Let $u$ be a Schwartz function as defined here https://en.wikipedia.org/wiki/Schwartz_space.
Let $D = (D_1, ..., D_n)$ where $D_j u = - i \frac{\partial u}{\partial x_j}$.
Let $A$ be a real quadratic form s.t. $A(x_1, .., x_n) \geq 0$ for all $x \in \mathbb{R}^n$.
I want to prove that $$ \| e^{-A(D)} (D^{\alpha} u) \|_2 = \| D^{\alpha} (e^{-A(D)} u) \|_2. $$ Here $\alpha = (\alpha_1, ..., \alpha_n)$. This is stated without any explanation in the book I am reading. Any hint on how I can get this would be appreciated. Thank you!
I try giving a proof that does not use the spectral theorem:
Let start by noticing that the operator $A(D): \mathcal{S} \to \mathcal{S}$ is bounded by the closed graph theorem. Indeed the Schwarz space is a Frechet space (see e.g. this)
In particular, given a set of functions $f_n \to 0$ in $\mathcal{S}$ we have $$ A(D)f_n \to 0 $$ in $\mathcal{S}$. So the graph of $A(D)$ is closed. As $A(D)$ is bounded we can define (see this) $$ e^{-A(D)}=\sum_{i=0}^\infty \frac{(-A(D))^n}{n!} $$ as a bounded operator without the need of the spectral theorem
By Parseval theorem $$ \| e^{-A(D)} (D^{\alpha} u) \|_2 =\|\widehat{e^{-A(D)} (D^{\alpha} u)}\|_2=\|e^{-A(\|x\|)}x^\alpha \hat{u}\|_2=\|x^\alpha e^{-A(\|x\|)} \hat{u}\|_2= \| D^{\alpha} (e^{-A(D)} u) \|_2 $$ (probably you don't even need Parseval, you can simply use the fact that $[D^\alpha, D^\beta]=0$)