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 v+ w,

The trick now is to choose the scalars just right. Some clever person - perhaps
Hermann Amandus Schwarz - figured out to choose = ||w||^{2} and
= - <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} .

QED

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