Approximation of Functions of Several Variables and Imbedding Theorems
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Herausgegeben von S. S. Chern J. L. Doob J. Douglas, jr. A. Grothendieck E. Heinz F. Hirzebruch E. Hopf W. Maak S. MacLane W. Magnus M. M. Postnikov F. K. Schmidt W. Schmidt D. S. Scott K. Stein Geschaftsjiihrende Herausgeber B. E 12 E Lp(~), i.e. functions differing on a set of measure zero. We shall take them to be equal to one and the same element of the functional space Lp(~) and write 11 = 12· In particular, if 1 E Lp(t$) is equal to zero for almost all x E 1$, we will write 1 = 0, in this way identifying this function to the by
6
1. Preparatory information
function identically zero on (g. Thus the equation 11/1 - 121IL p (6") = 0 implies that 11 - 12 = 0 and 11 = 12· The set (g may have a dimension m less than n, and then the integral entering into equation (3) will be understood in the sense of the natural (m-dimensional) Lebesgue measure defined on the set (g. We will not be concerned with sets (g of complicated structure. Often (g will coincide with the entire space JR n or with an m-dimensional subspace of JR n , or an m-dimensional cube or ball lying in JR n • Finally, (g might be a smooth or piecewise smooth hypersurface consisting of sufficiently smooth pieces, and then the measure of the measurable subset (g on the basis of which the integral appearing on the right side of (3) is defined is a generalization (extension) of the usual concept of area of a hypersurface. It is natural to extend the definition (3) to the case p = 00. Indeed, if the function I(x) is measurable and essentially bounded on the bounded set (g, i.e. if the quantity sup vrai
If(x) I = MJ ,
IlJE6"
which we call the essential maximuml of I/(x) I on (g, exists, then the equation (4)
lim 1I/IIL p (6") = M J
p""oo
holds. This is proved as follows. Let p,e denote the measure of e. If M J = 0 or p,(g = 0, then (4) is obvious. Suppose that 0 < M J < 00. If (g is a bounded measurable set, then
Accordingly,
(5 )
lim
p->oo
1I/IIL (6") p
~
Mf •
If (g is an unbounded measurable set, then equation (5) is in general not satisfied, as shown by the example (g = JR n , I(x) = 1. However one can prove this inequality under the hypotheses that I(x) E Lp«(g) for all sufficiently large p and that lim IlfIlL p (6") oo
1 M f is the smallest of the numbers 1'v[ having the property that the set of all x E (g for which [f(x) [ > M is of measure zero. It is easy to see that such a number exists.
1.1. The spaces
so that lim p~
C(,c) and Lp(.cl
7
(f,c If(x) IP dX)l/P ;;;; M}f2 [lim (f,c It(x) IP dX)1/P]1/2, p->oo
which implies inequality (5). On the other hand, from the definition of essential maximum of a function it follows that there exists a bounded set ~1 of positive measure such that for all of its points the inequality
It(x) I > M f is satisfied, where 0
IltIIL~(,c) ~
