Minimum Norm Extremals in Function Spaces: With Applications by Prof. Stephen D. Fisher, Prof. Joseph W. Jerome (auth.)

By Prof. Stephen D. Fisher, Prof. Joseph W. Jerome (auth.)

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In particular, sional and U is a closed flat, Then T U is a if N is finite dimen- then U + N is closed and in this case TU is closed and convex. Proof: in X. Let U be a c o n v e x subset of X and suppose that U + N is closed If J denotes the c a n o n i c a l m a p p i n g f r o m X onto X/N then the m a p p i n g T O : J(X) -* Y defined by T o J X = Tx is linear, theorem, c o n t i n u o u s and bijective. the set J(U+N) = J(U) Now, by the open m a p p i n g is closed, and its c o m p l e m e n t u n d e r J are disjoint.

M. in]E, Let I 0 , . . 2) -- inf[llh11 : (h,P) ~ u]. Then U contains at least one pair (h,P) for which llhll = a; the set S of such pairs is convex and compact in the weak-, topology. ,r and Z 0r Icjl = ~. Proof. Let {g~ = (hv, Pv)] be a sequence in U with IIkvlI -~ a. 3) since is also uniformly bounded l~i(Pv) I ~ C, i = 0 .... ,m and ~ = 1,2,... 3) that the sequence P is bounded in norm. ,m. Likewise, the measures point k 0 with Ilhoii~ hence IIkolI = a. ~. [k]. as j - ~ for have a weak* accumulation it is easy to check that (ho, P O) E U and Thus we have shown that S is nonempty.

F N are real analytic points. on 0 except then for example, at a finite n u m b e r of Let r E R N and let U = U(~) = [u E W2m(D) N W~O(~) : ~ LuF i = rl, i = i, .... N and Lu ~ L~(~)}. 1. Let $ E L~(D). 1) problem = = inf[llLu-~llL~ : u E U} has a unique modulus The m i n i m i z a t i o n on ~. solution ~ E U; the difference Furthermore, there are disjoint whose dense union omits only a set of measure Im - ~ has constant open sets D+ and O-, zero in ~, such that Im - ~ = a on D+ and Im - ~ = -a on O-.

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