Wednesday, 7 May 2014

real analysis - Limit of Lp norm



Could someone help me prove that given a finite measure space (X,M,σ) and a measurable function f:XR in L and some Lq, limpfp=f?



I don't know where to start.



Answer



Fix δ>0 and let Sδ:={x,|f(x)|fδ} for δ<f. We have
fp(Sδ(fδ)pdμ)1/p=(fδ)μ(Sδ)1/p,
since μ(Sδ) is finite and positive.
This gives
lim infp+fpf.
As |f(x)|f for almost every x, we have for p>q, fp(X|f(x)|pq|f(x)|qdμ)1/pfpqpfq/pq,
giving the reverse inequality.


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