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Sample text

2) and the Gauss -Green *From Lemma 23 it follows that the solution will have a finite integral of energy. )== ~ ~ ~ E uv't;: v }E (d~)== hu vt;~t> d~. E By virtue of the condition :E~i) 0, the last integral equals zero for all p E R" \E.. t. 4 is a function = which is harmonic in E. and continuous in E. , ~ ... ons t in E. The proof of the uniqueness of the exterior Neumann problem is completely analogous. R em ark. l: l') ('~::. ~·> ) belongs to the class

And Q'. Let and & , lies wholly in n. )~ ~Sl. '). k (n) " In\- RQ (eJ) A& < 2 e. If ~n (

Note that the sets do not intersect. Indeed, let there exist a point :x: ~ o" Q common to E A•n a*Q A* and B• . Then SPACE OF FUNCTIONS WHOSE DERIVATIVES ARE MEASURES 52 the volume density of each of the sets E. and Q \ E at the point ~equals ~/:~.. The latter is impossible since x E ~"Q. ,.. 1 (B·n~·Q)=Hn_, (P>*r\'()Q)=PcQ (Q\E) (see Proposition 4). The lemma is proved. THEOREM 3. If r-.. ~r. , P r o of . 3)- oo. Let, for definiteness, B:n (E)~ PcQ. (0 \E) and let the set E:> be such that & (1 Q= E, PcQ( &)= :11..

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