Creating a bounded operator from an unbounded symbol ? - discussion around the Calderon-Vaillancourt theorem

Question

Calderon-Vaillancourt theorem states that any quantization of a bounded symbol (in the sense of this article for example) is a bounded operator on the $L^2(\mathbb{R^n})$ space.

I was wondering if there was any way of going back from a $L^2(\mathbb{R^n})-$bounded pseudo-differential operator to one of its symbols. More precisely, I wondered if it is true that, for such operator $P$, there exists a bounded symbol $p$ such that $Op(p)=P$.

I've heard this is a hard task to go from operators to symbols in the general case. Beals theorem, which is a theorem proving a given operator is pseudo-differential, answers my question for some specific cases (hypothesis on commutators appear) but not in the general case.

Thus, I kind of assumed the answer to that question is : no.

So, my question here is :

Is it possible to find a counter-example, that is to say an unbounded symbol $p$ (a smooth function that has at least one of its derivatives not in $L^\infty(\mathbb{R^n})$) which quantization $P = Op(p)$ is bounded on $L^2(\mathbb{R^n})$ ?

Thanks in advance to anyone interested in this question.

Answer 1

Two thoughts:

1. You can go from an operator to its symbol using the notion of oscillatory testing for appropriate symbol classes. See, for example, Theorem 4.19 in Zworski's text.
2. I'm not sure exactly what you mean when you say a derivative of the symbol is not in $L^\infty(\mathbb{R}^n)$ since symbols are defined by $\mathbb{R}^{2n}.$ Let's assume that you meant $\mathbb{R}^{2n}$. Then, the answer is yes. In fact, the Calderon-Vaillancourt is even stronger than the theorem stated in your link; it holds for symbols in $S^{0}_{\rho,\rho}(\mathbb{R}^{2n})$ provided that $\rho\in [0,1).$ So, if $p\in S^{0}_{\rho,\rho}(\mathbb{R}^{2n})$ and $\rho\in [0,1),$ then $Op(p)$ is bounded on $L^2(\mathbb{R}^n)$, but one is only guaranteed that $$|D^\alpha_ x p(x,\xi)|\leq C_\alpha \langle\xi\rangle^{\rho|\alpha|}.$$ So, the $x$ derivatives do not need to be bounded.