12.07.2015 Views

Topics In Linear Algebra and Its Applications - STEM2

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So Bessel’s inequality is an equality if <strong>and</strong> only if x is a linear combinationof elements of S.The Cauchy-Schwarz inequality can be proved from the Bessel inequality.<strong>In</strong>deed, given x <strong>and</strong> y, letx 1 = y‖y‖ .Then the set containing just this vector is an orthonormal set, so we have bythe Bessel inequalityThus‖x‖ 2 ≥ |(x, x 1 )| 2 = |(x, y/‖y‖ 2 = |(x, y)| 2 /‖y‖ 2 .|(x, y)| ≤ ‖x‖‖y‖.13.3 The Triangle <strong>In</strong>equality‖x + y‖ 2 = (x + y, x + y)So= ‖x‖ 2 + (x, y) + (y, x) + ‖y‖ 2= ‖x‖ 2 + 2RE(x, y) + ‖y‖ 2≤ ‖x‖ 2 + 2|(x, y)| + ‖y‖ 2≤ ‖x‖ 2 + 2‖x‖‖y‖ + ‖y‖ 2= (‖x‖ + ‖y‖) 2 .‖x + y‖ ≤ ‖x‖ + ‖y‖√So (x, x) is a norm. Alsod(x, y) = ‖x − y‖is a metric, distance function. <strong>In</strong>deed we have‖x − y‖ = ‖x − z + z − y‖ ≤ ‖x − z‖ + ‖z − y‖.17

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