By Richard A. Silverman

ISBN-10: 0486647625

ISBN-13: 9780486647623

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2 (Null Function). is integrable and f I! I= 0. A function f is called a null function' iff Two functions f and g will be called equivalent iff- g is a null function. It is easy to check that the defined relation is actually an equivalence relation. Now we define the space £'\R) as the spa(;e of equivalence classes of Lebesgue integrable functions. , 1 (f]= { gE L (R): I If- g)=O }· With the usual definitions (f] + [g] = [f + g ], A(f] IIUJII = [Af], =I IJI, ( £' 1 (R), 11·11) becomes a normed space.

2+· ··and L~= 1 JIJ,,Isflfl+c:. Proof. Letf = g 1 + g 2 + · · ·be an arbitrary expansion off Then there exists n0 E N such that L~=no+I Ifni< c:/2. Define J for n 2:2. h +/ 2 + · · ·. Moreover, since we have The proof is complete. 1. If {fn} is a sequence of integrable functions and f=fl+h+· .. , then f is integrable and Proof. ft = ft1 + fh +·· ·. 1, there exist step functions J,,k ( n, kEN) such that fn = fn, I +fn,2 + ' ' ' and n = 1, 2, .... 51 The Lebesgue Integral Let {hn} be a sequence arranged from all the functions gn,k.

Since F is a closed subset of a complete space, there exists z E F such that xn -i> z as n -i> oc. We are going to show that z is a unique point such that f( z) = z. Indeed, since llf(z)-zll::; llf(z)-xnll + llxn -zll = IIJ(z)-J(xn-1)11 + llxn -zll ::; a liz -xn-111 + llxn- zll-i> 0 as n -i> oc, we have llf(z)- zll = 0, and thus f(z) = z. Suppose now f(w) wEF Then liz- wll = llf(z) -f(w)ll sa liz- wll. Since 0

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