# 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??

• You misunderstood something with the Dirac delta. Suppose that $\int_{-1}^1 |\delta(x)| dx$ exists and is $=1$ then $\delta$ is still not in $L^1([-1,1])$ because there is no sequence $(f_n)$ of continuous functions $\in L^p([-1,1])$ such that $\|\delta-f_n\|_1 \to 0$ as $n \to \infty$. – reuns May 30 '16 at 2:02
• another way to say this is that if a sequence of functions $f_n$ converges (in the sense of distributions !) to $\delta$, then $f_n$ is not a Cauchy sequence in $L^1$ (nor in $L^p$ for any $p \ge 1$) – reuns May 30 '16 at 2:05
• $L^2(\mathbb R)$ is a set of (equivalence classes) of functions defined on $\mathbb R$. The Dirac $\delta$ is not a function. Hence, $\delta$ is not in $L^2(\mathbb R)$. – Math1000 May 30 '16 at 13:21
• @Math1000: "$\delta$ is not a function" is a statement that requires proof - we need to show that there is no function $g \in L^2$ such that $\langle \delta, \phi \rangle = \int g \phi$. – Anthony Carapetis May 30 '16 at 13:29
• The proof is trivial: $\delta$ is zero almost everywhere with respect to Lebesgue measure and yet has a positive integral. This is impossible for a measurable function. – Math1000 May 30 '16 at 14:41

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.

• I found this document leetspeak.org/math/notes/dirac_delta_not_in_lp.pdf , but the document prove that Dirac's delta not belong in $L^p(0,2)$, then how could we pass the case $L^p(0,2)$ to the case $L^p(\mathbb{R})$ ? – Alex Pozo Jun 5 '16 at 3:28
• @AlexPozo You really can't use their result as a black box, you need to re-do the argument. But I already said an easy way to do that. For a concrete example, let $f$ be any continuous function with $f(0) \neq 0$ and $\int_{\mathbb{R}} f(x)^2 dx = 1$, then let $f_n(x)=\sqrt{n} f(nx)$. – Ian Jun 5 '16 at 3:54
• If I give $f_n(x)=\frac{ne^{-n^2 x^2}}{\sqrt{\pi}}$, and f(x) is the Dirac's delta, Have I to prove that $||f-fn||$ no converge to zero? – Alex Pozo Jun 5 '16 at 5:28
• If you had $\delta \in L^2$ then you would have $\int \delta (x) f (x) dx \leq \| \delta \|_2 \| f\|_2$ for any $L^2$ $f$. But by definition this integral is $f (0)$ for continuous $f$. So it is enough to show that there is no $C$ such that $f (0) \leq \| f \|_2$ for all continuous and $L^2$ functions $gf$. A sequence of the form $f_n (x)=\sqrt {n} f (nx)$ where $f$ is continuous and $L^2$ and $f (0)>0$ will show this, because $\| f_n \|_2$ will be constant but $f_n (0)$ will blow up. – Ian Jun 5 '16 at 5:47
• Typos: in the second sentence I should have $f(0) \leq C \| f \|_2$ and at the end of that sentence I should have $f$, not $gf$. – Ian Jun 5 '16 at 6:14

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.

• You are missing the constant $C$ that stands for "the norm of $\delta$" in the Cauchy-Schwartz inequality and in what follows. Still, the proof is easily adapted. – Jan Jul 24 '19 at 10:30

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 $\delta(x)$.