Keep in touch with English

Regole del forum

Consulta il nostro regolamento e la guida per scrivere le formule
Rispondi al messaggio

[EX] Another proof of the complex C-S inequality

30/10/2012, 22:50

The typical (in the Baire category sense :lol:) Math student should know that the Cauchy-Schwarz inequality, namely:
\[
\tag{CS}
\left| \sum_{n=1}^N z_n\ w_n \right|^2 \leq \left( \sum_{n=1}^N |z_n|^2\right)\ \left( \sum_{n=1}^N |w_n|^2\right)\; ,
\]
holds for any \(N\in \mathbb{N}\) and \(z_1,\ldots ,z_N,w_1,\ldots, w_N\in \mathbb{C}\); moreover, he should also know its "folkloristic proof" (i.e., the one obtained by minimizing the function \(\mathbb{C}\ni \lambda \mapsto \sum_{n=1}^N|z_n-\lambda \overline{w_n}|^2 \in \mathbb{R}\)).

Here we present a lesser known proof of (CS) which relies on a purely algebraic equality.

***

Exercise:

1. Let \(N\in \mathbb{N}\). Prove that equality:
\[
\tag{L} \left| \sum_{n=1}^N z_n w_n\right|^2 = \left( \sum_{n=1}^N |z_n|^2\right)\ \left( \sum_{n=1}^N |w_n|^2\right) - \sum_{n=1}^{N-1} \sum_{m=n+1}^N |z_n\overline{w_m} -z_m\overline{w_n}|^2
\]
holds for any \(z_1,\ldots ,z_N,w_1,\ldots ,w_N\in \mathbb{C}\).

2. Show that (CS) follows from (L).

3. Is it possible to characterize the case of equality in (CS) using (L)?
Rispondi al messaggio


Skuola.net News è una testata giornalistica iscritta al Registro degli Operatori della Comunicazione.
Registrazione: n° 20792 del 23/12/2010.
©2000— Skuola Network s.r.l. Tutti i diritti riservati. — P.I. 10404470014.