My Math Forum Cauchy Schwarz Inequality

 Real Analysis Real Analysis Math Forum

 July 8th, 2017, 04:12 AM #1 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100 Cauchy Schwarz Inequality Cauchy Schwarz Inequality in Rn: $\displaystyle |x\cdot y|\leq |x||y|$ Proof: $\displaystyle \left | x\cdot \frac{y}{|y|}\right |\leq |x|$ Explanation: $\displaystyle \frac{y}{|y|}$ is a unit vector which can be expanded to a basis. Then either x is in the direction of y and equality holds, or it has a component among the rest of the basis in which case it's component along y is less than |x|.
 July 9th, 2017, 04:18 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,165 Thanks: 867 What was your purpose in posting this? Do you have a question about it?
September 26th, 2017, 09:30 PM   #3
Senior Member

Joined: Sep 2016
From: USA

Posts: 379
Thanks: 205

Math Focus: Dynamical systems, analytic function theory, numerics
Quote:
 Originally Posted by zylo Cauchy Schwarz Inequality in Rn: $\displaystyle |x\cdot y|\leq |x||y|$ Proof: $\displaystyle \left | x\cdot \frac{y}{|y|}\right |\leq |x|$ Explanation: $\displaystyle \frac{y}{|y|}$ is a unit vector which can be expanded to a basis. Then either x is in the direction of y and equality holds, or it has a component among the rest of the basis in which case it's component along y is less than |x|.
This is completely circular. Specifically "or it has a component among the rest of the basis in which case it's component along y is less than |x|" holds only if the Cauchy-Schwarz inequality is true.

 Tags cauchy, inequality, schwarz

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post zylo Real Analysis 1 June 30th, 2017 06:26 AM mozganutyj Algebra 0 January 12th, 2014 04:45 AM jugger3 Linear Algebra 2 August 21st, 2013 02:00 PM aaron-math Calculus 3 February 5th, 2012 08:50 AM ElMarsh Algebra 8 September 21st, 2009 01:24 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top