Proof Let ( ρ ε ) ε > 0 {\displaystyle \left(\rho _{\varepsilon }\right)_{\varepsilon >0}} be the standard mollifier .
Fix a compact set Ω ′ ⋐ Ω {\displaystyle \Omega ^{\prime }\Subset \Omega } and put ε 0 = dist ( Ω ′ , ∂ Ω ) {\displaystyle \varepsilon _{0}=\operatorname {dist} \left(\Omega ^{\prime },\partial \Omega \right)} be the distance between Ω ′ {\displaystyle \Omega ^{\prime }} and the boundary of Ω {\displaystyle \Omega } .
For each x ∈ Ω ′ {\displaystyle x\in \Omega ^{\prime }} and ε ∈ ( 0 , ε 0 ) {\displaystyle \varepsilon \in \left(0,\varepsilon _{0}\right)} the function
y ⟼ ρ ε ( x − y ) {\displaystyle y\longmapsto \rho _{\varepsilon }(x-y)} belongs to test functions D ( Ω ) {\displaystyle {\mathcal {D}}(\Omega )} and so we may consider
⟨ u , ρ ε ( x − ⋅ ) ⟩ {\displaystyle \left\langle u,\rho _{\varepsilon }(x-\cdot )\right\rangle } We assert that it is independent of ε ∈ ( 0 , ε 0 ) {\displaystyle \varepsilon \in \left(0,\varepsilon _{0}\right)} . To prove it we calculate d d ε ρ ε ( x − y ) {\displaystyle {\frac {\mathrm {d} }{\mathrm {d} \varepsilon }}\rho _{\varepsilon }(x-y)} for x , y ∈ R n {\displaystyle x,y\in \mathbb {R} ^{n}} .
Recall that
ρ ε ( x − y ) = ε − n ρ ( x − y ε ) {\displaystyle \rho _{\varepsilon }(x-y)=\varepsilon ^{-n}\rho \left({\frac {x-y}{\varepsilon }}\right)} where the standard mollifier kernel ρ {\displaystyle \rho } on R n {\displaystyle \mathbb {R} ^{n}} was defined at Mollifier#Concrete_example . If we put
θ ( t ) = { 1 c n e 1 t − 1 if t < 1 , 0 if t ⩾ 1 , {\displaystyle \theta (t)={\begin{cases}{\frac {1}{c_{n}}}\mathrm {e} ^{\frac {1}{t-1}}&{\text{ if }}t<1,\\0&{\text{ if }}t\geqslant 1,\end{cases}}} then ρ ( x ) = θ ( | x | 2 ) {\displaystyle \rho (x)=\theta \left(|x|^{2}\right)} .
Clearly θ ∈ C ∞ ( R ) {\displaystyle \theta \in \mathrm {C} ^{\infty }(\mathbb {R} )} satisfies θ ( t ) = 0 {\displaystyle \theta (t)=0} for t ⩾ 1 {\displaystyle t\geqslant 1} . Now calculate
d d ε ( ε − n ρ ( x − y ε ) ) = − n ε − n − 1 ρ ( x − y ε ) − ε − n ∇ ρ ( x − y ε ) ⋅ x − y ε 2 = − 1 ε n + 1 ( n ρ ( x − y ε ) + ∇ ρ ( x − y ε ) ⋅ x − y ε ) {\displaystyle {\begin{aligned}{\frac {\mathrm {d} }{\mathrm {d} \varepsilon }}\left(\varepsilon ^{-n}\rho \left({\frac {x-y}{\varepsilon }}\right)\right)&=-n\varepsilon ^{-n-1}\rho \left({\frac {x-y}{\varepsilon }}\right)-\varepsilon ^{-n}\nabla \rho \left({\frac {x-y}{\varepsilon }}\right)\cdot {\frac {x-y}{\varepsilon ^{2}}}\\&=-{\frac {1}{\varepsilon ^{n+1}}}\left(n\rho \left({\frac {x-y}{\varepsilon }}\right)+\nabla \rho \left({\frac {x-y}{\varepsilon }}\right)\cdot {\frac {x-y}{\varepsilon }}\right)\end{aligned}}} Put K ( x ) = − n ρ ( x ) − ∇ ρ ( x ) ⋅ x {\displaystyle K(x)=-n\rho (x)-\nabla \rho (x)\cdot x} so that
d d ε ( ε − n ρ ( x − y ε ) ) = 1 ε n + 1 K ( x − y ε ) . {\displaystyle {\frac {\mathrm {d} }{\mathrm {d} \varepsilon }}\left(\varepsilon ^{-n}\rho \left({\frac {x-y}{\varepsilon }}\right)\right)={\frac {1}{\varepsilon ^{n+1}}}K\left({\frac {x-y}{\varepsilon }}\right).} In terms of ρ ( x ) = θ ( | x | 2 ) {\displaystyle \rho (x)=\theta \left(|x|^{2}\right)} we get
K ( x ) = − div ( ρ ( x ) x ) = − div ( θ ( | x | 2 ) x ) {\displaystyle K(x)=-\operatorname {div} (\rho (x)x)=-\operatorname {div} \left(\theta \left(|x|^{2}\right)x\right)} and if we set
Θ ( t ) = 1 2 ∫ t ∞ θ ( s ) d s {\displaystyle \Theta (t)={\frac {1}{2}}\int _{t}^{\infty }\theta (s)\mathrm {d} s} then Θ ∈ C ∞ ( R ) {\displaystyle \Theta \in \mathrm {C} ^{\infty }(\mathbb {R} )} with Θ ( t ) = 0 {\displaystyle \Theta (t)=0} for t ⩾ 1 {\displaystyle t\geqslant 1} , and Θ ′ ( t ) = − 1 2 θ ( t ) {\displaystyle \Theta ^{\prime }(t)=-{\frac {1}{2}}\theta (t)} . Consequently
− θ ( | x | 2 ) x = ∇ ( Θ ( | x | 2 ) ) {\displaystyle -\theta \left(|x|^{2}\right)x=\nabla \left(\Theta \left(|x|^{2}\right)\right)} and so K ( x ) = div ∇ ( Θ ( | x | 2 ) ) = ( Δ Φ ) ( x ) {\displaystyle K(x)=\operatorname {div} \nabla \left(\Theta \left(|x|^{2}\right)\right)=(\Delta \Phi )(x)} , where Φ ( x ) = Θ ( | x | 2 ) {\displaystyle \Phi (x)=\Theta \left(|x|^{2}\right)} . Observe that Φ ∈ D ( B 1 ( 0 ) ¯ ) {\displaystyle \Phi \in {\mathcal {D}}\left({\overline {B_{1}(0)}}\right)} , and
− 1 ε n + 1 ( n ρ ( x − y ε ) + ∇ ρ ( x − y ε ) ⋅ x − y ε ) = 1 ε n + 1 Δ y ( Φ ( x − y ε ) ) = Δ y ( ε 1 − n Φ ( x − y ε ) ) . {\displaystyle {\begin{aligned}-{\frac {1}{\varepsilon ^{n+1}}}\left(n\rho \left({\frac {x-y}{\varepsilon }}\right)+\nabla \rho \left({\frac {x-y}{\varepsilon }}\right)\cdot {\frac {x-y}{\varepsilon }}\right)&={\frac {1}{\varepsilon ^{n+1}}}\Delta _{y}\left(\Phi \left({\frac {x-y}{\varepsilon }}\right)\right)\\&=\Delta _{y}\left(\varepsilon ^{1-n}\Phi \left({\frac {x-y}{\varepsilon }}\right)\right).\end{aligned}}} Here y ↦ ε 1 − n Φ ( x − y ε ) {\displaystyle y\mapsto \varepsilon ^{1-n}\Phi \left({\frac {x-y}{\varepsilon }}\right)} is supported in B ε ( x ) ¯ ⊂ Ω {\displaystyle {\overline {B_{\varepsilon }(x)}}\subset \Omega } , and so by assumption
⟨ u , Δ y ( ε 1 − n Φ ( x − y ε ) ) ⟩ = 0 {\displaystyle \left\langle u,\Delta _{y}\left(\varepsilon ^{1-n}\Phi \left({\frac {x-y}{\varepsilon }}\right)\right)\right\rangle =0} .Now by considering difference quotients we see that
d d ε ⟨ u , ρ ε ( x − ⋅ ) ⟩ = ⟨ u , d d ε ρ ε ( x − ⋅ ) ⟩ {\displaystyle {\frac {\mathrm {d} }{\mathrm {d} \varepsilon }}\left\langle u,\rho _{\varepsilon }(x-\cdot )\right\rangle =\left\langle u,{\frac {\mathrm {d} }{\mathrm {d} \varepsilon }}\rho _{\varepsilon }(x-\cdot )\right\rangle } .Indeed, for ε , ε ′ > 0 {\displaystyle \varepsilon ,\varepsilon ^{\prime }>0} we have
ρ ε + ε ′ ( x − y ) − ρ ε ( x − y ) ε ′ = F T C ∫ 0 1 d d t ρ ε + t ε ′ ( x − y ) d t ⟶ ε ′ ↘ 0 d d s | s = ε ρ s ( x − y ) {\displaystyle {\begin{aligned}{\frac {\rho _{\varepsilon +\varepsilon ^{\prime }}(x-y)-\rho _{\varepsilon }(x-y)}{\varepsilon ^{\prime }}}&{\stackrel {\mathrm {FTC} }{=}}\int _{0}^{1}{\frac {\mathrm {d} }{\mathrm {d} t}}\rho _{\varepsilon +t\varepsilon ^{\prime }}(x-y)\mathrm {d} t\\&\left.{\underset {\varepsilon ^{\prime }\searrow 0}{\longrightarrow }}{\frac {\mathrm {d} }{\mathrm {d} s}}\right|_{s=\varepsilon }\rho _{s}(x-y)\end{aligned}}} in D ′ ( Ω ) {\displaystyle {\mathcal {D}}^{\prime }(\Omega )} with respect to y {\displaystyle y} , provided x ∈ Ω ′ {\displaystyle x\in \Omega ^{\prime }} and 0 < ε < ε 0 {\displaystyle 0<\varepsilon <\varepsilon _{0}} (since we may differentiate both sides with respect to y ) {\displaystyle y)} . But then d d ε ⟨ u , ρ ε ( x − ⋅ ) ⟩ = 0 {\displaystyle {\frac {\mathrm {d} }{\mathrm {d} \varepsilon }}\left\langle u,\rho _{\varepsilon }(x-\cdot )\right\rangle =0} , and so ⟨ u , ρ ε ( x − ⋅ ) ⟩ = ⟨ u , ρ ε 1 ( x − ⋅ ) ⟩ {\displaystyle \left\langle u,\rho _{\varepsilon }(x-\cdot )\right\rangle =\left\langle u,\rho _{\varepsilon _{1}}(x-\cdot )\right\rangle } for all ε ∈ ( 0 , ε 0 ) {\displaystyle \varepsilon \in \left(0,\varepsilon _{0}\right)} , where ε 1 ∈ ( 0 , ε 0 ) {\displaystyle \varepsilon _{1}\in \left(0,\varepsilon _{0}\right)} . Now let φ ∈ D ( Ω ′ ) {\displaystyle \varphi \in {\mathcal {D}}\left(\Omega ^{\prime }\right)} . Then, by the usual trick when convolving distributions with test functions ,
∫ Ω ′ ⟨ u , ρ ε ( x − ⋅ ) ⟩ φ ( x ) d x = ⟨ u , ∫ Ω ′ ρ ε ( x − ⋅ ) φ ( x ) d x ⟩ = ⟨ u , ρ ε ∗ φ ⟩ {\displaystyle {\begin{aligned}\int _{\Omega ^{\prime }}\left\langle u,\rho _{\varepsilon }(x-\cdot )\right\rangle \varphi (x)\mathrm {d} x&=\left\langle u,\int _{\Omega ^{\prime }}\rho _{\varepsilon }(x-\cdot )\varphi (x)\mathrm {d} x\right\rangle \\&=\left\langle u,\rho _{\varepsilon }*\varphi \right\rangle \end{aligned}}} and so for ε ∈ ( 0 , ε 1 ) {\displaystyle \varepsilon \in \left(0,\varepsilon _{1}\right)} we have
⟨ u , ρ ε ∗ φ ⟩ = ∫ Ω ′ ⟨ u , ρ ε 1 ( x − ⋅ ) ⟩ φ ( x ) d x {\displaystyle \left\langle u,\rho _{\varepsilon }*\varphi \right\rangle =\int _{\Omega ^{\prime }}\left\langle u,\rho _{\varepsilon _{1}}(x-\cdot )\right\rangle \varphi (x)\mathrm {d} x} .Hence, as ρ ε ∗ φ → φ {\displaystyle \rho _{\varepsilon }*\varphi \rightarrow \varphi } in D ( Ω ) {\displaystyle {\mathcal {D}}(\Omega )} as ε ↘ 0 {\displaystyle \varepsilon \searrow 0} , we get
⟨ u , φ ⟩ = ∫ Ω ′ ⟨ u , ρ ε 1 ( x − ⋅ ) ⟩ φ ( x ) d x {\displaystyle \langle u,\varphi \rangle =\int _{\Omega ^{\prime }}\left\langle u,\rho _{\varepsilon _{1}}(x-\cdot )\right\rangle \varphi (x)\mathrm {d} x} .Consequently u | Ω ′ ∈ C ∞ ( Ω ′ ) {\displaystyle \left.u\right|_{\Omega ^{\prime }}\in \mathrm {C} ^{\infty }\left(\Omega ^{\prime }\right)} , and since Ω ′ {\displaystyle \Omega ^{\prime }} was arbitrary, we are done.