# Showing $\sum_{j=1}^{n}\left(\frac{\partial}{\partial x_{j}}\right)^{m}$ is not elliptic for odd $m$

Let $$P:=\sum_{j=1}^{n}\left(\frac{\partial}{\partial x_{j}}\right)^{m}$$

Then I want to show that $P$ is an elliptic operator on $\mathbb{R}^{n}$ when $m$ is even, but not if $m$ is odd.

Definition: $P$ is elliptic if the top order symbol has no real zeros except $\xi=0$, or equivalently if the full symbol satisfies $|p(\xi)|\ge c|\xi|^{m}$ for $|\xi|\ge A$ for $c,A$ positive constants.

We see that

$$(Pu)^{\widehat{}}(\xi)=\left(-i\sum_{j=1}^{n}\xi_{j}\right)^{m}\hat{u}(\xi),$$ so that $$p(\xi)=\left(-i\sum_{j=1}^{n}\xi_{j}\right)^{m}=\begin{cases}\sum_{j=1}^{n}\xi^{m}_{j}&\mbox{if }m\text{ is even},\\ -i\sum_{j=1}^{n}\xi_{j}^{m}&\mbox{if }m\text{ is odd}.\end{cases}$$

However, $|p(\xi)|$ is the same for $m$ even and odd, and we get that

$$|p(\xi)|=\sum_{j=1}^{n}|\xi_{j}|^{m}\ge |\xi|^{m}$$

which means that $P$ is elliptic; but this isn't what I wanted. Where did I go wrong?

Indeed, the factor of $-i$ makes no difference: it's all about $\sum \xi_j^m$ being zero or not. Your proof goes wrong when you "distribute" the absolute value over a sum.
When $m$ is even, each $\xi_j^m$ is nonnegative, so the only way for their sum to be $0$ is if every term is $0$.
When $m$ is odd, the above no longer applies, and counterexamples such as $\xi=(1,-1,0,\dots,0)$ present themselves.