Reverse Triangle Inequalities for Riesz Potentials and Connections ...

More documents

Recommendations

Info

$Midterm Exam 2 (MATH 2163 Calculus III; Sections 1 and 3)$

$INTRODUCTION TO MODERN ALGEBRA MATH 3613 SECTION ...$

$Quiz 1: Vectors/Lines/Planes (MATH 2163 Calculus III; S 1/3)$

$Math 2123 Practice Exam III solutions 1. A car on a test track goes ...$

potentials as done in [22]. We do not consider the case N = α = 2 in this paper. For 0 < α
everywhere, and we may apply the Riesz Decomposition Theorem for Newtonian potentials.It states that there exists a unique Borel measure σ 2 such that∫d 2−NE(x) = |x − t| 2−N dσ 2 (t) + h(x),. Since d 2−NE(x) is positive and tendsis a Newtonianpotential. The complete proof of Theorem 2.2 is given in Section 4.In the Newtonian case, we can also give explicit examples of σ 2 . If B is the unit ball andd B (x) = |x| + 1, then the measure σ B is found using the Laplacian. Specifically, dσ B (x) =where h is the greatest harmonic minorant of d 2−NEto zero as x → ∞, h must be identically zero [1, p. 106] and hence d 2−NE∆d 2−NB(x) = c|x|N−2 d −NBdS where c is a constant and dS is the surface area measure onthe unit sphere. If L = [−1, 1] then ∆d 2−NL= 0 everywhere except on the hyperplanethat is the perpendicular bisector of the segment. It follows that σ L is supported on thathyperplane [23]. Its value can be calculated using the generalized Laplacian, and is given bydσ L = cd −NLdS where c is a constant and dS is the surface area measure on the hyperplane[23]. For 0 < α < 2, the measure σ E should be calculated using fractional Laplacians.We are now prepared to state a reverse triangle inequality for Riesz potentials.Theorem 2.3. Let E ⊂ R N be a compact set with the minimum α-energy W α (E) < ∞,where 0 < α ≤ 2. Suppose that ν k , k = 1, . . . , m, are positive compactly supported Borelmeasures, normalized so that ν := ∑ mk=1 ν k is a unit measure, with m ≥ 2. Thenm∑k=1infEU ν kα≥ C E (α, m) + infEm∑k=1U ν kα , (9)where∫C E (α, m) := minc k ∈Emin |x − c k| α−N dµ α (x) − W α (E)1≤k≤mcannot be replaced by a larger constant for each m ≥ 2. Furthermore, (9) holds with C E (α, m)replaced by∫C E (α) := d α−NE(x)dµ α (x) − W α (E),which does not depend on m.In the Newtonian case α = 2, the minimum principle holds and so the minimum inC E (2, m) is achieved on the boundary of E. Thus∫C E (2, m) = min min |x − c k| 2−N dµ 2 (x) − W 2 (E).c k ∈∂E 1≤k≤mA closed set S ⊂ E is called dominant ifd E (x) = max |x − t| for all x ∈ supp(µ α).t∈S5
Page 1 and 2: Reverse Triangle Inequalities for R
Page 3: We can calculate that M [−1,1]
Page 7 and 8: Example 2.8 (Unit ball B N in R N )
Page 9 and 10: ⎧2ζ(2 − α)−(2π) ⎪⎨2−
Page 11 and 12: Since supp(µ α ) ⊂ E, this impl
Page 13 and 14: The proof in the case of 0 < α < 2
Page 15 and 16: Proof of Theorem 2.3. For any posit
Page 17 and 18: Hence C E (α, m) is sharp. The alt
Page 19 and 20: ψ k−1 +ψ k2≤ θ ≤ ψ k+ψ k
Page 21: [19] I. E. Pritsker, Inequalities f

Reverse Triangle Inequalities for Riesz Potentials and Connections ...

You also want an ePaper? Increase the reach of your titles

Delete template?

Save as template?