Ch3 - Chapter 3 Orthogonal sets of functions In this chapter we investigate the following problem Given a fixed set of functions f n can we expand

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Unformatted text preview: Chapter 3: Orthogonal sets of functions In this chapter we investigate the following problem: Given a fixed set of functions { f n } , can we expand other functions f into a series f ( x ) = ∞ X n =0 c n f n ( x ) ? As we learned from the previous chapter, the convergence of such a series can be a delicate matter. Here we are interested in convergence in certain norms , rather than in pointwise convergence. 3.1. Mean square convergence, completeness of orthonormal sets Let a,b ∈ IR, a < b . (We also allow a =-∞ or b = + ∞ .) We denote by R ( a,b ) the set of all functions f : [ a,b ] → IC which are Riemann integrable on [ a,b ]. Assume that w ∈ R ( a,b ) is continuous on [ a,b ] and positive on the open interval ( a,b ). We set for any function f ∈ R ( a,b ) k f k := Z b a | f ( x ) | 2 w ( x ) dx ! 1 / 2 . (1) We call k f k the norm of f (w.r.t. the weight function w ). If f,g ∈ R ( a,b ) we set < f,g > := Z b a f ( x ) g ( x ) w ( x ) dx, (2) and we call the number < f,g > the inner product of f and g (w.r.t. the weight function w ). The weight function w will assumed to be fixed in all general results of this chapter, and will be specified only in the examples. Two functions f and g are said to be orthogonal to each other if < f,g > = 0. The following properties are well-known, we omit their proofs, ( f,g,h ∈ R ( a,b ), c,d ∈ IC): k f k ≥ , (3) < f,g > = < g,f >, (4) 16 < cf + dg,h > = c < f,h > + d < g,h >, (5) < h,cf + dg > = c < h,f > + d < h,g >, (6) k f k 2 = < f,f >, (7) | < f,g > | ≤ k f k · k g k , (Cauchy-Schwarz inequality) , (8) k f + g k ≤ k f k + k g k , (triangle inequality) . (9) Warning : Notice that if k f k = 0, then we do not have f ( x ) = 0 on the whole interval [ a,b ], in general ! However, this uncertainty does not bother too much if we restrict ourselves to piecewise continuous functions: If f ∈ PC ( a,b ), and k f k = 0, then we have that f ( x ) = 0 except perhaps at finitely many points. Finally, since we have | a + b | 2 = | a | 2 +2 Re ( a b )+ | b | 2 for any complex numbers a,b , there also holds k f + g k 2 = k f k 2 + 2 Re < f,g > + k g k 2 . (10) We will say that a sequence { f n } ⊂ R ( a,b ) converges to f in the norm – or, equivalently, it is mean square convergent to f – if k f- f n k → as n → ∞ . Notice that the norm and the inner product are continuous w.r.t. convergence in norm; that is, if f n → f in norm, then k f n k → k f k , < f n ,g > → < f,g > for all g. We have the following useful and simple result: Theorem 3.1. If f n → f uniformly on [ a,b ] then f n → f in the norm. Proof : Uniform convergence means that there is a sequence { M n } of positive constants with lim n →∞ M n = 0 such that | f ( x )- f n ( x ) | ≤ M n ∀ x ∈ [ a,b ], n = 1 , 2 ,... . But then k f- f n k 2 = Z b a | f- f n | 2 w dx ≤ M 2 n Z b a w dx → , as n → ∞ ....
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This note was uploaded on 02/26/2010 for the course MATH 212 taught by Professor Friedmannbrock during the Fall '08 term at American University of Beirut.

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Ch3 - Chapter 3 Orthogonal sets of functions In this chapter we investigate the following problem Given a fixed set of functions f n can we expand

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