## 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.