# Download Lecture notes and background materials on linear operators by Miller W. PDF

By Miller W.

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Then for every n there exists a vector vn ∈ DA such that ||un − vn || < n1 . Now ||vn || ≤ ||un || + ||vn − un || < b + 1 by the triangle inequality, so v 1 , v2 , · · · is a bounded sequence. Thus there is a convergent subsequence {Avnj : j = 1, 2, · · ·}. 1) by the triangle inequality. 1) is bounded by ||A||/nj , and the third term by ||A||/nk . Given any ǫ > 0 we can choose j, k so large that the middle term is less that ǫ. Thus ||Aunj − Aunk || → 0 as j, k → ∞. It follows that the sequence {Aunj } is Cauchy, hence convergent.

States that the solution of Au + u˙ = f is u = e−tA u0 + e−tA t 0 eτ A f dτ. Lemma 13 let x ∈ [a, b] and suppose 1. ∞ j=1 vj (x) = v(x) for all x ∈ [a, b]. 2. vj (x) ∈ C 1 [a, b], for all j. 3. ∞ j=1 vj′ (x) = g(x) where the convergence is pointwise uniform on [a, b]. Then v ∈ C 1 [a, b] and v ′ (x) = ∞ ′ j=1 vj (x) 60 for all x ∈ (a, b). PROOF OF THE LEMMA: Let sn (x) = nj=1 vj (x). Then sn ∈ C 1 [a, b] and sn (x) → v(x), s′n (x) → g(x) as n → ∞, where the convergence is uniform. By the Fundamental Theorem of Calculus, s n (x) − sn (a) = ax s′n (y)dy goes in the limit as n → ∞ to v(x) − v(a) = ax g(y)dy since the convergence is uniform.

33) o respectively. Here C 2 (Rn ) is the space of twice continuously differentiable functions with compact support in R n . Theorem 17 S-L operators A0 , and A1 (with some additional technical assumptions, see Hellwig, page 85) are symmetric in H. PROOF: (sketch) It is clear that DA0 = DA1 = H. v ∈ DA) , (Au, v) = Rn Au(x)v(x)k(x)dx = lim r→∞ |x|≤r Auvkdx. Now we integrate by parts on the ball |x| ≤ r, where |x|2 = x21 + · · · + x2n , νj (x) = 29 xj on |x| = r. |x| |x|≤r = x|≤r = [ Auvkdx = |x|≤r [ Dj (pjℓ Dℓ u)v + quv]dx pjℓ Dℓ uDj v + quv]dx + i |x|≤r uAvkdx + i Let Ψ(r) = |x|≤r Then |x|=r |x|=r xj pjℓ Dℓ u)vdS |x| ( xj [ pjℓ (vDℓ u + uDℓ v)dS.