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# The Vector Space L Squared

Module by: Don Johnson. E-mail the author

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, x2 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(i1)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  iTt<(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 etut 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: xti=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  0tT20  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 ( xy 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.

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