# Interpretation of Smoothing Operators as $\Psi$DO's

In most of the computations I've seen involving pseudodifferential operators (referred to as $\Psi$DO's from now on) we do not have true equality. Instead we often only have equivalence of operators, where we say that two $\Psi$DO's are equivalent if their difference is a smoothing operator.

That is to say, if $P(x,D)$ and $Q(x,D)$ are two $\Psi$DO's, then we say $P$ is equivalent to $Q$ (written $P\sim Q$) if $$P(x,D)-Q(x,D) = R(x,D) \in S^{-\infty} = \bigcap_{m}S^{m}.$$

This is an incredibly important aspect of $\Psi$DO calculus, since many of the fundamental theorems in the subject require us to express certain operators as "formal" sums:

If $P(x,D)\in S^{m}$, we say that $P \sim \sum_{\alpha}P_{\alpha},$ if $$P-\sum_{|\alpha|<N}P_{\alpha}\in S^{m-N}, \quad \text{for all N=1,2,3,\ldots}$$

My question is how does one interpret this? It is clear to me what equality of two operators means, but I'm not sure how to internalize what it means for two operators to be similar. I understand that a smoothing operator has a rapidly decreasing kernel, and so it can even "smooth out" tempered distributions. But what does a smoothing operator "do" qualitatively?

Since we're dealing with equivalence classes, intuitively two operators should be similar if they aren't "too different" from each other. Right? What properties do similar operators share? What property does a smoothing operator not introduce? Meaning, in what way does a smoothing operator not "mess things up" too much?

I hope this isn't too vague of a concept, but I really feel like I'm missing some big picture here. Everything makes sense algebraically, but I'm lost when it comes to the intuition. Hopefully you'll have some insight for me.

Thanks!

• The answer to most of your questions is "singularities". – Igor Khavkine Aug 19 '17 at 11:03
• That's quite vague, and not very enlightening. Can you possibly elaborate? – Patch Aug 20 '17 at 23:10
• If you want to be general, a "singularity" is any property of a distribution that distinguishes it from a smooth function. If there were no singularities, every distribution would be represented by integration against a smooth function. I'm sure you can think of your own examples, but the $\delta$-function is traditional one. If you want to think of operators acting on functions/distributions, it's just the same when you think of the distributional integral kernel corresponding to the operator (Schwartz kernel theorem). – Igor Khavkine Aug 21 '17 at 13:05