1.12 The vector space l squared

Special attention needs to be paid to the vector space $L^2(T_i, T_i)$ : the collection of functions $x(t)$ which are square-integrable over the interval $\left(T_i , T_f\right)$ : $$\int_{T_i}^{T_f} \left|x(t)\right|^2\,d t$$∞ An inner product can be defined for this space as:

There also exist orthonormal sets of functions that do not constitute a basis. For example, the set $\{{}_i(t)\}$ defined by:

While $L^2(0, T)$ is a separable space, examples can be given in which the representation of a vector in this space is not precisely equalto the vector. More precisely, let $x(t)\in L^2(0, T)$ and the set $\{{}_i(t)\}$ be defined by . The fact that $\{{}_i(t)\}$ constitutes a basis for the space implies: $$(x(t)-\sum_{i=1} )$$∞ x i i t 0 where $$x_i=\int_0^T x(t){}_i(t)\,d t$$ In particular, let $x(t)$ be: $$x(t)=\begin{cases}1 & \text{if $0\le t\le \frac{T}{2}$}\\ 0 & \text{if $\frac{T}{2}< t< T$}\end{cases}$$ Obviously, this function is an element of $L^2(0, T)$ . However, the representation of this function is not equal to 1at $t=\frac{T}{2}$ . In fact, the peak error never decreases as more terms are taken in the representation. In the special case ofthe Fourier series, the existence of this "error" is termed the Gibbs phenomenon . However, this "error" has zero norm in $L^2(0, T)$ ; consequently, the Fourier series expansion of thisfunction is equal to the function in the sense that the function and its expansion have zero distance between them. However, oneof the axioms of a valid inner product is that if $((e)=0)\implies (e=0)$ . The condition is satisfied, but the conclusion does not seem to be valid. Apparently, valid elements of $L^2(0, T)$ can be defined which are nonzero but have zero norm. An example is $$e=\begin{cases}1 & \text{if $t=\frac{T}{2}$}\\ 0 & \text{otherwise}\end{cases}$$ 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 $x$ and $y$ can differ from each other and be zero distance apart, the difference between them is"trivial." This difference has zero norm which, in $L^2$ , implies that the magnitude of ( $x-y$ ) integrates to zero. Consequently, the vectors are essentially equal. This notion of equality is usually writtenas $x=y\text{a.e.}$ ( $x$ equals $y$ almost everywhere ). With this convention, we have: $$((e)=0)\implies (e=0)\text{a.e.}$$ Consequently, the error between a vector and its representation is zero almosteverywhere.

Weakening the notion of equality in this fashion might seem to compromise the utility of the theory. However, if one suspectsthat 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 ashas been seen, this conclusion may not be valid). Usually, proving they are equal almost everywhere is much easier. Whilethis weaker notion of equality does not imply strict equality, one can be assured that any difference between them isinsignificant. The measure of "significance" for a vector space is expressed by the definition of the norm for the space.