I am looking at properties of the Glauber P-functions, which are distributions (in the sense of a dirac delta) on the complex plane, normalized so that $\int_{\mathbb{C}} d^2 \alpha P(\alpha) = 1$. On this wikipedia article on the Glauber representation, it says that
"By a theorem of Schwartz, distributions that are more singular than the Dirac delta function are always negative somewhere."
I am trying to understand this statement. The wiki article just links to a general page about distributions, which isn't very helpful. I understand "more singular than a Dirac delta" to mean something like $$P(\alpha) \sim \left(\frac{\partial}{\partial \alpha}\right)^n \delta(\alpha).$$ In what sense does a distribution like this have to "be negative somewhere"?
A precise statement of this "theorem of Schwartz" would be very welcome, as well as any intuition.
[Possibly related question here, but does not seem to contain the answer.]
[Edit: for posterity, to answer a comment below, you can get some basic intuition by regulating the Dirac delta as, for example, the limit $\epsilon \to 0$ of Gaussians of width $\epsilon$, and just taking derivatives. You clearly get negative behavior:
(I rescaled by powers of $2 \epsilon$ so the functions on fit on the graph).]
Best Answer
The idea is that for any distribution, you can assign it a “number” (a non-negative integer or $\infty$), called its order. The higher the order, the more singular it is. The theorem is then that a distribution is positive if and only if it is has order $0$ (and in fact it must be represented by a positive measure).
Now, let’s be more precise. Throughout, $\Omega$ denotes an open set in $\Bbb{R}^n$, $\mathcal{D}(\Omega)$ the space of test functions (with a suitable Frechet topology, generated by the seminorms of uniform convergence on compact sets of the various partial derivatives), and let $\mathcal{D}’(\Omega)$ be the topological dual space, i.e the space of distributions. To be very explicit, let us write out what continuity means in this context. A linear map $T:\mathcal{D}(\Omega)\to\Bbb{C}$ is continuous if and only if for each compact set $K\subset \Omega$, there is a constant $C_K>0$ and an integer $N_K>0$ such that for all $\phi\in\mathcal{D}_K(\Omega)$ (i.e $\phi$ is a smooth function $\Omega\to\Bbb{C}$ with support contained in $K$), we have \begin{align} |\langle T,\phi\rangle|&\leq C_K\|\phi\|_{K,N_K}:=C_K\sum_{|\alpha|\leq N_K}\sup_{x\in K}|\partial^{\alpha}\phi(x)|. \end{align}
Now, we’re ready to give the relevant definitions.
Of course, we have here the usual unfortunate terminological issue of positive vs non-negative. It is tradition to call these guys positive even though we only use the weak inequality $\geq 0$.
So, the point of this definition is that if you look back at the continuity condition above, the $N_K$ depends on the given compact set $K$. We are thus asking if it is possible to take $N$ independent of $K$. If it is, then the distribution is said to have finite order and the smallest such integer is called its order. If it is impossible to take $N_K$ independent of $K$, then we say $T$ has infinite order. So, the order is like “the smallest number of derivatives we need to control the distribution”.
Now we can ask for examples.
Let $\mu$ be a positive measure on $\Omega$ which is finite on compact sets. Then via integration $\phi\mapsto \int_{\Omega}\phi\,d\mu$, we get a distribution, $T_{\mu}$, on $\Omega$. To see the continuity condition is satisfied, note that if $K$ is a compact set, then for all $\phi\in\mathcal{D}_K(\Omega)$ we have \begin{align} |\langle T_{\mu},\phi\rangle|&=\left|\int_{\Omega}\phi\,d\mu\right|=\left|\int_K\phi\,d\mu\right|\leq \mu(K)\cdot \sup_{x\in K}|\phi(x)|. \end{align} This means we have satisfied the continuity condition with $C_K=\mu(K)$ and $N_K=0$, independent of $K$. Also, by virtue of $\mu$ being a positive measure, if we take a non-negative $\phi$ then the integral is non-negative. Hence $T_{\mu}$ is a positive distribution of order $0$. By the usual abuse of language, we then say the measure $\mu$ is a positive distribution of order $0$. For example, we can take the Lebesgue measure, or Dirac measure at a point.
The derivative $\partial^{\alpha}(\delta_p)$ of the Dirac measure/distribution is a distribution of order $|\alpha|$. So, in view of the theorem, if $|\alpha|>0$ then this is not a positive distribution. The other answer gives a rather concrete example in $1$ dimension. Unrelated for what follows, but as a general warning, note that differentiation doesn’t always increase the order. For example, differentiating smooth functions gives smooth functions (and their associated distributions can be identified) so it’s still order $0$. As a more ‘singular’ example (‘singular’ being used in the intuitive sense, not in the order sense), consider the Heaviside step function. This has a jump discontinuity, but it’s still a locally-integrable function so it gives an order $0$ distribution, and its derivative is the Dirac delta which is again order $0$.
With our first bullet point in mind, we can ask the follow up question “are there any more positive distributions other than the positive measures”, and the answer is no. To see this, suppose $T$ is a positive distribution on $\Omega$, and let $K\subset \Omega$ be any compact set, and fix a smooth non-negative bump function $\beta_K$ which is identically equal to $1$ on a neighborhood of $K$. Then, for any real-valued $\phi\in\mathcal{D}_K(\Omega)$, we have that the functions $\|\phi\|_{\infty}\beta_K\pm \phi$ are smooth and non-negative, so applying $T$ gives something non-negative: \begin{align} 0&\leq\langle T,\|\phi\|_{\infty}\beta_K\pm\phi\rangle=\|\phi\|_{\infty}\langle T,\beta_K\rangle \pm\langle T,\phi\rangle. \end{align} Rearranging gives $|\langle T,\phi\rangle|\leq \langle T,\beta_K\rangle\cdot\|\phi\|_{\infty}$. Now for a general complex-valued $\phi$, we apply this inequality to its real and imaginary parts to deduce that \begin{align} \left|\langle T,\phi\rangle\right|&\leq \langle T,\beta_K\rangle\cdot\left(\|\text{Re}(\phi)\|_{\infty}+\|\text{Im}(\phi)\|_{\infty}\right)\leq c\langle T,\beta_K\rangle\cdot\|\phi\|_{\infty}, \end{align} for some constant $c$, maybe $\sqrt{2}$ or something… I’m too lazy to work out the ‘best’ constant, but clearly it exists. So, taking the constant $C_K=c\langle T,\psi_K\rangle$, we see that we have managed to take $N_K=0$ regardless of the compact set $K$. Thus, we have shown that every positive distribution has order $0$ (this answers your question).
We can actually do a little more finessing by using some approximation arguments (bump functions, convolutions etc) to extend $T$ to a positive linear functional $\tilde{T}$ on $C_c(\Omega)$, and thus by Riesz’s representation theorem for positive functionals, there is a unique positive Radon measure $\mu$ such that for all $f\in C_c(\Omega)$, $\tilde{T}(f)=\int_{\Omega}f\,d\mu$, and hence for all $f\in C^{\infty}_c(\Omega)$, we have $T(f)=\tilde{T}(f)=\int_{\Omega}f\,d\mu$.
You can refer to Rudin’s functional analysis text (chapter 6) for more about distributions. The fact that every positive distribution gives rise to a positive measure is one of the exercises.
Also, for intuition, there’s a text by Strichartz on distributions and Fourier transforms. Chapter 6, The Structure of Distributions you may find helpful (this text of his is written in a more conversational manner than most mast texts).