A. F. Lavriks truncated equations by Kaufman R. M.

By Kaufman R. M.

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Proof Note first that u is real-valued. It will be shown now that the hypotheses imply that u is continuous on σ. To see this, consider any x ∈ σ and any ε > 0 such that − − − Bx,2ε ≥ σ. For any y ∈ Bx,ε , By,ε ≥ Bx,2ε . Since Bx,ε By,ε ≥ Bx,2ε and u is − integrable on Bx,2ε , |u(x) − u(y)| ≤ 1 νn ε n |u(z)| dz ⊂ 0 Bx,ε By,ε as y ⊂ x by the absolute continuity of the Lebesgue integral, which proves that u is continuous at x. Suppose there is a point x0 ∈ σ such that u(x0 ) = inf σ u. Letting = {y; u(y) = inf σ u}, is a relatively closed subset of σ by continuity of u.

To see this, let x be the projection of x = (x1 , . . , xn ) onto φσ and let z ∈ φσ. Then |z − x|2 = xn2 + |z − x|2 , and φσ 1 dz ≤ |z − x|n φσ∼{|z−x|<1} 1 dz + xnn φσ∼{|z−x|∇1} 1 dz. |z − x|n Since the first integral on the right is finite, only the second need be considered. Transforming to spherical coordinates in the (n − 1)-dimensional space φσ relative to the pole x, φσ∼{|z−x|∇1} 1 dz = |z − x|n ∅ |π|=1 1 1 n−2 r dπdr = ρn−1 < +∅. rn It will be shown now that PI(0, 1, σ) = 1 on σ. By Eq.

12 that f ∗ is ∞ harmonic on σ∗ . 1 Exercises for Sect. 8 1. Show that the circle C ≥ R 2 is invariant under an inversion with respect to the circle φ B0,θ if and only if it is orthogonal to φ B0,θ , that is, the normals to the two curves at the points of intersection must be orthogonal. Formulate the same proposition for spheres in R n . ) 2. Show that the Jacobian J (x ∗ , x), x ∈ R 2 , of the inversion x ∗ ⊂ x relative to φ B0,θ is −θ4 /|x|4 . 3. Show that the Jacobian J (x ∗ , x), x ∈ R 3 , of the inversion x ∗ ⊂ x relative to φ B0,θ is −θ6 /|x|6 .

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