# Inequality for multi-dimensional holomorphic functions

Let $B$ be the unit ball in $\mathbb{R}^n$ and define $$\widetilde{B} = \{\lambda x : x\in B, \lambda \in \mathbb{C}, \lvert \lambda \rvert \leq 1\}.$$ Now, suppose that $u$ is holomorphic in some neighbourhood of $\widetilde{B}$ such that $\sup\limits_{\widetilde{B}} \lvert u(z)\rvert \leq 1$.

For the case $n=1$, I was able to show (for any $N\in\mathbb{N}$) that: $$\left\lvert u(z) - \sum_0^{2N-1}\frac{u^{(k)}(0)z^k}{k!} \right\rvert \leq (2N+1)\lvert{z}\rvert^{2N}$$ and $$\left\lvert \frac{u^{(k)}(0)z^k}{k!} \right\rvert \leq \lvert{z}\rvert^{k}.$$

According to the notes I am reading, "apply the preceding inequalities to all complex lines passing through $0$, and we obtain for all $z\in\widetilde{B}$": $$\left\lvert u(z) - \sum_{\lvert\alpha\rvert \leq 2N-1}\frac{u^{(\alpha)}(0)z^\alpha}{\alpha!} \right\rvert \leq (2N+1)\lvert{z}\rvert^{2N}$$ and $$\left\lvert\sum_{\lvert\alpha\rvert=k}\frac{u^{(\alpha)}(0)z^\alpha}{\alpha!}\right\rvert \leq \lvert{z}\rvert^{k}.$$

I do not see how these inequalities are obtained. I am relatively new to multi-dimensional holomorphic functions and would appreciate if anyone could shed some light on how to do this.

• Those inequalities can't hold unless we have more hypotheses on $u.$ Is $u$ bounded in some way? – zhw. Mar 18 '18 at 3:02
• Yes sorry, I edited my question. We also have the assumption $\sup_{\tilde{B}}\lvert u(z)\rvert \leq 1$ – Quoka Mar 18 '18 at 3:08

$\def\d{\mathrm{d}}\def\peq{\mathrel{\phantom{=}}{}}$For any $x = (x_1, \cdots, x_n) \in \partial B$, define $f_x(λ) = u(λx)$ for $λ \in \mathbb{C}$, $|λ| < 1$. It will be proved by induction that$$f_x^{(m)}(λ) = \sum_{|α| = m} \frac{m!}{α!} u^{(α)}(λx) x^α. \tag{1}$$ The base case, i.e. $m = 0$, is trivial. Suppose that (1) holds for $m$. Note that the mapping $λ \mapsto λx$ is holomorphic, then by the chain rule,\begin{align*} f_x^{(m + 1)}(λ) &= \sum_{|α| = m} \frac{m!}{α!} x^α \frac{\d}{\d λ}(u^{(α)}(λx)) = \sum_{|α| = m} \frac{m!}{α!} x^α \sum_{|β| = 1} u^{(α + β)}(λx) x^β\\ &= \sum_{\substack{|α| = m\\|β| = 1}} \frac{m!}{α!} u^{(α + β)}(λx) x^{α + β} = \sum_{|α| = m + 1} u^{(α)}(λx) x^α \sum_{\substack{|β| = 1\\β \preccurlyeq α}} \frac{m!}{(α - β)!}\\ &= \sum_{|α| = m + 1} u^{(α)}(λx) x^α \sum_{α_k \geqslant 1} \frac{m!}{(α_k - 1)! \prod\limits_{j \neq k} α_j!}\\ &= \sum_{|α| = m + 1} u^{(α)}(λx) x^α · \frac{m! \sum\limits_{k = 1}^n α_k}{α!} = \sum_{|α| = m + 1} u^{(α)}(λx) x^α · \frac{m! · |α|}{α!}\\ &= \sum_{|α| = m + 1} \frac{(m + 1)!}{α!} u^{(α)}(λx) x^α. \end{align*} End of induction.
Note that$$\widetilde{B} = \{λx \mid x \in B,\ λ \in \mathbb{C},\ |λ| \leqslant 1\} = \{λx \mid x \in \partial B,\ λ \in \mathbb{C},\ |λ| < 1\}.$$ Now for any $z \in \widetilde{B}$, suppose $z = λx$, where $λ \in \mathbb{C}$, $x \in \partial B$. Because $z^α = λ^{|α|} x^α$ and $|x| = 1$, then\begin{align*} &\peq \left| u(z) - \sum_{|α| \leqslant 2m - 1} \frac{z^σ}{α!} u^{(α)}(0) \right| = \left| u(z) - \sum_{k = 0}^{2m - 1} \sum_{|α| = k} \frac{z^α}{α!} u^{(α)}(0) \right|\\ &= \left| f_x(λ) - \sum_{k = 0}^{2m - 1} \frac{λ^k}{k!} f_x^{(k)}(0) \right| \leqslant (2m + 1) |λ|^{2m} = (2m + 1) |z|^{2m}, \end{align*} and$$\left| \sum_{|α| = k} \frac{z^α}{α!} u^{(α)}(0) \right| = \left| \frac{λ^k}{k!} f_x^{(k)}(0) \right| \leqslant |λ|^k = |z|^k.$$
• Thanks for the answer! One question: how can I guarantee that $u$ has a power power series representation? I know this can be done if $u$ is holomorphic in the open poly-disk, but I'm not sure what to do in this case – Quoka Mar 24 '18 at 18:05