Sia
Ω
{\displaystyle \Omega }
R
n
{\displaystyle {\mathbb {R}}^{n}}
u
∈
C
0
(
Ω
)
{\displaystyle u\in C^{0}(\Omega )}
u
(
x
)
=
∮
∂
B
(
x
,
r
)
u
(
y
)
d
σ
(
y
)
∀
B
(
x
,
r
)
⊂
Ω
{\displaystyle u(x)=\oint _{\partial B(x,r)}u(y)d\sigma (y)\quad \forall B(x,r)\subset \Omega }
Allora
u
∈
C
∞
(
Ω
)
{\displaystyle u\in C^{\infty }(\Omega )}
Consideriamo
ϵ
{\displaystyle \epsilon }
0
<
ϵ
<
d
i
s
t
(
x
,
∂
Ω
)
{\displaystyle 0<\epsilon <dist(x,\partial \Omega )}
u
{\displaystyle u}
η
{\displaystyle \eta }
u
∗
η
ϵ
=
∫
Ω
u
(
y
)
η
ϵ
(
x
−
y
)
d
y
=
u
ϵ
(
x
)
{\displaystyle u*\eta _{\epsilon }=\int _{\Omega }u(y)\eta _{\epsilon }(x-y)dy=u_{\epsilon }(x)}
Dimostriamo che:
u
ϵ
∈
C
∞
(
Ω
)
{\displaystyle u_{\epsilon }\in C^{\infty }(\Omega )}
u
ϵ
=
u
(
x
)
∀
x
∈
Ω
{\displaystyle u_{\epsilon }=u(x)\;\forall x\in \Omega }
Prendiamo
u
ϵ
=
∫
Ω
η
ϵ
(
x
−
y
)
u
(
y
)
d
y
{\displaystyle u_{\epsilon }=\int _{\Omega }\eta _{\epsilon }(x-y)u(y)dy}
x
{\displaystyle x}
x
{\displaystyle x}
η
{\displaystyle \eta }
∂
u
ϵ
(
x
)
∂
x
=
∫
Ω
∂
∂
x
η
ϵ
(
x
−
y
)
u
(
y
)
d
y
{\displaystyle {\frac {\partial {u_{\epsilon }(x)}}{\partial {x}}}=\int _{\Omega }{\frac {\partial {}}{\partial {x}}}\eta _{\epsilon }(x-y)u(y)dy}
Essendo
η
ϵ
{\displaystyle \eta _{\epsilon }}
C
∞
(
Ω
)
{\displaystyle C^{\infty }(\Omega )}
u
ϵ
{\displaystyle u_{\epsilon }}
Ora calcoliamo
u
ϵ
{\displaystyle u_{\epsilon }}
η
ϵ
{\displaystyle \eta _{\epsilon }}
D
(
η
ϵ
(
x
−
y
)
)
=
{
y
|
|
x
−
y
|
≤
ϵ
}
=
B
(
x
,
ϵ
)
{\displaystyle {\mathcal {D}}\left(\eta _{\epsilon }(x-y)\right)=\{y|\;|x-y|\leq \epsilon \}=B(x,\epsilon )}
u
ϵ
(
x
)
=
∫
B
(
x
,
ϵ
)
η
ϵ
(
x
−
y
)
u
(
y
)
d
y
=
1
ϵ
n
∫
B
(
x
,
ϵ
)
η
(
|
x
−
y
|
ϵ
)
u
(
y
)
d
y
=
1
ϵ
n
∫
0
ϵ
(
∫
∂
B
(
x
,
ϵ
)
η
(
|
x
−
y
|
ϵ
)
u
(
y
)
d
σ
(
y
)
)
d
r
=
1
ϵ
n
∫
0
ϵ
η
(
r
ϵ
)
(
∫
∂
B
(
x
,
ϵ
)
u
(
y
)
d
σ
(
y
)
)
d
r
=
1
ϵ
n
∫
0
ϵ
η
(
r
ϵ
)
ω
(
n
)
r
n
−
1
(
∮
∂
B
(
x
,
ϵ
)
u
(
y
)
d
σ
(
y
)
)
⏟
=
u
(
x
)
d
r
{\displaystyle {\begin{aligned}u_{\epsilon }(x)&=\int _{B(x,\epsilon )}\eta _{\epsilon }(x-y)u(y)dy\\&={\frac {1}{\epsilon ^{n}}}\int _{B(x,\epsilon )}\eta \left({\frac {|x-y|}{\epsilon }}\right)u(y)dy\\&={\frac {1}{\epsilon ^{n}}}\int _{0}^{\epsilon }\left(\int _{\partial B(x,\epsilon )}\eta \left({\frac {|x-y|}{\epsilon }}\right)u(y)d\sigma (y)\right)dr\\&={\frac {1}{\epsilon ^{n}}}\int _{0}^{\epsilon }\eta \left({\frac {r}{\epsilon }}\right)\left(\int _{\partial B(x,\epsilon )}u(y)d\sigma (y)\right)dr\\&={\frac {1}{\epsilon ^{n}}}\int _{0}^{\epsilon }\eta \left({\frac {r}{\epsilon }}\right)\omega (n)r^{n-1}{\underset {=u(x)}{\underbrace {\left(\oint _{\partial B(x,\epsilon )}u(y)d\sigma (y)\right)} }}dr\end{aligned}}}
Otteniamo quindi da un lato
u
(
x
)
{\displaystyle u(x)}
ϵ
{\displaystyle \epsilon }
=
u
(
x
)
1
ϵ
n
∫
0
ϵ
η
(
r
ϵ
)
∫
∂
B
(
x
,
ϵ
)
1
⋅
d
σ
=
u
(
x
)
[
1
ϵ
n
∫
B
(
x
,
ϵ
)
η
(
r
ϵ
)
d
r
]
⏟
=
1
→
u
(
x
)
{\displaystyle {\begin{aligned}&=u(x){\frac {1}{\epsilon ^{n}}}\int _{0}^{\epsilon }\eta \left({\frac {r}{\epsilon }}\right)\int _{\partial B(x,\epsilon )}1\cdot d\sigma \\&=u(x){\underset {=1}{\underbrace {\left[{\frac {1}{\epsilon ^{n}}}\int _{B(x,\epsilon )}\eta \left({\frac {r}{\epsilon }}\right)dr\right]} }}\rightarrow u(x)\end{aligned}}}
Nell'ultimo integrale abbiamo utilizzato la normalizzazione del mollificatore.
Quindi abbiamo dimostrato che
u
ϵ
=
u
(
x
)
{\displaystyle u_{\epsilon }=u(x)}
u
∈
C
∞
{\displaystyle u\in C^{\infty }}
Osservazione: se consideriamo
u
∈
C
2
{\displaystyle u\in C^{2}}
u
{\displaystyle u}
![{\displaystyle {\begin{aligned}&=u(x){\frac {1}{\epsilon ^{n}}}\int _{0}^{\epsilon }\eta \left({\frac {r}{\epsilon }}\right)\int _{\partial B(x,\epsilon )}1\cdot d\sigma \\&=u(x){\underset {=1}{\underbrace {\left[{\frac {1}{\epsilon ^{n}}}\int _{B(x,\epsilon )}\eta \left({\frac {r}{\epsilon }}\right)dr\right]} }}\rightarrow u(x)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/deaea11dda834557041f59f1f96617693a7d55e6)
