Read e-book online A ''sup+ c inf'' inequality for the equation -delta u = [ PDF

By Bartolucci D.

Show description

Read Online or Download A ''sup+ c inf'' inequality for the equation -delta u = [ V(x2a)]eu PDF

Similar mathematics books

Download e-book for iPad: Indefinite-quadratic estimation and control: a unified by Babak Hassibi, Ali H. Sayed, Thomas Kailath

This monograph provides a unified mathematical framework for a variety of difficulties in estimation and keep watch over. The authors speak about the 2 most typically used methodologies: the stochastic H2 procedure and the deterministic (worst-case) H technique. regardless of the elemental adjustments within the philosophies of those ways, the authors have chanced on that, if indefinite metric areas are thought of, they are often taken care of within the similar method and are primarily a similar.

Download e-book for iPad: Metodos de Bezier y B-splines Spanish by Marco Paluszny, Hartmut Prautzsch, Wolfgang Boehm

Este libro provee una base sólida para l. a. teoría de curvas de Bézier y B-spline, revelando su elegante estructura matemática. En el texto se hace énfasis en las nociones centrales del Diseño Geométrico Asistido por Computadora con los angeles intención de dar un tratamiento analíticamente claro y geométricamente intuitivo de los principios básicos del área.

Additional resources for A ''sup+ c inf'' inequality for the equation -delta u = [ V(x2a)]eu

Sample text

Since u{V qei, : Q E Q} = R, X(R) > 0 + and X is R-invariant, we must have the inequality X(V) > 0. On the other hand, for any nonzero q E Q, we can write Taking q arbitrarily small, we see that the set V does not have the Steinhaus property (see Exercise 8 of Chapter 1). This circumstance immediately yields a contradiction with X(V) > 0. The contradiction obtained finishes the proof. Remark 1. It can easily be observed that V turns out to be a Vitali type set for the group I? = Qei,. This implies at once that the set V is not Lebesgue measurable (see Theorem 1 from Chapter 1).

We now assert that the sets A and B are nonmeasurable in the Lebesgue sense. Indeed, suppose otherwise. Then at least one of these sets is Lebesgue measurable and, in view of the relation -A = B , we claim that both these sets must be Lebesgue measurable. Since we derive that X(A) = X(B) > 0. On the other hand, the metrical transitivity of the Lebesgue measure (see Exercise 7 from Chapter 1) implies which leads to a contradiction. We thus conclude that each of the sets A and B is nonmeasurable in the Lebesgue sense.

Proof. Let {ei : i E I ) be a Hamel basis of R. Fix an index io from I and consider the vector subspace of R (over Q, of course) generated by the 36 CHAPTER 3 partial family {ei : i E I\ {io}}. We denote this subspace by V. Actually, V is a vector hyperplane in R regarded as a vector space over Q. Our purpose is to show that V cannot be Lebesgue measurable. Suppose to the contrary that V E dom(X) where X denotes, as usual, the standard Lebesgue measure on R. Since u{V qei, : Q E Q} = R, X(R) > 0 + and X is R-invariant, we must have the inequality X(V) > 0.

Download PDF sample

A ''sup+ c inf'' inequality for the equation -delta u = [ V(x2a)]eu by Bartolucci D.


by Christopher
4.4

Rated 4.73 of 5 – based on 42 votes