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HW7_Solutions

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

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Problem 1. By Cauchy-Shwartz inequality | 1 0 f ( t ) sin πtdt | ≤ 1 0 | f ( t ) sin πt | dt 1 0 | f ( t ) | 2 dt 1 0 | sinπt | 2 dt = 1 2 1 0 | f ( t ) | 2 dt . 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 ) = f 2 + t 2 g 2 + 2 t ( f, g ) 0 . Now, if we put t = - ( f, g ) / g 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( f + g 2 - f - g 2 ). So, we have ( f, g ) = 1 2 1 - 1 | x | ( f + g ) 2 + 3( f + g ) 2 dx - 1 - 1 | x | ( f - g ) 2 + 3( f - g ) 2 dx = 1 - 1 | x | ( fg ) + 3( f g ) dx. Now, apply Cauchy-Shwartz inequality to functions f and x 2 : | ( f, x 2 ) | = | 1 - 1 | x | 3 f + 6 xf dx | ≤ f x 2 = 1 - 1 | x | f 2 + 3 | f | 2 dx 1 / 2 1 - 1 | x | 3 + 12 x 2 dx 1 / 2 = 1 - 1 | x | f 2 + 3 | f | 2 dx 1 / 2 5 3 . Problem 3. Since continuous functions are dense in L 2 [0 , 1] and polynomials are dense in C [0 , 1] we conclude that polynomials are dense in L 2 [0 , 1]. Since L 2 [0 , 1] is sepa- rable, we can choose a countable dense set of polynomials. From this set we can choose a linear independent set of polynomials and apply the Gramm-Schmidt pro- cess. In the process we use only addition and multiplication by a constant, so the result will by also a set of polynomials.

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