My Fourier analysis materials define the energy of a function to be its $L^2$ norm over the real numbers. So far the reader has been lead to believe that if something does not belong to the $L^2$ space, then its energy is infinite. The reference gives a quick argument on why the Dirac's delta distribution does not belong to any $L^p$ space: Elements of $L^p$ spaces are functions in the classical sense and Dirac's delta is not a function. Hence $\delta \not\in L^p(\mathbb{R})$.
Later on the reference says casually that: *The signal $f \equiv \delta(b)$ (Dirac's delta at $b \in \mathbb{R}$) has an infinite energy.*At this point norm has not been defined for Schwartz's tempered distributions.
(Question:) Therefore, do you think that it is reasonable to assume that the reason for the comment that $\delta(b)$ has an infinite energy is that $\delta \not\in L^2(\mathbb{R})$, or is there a compelling argument why $\delta$ has an infinite norm in some sense?
Best Answer
There is an unfortunate issue in that to calculate the $L^2$ norm of $\delta$ one would have to somehow integrate its square, but there is no (obvious) way to make sense of $\delta^2$. We can instead try to approximate $\delta$ by a sequence of, say, smooth functions, and then we'll find that no matter what sequence we choose the $L^2$ norm of the sequence diverges.
But here is maybe a cleaner and more conceptual argument: the Fourier transform of $\delta$ is the constant function $1$, and the Fourier transform ought to be unitary. So the $L^2$ norm of $\delta$ ought to be the $L^2$ norm of the constant function $1$, which clearly diverges.