Home > Measure Theory, Real Analysis > Pointwise convergence implies other type of convergence

Pointwise convergence implies other type of convergence


Let (X,\mathcal M,\mu) be a measure space for which \mu(X)<\infty. Let 1< p< \infty. Suppose that \{f_k\} is a sequence in L^p(X) such that sup_k \|f_k\|_p<\infty and \lim_{n \to \infty}f_k(x)=f(x) exists for \mu-a.e. x. Prove that \lim_{k \to \infty} \|f_k-f\|_1 =0.

PHD 4324 (Indiana)

Proof: If we assume that \lim_{ k \to \infty} \|f_k-f\|_1 \neq 0 then there exists a subsequence denoted also \{f_k\} such that \lim_{k \to \infty } \|f_k-f\|_1 =\alpha >0. Then \{|f_k|\} is bounded in L^p(X) and L^p(X) is reflexive there exists a subsequence \{|f_k|\} converges weakly to |f| in L^p(X)

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  1. Davide Giraudo
    January 29, 2012 at 11:14 pm

    There is an other proof which uses Egoroff’s theorem.

    • Davide Giraudo
      January 30, 2012 at 12:50 pm

      Here are the arguments: let M:=\sup_{n\in\mathbb N}\lVert f_n\rVert_{L^p}. We fix \delta>0 and A_{\delta}\in\mathcal M such that f_n\to f uniformly on \delta and \mu(X\setminus A_{\delta})\leq \delta. We have
      \displaystyle \int_X|f_n-f|d\mu \leq \mu(X)\sup_{x\in A_{\delta}}|f_n(x)-f(x)|+\int_{A_{\delta}^c}(|f|+|f_n|)d\mu\leq \mu(X)\sup_{x\in A_{\delta}}|f_n(x)-f(x)|+2M\mu(A_{\delta}^c)^{\frac p{p-1}},
      so \limsup_n\int_X|f_n-f|d\mu  \leq 2M\delta^{\frac p{p-1}}, which gives the result.

      • January 30, 2012 at 12:58 pm

        Thank you very much for your solution. 🙂

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