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 $L^p(X)$

1. January 29, 2012 at 11:14 pm

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

• 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. 🙂