Conditions on the function Fourier inversion theorem




1 conditions on function

1.1 schwartz functions
1.2 integrable functions integrable fourier transform
1.3 integrable functions in 1 dimension
1.4 square integrable functions
1.5 tempered distributions





conditions on function

when used in physics , engineering, fourier inversion theorem used under assumption behaves nicely . in mathematics such heuristic arguments not permitted, , fourier inversion theorem includes explicit specification of class of functions being allowed. however, there no best class of functions consider several variants of fourier inversion theorem exist, albeit compatible conclusions.


schwartz functions

the fourier inversion theorem holds schwartz functions (roughly speaking, smooth functions decay , derivatives decay quickly). condition has benefit elementary direct statement function (as opposed imposing condition on fourier transform), , integral defines fourier transform , inverse absolutely integrable. version of theorem used in proof of fourier inversion theorem tempered distributions (see below).


integrable functions integrable fourier transform

the fourier inversion theorem holds continuous functions absolutely integrable (i.e.




l

1


(


r


n


)


{\displaystyle l^{1}(\mathbb {r} ^{n})}

) absolutely integrable fourier transform. includes schwartz functions, strictly stronger form of theorem previous 1 mentioned. these conditions have benefit integrals define fourier transform , inverse absolutely integrable. condition 1 used above in statement section.


a slight variant drop condition function



f


{\displaystyle f}

continuous still require , fourier transform absolutely integrable.



f
=
g


{\displaystyle f=g}

everywhere g continuous function, ,






f




1


(


f


f
)
(
x
)
=
g
(
x
)


{\displaystyle {\mathcal {f}}^{-1}({\mathcal {f}}f)(x)=g(x)}

every



x



r


n




{\displaystyle x\in \mathbb {r} ^{n}}

.


integrable functions in 1 dimension

piecewise smooth; 1 dimension

if function absolutely integrable in 1 dimension (i.e.



f


l

1


(

r

)


{\displaystyle f\in l^{1}(\mathbb {r} )}

) , piecewise smooth version of fourier inversion theorem holds. in case define










f




1


g
(
x
)
:=

lim

r








r


r



e

2
π
i
x
ξ



g
(
ξ
)

d
ξ
.


{\displaystyle {\mathcal {f}}^{-1}g(x):=\lim _{r\to \infty }\int _{-r}^{r}e^{2\pi ix\xi }\,g(\xi )\,d\xi .}



then



x


r



{\displaystyle x\in \mathbb {r} }










f




1


(


f


f
)
(
x
)
=


1
2


(
f
(

x




)
+
f
(

x

+


)
)
,


{\displaystyle {\mathcal {f}}^{-1}({\mathcal {f}}f)(x)={\frac {1}{2}}(f(x_{-})+f(x_{+})),}



i.e.






f




1


(


f


f
)
(
x
)


{\displaystyle {\mathcal {f}}^{-1}({\mathcal {f}}f)(x)}

equals average of left , right limits of



f


{\displaystyle f}

@



x


{\displaystyle x}

. note @ points



f


{\displaystyle f}

continuous equals



f
(
x
)


{\displaystyle f(x)}

.


a higher-dimensional analogue of form of theorem holds, according folland (1992) rather delicate , not terribly useful .



piecewise continuous; 1 dimension

if function absolutely integrable in 1 dimension (i.e.



f


l

1


(

r

)


{\displaystyle f\in l^{1}(\mathbb {r} )}

) merely piecewise continuous version of fourier inversion theorem still holds. in case integral in inverse fourier transform defined aid of smooth rather sharp cut off function; define










f




1


g
(
x
)
:=

lim

r








r



φ
(
ξ

/

r
)


e

2
π
i
x
ξ



g
(
ξ
)

d
ξ
,

φ
(
ξ
)
:=

e



ξ

2




.


{\displaystyle {\mathcal {f}}^{-1}g(x):=\lim _{r\to \infty }\int _{\mathbb {r} }\varphi (\xi /r)\,e^{2\pi ix\xi }\,g(\xi )\,d\xi ,\qquad \varphi (\xi ):=e^{-\xi ^{2}}.}



the conclusion of theorem same piecewise smooth case discussed above.



continuous; number of dimensions

if



f


{\displaystyle f}

continuous , absolutely integrable on





r


n




{\displaystyle \mathbb {r} ^{n}}

fourier inversion theorem still holds long again define inverse transform smooth cut off function i.e.










f




1


g
(
x
)
:=

lim

r









r


n




φ
(
ξ

/

r
)


e

2
π
i
x

ξ



g
(
ξ
)

d
ξ
,

φ
(
ξ
)
:=

e


|
ξ

|

2




.


{\displaystyle {\mathcal {f}}^{-1}g(x):=\lim _{r\to \infty }\int _{\mathbb {r} ^{n}}\varphi (\xi /r)\,e^{2\pi ix\cdot \xi }\,g(\xi )\,d\xi ,\qquad \varphi (\xi ):=e^{-\vert \xi \vert ^{2}}.}



the conclusion



x



r


n




{\displaystyle x\in \mathbb {r} ^{n}}










f




1


(


f


f
)
(
x
)
=
f
(
x
)
.


{\displaystyle {\mathcal {f}}^{-1}({\mathcal {f}}f)(x)=f(x).}




no regularity condition; number of dimensions

if drop assumptions (piecewise) continuity of



f


{\displaystyle f}

, assume merely absolutely integrable, version of theorem still holds. inverse transform again define smooth cut off, conclusion that










f




1


(


f


f
)
(
x
)
=
f
(
x
)


{\displaystyle {\mathcal {f}}^{-1}({\mathcal {f}}f)(x)=f(x)}



for every



x



r


n




{\displaystyle x\in \mathbb {r} ^{n}}


square integrable functions

in case fourier transform cannot defined directly integral since may not absolutely convergent, instead defined density argument (see fourier transform article). example, putting








g

k


(
ξ
)
:=



{
y



r


n


:

|
y
|


k
}



e


2
π
i
y

ξ



f
(
y
)

d
y
,

k


n

,


{\displaystyle g_{k}(\xi ):=\int _{\{y\in \mathbb {r} ^{n}:\left\vert y\right\vert \leq k\}}e^{-2\pi iy\cdot \xi }\,f(y)\,dy,\qquad k\in \mathbb {n} ,}



we can set






f


f
:=

lim

k





g

k





{\displaystyle \textstyle {\mathcal {f}}f:=\lim _{k\to \infty }g_{k}}

limit taken in




l

2




{\displaystyle l^{2}}

-norm. inverse transform may defined density in same way or defining terms of fourier transform , flip operator. have







f
(
x
)
=


f


(



f




1


f
)
(
x
)
=



f




1


(


f


f
)
(
x
)


{\displaystyle f(x)={\mathcal {f}}({\mathcal {f}}^{-1}f)(x)={\mathcal {f}}^{-1}({\mathcal {f}}f)(x)}



for every x∈ℝ.


tempered distributions

the fourier transform may defined on space of tempered distributions






s




(


r


n


)


{\displaystyle {\mathcal {s}} (\mathbb {r} ^{n})}

duality of fourier transform on space of schwartz functions.



f




s




(


r


n


)


{\displaystyle f\in {\mathcal {s}} (\mathbb {r} ^{n})}

, test functions



φ



s


(


r


n


)


{\displaystyle \varphi \in {\mathcal {s}}(\mathbb {r} ^{n})}

set










f


f
,
φ

:=

f
,


f


φ

,


{\displaystyle \langle {\mathcal {f}}f,\varphi \rangle :=\langle f,{\mathcal {f}}\varphi \rangle ,}



where





f


φ


{\displaystyle {\mathcal {f}}\varphi }

defined using integral formula. if



f


l

1


(


r


n


)


l

2


(


r


n


)


{\displaystyle f\in l^{1}(\mathbb {r} ^{n})\cap l^{2}(\mathbb {r} ^{n})}

agrees usual definition. may define inverse transform






f




1


:



s




(


r


n


)




s




(


r


n


)


{\displaystyle {\mathcal {f}}^{-1}\colon {\mathcal {s}} (\mathbb {r} ^{n})\to {\mathcal {s}} (\mathbb {r} ^{n})}

, either duality inverse transform on schwartz functions in same way, or defining in terms of flip operator (where flip operator defined duality). have









f





f




1


=



f




1




f


=

id




s




(


r


n


)


.


{\displaystyle {\mathcal {f}}{\mathcal {f}}^{-1}={\mathcal {f}}^{-1}{\mathcal {f}}=\operatorname {id} _{{\mathcal {s}} (\mathbb {r} ^{n})}.}








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