# Sharpest version of semiclassical Calderon-Vaillancourt theorem

Let $$S$$ be the space of symbols defined by $$S:=\{a\in C^{\infty}(T^*\mathbb{R}^d):\forall \alpha,\beta\in\mathbb{Z}^d,\, |\partial_x^{\alpha}\partial_{\xi}^{\beta}a(x,\xi)|\le C_{\alpha\beta}\},$$ and recall that the Weyl quantization of a symbol is given by $$(\mathrm{Op}_h(a)u)(x):=\frac{1}{(2\pi h)^d}\iint e^{\frac{i}{h}\langle x-y,\xi\rangle}a\left(\frac{x+y}{2},\xi\right)u(y)\,dy\,d\xi.$$ The Calderon-Vaillancourt theorem says that $$\mathrm{Op}_h(a)$$ is bounded on $$L^2(\mathbb{R}^d)$$ uniformly in $$h$$, with bound given by $$\|\mathrm{Op}_h(a)\|_{L^2\to L^2}\le C\sum_{|\alpha|\le Kd}h^{\frac{|\alpha|}{2}}\|\partial^{\alpha}a\|_{L^{\infty}}.$$ In particular, this implies a uniform bound even when $$a=a_h$$ is $$h$$-dependent, as long as it is in the class $$S_{\delta}:=\{a\in C^{\infty}(T^*\mathbb{R}^d):\forall \alpha,\beta\in\mathbb{Z}^d,\, |\partial_x^{\alpha}\partial_{\xi}^{\beta}a(x,\xi)|\le C_{\alpha\beta}h^{-\delta|\alpha+\beta|}\}$$ with $$\delta\le\frac{1}{2}$$. Specifically, if $$a\in S_{\delta}$$, then $$\|\mathrm{Op}_h(a)\|_{L^2\to L^2}\le\|a\|_{L^{\infty}}+O(h^{½-\delta}).$$

All of the above is standard, and spelled out in the books by Zworski, Martinez, and Dimassi–Sjöstrand. But in Chapter 13 of Zworski, we learn that it is not the full story. In fact (Theorem 13.13 in Zworski), it holds that for $$a\in S$$, $$\|\mathrm{Op}_h(a)\|_{L^2\to L^2}\le \|a\|_{L^{\infty}}+O(h).$$ The proof is fairly complicated, using the FBI transform. My questions are of the derivative dependence for this result. What powers of $$h$$ accompany which derivatives of $$a$$? In particular, can the error bound for $$S_{\delta}$$ symbols be strengthened to something better than $$O(h^{½-\delta})$$? And whatever the optimal bound is, has it ever been shown to be sharp?

It is a (relatively) simple computation to prove that given a symbol a, one can find another, b, so that $$Op^{weyl}(a) = Op^{toeplitz}(b) + O(h^\infty)$$. This is proved for example in Zworski's book (theorem 13.10). The main point being that $$b = a - h \Delta a + O(h^2)$$, where $$\Delta$$ is a suitably normalized Laplacian, and the $$O(h^2)$$ term contains derivatives of order at least 4 of a.
In particular, the formula $$\| Op^{weyl}(a)\| = \|a\|_{L^\infty} + \mathcal{O}(h)$$ you are referring to only removes odd derivatives from the Calderon Vaillancourt estimate, but does not change qualitatively the result (two derivatives = one power of $$h$$).
Following the proof in Zworski's book, will provide you with the estimate in $$S_\delta$$: $$\|Op^{weyl}(a)\|_{L^2} = \|a \|_{L^\infty} + \mathcal{O}(h^{1- 2\delta}).$$
edit: Actually, we can use the Fefferman-Phong inequality: we have $$Op^{weyl}(a \pm |a|_{L^\infty}) \geq - Ch^2,$$ as a quadratic form. In particular, we get the improved bound $$\|Op^{weyl}(a)\|_{L^2} \leq \| a \|_{L^\infty} + O(h^2).$$ To pass from this to the exotic statement, just rescale the symbols, and obtain $$\| Op^{weyl}(a) \|_{L^2} \leq \| a\|_{L^\infty} + O( h^{2-4\delta} ).$$
As to the question of sharpness. I only know of examples proving that not all symbols in $$S^{1,1}$$ give rise to $$L^2$$ bounded operators. See the works of Ching, Bourdaud. Maybe one can cook a semi-classical example from this, but I do not have a reference.