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I am trying to prove the following theorem:

Let $f:\mathbb{C}^{n}\to \mathbb{C}$ be a holomorphic function, and let, for $z\in \mathbb{C}^{n}$, there exists $A$ and $k$ constants, such that, $$ |f(z)|\leq A\|z\|_{2}^{k}. $$ Then, we have that $f$ is a polynomial of degree lesser or equal than $k$.

I don't know how to prove this theorem... I suppose I can use the Liouville Theorem in one variable taking on account that a holomorphic function $f:\mathbb{C}^{n}\to \mathbb{C}$ can be characterized as being holomorphic in each variable... Nevertheless, I have no more clues on this...

I would appreciate, or some bibliography, or some hints, on how to solve this problem. Thanks in advanced!

UPDATE: As suggested in the comments, I am trying to use an inductive argument: I suppose case $n=1$ is already proven. Then, take the $n$-th case. By Taylor (in several complex variables) $f(z)=\sum_{\alpha\in \mathbb{N}^{n}}\frac{1}{\alpha !}(D^{\alpha}f)(0,\dots,0)z^{\alpha}$ for all $z\in\mathbb{C}^{n}$. We fix $z_{n}=z_{0}\in\mathbb{C}$. Now, define $f_{1}:\mathbb{C}^{n-1}\to \mathbb{C}$ as follows: $$ \begin{split} f_{1}(z_{1},\dots,z_{n-1})& \triangleq f(z_{1},\dots,z_{n-1},z_{0}) \\ & =\sum_{\alpha\in \mathbb{N}^{n-1}}\frac{1}{\alpha !}\left(\sum_{j\in\mathbb{N}} \frac{1}{j!}(D^{(\alpha,j)}f)(0,\dots,0)z_{0}^{j}\right)\prod_{i=1}^{n-1}z_{i}^{\alpha_{i}} \\ & =\sum_{\alpha\in \mathbb{N}^{n-1}}c_{\alpha}\prod_{i=1}^{n-1}z_{i}^{\alpha_{i}}. \end{split} $$ Now, I would have to show by the inductive hypothesis that $f_{1}$ is a multivariable polynomial in $n-1$ variables. My problem now is:

To use the inductive hypothesis, I should have (with $A'$ positive constant not neccesarily equal to previous $A$) $$ |f_{1}(z_{1},\dots,z_{n-1})|\leq A'\|(z_{1},\dots,z_{n-1})\|_{2}^{k}.$$ I should deduce this from the fact that: $$|f_{1}(z_{1},\dots,z_{n-1})|\leq A'\|(z_{1},\dots,z_{n-1},z_{0})\|_{2}^{k}.$$ And in this previous argument where I am lost!

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    $\begingroup$ what about induction? eg $f(z,w)=\sum a_{n,m}z^nw^m$ and then fix $w$ and then $\sum_m a_{n,m}w^m=0$ for all $n >k$ by the usual Liouville and hence $a_{n,m}=0, n > k, m$ arbitrary and so $f(z,w)=\sum_{n \le k, m}a_{n,m}z^nw^m$ and now fix $z$ and apply Liouville in $w$ and repeat to get $a_{n,m}=0, m >k$ and then since $f$ is now already a polynomial, it's easy to show that actually $a_{n,m}=0, n+m >k$ so you are done; then the inductive step follows clearly to go to $3$ or higher variables $\endgroup$ Commented Jan 19, 2023 at 0:06
  • $\begingroup$ @Conrad Thanks for answering! I am trying your way, by induction. My only doubt is: why can I apply Liouville and get $ \sum_{m}a_{n,m}w^{m}$? I mean, I would requiere that, with fixed $w_{0}$, $|f(z,w_{0})|\leq A'|z|^{k}$ (for $A'$, not neccesarily equal to $A$), and I do not see why from $|f(z,w_{0})|\leq A(|z|^{2}+|w_{0}|^{2})^{k/2}$ I can get to that... $\endgroup$ Commented Jan 21, 2023 at 21:46
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    $\begingroup$ For large $z$ say greater than $2|w_0|$ that is obvious and for lower $z$ use continuity to get a bound $\endgroup$ Commented Jan 22, 2023 at 0:40

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Integral formulas are your friends when you want to deal with estimates. This is because for many purposes, integrals behave better than derivatives. Let us recall how we can prove such a statement in single-variable complex analysis:

Suppose $f$ is entire and satisfies the bound $|f(z)|\leq A|z|^k$ (by modifying the proof below, you can consider bounds like $|f(z)|\leq \phi(|z|)$, and depending on the properties of the radial function $\phi$, deduce corresponding properties of $f$). We have a Taylor expansion $f(z)=\sum\limits_{n=0}^{\infty}c_nz^n$ on $\Bbb{C}$. Now, for each $n\geq 0$, and each $R>0$ we have that \begin{align} c_n&=\frac{1}{2\pi i}\int_{|z|=R}\frac{f(z)}{z^{n+1}}\,dz. \end{align} This immediately implies the bound \begin{align} |c_n|\leq \frac{1}{2\pi}\cdot\frac{AR^k}{R^{n+1}}\cdot 2\pi R=\frac{A}{R^{n-k}}. \end{align} As soon as $n-k>0$, we can take the limit $R\to\infty$ to find that $c_n=0$. Thus, we see that $c_{k+1}=c_{k+2}=\dots =0$, meaning that $f$ is a polynomial of degree at most $k$.

I hope you see how to adapt this to $n$-dimensions: write a Taylor expansion $f(z)=\sum_{\alpha}c_{\alpha}z^{\alpha}$, and write \begin{align} c_{\alpha}&=\frac{1}{(2\pi i)^n}\int_{|z_1|=\dots=|z_n|=R}\frac{f(z)}{z^{\alpha+1}}\,dz_1\dots\,dz_n, \end{align} where $z^{\alpha+1}$ is the shorthand multindex notation for $(z_1)^{\alpha_1+1}(z_2)^{\alpha_2+1}\cdots(z_n)^{\alpha_n+1}$. Now, bounding this will give you (letting $|\alpha|=\alpha_1+\dots+\alpha_n$) \begin{align} |c_{\alpha}|\leq \frac{1}{(2\pi)^n}\cdot\frac{AR^k}{R^{|\alpha|+n}}\cdot(2\pi R)^n=\frac{A}{R^{|\alpha|-k}}. \end{align} So, as soon as the indices are large enough: $|\alpha|-k>0$, we can take the limit $R\to\infty$ to conclude $c_{\alpha}=0$. Hence, $f$ is again a polynomial of degree at most $k$.

In fact, if you only assumed that $f$ was holomorphic on a region like $\{|z|>r_0\}$, then you could write a Laurent expansion centered at the origin, and the proof would show that if $f$ satisfied such a bound, then its Laurent expansion only contains terms up to power $k$ (by working in the limit $R\to 0^+$, you can prove that if you have bounds on $f$ near the origin, then you can place restrictions on the negative part of the Laurent coefficients).

So, purely from the integral representation formula of holomorphic functions, we are able to deduce how the Laurent-coefficients behave as soon as we know bounds on $f$ in the appropriate region (infinity or near the origin).

Going further, harmonic functions have the mean-value property, which is a form of an integral representation. Try to see if you can generalize these arguments for harmonic functions too (very similar statements hold). Also, the fact that $f$ is $\Bbb{C}$-valued is irrelevant here; you can replace it with any complex Banach space $V$ (assume finite-dimensional for comfort if you like, and now the coefficients $c_{\alpha}$ will be elements of $V$).

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    $\begingroup$ Radiating intelligence $\endgroup$ Commented Jan 22, 2023 at 0:26
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    $\begingroup$ @TrystwithFreedom this is not intelligence; these are well-known classical facts: “integrals are nice” and just doing the basic single-variable argument $n$-times to get the $n$-dimensional argument (if we use the complex-analysis route rather than the elliptic PDE route). $\endgroup$ Commented Jan 22, 2023 at 0:29

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