By Della Dora J.

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We abbreviate this statement by the symbols: f 2 lipK . f ( ) K for all > 0. f ( ): Lemma. Let f be a bounded function on a b ] and let > 0. f ( ) for n = 1 2 : : :. f ( ) for any > 0. Proof. Given x < y with jx ; y j n , split the interval x y ] into n pieces, each of length at most . f ( ). The second assertion follows from the rst (and one of our exercises). Given > 0, choose an integer n so that n ; 1 < n. f ( ): We next repeat the proof of Bernstein's theorem, making a few minor adjustments here and there.

Thus, we may choose even trig polynomials T1 and T2 such that f (x) + f (;x) T1(x) and f (x) ; f (;x)] sin x T2(x): Multiplying the rst expression by sin2 x, the second by sin x, and adding, we get 2f (x) sin2 x T1 (x) sin2 x + T2(x) sin x T3(x) where T3 (x) is still a trig polynomial, and where \ " now means \within 2"" (since j sin x j 1). Step 3. " Repeat Step 2 for f (x ; =2) and translate: We rst choose a trig polynomial T4 (x) such that 2f x ; 2 sin2 x T4 (x): Trig Polynomials That is, 44 2f (x) cos2 x T5(x) where T5(x) is a trig polynomial.

If f ; p has an alternating set containing n + 2 (or more) points, then p is the best approximation to f out of Pn. Best Approximation 53 Proof. Let x0 x1 : : : xn+1 be an alternating set for f ; p, and suppose that some q 2 Pn is a better approximation to f than p that is, kf ; qk < kf ; pk. In particular, then, we must have jf (xi ) ; p(xi )j = kf ; pk > kf ; qk jf (xi ) ; q(xi )j for each i = 0 1 : : : n + 1. ), hence q ; p = (f ; p) ; (f ; q) alternates in sign n + 2 times (because f ; p does).

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Algorithmes de recherche de valeurs propres by Della Dora J.

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