Theorem 12.4.13.label Let $L=\sums{}{\abs{\al}\leq k}c_{\al}(x)\pd^{\al}$ be a linear differential operator on $\R^{n}$ then $L$ commutes with rotation and translation if and only if $L$ is a polynomial of $\la$, i.e. $L=\sums{}{}a_{j}(\la)^{j}$ for $a_{j}\in\R$.
Proof. $L$ commute with translations then $c_{\al}$ are constants since $\pd^{\al}$ are invariant under translations. $(\ha{Lu})(\xi)=P(\xi)\ha{u}$ where $P(\xi)=\sums{}{\abs{\al}\leq k }c_{\al}(2\pi i\xi)^{\al}$. Since Fourier Transform commute with rotations, $P(\xi)$ must be radial in $\xi$ so $P$ is a function of $\abs{\xi}$. So $P(\xi)=\sums{}{}c_{,}m(2\pi i\abs{\xi})^{m}=\sums{}{j}c_{j}(2\pi i\abs{\xi})^{2j}=\sums{}{j}c_{j}(2\pi i\abs{\xi}^{2})^{j}=\sums{}{}a_{j}(\la)^{j}$$\square$
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