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A=(A-B)(AB)B=(B-A)(AB)@ɲַքe,Կҵһһһ.

f(0)=1/2 ,f(1)=1/3 f(1/n)=1/(2+n),n=2 3 4..f(x)=x,.@һ[0 1](0 1)һһӳ.

=}Ŀ???һ}???㵹ǽocʾ..@}Ŀ]..

f(x)]^gabϿ΢ ΢Ҳ䌧 Ҍн cBmPϵ ɌضBm.f(x)]^gabϽ^Bm

^һcƽ,ÈDΘһһӳ,ɵC!

@ܺ,ֻҪһһһPϵͿ˱@,҂OƽȥǗlֱx=0ô҂ڌƽϵc@ӵČ(x,y),x,҂͌@ӵ(x,y)ԭƽͬһc(x,y)҂ҎxǴ0,҂cԭƽϵ(x-1,y)xС0,ô҂Ҳԭƽͬһc(x,y)@ӌƽϵκһcԭƽ϶Čc,ԭƽϵһc,ҲҵƽcǂΨһc,ƽȥһlcƽnjȵ

BC=(B-C) (C-B) A(BC)=A((B-C) (C-B))= (A - ((B-C) (C-B))) ((B-C) (C-B) - A)= ( A - BC) (ABC) ((B-CA) (C-BA) )=( A - BC) (B-CA) (C-BA) ( ABC ) ʽӵČQ,֪,߅ҲȻ,ԽYՓ.

Yes, try to prove the 1/p power of the integration of |f|^p over E is a norm. What you want is just an example of triangular inequality

(0,1)cZ+,H(0,1)cR

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