I need to prove that Dirac's delta does not belong in $L^2(\mathbb{R})$.
First, I found the next definition of Dirac's delta
$$\delta :D(\mathbb R)\to \mathbb R$$
is defined by:
$$\langle \delta,\varphi \rangle=\int_{-\infty}^{+\infty}\varphi(x)\delta(x)\,\mathrm{d}x = \varphi(0),$$
and
$$\delta(x)= \begin{cases} 1,& x= 0\\ 0 ,& x\ne 0. \end{cases} \\$$
The space $L^2(\mathbb{R})=\{f:f \text{ is measurable and } \|f\|_{2}<+\infty \}$.
I'm thinking suppose otherwise, i.e, Dirac's delta in $L^2$, but I have problems to prove that Dirac's delta is measurable, but I suspect that in calculating of $\|f\|_2$ I'll find the contradiction.
Could you give me any suggestions??
Best Answer
If it were $L^2$ then it would satisfy Cauchy-Schwarz, i.e. you would have $|f(0)| \leq C \| f \|_{L^2}$ for some $C$. Construct a sequence of functions $f_n$ such that $|f_n(0)|>n \| f_n \|_{L^2}$ to contradict this.