# Symbol of a partial differential operator.

We have $$P(fu)=\sum_{|\alpha|\le m}a_{\alpha}(x)\sum_{\beta+\gamma=\alpha}\frac{\alpha!}{\beta!\gamma!}\partial^{\beta}f\partial^{\gamma}u$$ The author claimed it is `obvious' we can put it into the form $$P(fu)=\sum_{\alpha\ge 0}\frac{\partial^{\alpha}f}{\alpha!}P^{\alpha}(u)=\sum_{\alpha\ge 0}\frac{\partial^{\alpha}u}{\alpha!}P^{\alpha}(f)$$

In here $P^{\alpha}$'s symbol is $\partial^{\alpha}_{\zeta} P(x,\chi)$.

I am wondering why this would work because in the most trivial case $P=a(x)\partial^{\alpha}$, we have $P(x,\zeta)=a(x)\zeta^{\alpha}$. Then $P^{\alpha}$'s symbol is $$\partial^{\alpha}_{\zeta} (a(x)\zeta^{\alpha})=\alpha!a(x)$$ In other words $P^{\alpha}(\phi)$ is just multiplication by $\alpha!a(x)$ to $\phi$ now. And applying this to $fu$above the result obviously does not match. So something elementary must involved or I made a computational mistake somewhere.

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