L
1
(
R
)
{\displaystyle L^{1}(\mathbb {R} )\!}
modifica
Sia
u
∈
L
1
(
R
)
:
x
→
u
(
x
)
{\displaystyle u\in L^{1}(\mathbb {R} ):x\to u(x)\!}
trasformata di Fourier della funzione
u
{\displaystyle u\!}
u
^
(
ξ
)
:=
∫
R
e
−
i
ξ
x
u
(
x
)
d
x
∀
ξ
∈
R
{\displaystyle {\hat {u}}(\xi ):=\int _{\mathbb {R} }e^{-i\xi x}u(x)\,dx\qquad \forall \xi \in \mathbb {R} \!}
indichiamo l'operazione con la lettera F calligrafica, perciò:
F
:
u
→
u
^
{\displaystyle {\mathcal {F}}:u\to {\hat {u}}\!}
Si può estendere questa definizione anche per funzioni
u
(
x
)
∈
L
1
(
R
n
)
{\displaystyle u(\mathbf {x} )\in L^{1}(\mathbb {R} ^{n})\!}
u
^
(
ξ
)
:=
∫
R
n
e
−
i
ξ
⋅
x
u
(
x
)
d
x
∀
ξ
∈
R
n
{\displaystyle {\hat {u}}({\boldsymbol {\xi }}):=\int _{\mathbb {R} ^{n}}e^{-i{\boldsymbol {\xi }}\cdot \mathbf {x} }u(\mathbf {x} )\,d\mathbf {x} \qquad \forall {\boldsymbol {\xi }}\in \mathbb {R} ^{n}\!}
Dove
ξ
⋅
x
{\displaystyle {\boldsymbol {\xi }}\cdot \mathbf {x} \!}
Sia
u
(
x
)
=
χ
[
−
1
,
+
1
]
(
x
)
{\displaystyle u(x)=\chi _{[-1,+1]}(x)\!}
u
^
(
ξ
)
=
∫
R
e
−
i
ξ
x
χ
[
−
1
,
+
1
]
(
x
)
d
x
=
∫
−
1
+
1
e
−
i
ξ
x
d
x
=
[
e
−
i
ξ
x
−
i
ξ
]
−
1
+
1
=
e
i
ξ
−
e
−
i
ξ
i
ξ
=
2
sin
ξ
ξ
{\displaystyle {\hat {u}}(\xi )=\int _{\mathbb {R} }e^{-i\xi x}\chi _{[-1,+1]}(x)\,dx=\int _{-1}^{+1}e^{-i\xi x}\,dx={\left[{\frac {e^{-i\xi x}}{-i\xi }}\right]}_{-1}^{+1}={\frac {e^{i\xi }-e^{-i\xi }}{i\xi }}=2{\frac {\sin \xi }{\xi }}\!}
Sia
u
(
x
)
=
1
1
+
x
2
{\displaystyle u(x)={\frac {1}{1+x^{2}}}\!}
u
^
(
ξ
)
=
∫
R
e
−
i
ξ
x
u
(
x
)
d
x
=
v
.
p
.
∫
R
e
−
i
ξ
x
u
(
x
)
d
x
=
lim
R
→
+
∞
∫
−
R
+
R
e
−
i
ξ
x
u
(
x
)
d
x
{\displaystyle {\hat {u}}(\xi )=\int _{\mathbb {R} }e^{-i\xi x}u(x)\,dx=v.p.\int _{\mathbb {R} }e^{-i\xi x}u(x)\,dx=\lim _{R\to +\infty }\int _{-R}^{+R}e^{-i\xi x}u(x)\,dx}
Ora applicando il principio del prolungamento analitico e il lemma di Jordan otteniamo:
lim
R
→
+
∞
∫
−
R
+
R
e
−
i
ξ
x
u
(
x
)
d
x
=
{
π
e
ξ
ξ
<
0
π
e
−
ξ
ξ
>
0
{\displaystyle \lim _{R\to +\infty }\int _{-R}^{+R}e^{-i\xi x}u(x)\,dx={\begin{cases}\pi e^{\xi }&\xi <0\\\pi e^{-\xi }&\xi >0\end{cases}}}
Mettendo insieme le due cose otteniamo:
u
^
(
ξ
)
=
∫
R
e
−
i
ξ
x
1
+
x
2
d
x
=
π
e
−
|
ξ
|
{\displaystyle {\hat {u}}(\xi )=\int _{\mathbb {R} }{\frac {e^{-i\xi x}}{1+x^{2}}}\,dx=\pi e^{-|\xi |}\!}
Elenchiamo le principali proprietà della trasformata di Fourier , utili per lo studio a priori e il calcolo di esse:
La trasformata di Fourier è un operatore lineare :
F
{
α
u
+
β
v
}
(
ξ
)
=
α
F
{
u
}
(
ξ
)
+
β
F
{
v
}
(
ξ
)
∀
α
,
β
∈
R
,
∀
u
,
v
∈
L
1
(
R
)
{\displaystyle {\mathcal {F}}\left\{\alpha u+\beta v\right\}(\xi )=\alpha {\mathcal {F}}\{u\}(\xi )+\beta {\mathcal {F}}\{v\}(\xi )\qquad \forall \alpha ,\beta \in \mathbb {R} ,\,\forall u,v\in L^{1}(\mathbb {R} )\!}
Presa una funzione
u
∈
L
1
(
R
)
{\displaystyle u\in L^{1}(\mathbb {R} )\!}
u
^
(
ξ
)
=
F
{
u
}
(
ξ
)
{\displaystyle {\hat {u}}(\xi )={\mathcal {F}}\{u\}(\xi )\!}
Se
u
∈
L
1
(
R
)
{\displaystyle u\in L^{1}(\mathbb {R} )\!}
pari
⟹
u
^
(
ξ
)
{\displaystyle \implies {\hat {u}}(\xi )\!}
pari .
Se
u
∈
L
1
(
R
)
{\displaystyle u\in L^{1}(\mathbb {R} )\!}
dispari
⟹
u
^
(
ξ
)
{\displaystyle \implies {\hat {u}}(\xi )\!}
dispari .
Se
u
∈
L
1
(
R
)
{\displaystyle u\in L^{1}(\mathbb {R} )\!}
reale pari
⟹
u
^
(
ξ
)
{\displaystyle \implies {\hat {u}}(\xi )\!}
reale pari .
Se
u
∈
L
1
(
R
)
{\displaystyle u\in L^{1}(\mathbb {R} )\!}
reale dispari
⟹
u
^
(
ξ
)
{\displaystyle \implies {\hat {u}}(\xi )\!}
immaginaria pura dispari .
Scaling :
F
{
u
(
a
x
)
}
(
ξ
)
=
1
|
a
|
F
{
u
(
x
)
}
(
ξ
a
)
{\displaystyle {\mathcal {F}}\{u(ax)\}(\xi )={\frac {1}{|a|}}{\mathcal {F}}\{u(x)\}\left({\frac {\xi }{a}}\right)\!}
Shifting :
F
{
u
(
x
−
a
)
}
(
ξ
)
=
e
i
a
ξ
F
{
u
(
x
)
}
(
ξ
)
{\displaystyle {\mathcal {F}}\{u(x-a)\}(\xi )=e^{ia\xi }{\mathcal {F}}\{u(x)\}(\xi )\!}
Modulazione complessa :
F
{
e
i
a
x
u
(
x
)
}
(
ξ
)
=
F
{
u
(
x
)
}
(
ξ
−
a
)
{\displaystyle {\mathcal {F}}\{e^{iax}u(x)\}(\xi )={\mathcal {F}}\{u(x)\}(\xi -a)\!}
Modulazione reale :
F
{
cos
(
ξ
0
x
)
u
(
x
)
}
(
ξ
)
=
F
{
u
(
x
)
}
(
ξ
−
ξ
0
)
+
F
{
u
(
x
)
}
(
ξ
+
ξ
0
)
2
{\displaystyle {\mathcal {F}}\{\cos(\xi _{0}x)u(x)\}(\xi )={\frac {{\mathcal {F}}\{u(x)\}(\xi -\xi _{0})+{\mathcal {F}}\{u(x)\}(\xi +\xi _{0})}{2}}\!}
Prendiamo la funzione
u
(
x
)
=
χ
[
−
1
,
+
1
]
(
x
)
{\displaystyle u(x)=\chi _{[-1,+1]}(x)\!}
u
^
(
ξ
)
=
2
sin
ξ
ξ
{\displaystyle {\hat {u}}(\xi )=2{\frac {\sin {\xi }}{\xi }}\!}
Siccome
u
(
x
)
{\displaystyle u(x)\!}
reale pari , anche
u
^
(
ξ
)
{\displaystyle {\hat {u}}(\xi )\!}
Calcoliamo
F
{
u
(
x
n
)
}
(
ξ
)
{\displaystyle {\mathcal {F}}\left\{u\left({\frac {x}{n}}\right)\right\}(\xi )\!}
scaling otteniamo:
F
{
u
(
x
n
)
}
(
ξ
)
=
|
n
|
F
{
u
(
x
)
}
(
n
ξ
)
=
2
|
n
|
sin
n
ξ
n
ξ
{\displaystyle {\mathcal {F}}\left\{u\left({\frac {x}{n}}\right)\right\}(\xi )=|n|{\mathcal {F}}\{u(x)\}(n\xi )=2|n|{\frac {\sin {n\xi }}{n\xi }}\!}
Calcoliamo
F
{
|
n
|
2
u
(
n
x
)
}
(
ξ
)
{\displaystyle {\mathcal {F}}\left\{{\frac {|n|}{2}}u\left(nx\right)\right\}(\xi )\!}
linearità della trasformata di Fourier e il teorema per lo scaling :
F
{
|
n
|
2
u
(
n
x
)
}
(
ξ
)
=
|
n
|
2
1
|
n
|
F
{
u
(
x
)
}
(
ξ
n
)
=
sin
ξ
n
ξ
n
{\displaystyle {\mathcal {F}}\left\{{\frac {|n|}{2}}u\left(nx\right)\right\}(\xi )={\frac {|n|}{2}}{\frac {1}{|n|}}{\mathcal {F}}\{u(x)\}\left({\frac {\xi }{n}}\right)={\frac {\sin {\frac {\xi }{n}}}{\frac {\xi }{n}}}\!}
Calcoliamo
F
{
cos
(
ξ
0
x
)
u
(
x
)
}
(
ξ
)
{\displaystyle {\mathcal {F}}\left\{\cos {(\xi _{0}x)}u(x)\right\}(\xi )\!}
modulazione reale :
F
{
cos
(
ξ
0
x
)
u
(
x
)
}
(
ξ
)
=
sin
(
ξ
−
ξ
0
)
ξ
−
ξ
0
+
sin
(
ξ
+
ξ
0
)
ξ
+
ξ
0
{\displaystyle {\mathcal {F}}\left\{\cos {(\xi _{0}x)}u(x)\right\}(\xi )={\frac {\sin {(\xi -\xi _{0})}}{\xi -\xi _{0}}}+{\frac {\sin {(\xi +\xi _{0})}}{\xi +\xi _{0}}}\!}
Sia
u
∈
L
1
(
R
n
)
{\displaystyle u\in L^{1}(\mathbb {R} ^{n})\!}
u
^
=
F
{
u
}
{\displaystyle {\hat {u}}={\mathcal {F}}\{u\}}
u
^
∈
C
0
(
R
n
)
∩
L
∞
(
R
n
)
{\displaystyle {\hat {u}}\in C^{0}(\mathbb {R} ^{n})\cap L^{\infty }(\mathbb {R} ^{n})\!}
‖
u
^
‖
L
∞
(
R
n
)
≤
‖
u
‖
L
1
(
R
n
)
{\displaystyle {\left\|{\hat {u}}\right\|}_{L^{\infty }(\mathbb {R} ^{n})}\leq {\left\|u\right\|}_{L^{1}(\mathbb {R} ^{n})}\!}
lim
‖
ξ
‖
→
∞
u
^
(
ξ
)
=
0
{\displaystyle \lim _{\|{\boldsymbol {\xi }}\|\to \infty }{\hat {u}}({\boldsymbol {\xi }})=0\!}
Siccome la variabile
ξ
{\displaystyle {\boldsymbol {\xi }}\!}
trasformata di Fourier , appartiene a
R
n
{\displaystyle \mathbb {R} ^{n}\!}
{
ξ
n
}
{\displaystyle \{{\boldsymbol {\xi }}_{n}\}\!}
ξ
∈
R
n
{\displaystyle {\boldsymbol {\xi }}\in \mathbb {R} ^{n}\!}
{
ξ
n
}
∈
R
n
:
ξ
n
→
ξ
{\displaystyle \{{\boldsymbol {\xi }}_{n}\}\in \mathbb {R} ^{n}:{\boldsymbol {\xi }}_{n}\to {\boldsymbol {\xi }}\!}
Allora anche la successione
e
−
i
ξ
n
⋅
x
u
(
x
)
{\displaystyle e^{-i{\boldsymbol {\xi }}_{n}\cdot \mathbf {x} }u(\mathbf {x} )\!}
e
−
i
ξ
n
⋅
x
u
(
x
)
→
e
−
i
ξ
⋅
x
u
(
x
)
{\displaystyle e^{-i{\boldsymbol {\xi }}_{n}\cdot \mathbf {x} }u(\mathbf {x} )\to e^{-i{\boldsymbol {\xi }}\cdot \mathbf {x} }u(\mathbf {x} )\!}
Ora la nostra idea, per dimostrare la continuità dalla funzione trasformata, si basa sul teorema di Lebesgue, dovremo quindi dimostrare la convergenza dell'integrale:
u
^
(
ξ
n
)
=
∫
R
n
e
−
i
ξ
n
⋅
x
u
(
x
)
d
x
→
∫
R
n
e
−
i
ξ
⋅
x
u
(
x
)
d
x
=
u
^
(
ξ
)
{\displaystyle {\hat {u}}({\boldsymbol {\xi }}_{n})=\int _{\mathbb {R} ^{n}}e^{-i{\boldsymbol {\xi }}_{n}\cdot \mathbf {x} }u(\mathbf {x} )\,d\mathbf {x} \to \int _{\mathbb {R} ^{n}}e^{-i{\boldsymbol {\xi }}\cdot \mathbf {x} }u(\mathbf {x} )\,d\mathbf {x} ={\hat {u}}({\boldsymbol {\xi }})\!}
Per fare ciò dovremo trovare una funzione
M
(
x
)
{\displaystyle M(x)\!}
|
e
−
i
ξ
n
⋅
x
u
(
x
)
|
≤
M
(
x
)
{\displaystyle |e^{-i{\boldsymbol {\xi }}_{n}\cdot \mathbf {x} }u(\mathbf {x} )|\leq M(x)\!}
R
n
{\displaystyle \mathbb {R} ^{n}\!}
|
e
−
i
ξ
n
⋅
x
|
=
1
{\displaystyle |e^{-i{\boldsymbol {\xi }}_{n}\cdot \mathbf {x} }|=1\!}
u
(
x
)
∈
L
1
(
R
n
)
{\displaystyle u(\mathbf {x} )\in L^{1}(\mathbb {R} ^{n})\!}
u
^
(
ξ
)
∈
C
0
(
R
n
)
{\displaystyle {\hat {u}}({\boldsymbol {\xi }})\in C^{0}(\mathbb {R} ^{n})\!}
La dimostrazione di
u
^
(
ξ
)
∈
L
∞
(
R
n
)
{\displaystyle {\hat {u}}({\boldsymbol {\xi }})\in L^{\infty }(\mathbb {R} ^{n})\!}
La dimostrazione del seguente punto avviene attraverso una banale minorazione.
|
u
^
(
ξ
)
|
=
|
∫
R
n
e
−
i
ξ
⋅
x
u
(
x
)
d
x
|
≤
∫
R
n
|
e
−
i
ξ
⋅
x
|
⋅
|
u
(
x
)
|
d
x
=
∫
R
n
|
u
(
x
)
|
d
x
=
‖
u
‖
L
1
(
R
n
)
{\displaystyle |{\hat {u}}({\boldsymbol {\xi }})|=\left|\int _{\mathbb {R} ^{n}}e^{-i{\boldsymbol {\xi }}\cdot \mathbf {x} }u(\mathbf {x} )\,d\mathbf {x} \right|\leq \int _{\mathbb {R} ^{n}}\left|e^{-i{\boldsymbol {\xi }}\cdot \mathbf {x} }\right|\cdot \left|u(\mathbf {x} )\right|\,d\mathbf {x} =\int _{\mathbb {R} ^{n}}\left|u(\mathbf {x} )\right|\,d\mathbf {x} ={\|u\|}_{L^{1}(\mathbb {R} ^{n})}\!}
Quindi:
‖
u
^
‖
L
∞
(
R
n
)
≤
‖
u
‖
L
1
(
R
n
)
{\displaystyle {\|{\hat {u}}\|}_{L^{\infty }(\mathbb {R} ^{n})}\leq {\|u\|}_{L^{1}(\mathbb {R} ^{n})}\!}
e non è migliorabile se
u
≥
0
{\displaystyle u\geq 0\!}
|
u
^
(
0
)
|
=
|
∫
R
n
e
0
u
(
x
)
d
x
|
=
∫
R
n
|
u
(
x
)
|
d
x
=
‖
u
(
x
)
‖
L
1
(
R
n
)
{\displaystyle |{\hat {u}}(0)|=\left|\int _{\mathbb {R} ^{n}}e^{0}u(\mathbf {x} )\,d\mathbf {x} \right|=\int _{\mathbb {R} ^{n}}|u(\mathbf {x} )|\,d\mathbf {x} =\|u(x)\|_{L^{1}(\mathbb {R} ^{n})}\!}
Consideriamo la funzione
χ
[
a
;
b
]
(
x
)
{\displaystyle \chi _{[a;b]}(x)\!}
X
^
(
ξ
)
=
∫
R
e
−
i
ξ
x
χ
[
a
;
b
]
(
x
)
d
x
=
∫
a
b
e
−
i
ξ
x
d
x
=
[
e
−
i
ξ
x
−
i
ξ
]
a
b
=
e
−
i
ξ
a
−
e
−
i
ξ
b
i
ξ
{\displaystyle {\hat {\mathrm {X} }}(\xi )=\int _{\mathbb {R} }e^{-i{\boldsymbol {\xi }}x}\chi _{[a;b]}(x)\,dx=\int _{a}^{b}e^{-i\xi x}\,dx={\left[{\frac {e^{-i\xi x}}{-i\xi }}\right]}_{a}^{b}={\frac {e^{-i\xi a}-e^{-i\xi b}}{i\xi }}\!}
Notare che
ξ
=
0
{\displaystyle \xi =0\!}
X
^
(
ξ
)
→
0
,
|
ξ
|
→
∞
{\displaystyle {\hat {\mathrm {X} }}(\xi )\to 0,\,|\xi |\to \infty \!}
s
n
=
∑
k
=
0
n
c
k
χ
I
k
{\displaystyle s_{n}=\sum _{k=0}^{n}c_{k}\chi _{I_{k}}\!}
Siccome la trasformata della funzione
χ
[
a
;
b
]
{\displaystyle \chi _{[a;b]}\!}
u
∈
L
1
(
R
n
)
{\displaystyle u\in L^{1}(\mathbb {R} ^{n})\!}
∃
ϵ
>
0
{\displaystyle \exists \epsilon >0\!}
∃
u
n
{\displaystyle \exists u_{n}\!}
u
n
=
∑
k
=
0
n
c
k
χ
I
k
{\displaystyle u_{n}=\sum _{k=0}^{n}c_{k}\chi _{I_{k}}\!}
tale che
u
n
→
u
{\displaystyle u_{n}\to u\!}
‖
u
−
u
n
‖
L
1
(
R
n
)
<
ϵ
{\displaystyle \|u-u_{n}\|_{L^{1}(\mathbb {R} ^{n})}<\epsilon \!}
Inoltre
∃
R
>
0
{\displaystyle \exists R>0\!}
|
u
^
n
(
ξ
)
|
<
ϵ
{\displaystyle |{\hat {u}}_{n}(\xi )|<\epsilon \!}
‖
ξ
‖
>
R
{\displaystyle \|\xi \|>R\!}
u
^
(
ξ
)
=
u
^
(
ξ
)
−
u
^
n
(
ξ
)
+
u
^
n
(
ξ
)
≤
|
u
^
(
ξ
)
−
u
^
n
(
ξ
)
|
+
|
u
^
n
(
ξ
)
|
{\displaystyle {\hat {u}}(\xi )={\hat {u}}(\xi )-{\hat {u}}_{n}(\xi )+{\hat {u}}_{n}(\xi )\leq |{\hat {u}}(\xi )-{\hat {u}}_{n}(\xi )|+|{\hat {u}}_{n}(\xi )|\!}
Per
‖
ξ
‖
>
R
{\displaystyle \|\xi \|>R\!}
|
u
^
(
ξ
)
−
u
^
n
(
ξ
)
|
+
|
u
^
n
(
ξ
)
|
≤
‖
u
^
(
ξ
)
−
u
^
n
(
ξ
)
‖
L
(
R
n
)
+
ϵ
≤
2
ϵ
{\displaystyle |{\hat {u}}(\xi )-{\hat {u}}_{n}(\xi )|+|{\hat {u}}_{n}(\xi )|\leq \|{\hat {u}}(\xi )-{\hat {u}}_{n}(\xi )\|_{L^{(}\mathbb {R} ^{n})}+\epsilon \leq 2\epsilon \!}
Riassumendo, per
‖
ξ
‖
>
R
{\displaystyle \|\xi \|>R\!}
|
u
^
(
ξ
)
|
<
2
ϵ
{\displaystyle |{\hat {u}}(\xi )|<2\epsilon \!}
Siano
u
(
x
)
,
x
⋅
u
(
x
)
∈
L
1
(
R
)
{\displaystyle u(x),\,x\cdot u(x)\in L^{1}(\mathbb {R} )\!}
F
{
u
}
∈
C
1
(
R
)
{\displaystyle {\mathcal {F}}\{u\}\in C^{1}(\mathbb {R} )\!}
F
{
x
⋅
u
}
=
i
⋅
d
d
ξ
F
{
u
}
(
ξ
)
{\displaystyle {\mathcal {F}}\{x\cdot u\}=i\cdot {\frac {d}{d\xi }}{\mathcal {F}}\{u\}(\xi )\!}
u
^
(
ξ
+
h
)
−
u
^
(
ξ
)
h
=
∫
R
e
−
i
ξ
x
(
e
−
i
h
x
−
1
)
h
u
(
x
)
d
x
→
−
i
∫
R
e
−
i
ξ
x
x
u
(
x
)
d
x
=
−
i
F
{
x
⋅
u
(
x
)
}
{\displaystyle {\frac {{\hat {u}}(\xi +h)-{\hat {u}}(\xi )}{h}}=\int _{\mathbb {R} }{\frac {e^{-i\xi x}\left(e^{-ihx}-1\right)}{h}}u(x)\,dx\to -i\int _{\mathbb {R} }e^{-i\xi x}xu(x)\,dx\!=-i{\mathcal {F}}\{x\cdot u(x)\}}
Più in generale per funzioni in
R
n
{\displaystyle \mathbb {R} ^{n}\!}
u
(
x
)
,
‖
x
‖
u
(
x
)
∈
L
1
(
R
n
)
⟹
u
^
(
ξ
)
∈
C
1
{\displaystyle u(\mathbf {x} ),\left\|\mathbf {x} \right\|u(\mathbf {x} )\in L^{1}(\mathbb {R} ^{n})\implies {\hat {u}}({\boldsymbol {\xi }})\in C^{1}}
F
{
x
j
u
(
x
)
}
=
i
∂
u
^
∂
ξ
j
{\displaystyle {\mathcal {F}}\{x_{j}u(\mathbf {x} )\}=i{\frac {\partial {\hat {u}}}{\partial \xi _{j}}}}
Siano
u
(
x
)
,
u
′
(
x
)
∈
L
1
(
R
)
{\displaystyle u(x),\,u'(x)\in L^{1}(\mathbb {R} )\!}
F
{
u
′
}
=
i
ξ
F
{
u
}
{\displaystyle {\mathcal {F}}\{u'\}=i\xi {\mathcal {F}}\{u\}\!}
Più in generale se
u
(
x
)
,
∂
u
(
x
)
∂
x
j
∈
L
1
(
R
n
)
{\displaystyle u(\mathbf {x} ),\,{\frac {\partial \,u(\mathbf {x} )}{\partial x_{j}}}\in L^{1}(\mathbb {R} ^{n})\!}
F
{
∂
u
(
x
)
∂
x
j
}
=
i
ξ
j
F
{
u
}
{\displaystyle {\mathcal {F}}\left\{{\frac {\partial \,u(\mathbf {x} )}{\partial x_{j}}}\right\}=i\xi _{j}{\mathcal {F}}\{u\}\!}
Nota bene : con la trasformata di Fourier l'operatore differenziale diventa una moltiplicazione per una potenza di
i
ξ
{\displaystyle i\xi \!}
u
,
u
′
,
.
.
.
,
u
(
k
)
∈
L
1
(
R
n
)
{\displaystyle u,u',...,u^{(k)}\in L^{1}(\mathbb {R} ^{n})\!}
F
{
∂
k
u
(
x
)
∂
x
j
k
}
=
(
i
ξ
j
)
k
F
{
u
}
{\displaystyle {\mathcal {F}}\left\{{\frac {\partial ^{k}\,u(\mathbf {x} )}{\partial x_{j}^{k}}}\right\}=(i\xi _{j})^{k}{\mathcal {F}}\{u\}\!}
Per la dimostrazione di questo teorema citiamo prima un lemma.
u
,
u
′
∈
L
1
⟹
∃
lim
x
→
∞
u
(
x
)
=
0
{\displaystyle u,u'\in L^{1}\implies \exists \lim _{x\to \infty }u(x)=0\!}
Calcoliamo semplicemente la trasformata della derivata della funzione
u
{\displaystyle u\!}
u
′
^
(
ξ
)
=
∫
R
e
−
i
ξ
x
u
′
(
x
)
d
x
=
lim
R
→
+
∞
∫
−
R
+
R
e
−
i
ξ
x
u
′
(
x
)
d
x
{\displaystyle {\hat {u'}}(\xi )=\int _{\mathbb {R} }e^{-i\xi x}u'(x)\,dx=\lim _{R\to +\infty }\int _{-R}^{+R}e^{-i\xi x}u'(x)\,dx\!}
Integriamo per parti:
lim
R
→
+
∞
∫
−
R
+
R
e
−
i
ξ
x
u
′
(
x
)
d
x
=
lim
R
→
+
∞
∫
−
R
+
R
i
ξ
e
−
i
ξ
x
u
(
x
)
d
x
+
[
e
−
i
ξ
x
u
(
x
)
]
−
∞
+
∞
{\displaystyle \lim _{R\to +\infty }\int _{-R}^{+R}e^{-i\xi x}u'(x)\,dx=\lim _{R\to +\infty }\int _{-R}^{+R}i\xi e^{-i\xi x}u(x)\,dx+{\left[e^{-i\xi x}u(x)\right]}_{-\infty }^{+\infty }\!}
Per il lemma:
[
e
−
i
ξ
x
u
(
x
)
]
−
∞
+
∞
=
0
{\displaystyle {\left[e^{-i\xi x}u(x)\right]}_{-\infty }^{+\infty }=0\!}
Concludendo:
lim
R
→
+
∞
∫
−
R
+
R
i
ξ
e
−
i
ξ
x
u
(
x
)
d
x
=
i
ξ
u
^
(
ξ
)
{\displaystyle \lim _{R\to +\infty }\int _{-R}^{+R}i\xi e^{-i\xi x}u(x)\,dx=i\xi {\hat {u}}(\xi )\!}
u
(
x
)
{\displaystyle u(x)\!}
u
^
(
ξ
)
{\displaystyle {\hat {u}}(\xi )\!}
1
1
+
x
2
{\displaystyle {\frac {1}{1+x^{2}}}\!}
π
e
−
|
ξ
|
{\displaystyle \pi e^{-|\xi |}\!}
e
−
|
x
|
{\displaystyle e^{-|x|}\!}
2
1
+
ξ
2
{\displaystyle {\frac {2}{1+\xi ^{2}}}\!}
χ
[
−
1
;
1
]
{\displaystyle \chi _{[-1;1]}\!}
2
sin
ξ
ξ
{\displaystyle {\frac {2\sin {\xi }}{\xi }}\!}
sin
x
π
x
{\displaystyle {\frac {\sin {x}}{\pi x}}\!}
χ
[
−
1
;
1
]
{\displaystyle \chi _{[-1;1]}\!}
(
1
−
|
x
|
)
χ
[
−
1
;
1
]
{\displaystyle (1-|x|)\chi _{[-1;1]}\!}
4
sin
(
ξ
2
2
)
ξ
2
{\displaystyle {\frac {4\sin {\left({\frac {\xi ^{2}}{2}}\right)}}{\xi ^{2}}}\!}
Funzioni assolutamente continue
modifica
Di fondamentale importanza per lo studio delle trasformate di Fourier è la conoscenza delle funzioni gaussiane e le loro principali proprietà. Cominciamo prendendo la più semplice
u
(
x
)
=
e
−
x
2
{\displaystyle u(x)=e^{-x^{2}}\!}
La prima cosa che si deve osservare è
u
(
x
)
∈
L
1
(
R
)
{\displaystyle u(x)\in L^{1}(\mathbb {R} )\!}
R
{\displaystyle \mathbb {R} \!}
v
(
x
)
=
e
−
(
x
2
+
y
2
)
=
e
−
x
2
e
−
y
2
{\displaystyle v(x)=e^{-(x^{2}+y^{2})}=e^{-x^{2}}e^{-y^{2}}\!}
e ne calcoliamo l'integrale su
R
2
{\displaystyle \mathbb {R} ^{2}\!}
∫
R
2
v
(
x
)
d
x
d
y
=
∫
R
2
e
−
(
x
2
+
y
2
)
d
x
d
y
=
{\displaystyle \int _{\mathbb {R} ^{2}}v(x)\,dx\,dy=\int _{\mathbb {R} ^{2}}e^{-(x^{2}+y^{2})}\,dx\,dy=\!}
=
∫
[
0
;
2
π
]
×
[
0
;
+
∞
]
e
−
ρ
2
ρ
d
ρ
d
θ
=
∫
0
2
π
d
θ
∫
0
+
∞
e
−
ρ
2
ρ
d
ρ
=
π
{\displaystyle =\int _{[0;2\pi ]\times [0;+\infty ]}e^{-\rho ^{2}}\rho \,d\rho \,d\theta =\int _{0}^{2\pi }\,d\theta \int _{0}^{+\infty }e^{-\rho ^{2}}\rho \,d\rho \ =\pi \!}
Ora, ritornando alla funzione iniziale:
∫
R
2
v
(
x
)
d
x
d
y
=
∫
R
2
e
−
(
x
2
+
y
2
)
d
x
d
y
=
∫
R
e
−
x
2
d
x
∫
R
e
−
y
2
d
y
=
(
∫
R
e
−
x
2
d
x
)
2
{\displaystyle \int _{\mathbb {R} ^{2}}v(x)\,dx\,dy=\int _{\mathbb {R} ^{2}}e^{-(x^{2}+y^{2})}\,dx\,dy=\int _{\mathbb {R} }e^{-x^{2}}\,dx\int _{\mathbb {R} }e^{-y^{2}}\,dy={\left(\int _{\mathbb {R} }e^{-x^{2}}\,dx\right)}^{2}\!}
Concludendo:
∫
R
e
−
x
2
d
x
=
π
{\displaystyle \int _{\mathbb {R} }e^{-x^{2}}\,dx={\sqrt {\pi }}\!}
Di facile dimostrazione (applicando un cambio di variabile) anche:
∫
R
e
−
(
x
+
i
b
)
2
d
x
=
π
b
∈
R
{\displaystyle \int _{\mathbb {R} }e^{-(x+i\,b)^{2}}\,dx={\sqrt {\pi }}\qquad b\in \mathbb {R} \!}
Ora calcoleremo
u
^
{\displaystyle {\hat {u}}\!}
u
^
(
ξ
)
=
∫
R
e
−
i
ξ
x
e
−
x
2
d
x
=
∫
R
e
−
(
i
ξ
x
+
x
2
)
d
x
=
{\displaystyle {\hat {u}}(\xi )=\int _{\mathbb {R} }e^{-i\xi \,x}e^{-x^{2}}\,dx=\int _{\mathbb {R} }e^{-(i\xi \,x+x^{2})}\,dx=\!}
=
∫
R
e
−
(
x
2
+
i
ξ
x
±
ξ
2
4
)
d
x
=
∫
R
e
−
(
x
+
i
ξ
2
)
2
e
−
ξ
2
4
d
x
=
π
e
−
ξ
2
4
{\displaystyle =\int _{\mathbb {R} }e^{-\left(x^{2}+i\xi \,x\pm {\frac {\xi ^{2}}{4}}\right)}\,dx=\int _{\mathbb {R} }e^{-\left(x+i{\frac {\xi }{2}}\right)^{2}}e^{-{\frac {\xi ^{2}}{4}}}\,dx={\sqrt {\pi }}e^{-{\frac {\xi ^{2}}{4}}}\!}
Più in generale:
e
−
a
x
2
→
π
a
e
−
ξ
2
4
a
{\displaystyle e^{-ax^{2}}\to {\sqrt {\frac {\pi }{a}}}e^{-{\frac {\xi ^{2}}{4a}}}\!}
e
−
a
‖
x
‖
2
→
(
π
a
)
n
2
e
−
‖
ξ
‖
2
4
a
x
,
ξ
∈
R
n
{\displaystyle e^{-a\|\mathbf {x} \|^{2}}\to {\left({\frac {\pi }{a}}\right)}^{\frac {n}{2}}e^{-{\frac {\|{\boldsymbol {\xi }}\|^{2}}{4a}}}\qquad \mathbf {x} ,\,{\boldsymbol {\xi }}\in \mathbb {R} ^{n}\!}
La dimostrazione di quest'ultima avviene in modo analogo alla precedente:
∫
R
n
e
−
i
ξ
⋅
x
e
−
‖
x
‖
2
d
x
=
∫
R
n
e
−
∑
j
=
0
n
(
i
ξ
j
x
j
+
x
j
2
)
d
x
=
{\displaystyle \int _{\mathbb {R} ^{n}}e^{-i{\boldsymbol {\xi }}\cdot \mathbf {x} }e^{-\|\mathbf {x} \|^{2}}\,d\mathbf {x} =\int _{\mathbb {R} ^{n}}e^{-\sum _{j=0}^{n}\left(i\xi _{j}x_{j}+x_{j}^{2}\right)}\,d\mathbf {x} =\!}
=
∫
R
n
∏
j
=
0
n
e
−
i
ξ
j
x
j
+
x
j
2
d
x
=
∏
j
=
0
n
(
∫
R
e
−
i
ξ
j
x
j
+
x
j
2
d
x
j
)
=
{\displaystyle =\int _{\mathbb {R} ^{n}}\prod _{j=0}^{n}e^{-i\xi _{j}x_{j}+x_{j}^{2}}\,d\mathbf {x} =\prod _{j=0}^{n}\left(\int _{\mathbb {R} }e^{-i\xi _{j}x_{j}+x_{j}^{2}}\,dx_{j}\right)=\!}
=
∏
j
=
0
n
π
e
−
ξ
j
2
4
=
π
n
e
−
‖
ξ
‖
2
4
{\displaystyle =\prod _{j=0}^{n}{\sqrt {\pi }}e^{\frac {-\xi _{j}^{2}}{4}}={\sqrt {\pi }}^{n}e^{-{\frac {\|{\boldsymbol {\xi }}\|^{2}}{4}}}\!}
Siano
f
,
g
∈
L
1
(
R
n
)
{\displaystyle f,\,g\in L^{1}(\mathbb {R} ^{n})\!}
convoluzione come:
(
f
∗
g
)
(
x
)
:=
∫
R
n
f
(
y
)
g
(
x
−
y
)
d
x
{\displaystyle (f*g)(\mathbf {x} ):=\int _{\mathbb {R} ^{n}}f(\mathbf {y} )g(\mathbf {x} -\mathbf {y} )\,d\mathbf {x} \!}
L'operazione di convoluzione è lineare (dalla proprietà degli integrali) ed è commutativa (per sostituzione):
(
(
α
f
+
β
g
)
∗
h
)
=
α
(
f
∗
h
)
+
β
(
g
∗
h
)
∀
f
,
g
,
h
∈
L
1
(
R
n
)
,
∀
α
,
β
∈
R
{\displaystyle ((\alpha f+\beta g)*h)=\alpha (f*h)+\beta (g*h)\qquad \forall f,\,g,\,h\in L^{1}(\mathbb {R} ^{n}),\,\forall \alpha ,\,\beta \in \mathbb {R} \!}
f
∗
g
=
g
∗
f
∀
f
,
g
∈
L
1
(
R
n
)
{\displaystyle f*g=g*f\qquad \forall f,\,g\in L^{1}(\mathbb {R} ^{n})\!}
Ed è anche associativa (per Fubini - Tonelli):
(
f
∗
g
)
∗
h
=
f
∗
(
g
∗
h
)
∀
f
,
g
,
h
∈
L
1
(
R
n
)
{\displaystyle (f*g)*h=f*(g*h)\qquad \forall f,\,g,\,h\in L^{1}(\mathbb {R} ^{n})\!}
Siano
f
,
g
∈
L
1
{\displaystyle f,\,g\in L^{1}\!}
f
∗
g
∈
L
1
{\displaystyle f*g\in L^{1}\!}
∫
R
n
|
(
f
∗
g
)
(
x
)
|
d
x
=
∫
R
n
|
∫
R
n
f
(
y
)
g
(
x
−
y
)
d
y
|
d
x
≤
∫
R
n
∫
R
n
|
f
(
y
)
|
|
g
(
x
−
y
)
|
d
y
d
x
=
{\displaystyle \int _{\mathbb {R} ^{n}}\left|(f*g)(\mathbf {x} )\right|\,d\mathbf {x} =\int _{\mathbb {R} ^{n}}\left|\int _{\mathbb {R} ^{n}}f(\mathbf {y} )g(\mathbf {x} -\mathbf {y} )\,d\mathbf {y} \right|\,d\mathbf {x} \leq \int _{\mathbb {R} ^{n}}\int _{\mathbb {R} ^{n}}\left|f(\mathbf {y} )\right|\left|g(\mathbf {x} -\mathbf {y} )\right|\,d\mathbf {y} \,d\mathbf {x} =\!}
per il teorema di Fubini - Tonelli:
=
∫
R
n
∫
R
n
|
f
(
y
)
|
|
g
(
x
−
y
)
|
d
x
d
y
=
∫
R
n
∫
R
n
|
f
(
y
)
|
(
∫
R
n
|
g
(
x
−
y
)
|
d
x
)
d
y
=
‖
f
‖
L
1
‖
g
‖
L
1
{\displaystyle =\int _{\mathbb {R} ^{n}}\int _{\mathbb {R} ^{n}}\left|f(\mathbf {y} )\right|\left|g(\mathbf {x} -\mathbf {y} )\right|\,d\mathbf {x} \,d\mathbf {y} =\int _{\mathbb {R} ^{n}}\int _{\mathbb {R} ^{n}}\left|f(\mathbf {y} )\right|\left(\int _{\mathbb {R} ^{n}}\left|g(\mathbf {x} -\mathbf {y} )\right|\,d\mathbf {x} \right)\,d\mathbf {y} =\|f\|_{L^{1}}\|g\|_{L^{1}}\!}
Quindi:
‖
f
∗
g
‖
L
1
≤
‖
f
‖
L
1
‖
g
‖
L
1
<
+
∞
{\displaystyle \|f*g\|_{L^{1}}\leq \|f\|_{L^{1}}\|g\|_{L^{1}}<+\infty \!}
Siano
f
,
g
∈
L
1
{\displaystyle f,\,g\in L^{1}\!}
F
{
f
∗
g
}
=
F
{
f
}
F
{
g
}
{\displaystyle {\mathcal {F}}\{f*g\}={\mathcal {F}}\{f\}{\mathcal {F}}\{g\}}
![{\displaystyle u(x)=\chi _{[-1,+1]}(x)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cda04aa331d5e7115d8fbe5c470c133d19d047a6)
![{\displaystyle {\hat {u}}(\xi )=\int _{\mathbb {R} }e^{-i\xi x}\chi _{[-1,+1]}(x)\,dx=\int _{-1}^{+1}e^{-i\xi x}\,dx={\left[{\frac {e^{-i\xi x}}{-i\xi }}\right]}_{-1}^{+1}={\frac {e^{i\xi }-e^{-i\xi }}{i\xi }}=2{\frac {\sin \xi }{\xi }}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1296a7dc0e5fadc707842aadaccb9f6c9ecf263f)
![{\displaystyle \chi _{[a;b]}(x)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/842b8832ac9eda0e52fd6720116b379a5d4fb8c0)
![{\displaystyle {\hat {\mathrm {X} }}(\xi )=\int _{\mathbb {R} }e^{-i{\boldsymbol {\xi }}x}\chi _{[a;b]}(x)\,dx=\int _{a}^{b}e^{-i\xi x}\,dx={\left[{\frac {e^{-i\xi x}}{-i\xi }}\right]}_{a}^{b}={\frac {e^{-i\xi a}-e^{-i\xi b}}{i\xi }}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/be39334f936d72e291360bfe98675b15b0c73192)
![{\displaystyle \chi _{[a;b]}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c6cfc23e3e7de262f9c2b7e27290c565581e4341)
![{\displaystyle \lim _{R\to +\infty }\int _{-R}^{+R}e^{-i\xi x}u'(x)\,dx=\lim _{R\to +\infty }\int _{-R}^{+R}i\xi e^{-i\xi x}u(x)\,dx+{\left[e^{-i\xi x}u(x)\right]}_{-\infty }^{+\infty }\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/42c0fb5af5c5024203a77d6b7e7ab25952a1a214)
![{\displaystyle {\left[e^{-i\xi x}u(x)\right]}_{-\infty }^{+\infty }=0\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e0fa541d5b148f57455cbba1b86c1538dd1da23f)
![{\displaystyle \chi _{[-1;1]}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5f4784aa81b0a5ecda8d3641ee32d85ebf72a078)
![{\displaystyle (1-|x|)\chi _{[-1;1]}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/378b1b84baa07bdf43cf1a12c98fab5129f92e36)
![{\displaystyle G(x)=(1-|x|)\chi _{[-1,+1]}(x)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6caa97ca8c963f446b1df205d340de34c4042d91)
![{\displaystyle g(x)=-{\mbox{sign}}(x)\chi _{[-1,+1]}(x)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e3c0901af824f2001229770f6385a480eb15624f)
![{\displaystyle G(x)=\chi _{[0,1]}(x)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/603e69973c6ea3d553fe0854beb22d8998bfc1ca)
![{\displaystyle =\int _{[0;2\pi ]\times [0;+\infty ]}e^{-\rho ^{2}}\rho \,d\rho \,d\theta =\int _{0}^{2\pi }\,d\theta \int _{0}^{+\infty }e^{-\rho ^{2}}\rho \,d\rho \ =\pi \!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/81f29787f484ada29ec26d24372fe6585ace108b)
