# $\int_{T^n}|P|=0$ implies $P=0$?

This is Exercise 1 of Chapter 8 in Rudin's Functional Analysis. We are asked to prove the following:

If $P$ is a polynomial in $\mathbb{C}^n$ and if $$\int_{T^n}|P|d\sigma_n=0,$$ then $P$ is identically zero.

Here $T_n$ is the n-dimensional torus and $\sigma_n$ is the Haar measure on $T_n$.

It should not be difficult but I guess I am missing a trick in complex analysis.

Thanks!

If you restrict $P$ as a function $\mathbb{T}^n$, then the Fourier series for $P$ is simply the polynomial itself (up to constant maybe). Therefore, by Plancherel's theorem, we have $$\int_{\mathbb{T}^n} |P| = 0 \Longrightarrow \int_{\mathbb{T}^n} |P|^2 = \sum_{j} |a_j|^2 = 0$$ again with possibly a constant $C > 0$ in front, and the $a_j$'s are the Fourier coefficients of $P$, which are the coefficients of the polynomial! Therefore $a_j = 0$ for all $j$, hence $P \equiv 0$.