Is every continuous microlocal operator a pseudo-differential operator?

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.

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 ′ (but in such a way that the formula for composition does not make sense).

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).