Let $P$ be a characteristic polynomial $\sum_{\mid \alpha\mid\leq m}a_\alpha (2\pi i\xi)^\alpha$ corresponding to some differential operator $L=\sum_{\mid\alpha\mid\leq m}a_\alpha\partial^\alpha_x$. Let $f$ be in Schwartz class and let $F$ be a distribution induced by $f$.
I want to know if $\hat{L(f)}=P\hat{f}$ then $\hat{L(F)}=P\hat{F}$ in a distribution sense. First, is it true? While I was studying distribution theory, this ought to be true in order to understand what was going on in the book.
So I was evaluating both distributions in some Schwartz class function $\phi$, but I am not sure how to evaluate $\hat{L(F)}(\phi)$ because of $(-1)^{\mid\alpha\mid}$ factor when we deal with differential operator. So how do I show that $\hat{L(F)}(\phi)=P\hat{F}(\phi)$?
Context: this comes up in observing that when solving PDE, it suffices to find $\hat{F}(\xi)=1/P(\xi)$
Yes. Differential operators are Fourier multipliers whose symbols are the polynomials corrresponding to replacing $D_\alpha$ by $\xi^\alpha$, so it is definitely true that $\widehat{Lf}=P\hat{f}.$ This naturally extends to making sense in distribution due to how operations on distributions are defined. In particular, if we let $M_\alpha f(x)=(ix)^\alpha f(x)$ and $\tilde{M}_\alpha f(x)=(-ix)^{\alpha}f(x),$ then \begin{align*} \langle \widehat{Lf},\varphi\rangle &=\langle Lf,\hat{\varphi}\rangle=\sum a_\alpha\langle \partial_\alpha f, \varphi\rangle=\sum (-1)^{|\alpha|}a_\alpha\langle f, \partial_\alpha\hat{\varphi}\rangle\\ &=\sum (-1)^{|\alpha|}a_\alpha\langle f,\widehat{\tilde{M}_\alpha\varphi}\rangle=\sum (-1)^{|\alpha|}a_\alpha\langle \hat{f},\tilde{M}_\alpha\varphi\rangle\\ &=\sum (-1)^{|\alpha|}a_\alpha\langle \tilde{M}_\alpha\hat{f},\varphi\rangle=\sum\limits a_\alpha\langle M_\alpha \hat{f},\varphi\rangle\\ &=\langle P\hat{f},\varphi\rangle. \end{align*}