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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:

Series of derivatives on Gaussian distributions

(I rescaled by powers of $2 \epsilon$ so the functions on fit on the graph).]

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    $\begingroup$ Have you explored this statement on Gaussians of arbitrary width? $\endgroup$ Aug 2, 2023 at 13:56
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    $\begingroup$ Yes! First thing I tried, but I couldn't figure out how to formally prove anything from this. I edited to include the result in the question above, for anyone reading this later. $\endgroup$
    – twoform
    Aug 2, 2023 at 15:36
  • $\begingroup$ You must be able to take the increasingly singular limits of the sequence.... $\endgroup$ Aug 2, 2023 at 15:40
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    $\begingroup$ Maybe I am not catching your point; if I just send $\epsilon \to 0$ the graphs will just rescale. The fact that all the derivatives give non-positive functions should commute with the limit. I know that's a pretty physics-proof-y way to look at it, so are you saying there's some subtlety in the $\epsilon \to 0$ limit, e.g., the derivatives don't commute with the limit and that's important? $\endgroup$
    – twoform
    Aug 2, 2023 at 15:44
  • $\begingroup$ Ach, you have rescaled wrongly. Work out the δ and δ' explicit analytical expressions, and forget about snazzy graphs. If you can't get that, you really don't need your question answered. $\endgroup$ Aug 2, 2023 at 15:55

2 Answers 2

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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.

Definition. (Positive distribution)

A distribution $T$ on $\Omega$ is said to be positive if for each non-negative test function $\phi\in\mathcal{D}(\Omega)$ we have $\langle T,\phi\rangle\geq 0$.

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 $\geq 0$.

Definition. (Order of a distribution)

Let $T$ be a distribution on $\Omega$. Let $E$ be the set of integers $N\geq 0$ such that for every compact set $K\subset \Omega$, there is a constant $C_K>0$ such that for all $\phi\in\mathcal{D}_K(\Omega)$, we have $|\langle T,\phi\rangle|\leq C_K\|\phi\|_{K,N}$.

  • if $E$ is empty (i.e no such integer exists) we say $T$ is a distribution of infinite order.
  • if $E$ is non-empty and $N\in E$, we say $T$ is a distribution of order at most $N$. We define $\min E$ to be the order of $T$.

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).

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    $\begingroup$ This is great! Thanks for writing it up. Time to dust off my old copy of Rudin... Some two-line summary of this is going into a paper footnote, and it would be great to acknowledge this comment--if you want to break anonymity so we can do it properly, my email is on the website in my profile :) $\endgroup$
    – twoform
    Aug 2, 2023 at 15:39
  • $\begingroup$ @twoform that’s fine; this is all “standard” stuff anyway so no need to credit me specifically. As you can see, proving a positive distribution has order zero is actually pretty straightforward (though it’s nice to realize that positivity (which is a monotonicity property because of linearity of distributions) is very closely related to continuity (cf the inequality proved)… this is a good general principle to keep in mind). It is only in proving that the result is a measure where we need to invoke the more advanced Riesz’s theorem. $\endgroup$
    – peek-a-boo
    Aug 2, 2023 at 22:23
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    $\begingroup$ … and the only reason I was able to answer this question is because I had this on my analysis final… (where I only answered it partially lol) $\endgroup$
    – peek-a-boo
    Aug 2, 2023 at 22:38
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While I am not clear on the specifics of your case, for example the distribution $\delta'(x)$ can be combined with the strictly positive $f(x)=\exp(-(x+1)^2)$ to produce $$\langle\delta', f\rangle = -2/e$$ and that's the intuition that I would use to make this statement rigorous.

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