By Miller W.
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Example text
Indeed, for k > ℓ, k 2 |Sk (x) − Sℓ (x)| = | k =| n=ℓ+1 λn (v, φn )φn (x)|2 ≤ (u, φn )φn (x)|2 n=ℓ+1 k n=ℓ+1 |(v, φn )|2 · k n=ℓ+1 |λn φn (x)|2 . Later we will show that kn=ℓ+1 |λn φn (x)|2 < β 2 < ∞ for all x ∈ D. Assuming this, we see that k |Sk (x) − Sℓ (x)| ≤ β n=ℓ+1 |(v, φn)|2 → 0 uniformly in x as k, ℓ → ∞. Thus lim k→∞ Sk (x) = w(x) and, since D is bounded, ||Sk − w|| → 0 as k → ∞. 52 2. Sk (x) → u(x), uniformly as k → ∞. Indeed ||u − w|| ≤ ||u − Sk || + ||Sk − w|| → 0 as k → ∞, so u = w in H and, since u, w are both continuous, we have u(x) ≡ w(x) for all x ∈ D.
If u ∈ RA , u = Av, v ∈ DA then u(x) = ∞ n=1 (u, φn )φn (x) for x ∈ D where the convergence is pointwise uniform absolute on D. PROOF: Assume there are an infinite number of nonzero eigenvalues. ) Recall that (φ i , φj ) = δij and Aφj = λj φj . For u ∈ RA with u = Av we have (u, φj ) = (Av, φj ) = (v, Aφj ) = λj (v, φj ). 1. Let Sk (x) = kn=1 (u, φn)φn (x) ∈ C 0 (D) for each integer k. We will show that {Sk (x)} is a uniformly pointwise convergent sequence of continuous functions on D, so that it converges to a continuous function w(x) in D.
NOTE: A kernel K with properties 1 and 2 is said to be weakly singular. 3. a(x, y) = a(y, x), for all x, y ∈ D. 4. H = L2c (D, k). 3) D dy ωn ≤ β |x − y| n−β nV ωn 1− β n if 0 < β < n, where ωn is the area of the unit n-sphere and V is the volume of D. Indeed, for fixed x we can write y − x in polar coordinates centered at x. Then dy is proportional to r n−1 dr dωn , where dωn is the differential of area measure on the unit n-sphere. Then an estimate for the integral n−1 n−β is 0B r rβ dr = Bn−β where D is contained in a ball of radius B about x.