# How to prove that the Dirac delta is not a function?

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!

• 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? – Steven Stadnicki Jan 23 '15 at 0:30
• 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 :/ – Mitscaype Jan 23 '15 at 0:34
• 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 $\alpha()$. – Steven Stadnicki Jan 23 '15 at 0:53

Suppose that we have a function $\psi(x)$ so that $$\delta(\phi)=\int_{\mathbb{R}}\psi(x)\phi(x)\,\mathrm{d}x\tag{1}$$ Let $0\le\eta(x)\le1$, $\eta(x)=0$ for $|x|\le\frac12$, and $\eta(x)=1$ for $|x|\ge1$.
Define $\phi_k(x)=\psi(x)\left(\eta(2^{-k} x)-\eta(2^{-k-1} x)\right)$. Then, since $\phi_k$ has compact support and is $0$ in a neighborhood of $0$, we have \begin{align} 0 &=\delta\left(\sum_{k=m}^n\phi_k\right)\\ &=\sum_{k=m}^n\int_{\mathbb{R}}\psi(x)\phi_k(x)\,\mathrm{d}x\\ &=\sum_{k=m}^n\int_{\mathbb{R}}\psi(x)^2\left(\eta(2^{-k} x)-\eta(2^{-k-1} x)\right)\mathrm{d}x\\ &=\int_{\mathbb{R}}\psi(x)^2\left(\eta(2^{-m} x)-\eta(2^{-n-1} x)\right)\mathrm{d}x\\ &\ge\int_{2^m\lt|x|\lt2^n}\psi(x)^2\,\mathrm{d}x\tag{2} \end{align} This 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 $$\delta(\phi)=0\tag{3}$$ for all $\phi$. Since $(3)$ is false, $(1)$ must be false.
• 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! – Mitscaype Jan 23 '15 at 18:18
• @Mitscaype: $\eta(x)$ is a function. Its existence is guaranteed by the $C^\infty$-Urysohn Lemma. We can give an explicit definition $$\eta(x)=\left\{\begin{array}{}0&\text{if }|x|\le\frac12\\\frac{\displaystyle e^{\frac1{1-2|x|}}}{\displaystyle e^{\frac1{1-2|x|}} + e^{\frac1{|x|-1}}} &\text{if }\frac12\lt|x|\lt1\\ 1&\text{if }|x|\ge1\end{array}\right.$$ – robjohn Jan 23 '15 at 18:36
• 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. – Mitscaype Jan 23 '15 at 18:46
• @Mitscaype: $\delta(\phi)=\phi(0)\le\|\phi\|_{C^\infty}$ – robjohn Jan 23 '15 at 19:00
• 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 $C_c^\infty(\Omega)$ – bunny Jun 23 '17 at 9:38