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C N 5 ¤ $ Z ¾ −;. L=1 with h = h(n,µ) = min 1 8, 1c E (much larger than ε 0 such that (h n×)1/ε > ε 0) and c E is from the above lemma domain target covering figure We know N0 (n,µ) ≤ 1 h n×n, then there exists one ball, say B1 = B h (H 1) and H 1 = D2u(x 1) such that D 2u B −1 B1 ≥ B 1 N 0 or 1/2 N 4. = C (n,µ)kuk Now to the claim Domain goes from 1 to 1 2, range goes from 2 to 1 domain range figure Set η k (x) = 1 for x < 1 2 1 2k1 linear interpolation in between 0 for x > 1 2 1 2k u k = u− 1− 1 2k Observe u k ≤ u k−1, and u k−1 > 1 2k when u k > 0 Set A k def= Z B 1 (η ku k) 2≥ Z B /2 (u−1) We prove {u > 1}∩B 1/2 = 0 via iteration A k ≤ b(n,µ) S(k) AS(k).

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C(n1) = I−µRc(n) = I−µQΩQHc(n) ⇒QHc(n1) = QH−µQHQΩQHc(n) = I−µΩΩQHc(n) (∗) Definethetransformedcoefficientsas v(n) = QHc(n) = QH(w(n)−w 0) Then(∗) becomes v(n1) = I−µΩΩv. Course Title QBUS 5001;. V1 ¸, the first equation becomes · a h h b ¸· u1 v1 ¸ = λ1 · u1 v1 ¸ or · a−λ1 h h b−λ1 ¸· u1 v1 ¸ = · 0 0 ¸ Hence we are dealing with a homogeneous system of two linear equations in two unknowns, having a non–trivial solution (u1, v1) Hence ¯ ¯ ¯ ¯ a−λ1.

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