Hello. I am trying to understand theorem 6.2.2 from the book Analytic Partial Differential Equations by author Treves. My first question at the moment is:
Question 1. What does vanishes identically mean?
Question 2. Why $\mathcal{L}\mathcal{O}'(\mathbb{C}^n)\subset \mathcal{O}(\mathbb{C}^n)$? (I have already solved this question )
Actualization.
I have already solved question 2, because $\partial_{\overline{\xi}}\mathcal{L}\mu(\xi)=0$ (Cauchy-Riemann operator) then $\mathcal{L}\mu(\xi)$ is holomorphic.
This uses that $e^{-z\cdot\zeta}$ is holomorphic to show that $\partial_{\bar\zeta}\mathcal{L}\mu(\zeta)$ is zero everywhere ("vanishes identically") and so $\mathcal{L}\mu$ is holomorphic, and $\mathcal{L}\mathcal{O}'(\mathbb{C}^n)\subset \mathcal{O}(\mathbb{C}^n)$.
The Laplace-Borel transform of $\mu$, $\mathcal{L}\mu(\zeta)=\mu(e^{-z\cdot\zeta})$, so $$ \partial_{\bar\zeta}\mathcal{L}\mu(\zeta) = \langle\;\mu,\partial_{\bar\zeta} e^{-z\cdot\zeta}\;\rangle $$
Since $e^{-z\cdot\zeta}$ is holomorphic, $\partial_{\bar\zeta} e^{-z\cdot\zeta}=0$ everywhere, so this is equivalent to $\langle\mu,0\rangle$, which is zero everywhere ("vanishes identically").
The second "vanishes identically" again means that $\mathcal{L}\mu=0$ everywhere, and if this is the case then the $\alpha$-th derivative with respect to $\zeta$, which is $\mu(z^\alpha)$, must then be zero for any $z$, so $\mu$ must be the zero functional.