1873_solutions

Observe that 2 0 2 f sin x dx 0 0 u

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Unformatted text preview: b  af a 0f 0  Þ0 f  Þ0 a fa f  1 Þ0 f  Þ0 f 1. a b The Case b  f a : The case b  f a is illustrated in the following figure. The case b  f a In this case, the number ab is the area of the yellow region plus the area of the turquoise region to the left of the vertical line at a. If we add on the area of the purple region then we obtain the sum of the two a b integrals Þ f and Þ f 1 and so 0 0 Þ0 f  Þ0 f 1. a ab b Now we say it precisely: It follows from the preceding exercise that ab  af a  a b Þ0 f  Þ0 a fa f  fa 1 Þ b f fa The case b  f a : The case b  f a is illustrated in the following figure. 282 Þ0 f  Þ0 a 1 fa y dy  f 1 Þ b a fa Þ0 f  Þ0 f 1. a b The case b  f a In this case the number ab is the area of the yellow region below the horizontal line at b plus the area of the turquoise region. If we add on the area of the purple region then we obtain the sum of the two a b integrals Þ f and Þ f 1 and so 0 0 Þ0 f  Þ0 f 1. a ab b Now we say it precisely: We write g  f 1 . Since a  g b we can apply the case just considered to the function g to obtain Þ0 g  Þ0 g 1 a ab b and this is exactly the desired result. 3. Given that a and b are positive numbers, that p and q are positive numbers satisfying the equation 1  1  1, q p prove that ap  bq . ab p q This inequality is known as W.H. Young’s inequality. Hint: Apply the preceding exercise to the function f defined by the equation f x  x p 1 for all x  0. The inequality Þ0 f  Þ0 f 1 a ab b gives us Þ 0 x p 1 dx  Þ 0 t 1/ p 1 dt a ab b which yields ab a p  b 11/ p 1 p 1  1/ p 1 p q a b p q 4. If f is an integrable function on an interval a, b then the p-norm f f p  Þ a |f | p b p of f is defined by the equation 1/p . Prove that if f and g are integrable on an interval a, b and if p and q are positive numbers satisfying the equation 1  1  1, p q then the following assertions are true: 283 a. For every number x a, b we have |f x g x | fpgq p |f x | fp 1 p q |g x | gq 1 q . This inequality follows at once from Exercise 3. b. Integrating both sides of the preceding inequality and applying additivity yields the inequality Þ a fg b f p q. g This inequality is known as Hölder’s inequality. We have Þa b |f x g x | dx fpgq Þa b 1 p p |f x | fp Þ a |f x b  1 pf p p  1 pf p p f p p Þa b dx  1 q 1 qg | p dx   1 qg q q g q |g x | gq Þ a |g x b q q q q dx q |dx  1  1 1 p q and so Þ a fg Þ a |fg | b b f p g q 5. This exercise makes use of the concept of a convex function that was introduced in earlier exercises. Suppose that f is integrable on the interval 0, 1 and that  fx  for every number x 0, 1 and that Þ 0 f  I. 1 Prove that if h is a convex function on an open interval that includes the interval ,  then the following assertions hold: a. There is a number k such that the inequality hy y k hy y holds whenever  hI I hI I k y  I and the inequality holds whenever I y . We suppose that h is a convex function on an interva...
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This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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