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I am currently taking a course on test functions and distributions and my task is to prove that the Dirac delta is not a function.

Furthermore, I would also like to prove that it is continuous as a distribution in the sense that the inequality : $|\delta_{a}(\phi)| \leq c\|{\phi}\|_{C^k}$, for $c>0$ and $\phi \in C_{0}^{\infty}$ holds.

Thank you!

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    $\begingroup$ What is your definition of the Dirac delta? There are several (equivalent) ones, but the proof would be different depending on what your starting point is. Also, what are your thoughts on the problem and what have you tried? $\endgroup$ Jan 23, 2015 at 0:30
  • $\begingroup$ The definition given is the point measure $\delta_{\alpha}$ such that: $\delta_{\alpha}(\phi):= \phi(\alpha)$ and $\phi \in C_{0}^{\infty}$. That is all we have. I have done a lot of online research, but the problem is that I do not really know where to begin from :/ $\endgroup$
    – Bazinga
    Jan 23, 2015 at 0:34
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    $\begingroup$ In that case, here's one approach: suppose that $\delta()$ were a function. Then first, can you prove it's 'positive' (i.e., $\delta()\geq 0$) everywhere? (Hint: try some test functions that 'expose' its negative values). Once you have that, you should be able to find a sequence of test functions that show you that $\delta_\alpha()$ takes on arbitrarily large values in arbitrarily small neighborhoods of $\alpha()$. $\endgroup$ Jan 23, 2015 at 0:53

2 Answers 2

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Suppose that we have a function $\psi(x)$ so that $$ \delta(\phi)=\int_{\mathbb{R}}\psi(x)\phi(x)\,\mathrm{d}x\tag{1} $$ Define $$ \begin{align} \eta(x)&=\left\{\begin{array}{} 0&\text{if }|x|\le\frac12\\ \frac{\displaystyle e^{\frac1{1-2|x|}}}{\displaystyle e^{\frac1{1-2|x|}}+e^{\frac1{|x|-1}}} &\text{if }\frac12\lt|x|\lt1\\ 1&\text{if }|x|\ge1 \end{array}\right.\tag{2a}\\[6pt] \eta_k(x)&=\eta\!\left(2^{-k} x\right)\tag{2b} \end{align} $$ Then $\eta_k\in C^\infty$, $\eta_k(x)=0$ for $|x|\le2^{k-1}$, and $\eta_k(x)=1$ for $|x|\ge2^k$. Since $C^\infty$ is dense in $L_\text{loc}^1$, for any $\epsilon>0$, we can choose a $\psi_\epsilon\in C^\infty$ so that $$ \begin{align} \|(\psi-\psi_\epsilon)(1-\eta)\|_{L^1}&\le\epsilon/2\tag{3a}\\ \|(\psi-\psi_\epsilon)(\eta_k-\eta_{k+1})\|_{L^1}&\le2^{-k-2}\epsilon\quad\text{for }k\ge0\tag{3b} \end{align} $$ which means $$ \|\psi-\psi_\epsilon\|_{L^1}\le\epsilon\tag{3c} $$ Furthermore, let $\sigma$ be the signum function, then since $\psi_\epsilon\in L_\text{loc}^\infty$, we can choose a $\sigma_\epsilon\in C^\infty$ so that $|\sigma_\epsilon|\le1$ and $$ \begin{align} \|(\sigma\circ\psi-\sigma_\epsilon)(1-\eta)\psi_\epsilon\|_{L^1}&\le\epsilon/2\tag{4a}\\ \|(\sigma\circ\psi-\sigma_\epsilon)(\eta_k-\eta_{k+1})\psi_\epsilon\|_{L^1}&\le2^{-k-2}\epsilon\quad\text{for }k\ge0\tag{4b} \end{align} $$ which means $$ \|(\sigma\circ\psi-\sigma_\epsilon)\psi_\epsilon\|_{L^1}\le\epsilon\tag{4c} $$ Since $\|\sigma\circ\psi-\sigma_\epsilon\|_{L^\infty}\le2$, we get $$ \begin{align} \|(\sigma\circ\psi-\sigma_\epsilon)\psi\|_{L^1} &\le\|(\sigma\circ\psi-\sigma_\epsilon)\psi_\epsilon\|_{L^1}+\|(\sigma\circ\psi-\sigma_\epsilon)(\psi-\psi_\epsilon)\|_{L^1}\tag{5a}\\ &\le\|(\sigma\circ\psi-\sigma_\epsilon)\psi_\epsilon\|_{L^1}+\|\sigma\circ\psi-\sigma_\epsilon\|_{L^\infty}\|\psi-\psi_\epsilon\|_{L^1}\tag{5b}\\ &\le3\epsilon\tag{5c} \end{align} $$

Define $\phi_k(x)=\sigma_\epsilon(x)\left(\eta(2^{-k} x)-\eta(2^{-k-1} x)\right)$. Then we have $$ \begin{align} 0 &=\delta\left(\sum_{k=m}^n\phi_k\right)\tag{6a}\\ &=\sum_{k=m}^n\int_{\mathbb{R}}\psi(x)\,\phi_k(x)\,\mathrm{d}x\tag{6b}\\ &=\sum_{k=m}^n\int_{\mathbb{R}}\psi(x)\,\sigma_\epsilon(x)\left(\eta(2^{-k} x)-\eta(2^{-k-1} x)\right)\mathrm{d}x\tag{6c}\\ &=\int_{\mathbb{R}}\psi(x)\,\sigma_\epsilon(x)\left(\eta(2^{-m} x)-\eta(2^{-n-1} x)\right)\mathrm{d}x\tag{6d}\\ &\ge\int_{\mathbb{R}}|\psi(x)|\left(\eta(2^{-m} x)-\eta(2^{-n-1} x)\right)\,\mathrm{d}x\\ &-\int_{\mathbb{R}}|(\sigma\circ\psi(x)-\sigma_\epsilon(x))\psi(x)|\left(\eta(2^{-m} x)-\eta(2^{-n-1} x)\right)\mathrm{d}x\tag{6e}\\ &\ge\int_{2^m\lt|x|\lt2^n}|\psi(x)|\,\mathrm{d}x-3\epsilon\tag{6f} \end{align} $$ Explanation:
$\text{(6a)}$: $\phi_k\in C_c^\infty$ and is $0$ in a neighborhood of $0$
$\text{(6b)}$: apply $(1)$
$\text{(6c)}$: apply the definition of $\phi_k$
$\text{(6d)}$: compute the sum of the $\eta_k$
$\text{(6e)}$: $\psi\sigma_\epsilon=|\psi|-(\sigma\circ\psi-\sigma_\epsilon)\psi$
$\text{(6f)}$: apply $(5)$

Since $\epsilon$ was arbitrary, $(6)$ implies that on any annulus centered at $0$, $\psi$ is $0$. That is, $\psi(x)=0$ for all $x\ne0$. Since the value of a function at a single point does not affect the integral of that function, $(1)$ would imply that $$ \delta(\phi)=0\tag7 $$ for all $\phi$. Since $(7)$ is false, $(1)$ must be false.

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    $\begingroup$ Seems to work, thank you so much! But what exactly is $n(x)$? A function? And how should I think of it? I mean what is the thinking way that you follow here? Thank you again! $\endgroup$
    – Bazinga
    Jan 23, 2015 at 18:18
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    $\begingroup$ @Mitscaype: $\eta(x)$ is a function. Its existence is guaranteed by the $C^\infty$-Urysohn Lemma. We can give an explicit definition $$\eta(x)=\left\{\begin{array}{}0&\text{if }|x|\le\frac12\\\frac{\displaystyle e^{\frac1{1-2|x|}}}{\displaystyle e^{\frac1{1-2|x|}} + e^{\frac1{|x|-1}}} &\text{if }\frac12\lt|x|\lt1\\ 1&\text{if }|x|\ge1\end{array}\right.$$ $\endgroup$
    – robjohn
    Jan 23, 2015 at 18:36
  • $\begingroup$ Thank you once again! I will study the Urysohn Lemma. One last question. How am I supposed to prove that $|\delta(\phi)|$ is bounded? In other words, I would like to prove that $\delta(\phi)$ is linear (obviously) and continuous, therefore a linear form. $\endgroup$
    – Bazinga
    Jan 23, 2015 at 18:46
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    $\begingroup$ @Mitscaype: $\delta(\phi)=\phi(0)\le\|\phi\|_{C^\infty}$ $\endgroup$
    – robjohn
    Jan 23, 2015 at 19:00
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    $\begingroup$ i have a doubt that how will you claim that your $\phi_{\lambda}$ has compact support as in order to define $\delta(\phi_{\lambda})$ our $\phi_{\lambda}$ must be in $C_c^\infty(\Omega)$ $\endgroup$
    – bunny
    Jun 23, 2017 at 9:38
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Although this question is very old and it already has an accepted answer that seems to be correct, I will add another one, much shorter and slightly easier. I will do so also because this question has many views and this basic fact seems to be easily overlooked in many standard texts providing an introduction to distributions (I guess that the two things are actually intertwined).

What follows is taken from Example 1.1 in G. Grubb, Distributions and operators, Graduated texts in mathematics, Springer Verlag (2009). This is a variant of the approach suggested in the comments below the OP.

Recall that $\delta$ is the distributional derivative of the Heaviside function $$H(x) := \begin{cases} 1 & x > 0,\\ 0 & x \leq 0.\end{cases}$$ In other words, $\delta(\phi) = -\int_{\mathbb R} H \phi' \,dx = \phi(0)$. The continuity of $\delta$ follows because $|\delta(\phi)| = |\phi(0)| \leq \|\phi\|_{C^0(\mathbb R)}$ (see section 3.1 in Grubb, or Theorem 6.8 in Rudin's Functional Analysis).

Suppose there exists $v \in L^1_{\rm loc}(\mathbb R)$ such that, for all $\phi \in C^\infty_c(\mathbb R)$, it holds $$\delta(\phi) = \int_{\mathbb R} v(x) \phi(x) \,dx. \quad \tag{$*$}$$ Let $\phi \in C^\infty_c(\mathbb R)$ be such that $\phi(0) = 1$ and, for $N \in \mathbb N$, set $\phi_N(x) := \phi(Nx)$. Note that, for all $N$, $\max |\phi(x)| = \max|\phi_N(x)|$, and that when $\phi$ is supported in $[-R,R]$, $\phi_N$ is supported in $[-R/N, R/N]$. By the theorem of Lebesgue,

$$\int_{\mathbb R} v \phi_N \,dx \to 0 \quad \mbox{as } N \to \infty.$$

On the other hand,

$$-\int_{\mathbb R} H\phi_N' \,dx = -\int_0^{+\infty} N \phi'(Nx) \,dx = -\int_0^{+\infty} \phi'(y) \,dy = \phi(0) = 1,$$

a contradiction. Hence, ($*$) does not hold for the sequence $\{\phi_N\} \subset C^\infty_c(\mathbb R)$, and in turn $v$ cannot exist.

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