Mollifiers
📂Partial Differential Equations Mollifiers Definition Let’s define the function η ∈ C ∞ ( R n ) \eta \in C^{\infty}(\mathbb{R}^{n}) η ∈ C ∞ ( R n )
η ( x ) : = { C exp ( 1 ∣ x ∣ 2 − 1 ) ∣ x ∣ < 1 0 ∣ x ∣ ≥ 1
\begin{equation}
\eta (x) := \begin{cases} C \exp \left( \dfrac{1}{|x|^2-1} \right) & |x|<1
\\ 0 & |x| \ge 1\end{cases}
\label{1}
\end{equation}
η ( x ) := ⎩ ⎨ ⎧ C exp ( ∣ x ∣ 2 − 1 1 ) 0 ∣ x ∣ < 1 ∣ x ∣ ≥ 1
Such η \eta η mollifier . In particular, when C > 0 C>0 C > 0 ∫ R n η d x = 1 \displaystyle \int_{\mathbb{R}^{n}} \eta dx=1 ∫ R n η d x = 1 η \eta η standard mollifier .
Let’s define η ϵ \eta_{\epsilon} η ϵ ϵ > 0 \epsilon>0 ϵ > 0
η ϵ ( x ) : = 1 ϵ n η ( x ϵ )
\eta_\epsilon (x) := \dfrac{1}{\epsilon^n}\eta\left( \dfrac{x}{\epsilon} \right)
η ϵ ( x ) := ϵ n 1 η ( ϵ x )
If η \eta η ∫ R n η ϵ d x = 1 \int_{\mathbb{R^n}} \eta_{\epsilon} dx=1 ∫ R n η ϵ d x = 1
Description A mollifier serves to smooth out functions that are not smooth through convolution. The word mollify means ’to soothe’, which expresses that η \eta η
Meanwhile, one may consider that the form of a mollifier does not necessarily have to be like ( def ) \eqref{def} ( def )
Generalization If φ ∈ C ∞ ( R n ) \varphi \in C^{\infty}(\mathbb{R}^{n}) φ ∈ C ∞ ( R n ) φ \varphi φ
Has a compact support . ∫ R n φ d x = 1 \displaystyle \int_{\mathbb{R^n}} \varphi dx =1 ∫ R n φ d x = 1 lim ϵ → 0 φ ϵ = lim ϵ → 0 1 ϵ n φ ( x ϵ ) = δ ( x ) \lim \limits_{\epsilon \rightarrow 0}\varphi_\epsilon = \lim \limits_{\epsilon \rightarrow 0} \dfrac{1}{\epsilon^n}\varphi \left( \dfrac{x}{\epsilon} \right) =\delta (x) ϵ → 0 lim φ ϵ = ϵ → 0 lim ϵ n 1 φ ( ϵ x ) = δ ( x ) δ ( x ) \delta (x) δ ( x ) Dirac delta function .The following condition is called a positive mollifier .
φ ( x ) ≥ 0 , ∀ x ∈ R n \varphi (x) \ge 0, \quad \forall x \in \mathbb{R^n} φ ( x ) ≥ 0 , ∀ x ∈ R n The following condition is called a symmetric mollifier .
For μ ∈ C ∞ \mu \in C^{\infty} μ ∈ C ∞ φ ( x ) = μ ( ∣ x ∣ ) \varphi (x)=\mu (|x|) φ ( x ) = μ ( ∣ x ∣ ) Meanwhile, it can be proved that η \eta η ( 1 ) \eqref{1} ( 1 )
Proof First, let’s define f f f g g g
f ( s ) : = { C exp ( 1 s − 1 ) s < 1 0 s ≥ 1 g ( x ) : = ∣ x ∣ 2 = x 1 2 + x 2 2 + ⋯ + x n 2 , x ∈ R n
\begin{align*}
f(s) &:= \begin{cases} C \exp\left( \dfrac{1}{s-1} \right) & s<1 \\ 0 & s \ge 1 \end{cases}
\\g(x) &:= |x|^2={x_{1}}^2+{x_{2}}^2+\cdots+{x_{n}}^2 , \quad x\in \mathbb{R}^{n}
\end{align*}
f ( s ) g ( x ) := ⎩ ⎨ ⎧ C exp ( s − 1 1 ) 0 s < 1 s ≥ 1 := ∣ x ∣ 2 = x 1 2 + x 2 2 + ⋯ + x n 2 , x ∈ R n
Then, it aims to show f ∈ C ∞ f\in C^\infty f ∈ C ∞ g ∈ C ∞ g\in C^\infty g ∈ C ∞ η = f ∘ g ∈ C ∞ \eta=f \circ g \in C^\infty η = f ∘ g ∈ C ∞
Part 1. f ∈ C ∞ f \in C^\infty f ∈ C ∞
First, by calculating the derivatives of f f f ( − ∞ , 1 ) (-\infty, 1) ( − ∞ , 1 )
f ′ ( s ) = C e 1 s − 1 − 1 ( s − 1 ) 2 f ′ ′ ( s ) = C e 1 s − 1 − 1 ( s − 1 ) 4 + C e 1 s − 1 2 ( s − 1 ) 3 f ′ ′ ′ ( s ) = C e 1 s − 1 − 1 ( s − 1 ) 6 + C e 1 s − 1 2 ( s − 1 ) 5 + C e 1 s − 1 − 6 ( s − 1 ) 4
\begin{align*}
f^{\prime}(s) =&\ Ce ^{ \frac{1}{s-1}} \frac{-1}{(s-1)^2}
\\ f^{\prime \prime}(s) =&\ Ce ^{ \frac{1}{s-1}} \frac{-1}{(s-1)^4} + Ce ^{ \frac{1}{s-1}} \frac{2}{(s-1)^3}
\\ f^{\prime \prime \prime}(s) =&\ Ce ^{ \frac{1}{s-1}} \frac{-1}{(s-1)^6} +Ce ^{ \frac{1}{s-1}} \frac{2}{(s-1)^5}+Ce ^{ \frac{1}{s-1}} \frac{-6}{(s-1)^4}
\end{align*}
f ′ ( s ) = f ′′ ( s ) = f ′′′ ( s ) = C e s − 1 1 ( s − 1 ) 2 − 1 C e s − 1 1 ( s − 1 ) 4 − 1 + C e s − 1 1 ( s − 1 ) 3 2 C e s − 1 1 ( s − 1 ) 6 − 1 + C e s − 1 1 ( s − 1 ) 5 2 + C e s − 1 1 ( s − 1 ) 4 − 6
Henceforth, it aims to prove via induction that for all k ≥ 1 k \ge 1 k ≥ 1 a 0 ( k ) , a 1 ( k ) , ⋯ , a 2 k ( k ) a^{(k)}_{0},\ a^{(k)}_{1},\ \cdots,\ a^{(k)}_{2k} a 0 ( k ) , a 1 ( k ) , ⋯ , a 2 k ( k )
f ( k ) ( s ) = { e 1 s − 1 ( a 0 ( k ) + a 1 ( k ) s − 1 + ⋯ + a 2 k ( k ) ( s − 1 ) 2 k ) s < 1 0 s ≥ 1
f^{(k)}(s) = \begin{cases} e^{\frac{1}{s-1}} \left( a^{(k)}_{0} + \frac{a^{(k)}_{1}}{s-1}+\cdots+\frac{a^{(k)}_{2k}}{(s-1)^{2k}} \right) & s<1
\\ 0 & s \ge 1 \end{cases}
f ( k ) ( s ) = ⎩ ⎨ ⎧ e s − 1 1 ( a 0 ( k ) + s − 1 a 1 ( k ) + ⋯ + ( s − 1 ) 2 k a 2 k ( k ) ) 0 s < 1 s ≥ 1
Part 1-1. k = 1 k=1 k = 1
f ′ ( s ) = { e 1 s − 1 − C ( s − 1 ) 2 s < 1 0 s > 1
f^{\prime}(s) = \begin{cases} e^{\frac{1}{s-1} }\frac{-C}{(s-1)^2} & s<1
\\ 0 & s >1\end{cases}
f ′ ( s ) = { e s − 1 1 ( s − 1 ) 2 − C 0 s < 1 s > 1
Therefore, a 0 ( 1 ) = a 1 ( 1 ) = 0 a^{(1)}_{0} =a^{(1)}_{1}=0 a 0 ( 1 ) = a 1 ( 1 ) = 0 a 2 ( 1 ) = − C a^{(1)}_2=-C a 2 ( 1 ) = − C
lim h → 0 + f ( 1 + h ) − f ( 1 ) h = lim h → 0 + C e 1 h h = lim h → 0 + C 1 h e − 1 h = 0
\lim \limits_{h \rightarrow 0^+} \dfrac{ f(1+h)-f(1) }{h}=\lim \limits_{h \rightarrow 0^+} \dfrac{Ce^{ \frac{1}{h} } }{h}=\lim \limits_{h \rightarrow 0^+} \dfrac{C\frac{1}{h}}{e^{-\frac{1}{h} } }=0
h → 0 + lim h f ( 1 + h ) − f ( 1 ) = h → 0 + lim h C e h 1 = h → 0 + lim e − h 1 C h 1 = 0
lim h → 0 − f ( 1 + h ) − f ( 1 ) h = lim h → 0 − C e 1 h h = lim h → 0 − C 1 h e − 1 h = 0
\lim \limits_{h \rightarrow 0^-} \dfrac{f(1+h)-f(1) }{h} = \lim \limits_{h \rightarrow 0^-} \dfrac{Ce^{ \frac{1}{h} } }{h}=\lim \limits_{h \rightarrow 0^-} \dfrac{C\frac{1}{h}}{e^{-\frac{1}{h} } }=0
h → 0 − lim h f ( 1 + h ) − f ( 1 ) = h → 0 − lim h C e h 1 = h → 0 − lim e − h 1 C h 1 = 0
The limit 0 0 0 f ′ ( 1 ) = 0 f^{\prime}(1)=0 f ′ ( 1 ) = 0 k = 1 k=1 k = 1
Part 1-2. Assuming it holds for an arbitrary k k k k + 1 k+1 k + 1
Assume it holds for any given k k k a 0 ( k ) , a 1 ( k ) , ⋯ , a 2 k ( k ) a^{(k)}_{0},\ a^{(k)}_{1},\ \cdots,\ a^{(k)}_{2k} a 0 ( k ) , a 1 ( k ) , ⋯ , a 2 k ( k )
f ( k ) ( s ) = { e 1 s − 1 ( a 0 ( k ) + a 1 ( k ) s − 1 + ⋯ + a 2 k ( k ) ( s − 1 ) 2 k ) s < 1 0 s ≥ 1
f^{(k)}(s) = \begin{cases} e^{\frac{1}{s-1}} \left( a^{(k)}_{0} + \frac{a^{(k)}_{1}}{s-1}+\cdots+\frac{a^{(k)}_{2k}}{(s-1)^{2k}} \right) & s<1
\\ 0 & s \ge 1 \end{cases}
f ( k ) ( s ) = ⎩ ⎨ ⎧ e s − 1 1 ( a 0 ( k ) + s − 1 a 1 ( k ) + ⋯ + ( s − 1 ) 2 k a 2 k ( k ) ) 0 s < 1 s ≥ 1
When calculating f ( k + 1 ) ( s ) f^{(k+1)}(s) f ( k + 1 ) ( s ) ( − ∞ , 1 ) (-\infty, 1) ( − ∞ , 1 )
f ( k + 1 ) ( s ) = e 1 s − 1 − 1 ( s − 1 ) 2 ( a 0 ( k ) + a 1 ( k ) s − 1 + ⋯ + a 2 k ( k ) ( s − 1 ) 2 k ) + e 1 s − 1 ( − a 1 ( k ) ( s − 1 ) 2 + ⋯ + − 2 k a 2 k ( k ) ( s − 1 ) 2 k + 1 ) = e 1 s − 1 ( a 0 ( k + 1 ) + a 1 ( k + 1 ) s − 1 + ⋯ + a 2 k + 2 ( k + 1 ) ( s − 1 ) 2 k + 2 )
\begin{align*}
f^{(k+1)}(s) =&\ e^{\frac{1}{s-1}} \dfrac{-1}{(s-1)^2}\left( a^{(k)}_{0} + \frac{a^{(k)}_{1}}{s-1}+\cdots+\frac{a^{(k)}_{2k}}{(s-1)^{2k}} \right) + e^{\frac{1}{s-1}}\left( \frac{-a^{(k)}_{1}}{(s-1)^2}+\cdots+\frac{-2k a^{(k)}_{2k}}{(s-1)^{2k+1}} \right)
\\ =&\ e^{\frac{1}{s-1}} \left( a^{(k+1)}_{0} + \frac{a^{(k+1)}_{1}}{s-1}+\cdots+\frac{a^{(k+1)}_{2k+2}}{(s-1)^{2k+2}} \right)
\end{align*}
f ( k + 1 ) ( s ) = = e s − 1 1 ( s − 1 ) 2 − 1 ( a 0 ( k ) + s − 1 a 1 ( k ) + ⋯ + ( s − 1 ) 2 k a 2 k ( k ) ) + e s − 1 1 ( ( s − 1 ) 2 − a 1 ( k ) + ⋯ + ( s − 1 ) 2 k + 1 − 2 k a 2 k ( k ) ) e s − 1 1 ( a 0 ( k + 1 ) + s − 1 a 1 ( k + 1 ) + ⋯ + ( s − 1 ) 2 k + 2 a 2 k + 2 ( k + 1 ) )
Here, each a 0 ( k + 1 ) , ⋯ , a 2 k + 2 ( k + 1 ) a^{(k+1)}_{0},\ \cdots,\ a^{(k+1)}_{2k+2} a 0 ( k + 1 ) , ⋯ , a 2 k + 2 ( k + 1 ) a 0 ( k ) , ⋯ , a 2 k ( k ) a^{(k)}_{0},\ \cdots,\ a^{(k)}_{2k} a 0 ( k ) , ⋯ , a 2 k ( k ) ( 1 , ∞ ) (1, \infty) ( 1 , ∞ ) f ( k + 1 ) ( s ) = 0 f^{(k+1)}(s)=0 f ( k + 1 ) ( s ) = 0 f ( k + 1 ) = 0 f^{(k+1)}=0 f ( k + 1 ) = 0
lim h → 0 + f ( 1 + h ) − f ( 1 ) h = lim h → 0 + e 1 s − 1 ( a 0 ( k ) + a 1 ( k ) h + ⋯ + a 2 k ( k ) ( h ) 2 k ) h = lim h → 0 + ( a 0 ( k ) h + a 1 ( k ) h 2 + ⋯ + a 2 k ( k ) ( h 2 k + 1 ) 2 k ) e − 1 h = 0
\begin{align*}
\lim \limits_{h \rightarrow 0^+} \dfrac{ f(1+h)-f(1)}{h}=&\ \lim \limits_{h \rightarrow 0^+} \dfrac{e^{\frac{1}{s-1}} \left( a^{(k)}_{0} + \frac{a^{(k)}_{1}}{h}+\cdots+\frac{a^{(k)}_{2k}}{(h)^{2k}} \right) }{h}
\\ =&\ \lim \limits_{h \rightarrow 0^+} \dfrac{ \left( \frac{a^{(k)}_{0}}{h} + \frac{a^{(k)}_{1}}{h^2}+\cdots+\frac{a^{(k)}_{2k}}{(h^{2k+1})^{2k}} \right) }{e^{-\frac{1}{h}} } =0
\end{align*}
h → 0 + lim h f ( 1 + h ) − f ( 1 ) = = h → 0 + lim h e s − 1 1 ( a 0 ( k ) + h a 1 ( k ) + ⋯ + ( h ) 2 k a 2 k ( k ) ) h → 0 + lim e − h 1 ( h a 0 ( k ) + h 2 a 1 ( k ) + ⋯ + ( h 2 k + 1 ) 2 k a 2 k ( k ) ) = 0
lim h → 0 − f ( 1 + h ) − f ( 1 ) h = lim h → 0 − e 1 s − 1 ( a 0 ( k ) + a 1 ( k ) h + ⋯ + a 2 k ( k ) ( h ) 2 k ) h = lim h → 0 − ( a 0 ( k ) h + a 1 ( k ) h 2 + ⋯ + a 2 k ( k ) ( h 2 k + 1 ) 2 k ) e − 1 h = 0
\begin{align*}
\lim \limits_{h \rightarrow 0^-} \dfrac{ f(1+h)-f(1)}{h}=&\ \lim \limits_{h \rightarrow 0^-} \dfrac{e^{\frac{1}{s-1}} \left( a^{(k)}_{0} + \frac{a^{(k)}_{1}}{h}+\cdots+\frac{a^{(k)}_{2k}}{(h)^{2k}} \right) }{h}
\\ =&\ \lim \limits_{h \rightarrow 0^-} \dfrac{ \left( \frac{a^{(k)}_{0}}{h} + \frac{a^{(k)}_{1}}{h^2}+\cdots+\frac{a^{(k)}_{2k}}{(h^{2k+1})^{2k}} \right) }{e^{-\frac{1}{h}} }=0
\end{align*}
h → 0 − lim h f ( 1 + h ) − f ( 1 ) = = h → 0 − lim h e s − 1 1 ( a 0 ( k ) + h a 1 ( k ) + ⋯ + ( h ) 2 k a 2 k ( k ) ) h → 0 − lim e − h 1 ( h a 0 ( k ) + h 2 a 1 ( k ) + ⋯ + ( h 2 k + 1 ) 2 k a 2 k ( k ) ) = 0
Repeated application of L’Hôpital’s rule ultimately yields 0 0 0 f ∈ C ∞ ( R ) f \in C^\infty(\mathbb{R}) f ∈ C ∞ ( R )
Part 2. g ∈ C ∞ ( R n ) g \in C^\infty( \mathbb{R^n}) g ∈ C ∞ ( R n )
This is trivial by the definition of g g g
Therefore, by Part 1., Part 2. , the following holds.
η = f ∘ g ∈ C ∞ ( R n )
\eta = f \circ g \in C^\infty ( \mathbb{R^n} )
η = f ∘ g ∈ C ∞ ( R n )
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