Prove $\int_0^1 | \Sigma_{k=1}^N a_k e^{2\pi i k \theta}|d \theta \geq 1$

Question

Suppose $a_1,...a_N$ are complex numbers with $|a_i|=1$ for $i=1,....N$. Prove that:

$\int_0^1 | \Sigma_{k=1}^N a_k e^{2\pi i k \theta}|d \theta \geq 1$

My attempt:

I know that $\int_0^1 e^{2 \pi i k \theta} d \theta = 1$ if $k=0$ and equals $0$ if $k \neq 0$ otherwise.

I'm trying to use the fact that if $f \in L^1 \cap L^{\infty}$ then $f \in L^2$ and $||f||_2^2 \leq ||f||_1||f||_{\infty}$

I'm stuck in tunnel vision.. Could use a new perspective!

Answer 1

I think you can also show this by the triangle inequality for integrals. First, I claim that

$$ | \sum_{k=1}^N a_k e^{2\pi i k \theta}|=| \sum_{k=1}^N a_k e^{2\pi i (k-1) \theta}|\cdot \vert e^{2\pi i\theta} \vert= | \sum_{k=1}^N a_k e^{2\pi i (k-1) \theta}| .$$

Thus our integral equals

$$ \int_0^1 | \sum_{k=1}^N a_k e^{2\pi i(k-1)\theta}|d \theta \geq \Big\vert \int_0^1 \sum_{k=1}^N a_k e^{2\pi i (k-1) \theta}d \theta \Big\vert =|a_1|=1. $$

I'm not completely sure about this argument, since it seems to work whenever one of the $a_i$-s have absolute value one.

Answer 2

Your idea is pretty good. Note that

\begin{align} \int_0^1 \left|\sum_{k=1}^N a_k e^{2\pi i k\theta} \right|^2\textrm{d}\theta &=\int_0^1 \left(\sum_{k=1}^N a_k e^{2\pi i k\theta}\right)\left(\sum_{j=1}^N \overline{a_j} e^{-2\pi i j\theta}\right)\textrm{d}\theta\\ &= \sum_{j,k=1}^N \int_0^1 a_k \overline{a_j} e^{i(k-j)\theta}\textrm{d}\theta \\ &= \sum_{k=1}^N |a_k|^2=N \end{align} By your observation. Furthermore, by your inequality, we have

$$ N=\int_0^1 \left|\sum_{k=1}^N a_k e^{2\pi i k\theta} \right|^2\textrm{d}\theta \leq \left\| \sum_{k=1}^N a_k e^{2\pi i k\theta} \right\|_{\infty} \int_0^1 \left|\sum_{k=1}^N a_k e^{2\pi i k\theta} \right|\textrm{d}\theta $$

Since $\left\| \sum_{k=1}^N a_k e^{2\pi i k\theta} \right\|_{\infty}\leq \sum_{k=1}^N |a_k|=N,$ you get your result.