Copy of s12 - -1 / 2. By the Schwarz inequality, we...

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Solutions to homework # 12. 1. Let f = 1 at rational points and - 1 at irrational points. Then f is real and bounded, and f 2 = 1 is Riemann integrable. But f itself is not integrable. Now suppose f 3 is Riemann integrable. Consider the function φ ( t ) := t 1 / 3 . This function is defined and coninuous over all of IR. By Theorem 6.11, the function f = φ ( f 3 ) is therefore integrable. 2. Suppose f , g ∈ R ( α ). First let us prove that k f + g k 2 ≤ k f k 2 + k g k 2 . Using the Schwarz inequality , we have k f + g k 2 2 = Z b a | f + g | 2 = Z b a | f | 2 + 2Re Z b a f g dα + Z b a | g | 2 Z b a | f | 2 + 2 ± ± ± ± ± Z b a f g dα ± ± ± ± ± + Z b a | g | 2 Z b a | f | 2 + 2 ² Z b a | f | 2 ! 1 / 2 ² Z b a | g | 2 ! 1 / 2 + Z b a | g | 2 = ² ( Z b a | f | 2 ) 1 / 2 + ( Z b a | g | 2 ) 1 / 2 ! 2 = ( k f k 2 + k g k 2 ) 2 . So, k f + g k 2 ≤ k f k 2 + k g k 2 . Now replace f by f - g and g by g - h where h is an arbitrary function in R ( α ), to get k f - h k 2 ≤ k f - g k 2 + k g - h k 2 . 3. Let F ( x ) := xf 2 ( x ). Since f is continuously differentiable, so is F . By the Fundamental Theorem of Calculus, Z b a ( f 2 ( x ) + 2 xf ( x ) f 0 ( x ))dx = F ( b ) - F ( a ) = 0 and, since R b a f 2 ( x )dx = 1, we have R b a ff ( x ) f 0 ( x )dx =
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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 Z b a | f ( x ) | 2 ! 1 / 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 dierential 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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Copy of s12 - -1 / 2. By the Schwarz inequality, we...

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