HW7_Solutions

HW7_Solutions - Problem 1 By Cauchy-Shwartz inequality 1 1...

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Problem 1. By Cauchy-Shwartz inequality | Z 1 0 f ( t ) sin πtdt | ≤ Z 1 0 | f ( t ) sin πt | dt ±Z 1 0 | f ( t ) | 2 dt ²±Z 1 0 | sinπt | 2 dt ² = 1 2 ±Z 1 0 | f ( t ) | 2 dt ² . Also, equality holds if and only if f ( t ) = C sin πt . Indeed, consider functions f and g . Then we have ( f + tg, f + tg ) = k f k 2 + t 2 k g k 2 + 2 t ( f, g ) 0 . Now, if we put t = - ( f, g ) / k g k 2 we get Cauchy-Shwartz inequality. So, for the equality to be true we need ( f + tg, f + tg ) = 0, which means that f = - tg for some t . Problem 2. We know that ( f, g ) = 1 / 4( k f + g k 2 - k f - g k 2 ). So, we have ( f, g ) = 1 2 ±Z 1 - 1 | x | ( f + g ) 2 + 3( f 0 + g 0 ) 2 dx - Z 1 - 1 | x | ( f - g ) 2 + 3( f 0 - g 0 ) 2 dx ² = Z 1 - 1 | x | ( fg ) + 3( f 0 g 0 ) dx. Now, apply Cauchy-Shwartz inequality to functions f and x 2 : | ( f, x 2 ) | = | Z 1 - 1 | x | 3 f + 6 xf 0 dx | ≤ k f kk x 2 k = ±Z 1 - 1 | x | f 2 + 3 | f 0 | 2 dx ² 1 / 2 ±Z 1 - 1 | x | 3 + 12 x 2 dx ² 1 / 2 = ±Z 1 - 1 | x | f 2 + 3 | f 0 | 2 dx ² 1 / 2 5 3 . Problem 3.
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HW7_Solutions - Problem 1 By Cauchy-Shwartz inequality 1 1...

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