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If $P$ and $P'$ are linear partial differential operators with constant complex coefficients on $U = \mathring U \subseteq \Bbb R^m$, we say that $P \sim P'$ if and only if $\dfrac {\tilde P} {\tilde {P '}}$ and $\dfrac {\tilde {P '}} {\tilde P}$ are bounded on $\Bbb R^m$, where $P = \sum_\alpha a_\alpha D^\alpha$, $D^\alpha = (-\Bbb i)^{|\alpha|} \partial _\alpha$, $\tilde P = \sqrt{ \sum _\alpha |P^\alpha|^2 }$ and $P^\alpha (v) = \partial ^\alpha P (v)$ (as usual, $\alpha$ is a multi-index). We say that $P$ and $P'$ "are of equal strength".

If $P \sim P'$ on $U$ and $Q \sim Q'$ on $V$, then is it true that $(P+Q) \sim (P' + Q')$ on $U \times V$? If this is not true in general, are there (necessary and) sufficient conditions under which it is?

By $P + Q$ I understand $P \otimes \text{id} + \text{id} \otimes Q$.

I am investigating the hypoellipticity of an operator and if the above were true, then it would allow me to split my problem into smaller, more tractable blocks.

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  • $\begingroup$ Could you please more explain on $P^{\alpha}$ and its norm?What does $\partial^{\alpha}P$ mean?Is it $P$ a differential operator itself? Could you explain your notations by a precise operator, say Laplacian on the plane and $\alpha=(2,0)$?How it act on a given $v\in \mathbb{r}^2$ ? $\endgroup$
    – Ali Taghavi
    Commented Jun 18, 2018 at 13:43
  • $\begingroup$ @AliTaghavi: $\alpha$ is a multi-index, i.e. $\alpha = (\alpha_1, \dots, \alpha_m) \in \mathbb N^m$. Then $\partial ^{\alpha} = \partial _{x_1} ^{\alpha_1} \dots \partial _{x_m} ^{\alpha_m}$. $\endgroup$
    – Alex M.
    Commented Jun 18, 2018 at 13:49
  • $\begingroup$ Yes I know but what does it mean $\partial^{\alpha}P$ while P is itself a differential operator. For example what is ${|P^{\alpha}|}^2$ where $P$ is Laplacian and $\alpha=(2,0)$. I really do not understand your question. $\endgroup$
    – Ali Taghavi
    Commented Jun 18, 2018 at 13:57
  • $\begingroup$ I am using the notations of Hörmander: If $P$ is a polynomial, then $P(D)$ is a differential operator, but $P(v)$ is its symbol. Therefore, $\partial^\alpha P(v)$ means $\partial _{v_1} ^{\alpha_1} \dots \partial _{v_m} ^{\alpha_m} (P(v_1, \dots, v_m))$. $\endgroup$
    – Alex M.
    Commented Jun 18, 2018 at 14:04

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