- Dirac Delta Function in L²[a, b] and Dirac Delta Functional in Dual Space
1. $L^2[a, b]$ as a Pure Vector Space
The set of all complex-valued functions that are square integrable on the real interval $[a, b]$ forms a complex vector space.
The meaning of a square integrable function $f$: $$ \int_a^b |f(x)|^2 dx < \infty $$
2. $L^2[a, b]$ as a Complete Inner Product Space, i.e., a Hilbert Space
Definition of the inner product: $$ (f | g) = \frac{1}{b-a} \int_a^b f^*(x) g(x) dx $$
The meaning of square integrable from the inner product perspective: $$ \int_a^b |f(x)|^2 dx = (f | f) = |f|^2 < \infty $$
That is, the squared norm of the function $f$ as a vector is less than infinity, meaning it has a finite value of a certain magnitude.
3. Fourier Series as Vectors in $L^2[a, b]$
Selection of orthonormal basis: $$ e_n(x) = e^{i \frac{2n \pi}{b-a} x} $$
Representation of square integrable function $f(x)$: $$ f(x) = \sum_{n=-\infty}^\infty c_n e^{i \frac{2n \pi}{b-a} x} $$
Square integrable $f$: $$ \sum_{n=-\infty}^\infty |c_n|^2 < \infty $$
Definition of inner product from the basis perspective: $$ f = \sum_{n=-\infty}^\infty c_n e_n, \quad g = \sum_{n=-\infty}^\infty d_n e_n $$
$$ \implies (f | g) = \sum_{n=-\infty}^\infty c_n^* d_n $$
4. Finding $c_n$ Using the Basis of the Dual Space of $L^2[a, b]$
$$ c_n = e^n(f) = (e_n | f) = \frac{1}{b-a} \int_a^b e^{-i \frac{2n \pi x}{b-a}} f(x) dx $$
5. Positive Real-valued Dirac Delta Function $\delta$ in $L^2[a, b]$ Inner Product Space and Dirac Delta Functional $\tilde{\delta}$ in the Dual Space
$$ f(0) = \tilde{\delta}(f) = (\delta | f) $$
$$ f(x) = \sum_{n=-\infty}^\infty c_n e_n, \quad e_n(0) = 1 $$
$$ \implies f(0) = \sum_{n=-\infty}^\infty c_n $$
Therefore, $$ (\delta | f) = \sum_{n=-\infty}^\infty c_n \implies \delta = \sum_{n=-\infty}^\infty e_n $$
However, $$ |\delta|^2 \to \infty $$
diverges, so it is not square integrable. Calculating this differently:
$$ |\delta|^2 = (\delta | \delta) = \tilde{\delta}(\delta) = \delta(0) $$
And since this value diverges to infinity: $$ \text{The value of } \delta(0) \text{ diverges to infinity.} $$
The meaning of this is that while finite-dimensional vector spaces and their dual spaces have a one-to-one correspondence, infinite-dimensional vector spaces and their dual spaces do not necessarily have such a correspondence.
As in this case, there are instances where the functional $\tilde{\delta}$ exists in the dual space, but the corresponding function $\delta$ does not exist in the vector space.