \mathrm{C}_c(ℝ^n)$ and so
\[
ϕ ↦⟨\frac{u}{T_t(u)} Δ u, ϕ⟩=⟨Δ u, \frac{u}{T_t(u)} ϕ⟩
\]
is a well-defined distribution of order 0. We record that we then have the continuity property: $⟨\frac{u}{T_t(u)} Δ u, ϕ_j⟩=⟨Δ u, \frac{u}{T_t(u)} ϕ_j⟩ → 0$ whenever $ϕ_j ∈ \mathrm{C}_c(ℝ^n)$ with $\supp(ϕ_j) ⊆$ fixed compact and $ϕ_j → 0$ uniformly.

In order to verify that the distribution $Δ T_t(u)-\frac{u}{T_t(u)} Δ u$ is positive we follow the hint and use mollification and (i): put $u_{ε}=ρ_{ε} * u$. Since $u$ is a continuous function, $u_{ε} → u$ locally uniformly on $ℝ^n$ as $ε ↘ 0$. Anticipating the calculation below we also record that for $ϕ ∈ 𝒟(ℝ^n)$ we have $\frac{u_{ε}}{T_t(u_{ε})} ϕ → \frac{u}{T_t(u)} ϕ$ uniformly as $ε ↘ 0$ and all supports are contained in $B_1(\supp(ϕ))$ when $ε ∈(0,1)$. Now for $ϕ ∈ 𝒟(ℝ^n)^+$ we then have \begin{aligned} ⟨Δ T_t(u)-\frac{u}{T_t(u)} Δ u, ϕ⟩ &=⟨T_t(u), Δ ϕ⟩-⟨\frac{u}{T_t(u)} Δ u, ϕ⟩ \\ &=\lim_{ε ↘ 0}(⟨T_t(u_{ε}), Δ ϕ⟩-⟨\frac{u_{ε}}{T_t(u_{ε})} Δ u_{ε}, ϕ⟩) \\ &=\lim_{ε ↘ 0}⟨Δ T_t(u_{ε})-\frac{u_{ε}}{T_t(u_{ε})}, ϕ⟩ ≥ 0 \end{aligned} where the last inequality follows from (i). - Proof. First note that for $ϕ ∈ 𝒟(ℝ^n)$, \[ ⟨Δ|u|, ϕ⟩=∫_{ℝ^n}|u| Δ ϕ \mathrm{d} x=\lim_{t↘0} ∫_{ℝ^n} T_t(u) Δ ϕ \mathrm{d} x. \] Fix a compact set $K$ in $ℝ^n$. By the zero order boundedness property for $Δ u$ proved in (a)(iii) we find a constant $c_K ≥ 0$ such that $|⟨Δ u, ψ⟩| ≤ c_K \max |ψ|$ holds for all $ψ ∈ \mathrm{C}_c(ℝ^n)$ with support in $B_2(K)$. In particular $ψ=\frac{u}{T_t(u)} ϕ $ we then have \[ |⟨Δ u, \frac{u}{T_t(u)} ϕ⟩| ≤ c_K \max |ϕ| \] for all $ϕ ∈ 𝒟(B_2(K))$. Now by the positivity proved in (i) we get for $ϕ ∈ 𝒟(B_2(K))^+:$ \[ ⟨Δ T_t(u), ϕ⟩ ≥⟨Δ u, \frac{u}{T_t(u)} ϕ⟩ ≥-c_K \max ϕ, \] and hence taking $t ↘ 0$ we get$$⟨Δ|u|, ϕ⟩ ≥-c_K \max ϕ$$Now let $ψ ∈ 𝒟(K)$ and put $χ=ρ * 𝟏_{B_1(K)}$. If $ϕ^±=χ(\max |ψ| ± ψ)$, then $ϕ^± ∈ 𝒟(B_2(K))^+$ and so from $⟨Δ|u|, ϕ^±⟩ ≥-c_K \max ϕ^±$ and the properties of $χ$ we get \[ |⟨Δ|u|, ψ⟩| ≤ C_0 \max |ψ| \] where $C_0=2 c_K+⟨Δ|u|, χ⟩$.