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I'm learning several complex variables. There is a corollary of Cauchy's integral formula that I don't know how to prove.

Let $X\subset\mathbb{C}^n$ be a domain. For each multi-index $\nu$, for each polydisk $P=P^n(a;\rho)\subset\subset X$ with distinguished boundary $T$, and for each holomorphic function $f$ on $X$, fix $w\in P$ and $m \in\mathbb{N}$; if, for every $z\in P$, $X$ includes the closed complex segment $\{(1-\lambda)w+\lambda z;\lambda\in\overline{P^1(1)}\}$, then $$\sup_{z\in P}\left|\sum\limits_{|\nu|=m}\frac{1}{\nu!}D^{\nu}f(w)(z-w)^{\nu}\right|\le \|f\|_X.$$ I have tried to solve this problem by induction,but I failed to solve it even by assuming $m=1$. I don't know how to use the condition "$X$ includes the closed complex segment". Can you give me an idea? Thanks in advance!

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For $z\in P$, consider the parametrisation $g_z \colon \mathbb{C} \to \mathbb{C}^n;\: g_z(\lambda) = w + \lambda(z-w)$ of a complex line in $\mathbb{C}^n$. The premise says that $g_z(\overline{\mathbb{D}}) \subset X$, hence $h = f \circ g_z$ is a holomorphic function in a neighbourhood of the closed unit disk in $\mathbb{C}$. Now the Cauchy integral formula for the $m$-th derivative of $h$ gives you

$$h^{(m)}(0) = \frac{m!}{2\pi i} \int_{\partial \mathbb{D}} \frac{h(\zeta)}{\zeta^{m+1}}\,d\zeta.$$

The standard estimate then yields

$$\left\lvert h^{(m)}(0)\right\rvert \leqslant m! \cdot \sup_{\zeta \in \partial \mathbb{D}} \lvert h(\zeta)\rvert.$$

Evidently we have $\sup\limits_{\zeta\in\partial\mathbb{D}} \lvert h(\zeta)\rvert \leqslant \lVert f\rVert_X$, and all that remains is to use the chain rule to obtain

$$h^{(m)}(0) = \sum_{\lvert \nu\rvert = m} \binom{m}{\nu} D^\nu f(w)(z-w)^\nu.$$

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  • $\begingroup$ I'm jealous of you.I have seen you solved many problems perfectly in this website.You do the math just like playing games!I dream of being somebody like you someday!Thank you very much! $\endgroup$ – user108005 Nov 12 '13 at 4:09

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