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Unformatted text preview: 1 / 2. By the Schwarz inequality, we therefore conclude 1 4 = ± ± ± ± ± Z b a xf ( x ) f ( x )dx ± ± ± ± ± ≤ ² Z b a  f ( x )  2 ! 1 / 2 ² Z b a  xf ( x )  2 dx ! 1 / 2 . (1) The equality holds in this inequality if and only if the integrands are proportional. That means f ( x ) = cxf ( x ) for some c 6 = 0, since f is not identically zero. All solutions to this diﬀerential equation have the form f ( x ) = de cx 2 / 2 , 1 where d is another nonzero constant. For such a function though we cannot have f (0) = f (1) = 0, so equality in (1) is impossible. Thus Z b a ( f ( x )) 2 dx · Z b a x 2 f 2 ( x )dx > 1 / 4 . 2...
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 Spring '09
 Fundamental Theorem Of Calculus, Binary relation, Riemann, Hölder's inequality

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