38 2. FOUR IMPORTANT LINEAR PDE Proof. 1. For each fixed x, the mapping y → G(x, y) is harmonic, except for y = x. As G(x, y) = G(y, x), x → G(x, y) is harmonic, except for x = y. Thus x → − ∂G ∂yn (x, y) = K(x, y) is harmonic for x ∈ R+, n y ∈ ∂R+.n 2. A direct calculation, the details of which we omit, verifies (34) 1 = ∂R+n K(x, y) dy for each x ∈ R+. n As g is bounded, u defined by (33) is likewise bounded. Since x → K(x, y) is smooth for x = y, we easily verify as well u ∈ C∞(R+),n with Δu(x) = ∂R+n ΔxK(x, y)g(y) dy = 0 (x ∈ R+).n 3. Now fix x0 ∈ ∂R+, n ε 0. Choose δ 0 so small that (35) |g(y) − g(x0)| ε if |y − x0| δ, y ∈ ∂R+.n Then if |x − x0| δ 2 , x ∈ R+,n (36) |u(x) − g(x0)| = ∂Rn + K(x, y)[g(y) − g(x0)] dy ≤ ∂R+∩B(x0,δ)n K(x, y)|g(y) − g(x0)| dy + ∂R+−B(x0,δ)n K(x, y)|g(y) − g(x0)| dy =: I + J. Now (34), (35) imply I ≤ ε ∂R+n K(x, y) dy = ε. Furthermore if |x − x0| ≤ δ 2 and |y − x0| ≥ δ, we have |y − x0| ≤ |y − x| + δ 2 ≤ |y − x| + 1 2 |y − x0| and so |y − x| ≥ 1 2 |y − x0|. Thus J ≤ 2 g L∞ ∂R+−B(x0,δ)n K(x, y) dy ≤ 2n+2 g L∞ xn nα(n) ∂R+−B(x0,δ)n |y − x0|−n dy → 0 as xn → 0+.
Purchased from American Mathematical Society for the exclusive use of nofirst nolast (email unknown) Copyright 2010 American Mathematical Society. Duplication prohibited. Please report unauthorized use to cust-serv@ams.org. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.
