Pointwise convergence implies other type of convergence

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Pointwise convergence implies other type of convergence

January 3, 2012 beni22sof Leave a comment Go to comments

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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Categories: Measure Theory, Real Analysis Tags: measure space

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Davide Giraudo

January 29, 2012 at 11:14 pm | #1

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

Davide Giraudo

January 30, 2012 at 12:50 pm | #2

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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.