1 statement
1.1 inverse fourier transform integral
1.2 fourier integral theorem
1.3 inverse transform in terms of flip operator
1.4 2 sided inverse
statement
in section assume
f
{\displaystyle f}
integrable continuous function. use convention fourier transform that
(
f
f
)
(
ξ
)
:=
∫
r
n
e
−
2
π
i
y
⋅
ξ
f
(
y
)
d
y
.
{\displaystyle ({\mathcal {f}}f)(\xi ):=\int _{\mathbb {r} ^{n}}e^{-2\pi iy\cdot \xi }\,f(y)\,dy.}
furthermore, assume fourier transform integrable.
inverse fourier transform integral
the common statement of fourier inversion theorem state inverse transform integral. integrable function
g
{\displaystyle g}
,
x
∈
r
n
{\displaystyle x\in \mathbb {r} ^{n}}
set
f
−
1
g
(
x
)
:=
∫
r
n
e
2
π
i
x
⋅
ξ
g
(
ξ
)
d
ξ
.
{\displaystyle {\mathcal {f}}^{-1}g(x):=\int _{\mathbb {r} ^{n}}e^{2\pi ix\cdot \xi }\,g(\xi )\,d\xi .}
then x∈ℝ have
f
−
1
(
f
g
)
(
x
)
=
g
(
x
)
.
{\displaystyle {\mathcal {f}}^{-1}({\mathcal {f}}g)(x)=g(x).}
fourier integral theorem
the theorem can restated as
f
(
x
)
=
∫
r
n
∫
r
n
e
2
π
i
(
x
−
y
)
⋅
ξ
f
(
y
)
d
y
d
ξ
.
{\displaystyle f(x)=\int _{\mathbb {r} ^{n}}\int _{\mathbb {r} ^{n}}e^{2\pi i(x-y)\cdot \xi }\,f(y)\,dy\,d\xi .}
if f real valued taking real part of each side of above obtain
f
(
x
)
=
∫
r
n
∫
r
n
cos
(
2
π
(
x
−
y
)
⋅
ξ
)
f
(
y
)
d
y
d
ξ
.
{\displaystyle f(x)=\int _{\mathbb {r} ^{n}}\int _{\mathbb {r} ^{n}}\cos(2\pi (x-y)\cdot \xi )\,f(y)\,dy\,d\xi .}
inverse transform in terms of flip operator
for function
g
{\displaystyle g}
define flip operator
r
{\displaystyle r}
by
r
g
(
x
)
:=
g
(
−
x
)
.
{\displaystyle rg(x):=g(-x).}
then may instead define
f
−
1
f
:=
r
f
f
=
f
r
f
.
{\displaystyle {\mathcal {f}}^{-1}f:=r{\mathcal {f}}f={\mathcal {f}}rf.}
it immediate definition of fourier transform , flip operator both
r
f
f
{\displaystyle r{\mathcal {f}}f}
,
f
r
f
{\displaystyle {\mathcal {f}}rf}
match integral definition of
f
−
1
f
{\displaystyle {\mathcal {f}}^{-1}f}
, , in particular equal each other , satisfy
f
−
1
(
f
f
)
(
x
)
=
f
(
x
)
{\displaystyle {\mathcal {f}}^{-1}({\mathcal {f}}f)(x)=f(x)}
.
note since
r
f
=
r
f
−
1
f
f
=
r
r
f
f
f
{\displaystyle rf=r{\mathcal {f}}^{-1}{\mathcal {f}}f=rr{\mathcal {ff}}f}
have
r
=
f
2
{\displaystyle r={\mathcal {f}}^{2}}
and
f
−
1
=
f
3
.
{\displaystyle {\mathcal {f}}^{-1}={\mathcal {f}}^{3}.}
two sided inverse
the form of fourier inversion theorem stated above, common, that
f
−
1
(
f
f
)
(
x
)
=
f
(
x
)
.
{\displaystyle {\mathcal {f}}^{-1}({\mathcal {f}}f)(x)=f(x).}
in other words,
f
−
1
{\displaystyle {\mathcal {f}}^{-1}}
left inverse fourier transform. right inverse fourier transform i.e.
f
(
f
−
1
f
)
(
ξ
)
=
f
(
ξ
)
.
{\displaystyle {\mathcal {f}}({\mathcal {f}}^{-1}f)(\xi )=f(\xi ).}
since
f
−
1
{\displaystyle {\mathcal {f}}^{-1}}
similar
f
{\displaystyle {\mathcal {f}}}
, follows fourier inversion theorem (changing variables
ζ
:=
−
ζ
{\displaystyle \zeta :=-\zeta }
):
f
=
f
−
1
(
f
f
)
(
x
)
=
∫
r
n
∫
r
n
e
2
π
i
x
⋅
ξ
e
−
2
π
i
y
⋅
ξ
f
(
y
)
d
y
d
ξ
=
∫
r
n
∫
r
n
e
−
2
π
i
x
⋅
ζ
e
2
π
i
y
⋅
ζ
f
(
y
)
d
y
d
ζ
=
f
(
f
−
1
f
)
(
x
)
.
{\displaystyle {\begin{aligned}f&={\mathcal {f}}^{-1}({\mathcal {f}}f)(x)\\&=\int _{\mathbb {r} ^{n}}\int _{\mathbb {r} ^{n}}e^{2\pi ix\cdot \xi }\,e^{-2\pi iy\cdot \xi }\,f(y)\,dy\,d\xi \\&=\int _{\mathbb {r} ^{n}}\int _{\mathbb {r} ^{n}}e^{-2\pi ix\cdot \zeta }\,e^{2\pi iy\cdot \zeta }\,f(y)\,dy\,d\zeta \\&={\mathcal {f}}({\mathcal {f}}^{-1}f)(x).\end{aligned}}}
alternatively, can seen relation between
f
−
1
f
{\displaystyle {\mathcal {f}}^{-1}f}
, flip operator , associativity of function composition, since
f
=
f
−
1
(
f
f
)
=
f
r
f
f
=
f
(
f
−
1
f
)
.
{\displaystyle f={\mathcal {f}}^{-1}({\mathcal {f}}f)={\mathcal {f}}r{\mathcal {f}}f={\mathcal {f}}({\mathcal {f}}^{-1}f).}
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