Special attention needs to be paid to the vector space
L
2
T
i
T
i
L
2
T
i
T
i
: the collection of functions
xt
x
t
which are square-integrable over the interval
T
i
T
f
T
i
T
f
:
∫
T
i
T
f
|xt|2dt<∞
t
T
i
T
f
x
t
2
An inner product can be defined for this space as:
〈x,y〉=∫
T
i
T
f
xtytdt
x
y
t
T
i
T
f
x
t
y
t
(1)
Consistent with this definition, the length of the vector
xt
x
t
is given by
∥x∥=∫
T
i
T
f
|xt|2dt
x
t
T
i
T
f
x
t
2
Physically,
∥x∥2
x
2
can be related to the energy contained in the signal over
T
i
T
f
T
i
T
f
. This space is a Hilbert space. If
T i
T i and
T
f T
f are both finite, an
orthonormal basis is easily found which spans it. For
simplicity of notation, let
T
i
=0
T
i
0
and
T
f
=T
T
f
T
. The set of functions defined by:
φ
2
i
-
1
t=2T1/2cos2π(i−1)tT
φ
2
i
-
1
t
2
T
1/2
2
i
1
t
T
(2)
φ
2
i
t=2T1/2sin2πitT
φ
2
i
t
2
T
1/2
2
i
t
T
is complete over the interval
0
T
0
T
and therefore constitutes a basis for
L
2
0T
L
2
0
T
. By demonstrating a basis, we conclude that
L
2
0T
L
2
0
T
is a separable vector space. The representations of
functions with respect to this basis corresponds to the
well-known Fourier series expansion of a function. As most
functions require an infinite number of terms in their Fourier
series representation, this space is infinite dimensional.
There also exist orthonormal sets of functions that do
not constitute a basis. For example, the
set
φ
i
t
φ
i
t
defined by:
∀i,i=01…:
φ
i
t={1T if iT≤t<(i+1)T0 otherwise
i
i
0
1
…
φ
i
t
1
T
i
T
t
i
1
T
0
(3)
over
L
2
0∞
L
2
0
. The members of this set are normal (unit
norm) and are mutually orthogonal (no member overlaps with any
other). Consequently, this set is an orthonormal set. However,
it does not constitute a basis for
L
2
0∞
L
2
0
.
Functions piecewise constant over intervals of length
TT are the only members of
L
2
0∞
L
2
0
which can be represented by this set. Other
functions such as
e−tut
t
u
t
cannot be represented by the
φ
i
t
φ
i
t
defined above. Consequently,
orthonormality
of a set of functions does not guarantee
completeness.
While
L
2
0T
L
2
0
T
is a separable space, examples can be given in which the
representation of a vector in this space is not precisely equal
to the vector. More precisely, let
xt∈
L
2
0T
x
t
L
2
0
T
and the set
φ
i
t
φ
i
t
be defined by Equation 2. The fact that
φ
i
t
φ
i
t
constitutes a basis for the space implies:
∥xt−∑i=1∞
x
i
φ
i
t∥=0
x
t
i
1
x
i
φ
i
t
0
where
x
i
=∫0Txt
φ
i
tdt
x
i
t
0
T
x
t
φ
i
t
In particular, let
xt
x
t
be:
xt={1 if 0≤t≤T20 if T2<t<T
x
t
1
0
t
T
2
0
T
2
t
T
Obviously, this function is an element of
L
2
0T
L
2
0
T
.
However, the representation of this function is not equal to 1
at
t=T2
t
T
2
. In fact, the peak error never decreases as more
terms are taken in the representation. In the special case of
the Fourier series, the existence of this "error" is termed the
Gibbs phenomenon. However, this "error" has zero
norm in
L
2
0T
L
2
0
T
; consequently, the Fourier series expansion of this
function is equal to the function in the sense that the function
and its expansion have zero distance between them. However, one
of the axioms of a valid inner product is that if
(∥e∥=0)⇒(e=0)
e
0
e
0
. The condition is satisfied, but the conclusion does
not seem to be valid. Apparently, valid elements of
L
2
0T
L
2
0
T
can be defined which are nonzero but have zero
norm. An example is
e={1 if t=T20 otherwise
e
1
t
T
2
0
So as not to destroy the theory, the most common method of
resolving the conflict is to weaken the definition of equality.
The essence of the problem is that while two vectors
xx and yy can differ from each other and
be zero distance apart, the difference between them is
"trivial." This difference has zero norm which, in
L
2
L
2
, implies that the magnitude of (
x−y
x
y
) integrates to zero. Consequently, the vectors are
essentially equal. This notion of equality is usually written
as
x=y
a.e.
x
y
a.e. (xx equals
yy almost
everywhere). With this convention, we have:
(∥e∥=0)⇒(e=0)
a.e.
e
0
e
0
a.e. Consequently, the error
between a vector and its representation is zero almost
everywhere.
Weakening the notion of equality in this fashion might seem to
compromise the utility of the theory. However, if one suspects
that two vectors in an inner product are equal
(e.g., a vector and its representation), it
is quite difficult to prove that they are strictly equal (and as
has been seen, this conclusion may not be valid). Usually,
proving they are equal almost everywhere is much easier. While
this weaker notion of equality does not imply strict equality,
one can be assured that any difference between them is
insignificant. The measure of "significance" for a vector space
is expressed by the definition of the norm for the space.
