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7330f02projops

7330f02projops - Math 7330 Functional Analysis Notes In the...

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Math 7330: Functional Analysis Fall 2002 Notes A. Sengupta In the following H is a complex Hilbert space. 1 Orthogonal Projections We shall study orthogonal projections onto closed subspaces of H . In summary, we show: If X is any closed subspace of H then there is a bounded linear operator P : H H such that P = X and each element x can be written unqiuely as a sum a + b , with a Im( P ) and b ker ( P ); explicitly, a = Px and b = x - Px . The point Px is the point on X closest to x . The operator P satisfies P 2 = P and is self-adjoint. Conversely, if P 1 is any bounded linear operator H H for which P 2 1 = P 1 then the following are equivalent: (i) P 1 is an orthogonal projection onto a closed subspace, (ii) P 1 is self-adjoint, (iii) P 1 is normal, i.e. commutes with its adjoint P * 1 . 1.1 Point in a convex set closest to a given point Let C be a closed convex subset of H . We will prove that there is a unique point in C which is closest to the origin . This will use convexity of C , that C is closed, the fact that the topology on H arises from an inner-product, and that this topology makes H a complete metric space. Let r = d (0 , C ) = inf {| x | : x C } The function f : H R : x d ( x, C ) = inf {| x - y | : y C } is continuous because it satisfies | f ( x ) - f ( y ) | ≤ d ( x, y ) (1) This inequality can be verified as follows: for any for any a C the triangle inequality implies f ( x ) d ( x, a ) d ( x, y ) + d ( y, a ) and so taking inf over a C gives f ( x ) d ( x, y ) + f ( y ); this implies f ( x ) - f ( y ) d ( x, y ), and then we can interchange x and y . 1
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Since C is closed, there is a sequence of points x n C with | x n | → r as n → ∞ . We will show that the sequence ( x n ) is necessarily Cauchy. This uses the parallelogram property: | x n - x m | 2 + | x n + x m | 2 = 2( | x n | 2 + | x m | 2 ) which gives | x n - x m | 2 = 2( | x n | 2 + | x m | 2 ) - 4 x n + x m 2 2 Since C is convex, the midpoint x n + x m 2 lies in C . Since r is the closes distance of C from 0, it follows that | x n + x m 2 | ≥ r and so | x n - x m | 2 2( | x n | 2 + | x m | 2 ) - 4 r 2 If we let n, m → ∞ the the right side approaches 2( r 2 + r 2 ) - 4 r 2 = 0, and so the sequence ( x n ) n is Cauchy. Since H is complete, this sequence has a
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