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Given the following definition of the Dirac's Delta:

$$\delta: \mathcal{D}(\mathbb{R}^n) \to \mathbb{R}: \varphi \mapsto \langle \delta,\varphi \rangle = \varphi(\mathbf{0})$$

where $\mathcal{D}(\mathbb{R}^n)$ is the space of bump functions over $\mathbb{R}^n$, i.e. the space of smooth functions

$$C_0^\infty(\Omega) := \{ f \in C^\infty(\Omega): \mathrm{supp}(f) \ \mathrm{is \ compact} \}$$

with a suitable topology, and given the following association between an ordinary function $u \in L^1_{loc}(\Omega) $ and the corresponding distribution $I_u \in \mathcal{D}'(\Omega)$

$$ I_u: \mathcal{D}(\Omega) \to \mathbb{R}^n : \varphi \mapsto \langle I_u,\varphi \rangle =\int_\Omega u\varphi $$

how to prove there is not $u \in L^1_{loc}(\mathbb{R}^n)$ such that $I_u = \delta$?

Edit. Or, how to prove that if such $u$ exists, $u \equiv 0$ in $\mathbb{R}^n\setminus\{0\}$ and $\int_{\mathbb{R}^n}u=1$? This leads to a contradiction because $\int_{\mathbb{R}^n}u = \int_{\mathbb{R}^n\setminus\{0\}}u$, so no such $u$ exists.

unlikely
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  • Apply $\delta$ to $u\chi_{[-1/n,1/n]}$, where $u$ is a possible candidate, and $\chi_{[-1/n,1/n]}$ is the indicator function of $[-1/n,1/n]$. Take limit on both sides as $n\rightarrow\infty$ and apply dominated convergence. – OR. Mar 28 '14 at 12:27
  • You mean $\varphi(0)$, not $\delta(0)$, in the first line, right? – fgp Mar 28 '14 at 12:34
  • Forgot the translation. It should be $u(x-a)\chi_{[-1/n,1/n]}(x-a)$ for constants $a$. So that you test all the points. – OR. Mar 28 '14 at 12:35
  • @fgp, yes, thanks, corrected – unlikely Mar 28 '14 at 12:35
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    @ABC What do you means with apply $\delta$ to $u\chi$? $u\chi \notin \mathcal{D}(\mathbb{R})$ because $u\chi \notin C_0^\infty(\mathbb{R})$... – unlikely Mar 28 '14 at 13:56

2 Answers2

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Intuitively, because there a bump functions $\varphi$ with $\varphi(0) = 1$ but which are zero except on an arbitrary small interval $[-\epsilon,\epsilon]$ around $0$. That, however, is incompatible with the requirement that for some fixed $u$, $$ \int_{\mathbb{R}^n} u(x)\varphi(x) \,dx = \varphi(0) = 1 $$ essentially because you can make that integral arbitrarily small by picking a suitably narrow $\varphi$.

For a formal proof, let $(\varphi_n)_{n\in\mathbb{N}}$ be a sequence of bump functions with $\varphi_n(0) = 1$, $\|\varphi_n\|_\infty = 1$ and $\textrm{supp}\, \varphi_n = [-\frac{1}{n},\frac{1}{n}]^k$ (i.e., $\varphi_n$ is zero outside the cube $[-\frac{1}{n},\frac{1}{n}]^k$).

You then on the one hand get $$ \lim_{n\to\infty} \int_{\mathbb{R}^k} u \varphi_n = \lim_{n\to\infty} \langle I_u, \varphi_n \rangle = \lim_{n\to\infty} \varphi(0) = 1 \text{.} $$ But on the other hand, since obviously $u \varphi_n \to 0 $ pointwise as $n \to \infty$, you get by dominated convergence (note that $|u(x)\varphi_n(x)| \leq |u(x)|\cdot\|\varphi_n\|_\infty \leq |u(x)|$ for all $x \in [-1,1]^k$ and that $|u|\in L^1_{[-1,1]^k} \subset L^1_{\textrm{loc}}$) that $$ \lim_{n\to\infty} \int_{\mathbb{R}^k} u \varphi_n = \lim_{n\to\infty} \int_{[-1,1]^k} u \varphi_n = 0 \text{.} $$ This obviously is a contradiction.

Cortizol
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fgp
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  • No sure to understand... ${f_n}$ converges almost everywhere to $f \equiv 0$, but $f_n$ cannot be used as bump function in place of $\varphi$... Maybe a convolution with a mollifier? – unlikely Mar 28 '14 at 14:30
  • @unlikely The way I had presented things previously, it wasn't really clear that this works, because I first picked a specific $\varphi$, but then chose an arbitrarily narrow $\varphi$. Hope things are clearer now... – fgp Mar 28 '14 at 16:10
  • OK, now I understood, thanks. I only suppose we need to change $\varphi_n$ to $\varphi_k$ and $1/n$ to $1/k$ to avoid confusion with the dimension of space $\mathbb{R}^n$. As dominating function we can use $g=|u| \chi_{[-1,1]}$ wich is not negative, $L^1$ and such that $g >= u\varphi_n$ almost everywhere? – unlikely Mar 28 '14 at 21:49
  • @unlikely Ups, yeah, calling both $n$ wasn't very smart ;-) I renamed the $n$ in $\mathbb{R}^n$, though - that was less work. Regarding the dominating function - I was again sloppy there. You need the additional requirement that all the $\varphi_n$ are bounded, for simplicity say $0 \leq \varphi_n(x) \leq $ for all $n$ and all $x$. Then $|u|$ is a dominating function. Will edit to make that clear. – fgp Mar 28 '14 at 22:21
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Proof by @fgp is very nice. Here it's alternative proof:

Suppose that exist function $u(x) \in L^1_{\text{loc}}(\mathbb{R}^n)$ such that $$\int_{\mathbb{R}^n} u(x)\varphi(x)\, dx = \varphi(0), \quad \text{for all } \varphi \in \mathcal{D}(\mathbb{R}^n).$$

If $\varphi \in \mathcal{D}(\mathbb{R}^n)$ then also $x_1 \varphi(x)=x_1 \varphi(x_1,x_2,\ldots,x_n)\in \mathcal{D}(\mathbb{R}^n)$, so then $$\int_{\mathbb{R}^n} u(x)x_1 \varphi(x) \, dx =x_1 \varphi(x){\Large{|}}_{x=0}=0, \quad \text{for all } \varphi \in \mathcal{D}(\mathbb{R}^n).$$

According to du Bois-Reymond lemma that implies $x_1 u(x)=0$ almost everywhere, and then $u(x)=0$ almost everywhere, so we get

$$\int_{\mathbb{R}^n} u(x)\varphi(x)\, dx = 0,$$

which is in contradiction with our initial assumption.

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