6.1 (16) 6.2(13) 6.3(3)

6.1 (16) 6.2(13) 6.3(3) - Section 6.1: 16: Showing...

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Section 6.1: 16: Showing that H is an inner product space is mostly a straightforward exercise except for the positivity condition. As in example 3, the continuity of the functions is an essential part of the proof. To see why, consider a function that is zero everywhere except at x = π where it is 1. This is a nonzero (noncontinuous) function, and its inner product with itself is 0. On the other hand, if f 6 = 0, then there is some x 0 with f ( x 0 ) 6 = 0. Since f ( x ) is continuous, there is some δ > 0 such that | f ( x ) | 2 > | f ( x 0 ) | 2 / 2 whenever | x - x 0 | < δ . Therefore, since | f ( x ) | 2 0, we have h f, f i = 1 2 π Z 2 π 0 | f ( x ) | 2 dx > 1 2 π δ | f ( x 0 ) | 2 > 0 . (Note: if the point had been in the interior, 2 δ could have been used rather than δ , I used δ to allow for the x 0 to be an end points.) For part b, we need to find a specific continuous function f ( x ) on [0 , 1] whose “inner product” with itself is zero. This will be the case if f ( x ) = 0 on [0 , 1 / 2]. Thus the example f ( x ) = ± 0 0 x 1 / 2 x - 1 / 2 1 / 2 < x 1 , which is a continuous function, shows that this is not an inner product. Section 6.2: 13c: Most of this problem requires direct application of the various definitions and theorems. Part c is a bit harder, and in fact, this statement that ( W ) = W is not true in general (seen problem 23 below). It is true when we are working with a finite dimensional
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This note was uploaded on 01/26/2012 for the course MATH 115A 262398211 taught by Professor Fuckhead during the Spring '10 term at UCLA.

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6.1 (16) 6.2(13) 6.3(3) - Section 6.1: 16: Showing...

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