By Martin Schechter
The innovations used to unravel nonlinear difficulties vary tremendously from these facing linear positive factors. Deriving all of the valuable theorems and ideas from first rules, this textbook offers higher undergraduates and graduate scholars a radical realizing utilizing as little historical past fabric as attainable.
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82) ˜k (x) = uk (x)/ρk → 0, and, consequently, u ˜(x) = 0. 82) holds as well. If x ∈ Ω3 , there are subsequences for which |uk (x)| → ∞ and subsequences for which |uk (x)| is bounded. 82) holds, and for the latter it holds as well since u˜(x) = 0. e. on the whole of I. e. 78). 18) F (x, uk )/ρ2k dx → 2 I β(x)˜ u(x)2 dx. 84) I Hence, 2G(uk )/ρ2k → 1 − β(x)˜ u(x)2 dx I = (1 − u ˜ 2 H) + u ˜ 2 [1 − β(x)]˜ u(x)2 dx + I = A + B + C. Since u ˜ H ≤ 1 and β(x) ≤ 1, the quantities A, B, C are each ≥ 0. The only way the sum can equal 0, is if each equals 0.
2 |k|≤n Consequently, |αk |2 ≤ u 2 . |k|≤n Since this is true for every n, we have ∞ |αk |2 ≤ u 2 . 54) k=−∞ Moreover, if m < n, then un − um 2 |αk |2 → 0 = m<|k|≤n as n → ∞. 4 Fourier series 19 Hence, {un } is a Cauchy sequence in L2 (I). It converges in L2 (I) to a function u ˜. Let k = 0, ±1, ±2, . . u, ϕ¯k ), α ˜ k = (˜ Then, α ˜ k = ( lim un , ϕ¯k ) = lim ( n→∞ n→∞ αj ϕj , ϕ¯k ) = lim n→∞ |j|≤n αj (ϕj , ϕ¯k ) = αk . |j|≤n Let f = u ˜ − u. 51). 19 that f ≡ 0. 45) holds. 17. Proof. Assume that there is a point x0 ∈ I such that f (x0 ) > 0.
Let M be the subspace of those functions in H which are orthogonal to N, that is, functions w ∈ H which satisfy (w, 1)H = w(x) dx = 0. 23. 72) for some constants a, b. We shall also prove these at the end of the section. 24. 67) there is a u in H such that G(u) = min G. 2) in the usual sense. Proof. Let α = inf G, H and let {uk } be a minimizing sequence, that is, a sequence satisfying G(uk ) α. Assume first that ρk = uk H ≤ C. 73) and uk (x) → u0 (x) uniformly in I. 74) Then F (x, uk )dx → I F (x, u0 )dx I by arguments given previously.