The Dirac delta does not belong in L2

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??

Solution 1:

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.

Solution 2:

I present a sketch show I made from suggestions @Ian:

Suppose that Dirac's delta belongs in $L^2(\mathbb{R})$, i.e., that has $\int\delta(x)f(x)dx\le||\delta||_{L^2}||f||_{L^2}$, for all $f\in L^2(\mathbb{R})$.

By definition Dirac's delta, $f(0)=\int\delta(x)f(x)dx\le||f||_{L^2}$

i.e, $f(0)\le||f||_{L^2}$, for all $f\in L^2(\mathbb{R})$.

Then we will give a sequence of continuous functions in $L^2(\mathbb{R})$ such that the inequality present some inconvenience.

Let $f_n(x)=\sqrt n e^{-nx^2}$, where $f_n$ are continuous and belong to $L^2(\mathbb{R})$.

Next, $f_n(0)=\sqrt n>0$, and $||f_n||_{L^2}=(\int_{\mathbb{R}}ne^{-2nx^2})^{1/2}\rightarrow 0$, when $n\rightarrow +\infty$, this part is by dominated convergence theorem.

On the one hand we see that $f_n(0)\rightarrow +\infty$ and by other hand $||f_n||_{L^2}\rightarrow 0$, when $n\rightarrow +\infty$. That is a contradiction.

Therefore must be that Dirac's delta doesn't belong to $L^2(\mathbb{R})$.

Any suggestion is welcome.

Solution 3:

Suggestion for a proof.

The Riesz-Fisher theorem states that $L^p(\mathbb{R})$ is complete. Riesz-Fisher theorem

Another theorem states that $C^\infty_c(\mathbb{R})$ is dense in $L^p(\mathbb{R})$. Check out denseness of smooth functions

This implies that for any $f\in L^p(\mathbb{R})$, there exists a sequence of functions $\{f_n\}\in C^\infty_c(\mathbb{R})$ such that $f_n\to f$ in $L^p(\mathbb{R})$

If $\delta(x)\in L^p(\mathbb{R})$, for some fixed $\epsilon >0$ there is an $N\in\mathbb{N}$ such that $||f_n(x) - \delta(x)||_p<\epsilon$, for $n\ge N$.

This results in a contradiction though, because at some point, $f_n$ would not be continuous.

Refer to reuns's first comment: no sequence of continuous functions exist that converge to $\delta(x)$.