# Estimating a finite sum of complex numbers (Rudin Lemma 6.3).

In Rudin's book, Real and Complex analysis, Lemma 6.3. states:

If $z_1,...,z_N$ are complex numbers then, there is a subset $S$ of $\{1,...,N\}$ for which $$\left|\sum_{k\in S} z_k\right|\ge \frac{1}{\pi}\sum _{k=1}^N |z_k|.$$

In the proof he claims the following: Let $\theta_0$ be the value for which $$\sum _{k=1}^N |z_k|\cos^{+}(\alpha_k-\theta),$$

attains it's maximum. Therefore

$$\sum _{k=1}^N |z_k|\cos^{+}(\alpha_k-\theta_0)\ge \frac{1}{2\pi}\int_{-\pi}^\pi\sum _{k=1}^N |z_k|\cos^{+}(\alpha_k-\eta)d\eta=\frac{1}{\pi}\sum _{k=1}^N |z_k|,\tag{1}$$

i.e. the maximum is bounded below by the avergae of the sum over $[-\pi,\pi]$.

Could someone help me to understand why this is true. I can see, for example, that there is a constant $0<C<1$ such that inequality $(1)$ is true for $C$ instead of $\pi$, however, I fail to see why his claim is true.

Moreover, is $1/\pi$ the best constant?

• You have a continuous function $f\colon [-\pi,\pi] \to \mathbb{R}$. Then the maximum of $f$ is not smaller than the average, $$\max \{ f(\theta) : -\pi \leqslant \theta \leqslant \pi\} \geqslant \frac{1}{2\pi}\int_{-\pi}^\pi f(\eta)\,d\eta.$$ In this case, $$f(\theta) = \sum_{k=1}^N \lvert z_k\rvert \cos^+(\alpha_k-\theta),$$ and the average of $\cos^+(\alpha-\theta)$ is $\frac{1}{\pi}$ for all $\alpha$. $1/\pi$ is, iirc, sharp. – Daniel Fischer Jun 11 '14 at 20:55
• @DanielFischer, sometimes I feel so stupid. I was considering for each $k$, the function $|z_k|\cos^{+}(\alpha_k-\theta)$ and trying to apply your idea. Anyway, what is iirc? – Tomás Jun 11 '14 at 21:00
• "If I recall/remember correctly" – Daniel Fischer Jun 11 '14 at 21:02

Yes, $1/\pi$ cannot be replaced by a larger constant. Observe that $$\sup_S\left|\sum_{k\in S } z_k\right| = \sup_{\theta \in [0,2\pi]} \sum_{k=1}^N |z_k|\cos^+ (\alpha_k-\theta) \tag1$$ (This relation is also used in the proof of the inequality). The point being that maximizing the modulus of some quantity $\Sigma$ amounts to maximizing $\operatorname{Re}(e^{-i\theta}\Sigma)$ over $\theta$. And the maximum of the latter is obtained simply by taking all $z_k$ which contribute positively.

Let $z_k$ be uniformly distributed on the unit circle, so the gap between them is $2\pi/N$. Then
$$\frac{2\pi}{N}\sum_{k=1}^N |z_k|\cos^+ (\alpha_k-\theta) \approx\int_{ 0 }^{2\pi } \cos^+ (t-\theta) \,dt = 2$$ no matter what $\theta$ is. (We have convergence of Riemann sum to the integral as $N\to\infty$). Hence , $$\frac{1}{N} \sum_{k=1}^N |z_k|\cos^+ (\alpha_k-\theta) \approx \frac{1}{\pi}$$

For completeness I copy Daniel Fischer's comment:

You have a continuous function $f\colon [-\pi,\pi] \to \mathbb{R}$. Then the maximum of $f$ is not smaller than the average, $$\max \{ f(\theta) : -\pi \leqslant \theta \leqslant \pi\} \geqslant \frac{1}{2\pi}\int_{-\pi}^\pi f(\eta)\,d\eta.$$ In this case, $$f(\theta) = \sum_{k=1}^N \lvert z_k\rvert \cos^+(\alpha_k-\theta),$$ and the average of $\cos^+(\alpha-\theta)$ is $\frac{1}{\pi}$ for all $\alpha$.