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Let $\mathcal S'=\mathcal S'(\mathbb R^n)$ be the Schwartz distribution space. Suppose $A\colon\mathcal S'\to\mathcal S'$ is linear, continuous and microlocal. By being microlocal I mean that the wave front sets satisfy $WF(Af)\subset WF(f)$ for all $f$. (For another version, one could consider the singular supports instead of wave fronts, but I assume the answer wouldn't be different.) Does it follow that $A$ is a pseudo-differential operator?

This is a variation on this question about continuous endomorphisms on the Schwartz space. There one only assumed linearity and continuity, and the answer was negative. The other question was about $\mathcal S$ instead of $\mathcal S'$, but I need distributions to allow singularities.

I realize that there are many classes of pseudo-differential operators. The question is whether such a microlocal operator is always a ΨDO of some kind. Any details on what kind would be very interesting, of course.

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    $\begingroup$ Thanks! For starters, mustn't such $A$ be induced from some continuous endomorphism of $\mathcal{S}$? If not then its a negative answer to the question, if yes then it doesn't hurt to assume it in the premise. $\endgroup$ – Saal Hardali Nov 3 '17 at 23:04
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    $\begingroup$ @SaalHardali Interesting question! It would be weird if the result held for extensions of operators on test functions but failed for operators on distributions, but we'll have to wait and see. $\endgroup$ – Joonas Ilmavirta Nov 3 '17 at 23:08
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This is more like a longish series of comments somewhat complementing Ilya Zakharevich's answers rather than an answer by itself. First of all, notice that since $\mathscr{S}(\mathbb{R}^n)$ embeds continuously (and densely) into $\mathscr{S}'(\mathbb{R}^n)$, any continuous linear map $A:\mathscr{S}'(\mathbb{R}^n)\rightarrow\mathscr{S}'(\mathbb{R}^n)$ induces by restriction a unique continuous linear map $A|_{\mathscr{S}(\mathbb{R}^n)}:\mathscr{S}(\mathbb{R}^n)\rightarrow\mathscr{S}'(\mathbb{R}^n)$, which we also denote by $A$ not to overburden our notation. By Schwartz's kernel theorem, we must have $$(A\phi_1)(\phi_2)=K_A(\phi_1\otimes\phi_2)\ ,\quad\forall\phi_1\ ,\,\phi_2\in\mathscr{S}(\mathbb{R}^n)$$ for a unique $K_A\in\mathscr{S}'(\mathbb{R}^{2n})$.

One sees from the calculus of wave front sets that microlocality of $A$ $$\mathrm{WF}(Au)\subset\mathrm{WF}(u)\ ,\quad\forall u\in\mathscr{S}'(\mathbb{R}^n)$$ is entailed by the condition $$\mathrm{WF}(K_A)\subset\{(x,x;\xi,-\xi)\ |\ (x;\xi)\in\mathbb{R}^{2n}\smallsetminus 0\}\ ,$$ which all pseudodifferential operators with (say) smooth symbols must satisfy by the stationary phase formula, but the converse (as far as I know) is not necessarily true.

To have a better idea of what the would-be symbol of $A$ should look like, we have that $$\sigma_A(x,\xi)=A(e^{-i\langle x-\cdot,\xi\rangle})(x)$$ defines for each $\xi\in\mathbb{R}^n$ a tempered distribution $\sigma_A$ in $x$ which is smooth and polynomially bounded in $\xi$ together with all its $x$- and $\xi$-derivatives. By microlocality, $\sigma_A$ should be smooth in $x$ as well. However, this alone does not guarantee that $\sigma_A$ fits into any decent symbolic calculus since the polynomial bounds on its derivatives in $x$ and $\xi$ may vary too wildly.

One can say more if we know that $A$ is translation invariant in the sense that $$A(u(\cdot+x))=(Au)(\cdot+x)$$ for all $x\in\mathbb{R}^n$, $u\in\mathscr{S}(\mathbb{R}^n)$. Due to microlocality, this can only be the case if $A$ is a convolution operator, that is, there is a unique $F_A\in\mathscr{S}'(\mathbb{R}^n)$ such that $$Au=F_A*u\ ,\quad\forall u\in\mathscr{S}'(\mathbb{R}^n)\ .$$ Equivalently, $K_A$ is the convolution kernel associated to $F_A$: $$K_A(x,y)=F_A(x-y)\ .$$ The Fourier transform $\sigma_A$ of $F_A$ is in this case the obvious candidate for the symbol of $A$ (in this case, $\sigma_A$ no longer depends on $x$, of course). The image of $\mathscr{S}(\mathbb{R}^n)$ under $A$ consists of smooth functions with at most polynomial growth together with all their derivatives, but that is pretty much it - the convolution with any $F\in\mathscr{S}'(\mathbb{R}^n)$ will have the same property. Microlocality of $A$ follows if in addition $\mathrm{WF}(F_A)\subset\{(0;\xi)\ |\ \xi\in\mathbb{R}^n\smallsetminus 0\}$, which is equivalent to the above condition on $\mathrm{WF}(K_A)$ in the translation invariant case, but again this (as far as I know) is not a necessary condition.

If we could show that:

  1. $(1+\|\xi\|^2)^{-k}\sigma_A\in L^\infty(\mathbb{R}^n)$ for some $k\in\mathbb{N}$, which amounts to saying that $(1-\Delta)^{-k}A$ is a bounded operator in $L^2(\mathbb{R}^n)$ by the Plancherel formula ($\Delta$ is the Laplacian in $\mathbb{R}^n$), and

  2. The commutators of $(1-\Delta)^{-k}A$ with (multiplication by) polynomials are also bounded operators in $L^2(\mathbb{R}^n)$,

one could appeal to the Beals-Cordes commutator characterization of pseudodifferential operators of order zero since $A$ commutes with derivatives in the translation invariant case (notice that (2.) and the latter form of (1.) still make sense in the non-translation-invariant case). In the general (non-translation-invariant) case, we need as well to show that

  1. The commutators of $(1-\Delta)^{-k}A$ with derivatives are also bounded operators in $L^2(\mathbb{R}^n)$

in order to use the Beals-Cordes criterion, which precisely guarantees the kind of polynomial bounds on the derivatives of $\sigma_A$ needed for it to be a symbol (of order $2k$). I have no idea if microlocality is enough to yield (1.)-(3.) (or even (1.)-(2.) in the translation-invariant case), but it seems a rather tall order to me.

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Preliminaries

Averaging arguments show that the question boils down to taking convolutions. So one needs to find a generalized function F smooth outside of 0 such that the Fourier transform ℱF(ξ) does not have asymptotic at ξ→∞ “suitable for ΨDSymbol”. As I already mentioned, until one provides a particular flavor of ΨDS to consider, the question does not make any sense. Below, I would consider ρ-flavored definition: ∂/∂ξ-derivatives of ℱF(ξ) should decrease as O(|ξ|ᵗ), with t=C-kρ; here ρ>0, and k is the order of derivative.

Note that for k=0 the estimate follows from the fact that F is a generalized function: it has a “finite degree of non-differentiability”: taking enough anti-derivatives, one can make it continuous, differential etc. Now, when we consider derivatives of ℱF, we are essentially replacing F by xᵏF(x); for F to be a ΨDS, the degree of differentiability of xᵏF(x) should increase linearly with k.

Summary: we need a generalized function F smooth outside of 0 (so behaviour far away from 0 is irrelevant) and such that the number of continuous derivatives of xᵏF(x) does not increase linearly with k. I do not want to give particular examples, just a framework which allows finding “arbitrary” examples of similar kind.¹⁾

    ¹⁾ So when one comes with a wider class of “what is a ΨDS”, one may use a framework to find something outside of this class — same as we did for the ρ-class.

Localizability

Existence of ρ>0 as above corresponds to Fourier transforms of x(xᵏF(x)) and xᵏF(x) differing by at least a factor |ξ|^{-ρ}. Equating the factor x and the factor |ξ|^{-ρ} shows that “the part of F(x) contributing most to value of ℱF(ξ) at ξ=ξ₀” is the part near |x| ∼ |ξ₀|^{-ρ}. Translating to our requirements: we need the value of ℱF(ξ) at ξ=ξ₀ to “be contributed” by behaviour of F in a region |x|∼x₀, and x₀ should depend on ξ₀ slower that a power law.

Summary

  • We need ℱF to be “localizable”: its behaviour near ξ=ξ₀ should depend on behaviour of F at a certain region |x|∼x₀.
  • x₀ should decay slowly when ξ₀→∞.

Stationary phase

A typical reason of localizability a possibility to calculate an integral transform (such as a Fourier transform) by a saddle point method. For simplicity, we assume that this is reducible to stationary phase; for this, we assume that F(x) = exp i φ(x); the method of stationary phase is applicable for a very wide class of functions φ. (I do not remember the details; convexity of φ, or maybe of φ' should be enough.) Below, I assume that this method is applicable.

Conclusion: we need to find φ(x) such that φ'(x₀) = ξ₀ has a solution x₀(ξ₀) decreasing to 0 slower than a power of ξ₀. In other words, φ' (or φ) should not be of tempered growth near 0.

The result

Consider the operator of convolution with the bounded function F(x) = exp i exp 1/|x| (cut off F far away from 0). This is an operator with symbol ℱF(ξ); however, this symbol does not satisfy ρ-estimate.

(Note that F is a derivative of a continuous function, so it is a generalized function.)

Defects

As far as I understand, there is no big deal to extend the theory of symbols such that it would also consider the symbol ℱF(ξ) as an “admissible” symbol. This is why I think the question does not make a major sense: different people have different needs, and take different classes of symbols as fits their problem domains.

Update. Somehow, while I explained in details the arguments which allow to construct the example above, I omitted the part “in the other direction”: why, indeed, the function F does not satisfy the (ρ,δ)-condition on ΨDS with ρ>0. This part is completely trivial; it does not even require stationary phase:

Lemma. If φ is real, and φ' is not tempered near 0, then the derivative of xⁿF(x) is unbounded near 0 for any n∈ℤ. Here F(x) ≔ x exp i φ(x).

Hence F∈C⁰, and F(x) is smooth for x≠0 (provided φ is), but xⁿF(x)∉C¹ for any n. This implies that the ρ-condition does not hold for the Fourier transform ℱF of F.

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    $\begingroup$ Question: what do you mean by x(xᵏF(x))? Comment: maybe a sensible question is whether there is at all a rigorously formulable property that distinguishes a proper subclass of microlocal operators which are definitely not pseudodifferentiable in any possible sense, then $\endgroup$ – მამუკა ჯიბლაძე Jan 6 '18 at 9:03
  • $\begingroup$ x(xᵏF(x)) is a product of x and of xᵏF(x). ;-) [And I said it many times: I do not see how it may make sense to have a notion of ΨDO with a “well-defined boundary”. With (ρ,δ), it is easy to say in which sense the composition is defined as “a sum” of a series. However, if a need arises, one could use other notions of “summation of a series” to handle more general classes.] $\endgroup$ – Ilya Zakharevich Jan 6 '18 at 22:57
  • $\begingroup$ So would you say that any microlocal operator is, in some generalized sense, pseudodifferential? $\endgroup$ – მამუკა ჯიბლაძე Jan 7 '18 at 6:32
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    $\begingroup$ Now this is a very productive question! In 1-dimensional case the answer is kind of tricky: I’m mentally ready to treat any such operator as a ΨDO, but I’m not ready to do this for two such operators simultaneously. In my mental picture, different operators may require different, incompatible, generalizations. $\endgroup$ – Ilya Zakharevich Jan 7 '18 at 20:24
  • $\begingroup$ In more than 1-dimensional case, all hell breaks loose. Def: a ΨD kernel K(x,y)dy is a kernel with a singularity on (conormal bundle to) Diag≔{x=y} such that after blowing up Diag, it has a singularity in codimension 1. (In other words, K is a direct image of a generalized function on Z with such a property; here Z → X×X is the blowup map.) This is my mental picture for ΨDO in dim>1. Omitting the condition of codim=1 gives the “preservation of wavefront” condition. One can see how far these notions are from each other. $\endgroup$ – Ilya Zakharevich Jan 7 '18 at 20:32
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Not really an answer: but you do realize how many different notions of a ΨDO are there? So pick up one of them, and the example would not be a ΨDO in the sense of another definition.

BTW: AFAIK, the (ε,δ)-class is tuned up to be closed under composition. It might be possible to take ε,δ∉[0,1] and get something which still acts in 𝒮 ′ (but in such a way that the formula for composition does not make sense).

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  • $\begingroup$ A possible candidate: start with multiplicative convolution with a Gaussian kernel on ℝ₊⊂ℝ (extended as id onto ℝ₋), and tune up the parameters. Something like exp(-(x-y)²/x²ˢ)/xˢ dy for an appropriate s (or an appropriate substitute for xˢ). $\endgroup$ – Ilya Zakharevich Nov 4 '17 at 3:58
  • $\begingroup$ Oups: the example above is acting in 𝒮, not in 𝒮 ′. Dualize! $\endgroup$ – Ilya Zakharevich Nov 4 '17 at 4:19
  • $\begingroup$ Welcome to MO, Ilya! Great to have you here! $\endgroup$ – მამუკა ჯიბლაძე Nov 4 '17 at 5:32
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    $\begingroup$ Why exactly is it pseudolocal at the origin? $\endgroup$ – მამუკა ჯიბლაძე Nov 4 '17 at 11:33
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    $\begingroup$ Thanks for the remark! I know that there are various classes of $\Psi$DOs, and the question is really whether a microlocal operator is a $\Psi$DO of any kind. I updated my question to reflect this. $\endgroup$ – Joonas Ilmavirta Nov 4 '17 at 16:10
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I do not remember how to prove the general statement I mentioned about gluing operators on ⟦-∞,0] and [0,∞⟧ which match on ∞-Jets at 0; neither can I prove micro-locality of the operator I described in my comment. However, when replacing Gauss kernel with Poisson kernel, it becomes much simpler.

So: given a generalized function f on ℝ, denote by F its harmonic extension into ℂ. Take a smooth curve γ⊂ℂ such that its vertical projection π to ℝ is a diffeomorphism; suppose that γ coincides with ⟦-∞,0] in the left half-plane. Since F is harmonic with tempered growth near ℝ, it also has tempered growth near γ; hence restriction φ of F to γ makes sense as a generalized function. Identify φ with a generalized function on ℝ via π; denote it by Af.

If γ is in the upper half-plane, then A is the required operator. Indeed, it sends a function f holomorphically extendable to upper half-plane to a function φ holomorphically extendable to the region above γ; hence it preserves “positivity” of wavefront. On the other hand, the operator A does not change if one replaces γ by its reflection in ℝ; hence the same argument works for “negativity”. Obviously, singular support cannot increase. This implies micro-locality.

Update: No, this is not a counterexample. Indeed, the symbol of this operator is just exp -|ℾ(x)ξ|; here the curve γ⊂ℂ is given by y=ℾ(x), x+yi∈ℂ.

So as “a counterexample to something”, it is good enough to show how weird the ΨDS may be which are “almost classical” ΨDS. Also, it is good enough to show that preservation of singular support is not equivalent to presentation of wavefront. Otherwise, it is a very plain ΨDS in the sense of (ρ,δ).

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    $\begingroup$ So A is not pseudodifferential in what sense? $\endgroup$ – მამუკა ჯიბლაძე Nov 6 '17 at 10:36
  • $\begingroup$ @მამუკაჯიბლაძე: In any sense I know. It it a ΨDO on positive and negative axes (well, here I mean “negative”≔ℝ∩γ), with symbols 0 and 1, so if it were globally ΨDO, the symbol would have a discontinuity. $\endgroup$ – Ilya Zakharevich Nov 6 '17 at 21:45
  • $\begingroup$ I asked not only to make sure, but also mainly because I believe that for a first-time reader without this argument in the answer it is not clear at all. $\endgroup$ – მამუკა ჯიბლაძე Nov 7 '17 at 4:37
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    $\begingroup$ @მამუკაჯიბლაძე: BTW, commuting this A with ∂, one gets a certain operator B which is “kinda ΨDO” but “with the symbol” B(x,ξ)≔δ(x). Have not thought about this before… $\endgroup$ – Ilya Zakharevich Nov 7 '17 at 6:09
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    $\begingroup$ If the trick with taking commutator with ∂/∂x works indeed (cannot invent a mechanism why it won’t, but the conclusion is kinda incredible), this would give something really surprising: negative answer to the equivalence of micro-locality with preservation of singular support. Indeed, composing B with reflection would give an operator which does not increase singular support, but it is inverting the wavefront over x=0… (?!) $\endgroup$ – Ilya Zakharevich Nov 8 '17 at 10:20

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