By Carothers N.L.
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Extra resources for A Short course on approximation theory
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 first 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 2ε” (since | sin x | ≤ 1). Step 3. ” Repeat Step 2 for f(x − π/2) and translate: We first choose a trig polynomial T4 (x) such that 2f x − π sin2 x ≈ T4 (x). 2 Trig Polynomials 44 That is, 2f(x) cos2 x ≈ T5 (x), where T5 (x) is a trig polynomial.
Bernstein’s theorem states that the sequence Bn (f) converges uniformly to f for each f ∈ C[ 0, 1 ]; the proof is rather simple once we have a few facts about the Bernstein polynomials at our disposal. For later reference, let’s write f0 (x) = 1, f1 (x) = x, and f2 (x) = x2 . Among other things, the following exercise establishes Bernstein’s theorem for these three polynomials. Curiously, these few special cases will imply the general result. ⊲ 30. (i) Bn (f0 ) = f0 and Bn (f1 ) = f1 . ] (ii) Bn (f2 ) = 1 − n k=0 (iii) k n −x 1 n f2 + n1 f1 , and hence (Bn (f2 )) converges uniformly to f2 .
Thus, ωf (nδ) ≤ n ωf (δ). The second assertion follows from the first (and one of our exercises). Given λ > 0, choose an integer n so that n − 1 < λ ≤ n. Then, ωf (λδ) ≤ ωf (n δ) ≤ n ωf (δ) ≤ (1 + λ) ωf (δ). We next repeat the proof of Bernstein’s theorem, making a few minor adjustments here and there. Theorem. For any bounded function f on [ 0, 1 ] we have f − Bn (f) ≤ 3 2 In particular, if f ∈ C[ 0, 1 ], then En (f) ≤ 1 3 ωf √ 2 n . ωf ( √1n ) → 0 as n → ∞. Proof. We first do some term juggling: n |f(x) − Bn (f)(x)| = k=0 n ≤ k=0 ωf k=0 ≤ ωf 1 √ n = ωf 1 √ n x− n k x (1 − x)n−k k k n f(x) − f n ≤ k n f(x) − f n k x (1 − x)n−k k k n n 1+ k=0 n k x (1 − x)n−k k √ k n x− n √ 1 + n n k=0 x− k n n k x (1 − x)n−k k n k x (1 − x)n−k , k 31 Algebraic Polynomials where the third inequality follows from our previous Lemma (by taking λ = and δ = √1 n √ n x− k n ).