Note that by taking the DTFT of a sequence we get a function defined on [π,π][π,π].
In vector space notation we can view the DTFT as an operator (transformation). In this context it is useful to consider the normalized
DTFT
F
(
x
)
:
=
X
(
e
j
ω
)
=
1
2
π
∑
n
=

∞
∞
x
[
n
]
e

j
ω
n
.
F
(
x
)
:
=
X
(
e
j
ω
)
=
1
2
π
∑
n
=

∞
∞
x
[
n
]
e

j
ω
n
.
(1)One can show that the summation converges for any x∈ℓ2(π)x∈ℓ2(π), and
yields a function X(ejω)∈L2[π,π]X(ejω)∈L2[π,π]. Thus,
F
:
ℓ
2
(
Z
)
→
L
2
[

π
,
π
]
F
:
ℓ
2
(
Z
)
→
L
2
[

π
,
π
]
(2)can be viewed as a linear operator!
Note: It is not at all obvious that FF can be defined for all x∈ℓ2(Z)x∈ℓ2(Z).
To show this, one can first argue that if x∈ℓ1(Z)x∈ℓ1(Z), then
X
(
e
j
ω
)
≤
1
2
π
∑
n
=

∞
∞
x
[
n
]
e

j
ω
n
≤
1
2
π
∑
n
=

∞
∞
x
[
n
]
e

j
w
n
=
1
2
π
∑
n
=

∞
∞
x
[
n
]
<
∞
X
(
e
j
ω
)
≤
1
2
π
∑
n
=

∞
∞
x
[
n
]
e

j
ω
n
≤
1
2
π
∑
n
=

∞
∞
x
[
n
]
e

j
w
n
=
1
2
π
∑
n
=

∞
∞
x
[
n
]
<
∞
(3)
For an x∈ℓ2(Z)∖ℓ1(Z)x∈ℓ2(Z)∖ℓ1(Z), one must show that it is always possible to construct a sequence
xk∈ℓ2(Z)∩ℓ1(Z)xk∈ℓ2(Z)∩ℓ1(Z) such that
lim
k
→
∞
∥
x
k

x
∥
2
=
0
.
lim
k
→
∞
∥
x
k

x
∥
2
=
0
.
(4)This means {xk}{xk} is a Cauchy sequence, so that since ℓ2(Z)ℓ2(Z) is a Hilbert space, the limit exists (and is xx). In this case
X
(
e
j
ω
)
=
lim
k
→
∞
X
k
(
e
j
ω
)
.
X
(
e
j
ω
)
=
lim
k
→
∞
X
k
(
e
j
ω
)
.
(5)So for any x∈ℓ2(Z)x∈ℓ2(Z), we can define F(x)=X(ejω)F(x)=X(ejω), where X(ejω)∈L2[π,π]X(ejω)∈L2[π,π].
Can we always get the original xx back? Yes, the DTFT is invertible
F

1
(
X
)
=
1
2
π
∫

π
π
X
(
e
j
ω
)
·
e
j
ω
n
d
ω
F

1
(
X
)
=
1
2
π
∫

π
π
X
(
e
j
ω
)
·
e
j
ω
n
d
ω
(6)To verify that F1(F(x))=xF1(F(x))=x, observe that
1
2
π
∫

π
π
1
2
π
∑
k
=

∞
∞
x
[
k
]
e

j
ω
k
e
j
ω
n
d
ω
=
1
2
π
∑
k
=

∞
∞
x
[
k
]
∫

π
π
e

j
ω
(
k

n
)
d
ω
=
1
2
π
∑
k
=

∞
∞
x
[
k
]
·
2
π
δ
[
n

k
]
=
x
[
n
]
1
2
π
∫

π
π
1
2
π
∑
k
=

∞
∞
x
[
k
]
e

j
ω
k
e
j
ω
n
d
ω
=
1
2
π
∑
k
=

∞
∞
x
[
k
]
∫

π
π
e

j
ω
(
k

n
)
d
ω
=
1
2
π
∑
k
=

∞
∞
x
[
k
]
·
2
π
δ
[
n

k
]
=
x
[
n
]
(7)
One can also show that for any X∈L2[π,π]X∈L2[π,π], F(F1(X))=XF(F1(X))=X.
Operators that satisfy this property are called unitary operators or unitary
transformations. Unitary operators are nice! In fact, if A=X→YA=X→Y is a unitary
operator between two Hilbert spaces, then one can show that
〈
x
1
,
x
2
〉
=
〈
A
x
1
,
A
x
2
〉
∀
x
1
,
x
2
∈
X
,
〈
x
1
,
x
2
〉
=
〈
A
x
1
,
A
x
2
〉
∀
x
1
,
x
2
∈
X
,
(8)i.e., unitary operators obey Plancherel's and Parseval's theorems!