# Proving a complex polynomial is identically $0$.

Let $f:\mathbb{C}^n\to\mathbb{C}$ be a complex polynomial of $n$ complex variables. Let $T^n:=\{(e^{i\theta_1},\dots,e^{i\theta_n}),\theta_j\in\mathbb{R}\}$ be the $n$-dimensional torus and $\sigma_n$ the Lebesgue measure divided by $(2\pi)^n$. Suppose that $$\int_{T^n}|f|\,d\sigma_n = 0.$$ Then $f\equiv0$.

Can we use induction on $n$ along with the identity theorem in 1 dimension to prove this?

I was thinking if $n= 1$ then this holds since $f= 0$ on the unit circle and is entire.

If we assume it holds for $n-1$ and we examine the map $z\mapsto f(e^{i\theta_1},\dots,e^{i\theta_{n-1}},z)$ then this must be $0$ for all $z\in \mathbb{C}$ and for all $(e^{i\theta_1},\dots,e^{i\theta_{n-1}})\in T^n$ since $f=0$ on $T^n$. Therefore for all $z$ $$\int_{T^{n-1}}\big|f(e^{i\theta_1},\dots,e^{i\theta_{n-1}},z)\big|\,d\sigma_{n-1}(\theta_1,\dots,\theta_{n-1}) = 0.$$ Then the induction hypothesis implies that $f(-,z)$is identically $0$ for in the '$-$' variable.

This is all that is needed to show that $f$ is $0$, correct? If it is, do we need the assumption that $f$ is a polynomial, or would it suffice for $f$ to be a holomorphic function on a connected open set containing $T^n$.

• The fact that $f$ is zero on the full integration domain is easy to obtain, the induction step also is fine. I'm just wondering about how exactly we get the claim for $n=1$... (it's been a while since I was buisy on $\mathbb{C}$ the last time) – Max Oct 19 '15 at 22:44
• the fundamental theorem of algebra it is of course – Max Oct 19 '15 at 22:55
• For $n=1$ it follows from the identity theorem – ItsNotMeItsYou Oct 20 '15 at 0:09

## 1 Answer

This looks fine. (And as you say, it's not necessary to assume that $f$ is polynomial. Assuming that $f$ is holomorphic on the polydisc $\mathbb{D}^n$ and continuous on $\bar{\mathbb{D}}^n$ is enough.

Another way to do it: $\mathbb{T}^n$ is the distingushed boundary of $\mathbb{D}^n$ (or the Shilov boundary). Note that $\mathbb{T}^n$ is just a small part of the topological boundary, but perhaps a little surprisingly $$|f(p)| \le \max_{z \in \mathbb{T}^n} |f(z)|$$ for all $f$ that are holomorphic on the polydisc $\mathbb{D}^n$ and continuous on $\bar{\mathbb{D}}^n$ and all $z \in \bar{\mathbb{D}}^n$. In other words, we have a version of the maximum modulus principle using only a very small part of the boundary. This can be proved for example using an induction argument similar to yours.

For your problem, the condition forces $|f|$ to vanish on $\mathbb{T}^n$, and by the "maximum modulus principle" above, $f$ vanishes on $\mathbb{D}^n$, thus everywhere.

Thinking a little harder about this, the reason for this is that the boundary of the polydisc $\mathbb{D}^n$ contains analytic structure (i.e. lower-dimensional complex manifolds). This cannot happen if we start with a strictly pseduoconvex domain such an the unit ball. Hopefully the last few lines give you a little motivation to learn more about several complex variables, which is a very fascinating subject.