实变函数与泛函分析基础第四章习题答案

? ? 1.?f(x)?E?r,?Ef r? ?Ef = r?f(x)? ?r,Ef r?,?rn? ?Ef a = S n=1 Ef rn,?Ef rn?Ef ?f(x)?E? ? ?r,Ef = r?f(x)?E = (,), z?(,) ?x z,f(x) = 3;x 6 z,f(x) =2, ?r,Ef = r = ? ?Ef 2 = z ?f? 2.?f(x),fn(x)(n = 1,2,)?a,b?k? k=1 lim n E ? | fn f | N ? | fn(x) f(x) | 1 k, ? x lim n E ? | fn f | 1 k ? . ?k? x k=1 lim n E ? | fn f | 1 k ? . ?x T k=1 lim n E ?| f n f | 0,?k0,? 1 k0 ,?x lim n E h | fn f | N ?x E h | fn f | 1 k0 i ,?| fn(x) f(x) | 1 k0 limfn. 1 ?E lim n fn= + E lim n fn= E lim n fn lim n fn? ? 4.?E?0,1? f(x) = ( x,x E, x,x 0,1 E. ?f(x)?0,1?| f(x) |? ?f(x)?0 E,?Ef 0 = E?0 6 E,?Ef 0 = E? ?f(x)? ?x 0,1 ?| f(x) |= x?| f(x) |?0,1? 5.?fn(x)(n = 1,2,)?E?a.e.?| fn|a.e.?f. ? 0?c?E0 E,m(EE0) ,?E0?n?| fn(x) | c. ?mE . ?E| fn|= ,Efn6 f?n = 0,1,2, .?E1= Efn6 f( S n=0 E| fn|= ),?mE1= 0.?EE1?fn(x)?f(x).?E2= EE1, ?x E2,sup n | fn(x) | .? E2= k=1 E2sup n | fn| k,E2sup n | fn| k E2sup n | fn| k + 1. ?mE2=lim k mE2sup n | fn| k.?k0?mE2 mE2sup n | fn| k0 .? E0= E2sup n | fn| k0,c = k0.?E0?n,| fn(x) | c,? m(E E0) = m(E E2) + m(E2 E0) c = S n=1 E2n 0,? E E,?m(E E) ?fn?E?f(x).?E0?E?fn? ?, E0 E E(?E?fn?),?mE0 m(E E0) 0,?E E? f(x)?E?m(E E) ,?f(x)?E?a.e.? ?1/n,?En E,?f(x)?En? m(E En) a ( S n=1 Enf a),?f?En?Enf a?m(E0f a) mE0= 0, ?E0f a?Ef a?f?f?En? ? S n=1 En?f(x)a.e.? 9.?fn?E?f,?fn(x) g(x)a.e.?E,n = 1,2, .? f(x) g(x)?E? ?fn(x) f(x),?fni fn,?fni(x)?E?a.e.?f(x).?E0? fni(x)?f(x)?En= Efn g.?mE0= 0,mEn= 0.m( S n=0 En) P n=0 mEn= 0. ?E S n=0 En?fni(x) g(x),fni(x)?f(x),?f(x) = limfni(x) g(x)?E S n=0 En ?f(x) g(x)?E? 10.?E?fn(x) f(x),?fn(x) fn+1(x)?n = 1,2, ,? ?fn(x)?f(x). ?fn(x) f(x),?fni fn,?fni(x)?E?a.e.?f(x).?E0 ?fni(x)?f(x)?En= Efn 0,E| f gn| ( S n=1 En) E| f fn| .? mE| f gn| m( n=1 En) + mE| f fn| = mE| f fn| . 3 ?fn(x) f(x),?0 limmE| f gn| limmE| f fn| = 0?gn(x) f(x). 12.?mE 0,?mE| fn f | 0?0 0,? ?fnk,? mE| fnk f | 0 0 0.(1) ?fnk?f(x)?fnkj ?E?a.e.?f,?mE +,?E?fnkj f(x),?(1)? 13.?mE 0,?k,mE| f | k 5 ?mE| g | k N ?mE| gn g | 0 5,mE| fn f | 0 5 ? ? E| gn| k + 1 E| g | k E| gn g | 1 E| g | k E| gn g | 0. mE| gn| k + 1 mE| g | k + mE| gn g | 0 5 + 5 = 2 5 . ? E h | gnfn gnf | 2 i E| gn| k+1E ? | fn f | 2(k + 1) ? E| gn| k+1E| fnf | 0. ? mE h | gnfn gnf | 2 i mE| gn| k + 1 + mE| fn f | 0 2 5 + 5 = 3 5 . ? E h | fgn fg | 2 i E| f | k+1E ? | gn g | 2(k + 1) ? E| f | kE| gng | 0. 4 ? mE h | fgn fg | 2 i mE| f | k + mE| gn g | 0 0,?N,?n N ?mE| gnfn gf | ,?gnfn gf. (2) E| (fn+ gn) (f + g) | E h | fn f | 2 i E h | gn g | 2 i ? mE| (fn+ gn) (f + g) | mE h | fn f | 2 i + mE h | gn g | 2 i , lim n mE| (fn+ gn) (f + g) | lim n mE h | fn f | 2 i + lim n mE h | gn g | 2 i . ?fn+ gn f + g. (3)?fn f,?| fn| f | .? E| fn f | E| fn| | f | . ?lim n mE| fn| | f | lim n mE| fn f | = 0,?| fn| f |. ?fn f,?a 6= 0,afn af.? E| afn af | = E ? | fn f | | a | ? , ? lim n mE| afn af | = lim n mE ? | fn f | | a | ? = 0. minfn(x),gn(x) = fn(x) + gn(x) | fn(x) gn(x) | 2 . ?(2), fn(x) + gn(x) f + g,fn(x) gn(x) f g. ? | fn(x) gn(x) | f(x) g(x) |, ?(2), fn(x) + gn(x) | fn(x) gn(x) | f(x) + g(x) | f(x) g(x) | . ? fn(x) + gn(x) | fn(x) g