Is ${L^p}({\mathbb R^n},B)$ complete for $p \ge 1$ where $B$ is a Banach space?
The completeness of ${L^p}({\mathbb R^n},B)$
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7It would be nice if you started showing at least a little bit of your own work and thoughts. Your questions always are one or two lines and don't show any research effort. Most of the results you ask about are standard and easily located in the literature. Where do you get the statements from? What did you try to do to prove them yourself? Where did you get stuck? – 2011-12-13
1 Answers
If $1\leq p<\infty$, let $\{f_n\}$ be a Cauchy sequence in $L^p(\Omega,B)$. We will show that there is a converging subsequence. We can assume that $\lVert f_n-f_m\rVert_{L^p(\Omega,B)}\leq 2^{-n}$ for $m>n$, after taking a subsequence, denoted in the same way. Put $A(n):=\left\{x\in\Omega\mid \lVert f_n(x)-f_{n+1}(x)\rVert_B\geq n^{-2}\right\}$. Then $\mathbf 1_{A(n)}n^{-2}\leq \lVert f_n(x)-f_{n+1}(x)\rVert_B$ for all $x$ hence $\mu(A_n)^pn^{-2p}\leq \int_{\Omega}\lVert f_n(x)-f_{n+1}(x)\rVert_B^pd\mu(x)\leq 2^{-np}$ so $\mu(A_n)\leq n^22^{-n}$. Put $B(n):=\bigcup_{m\geq n}A(n)$. The sequence $\{B(n)\}$ is decreasing and $\mu(B(n))\to 0$.
We define $N:=\bigcap_n B(n)$. If $x\notin N$, then for $n$ large enough we have $\lVert f_n(x)-f_{n+1}(x)\rVert_B\leq n^{-2}$, and since $B$ is complete, let $f(x):=\lim_n f_n(x)$. Put $f(x)=0$ if $x\in N$. Then we use Fatou lemma to get that $f\in L^p(\Omega,B)$ and $f_n\to f$ in $L^p(\Omega,B)$: $\int_{\Omega}\lVert f(x)\rVert^pd\mu(x)\leq \liminf_n\int_{\Omega}\lVert f_n(x)\rVert^pd\mu(x)<\infty$ since a Cauchy sequence is bounded, and $\int_{\Omega}\lVert f_n(x)-f(x)\rVert^pd\mu(x)\leq \liminf_m\int_{\Omega}\lVert f_n(x)-f_m(x)\rVert^pd\mu(x)\leq 2^{-np}.$ This concludes the proof in the case $p$ finite.
If $p=+\infty$ and $\{f_n\}$ is a Cauchy sequence in $L^{\infty}(\Omega,B)$, we have for almost every $x$ $\lVert f_n(x)-f_m(x)\rVert_B\leq \lVert f_n-f_m\rVert_{\infty}$. We use the completeness of $B$ to get a limit $f(x)$ for these $x$, and we put $f(x)=0$ for the other $x$. We get that $f\in L^{\infty}(\Omega,B)$ and $f_n\to f$ in $L^{\infty}(\Omega,B)$.