Proof of CSB Inequality

Linear Methods of Applied Mathematics

Evans M. Harrell II and James V. Herod*

*(c) Copyright 1994,1995,1996 by Evans M. Harrell II and James V. Herod. All rights reserved.

Theorem II.3. If V is an inner product space, then the CSB inequality (Property 5) and the triangle inequality (Property 6) hold.

proof (don't worry -it won't hurt you!): Because of the positivity property 4, the square length of any vector is >= 0, in particular for any linear combination alpha v+ beta w,

0<= ||alpha v + \beta w||^2 = |alpha|^2 ||v||^2 + |beta|^2 
||w||^2 + 2 Re alpha beta-bar <v,w>

The trick now is to choose the scalars just right. Some clever person - perhaps Hermann Amandus Schwarz - figured out to choose alpha = ||w||2 and beta = - <v,w>. The inequality then boils down to

    0 <= ||w||4 ||v||2 - ||w||2 |<v,w>|2.

If we collect terms and divide through by ||w||2, we get the CSB inequality. (If ||w||=0, the CSB inequality is automatic; otherwise we can divide it out.)

For the triangle inequality, we just calculate:

	||v+w||2 = <v+w,v+w> = <v,v> + <w,w> + 2 Re(<v,w>)
		<= ||v||2 + ||w||2 + 2 ||v|| ||w||
		<= (||v|| + ||w||)2 .


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