# Completeness of $(BMO(\Bbb R^n),||\cdot||_{\ast})$

Recall that $$BMO(\Bbb R^n):=\left\{f\in L_{loc}^1(\Bbb R^n)\;\mbox{modulo constant functions, such that}\\ \forall B\subseteq\Bbb R^n\;\mbox{ball}, \exists\alpha(B)\in\Bbb R\;\;\mbox{such that}\\ \frac1{|B|}\int_B|f(x)-\alpha(B)|\,dx\le C\;\;\;\;\exists C>0\right\}$$ Observe that $C$ is uniform in $B$. Given $f\in BMO$, the infimum of such $C$'s is the $BMO$ norm of $f$, denoted by $||f||_*$.

In particular, elements of $BMO$ are not functions, they are classes of functions.

I have to prove that $(BMO(\Bbb R^n),||\cdot||_{\ast})$ is complete.

So let's take a Cauchy sequence $\{f_n\}_{n\ge1}\subseteq BMO(\Bbb R^n)$; thus $$||f_n-f_m||_*\to0\;\;\;\mbox{as}\;\;\;n,m\to+\infty.$$ So we can extract a subsequence $\{f_{n_k}\}_{k\ge1}\subseteq \{f_n\}_{n\ge1}$ such that, given $\epsilon>0\;\;\;\exists N_{\epsilon}\ge1$, such that $$||f_{n_k}-f_{n_1}||_*<\epsilon 2^{-k}\;\;\;\forall k\ge1$$ from which clearly $$||f_{n_k}-f_{n_1}||_*\to0\;\;\;\mbox{as}\;\;k\to+\infty$$ i.e. $\{f_n\}_n$ admits a converging subsequence, thus $\{f_n\}_n$ itself converges, so we have done.

But this doesn't sounds good. The fact we are dealing with classes and not with functions must play some role, but I can't understand what is. Can you tell me if it's right or not please? Many thanks

It is well known that if $f \in \text{BMO}(\mathbb R^n)$, then for any $\epsilon > 0$ there exists a ball $B$ with center $x_0$ and radius $R$ such that \begin{equation}\label{3} R^\epsilon \int_{\mathbb R^n} \frac{|f(x) - f(B)|}{(R + |x - x_0|)^{n + \epsilon}} \, dx \leq C_{n,\epsilon} \|f\|_*\;. \end{equation} We can deduce from this theorem that any $\text{BMO}$-Cauchy sequence is Cauchy in $L^1$ on every compact subset of $\Bbb R^n$. From this we will deduce that every Cauchy sequence converges. \newline First of all, setting $\int_B^*:=\frac1{|B|}\int_B$ let's define $$\text{quasi-BMO}(\Bbb R^n):=\left\{f\in L_{loc}^1(\Bbb R^n)\;:\; \forall B\subseteq\Bbb R^n\;\mbox{ball}\\ \int_B^*|f(x)-f(B)|\,dx\le C\;\;\;\;\exists C>0\right\}$$%\end{align*} Given $f\in\text{quasi-BMO}$, let's define then $$\|f\|_q:=\sup_B\fint_B|f(x)-f(B)|\,dx$$ the problem is that $\|\cdot\|_q$ is actually a seminorm, that is $\|f\|_q = 0$ does not only occur when $f = 0$ but actually for all constants. $\text{BMO}$ space is built defining an equivalence relation on $\text{quasi-BMO}$ in order to making it properly a normed space.

So let $f \sim g$ if and only if $f - g$ is a constant. The normed space we call $\text{BMO}$ is thus the quotient space $\text{quasi-BMO}/\{\mbox{const. funct.}\}$ and the projection $\pi:\text{quasi-BMO}\to\text{BMO}$ that sends every function to its equivalence class, i.e. $\pi(f):=[f]$, is linear and continuous.

Also, we have a norm on this space $\|[f]\|_* = \inf_{c \in \mathbf{R}} \|f + c\|_q$. If we want to show completeness, we would have to build a function where a Cauchy sequence converges to but this doesn't seem smart. We rather use the following:

$\textbf{Theorem:}$ A normed space $(X,\|\cdot\|)$ is complete iff $\sum_m x_m$ converges in the norm for every sequence $\{x_m\}_m\subseteq X$ such that $\sum_m \|x_m\|<+\infty$.

First of all, note that $\|[f]\|_*=\|f\|_q$ for all $f\in\text{quasi-BMO}$: \begin{align*} \|[f]\|_* &= \inf_{c\in\R} \|f + c\|_q\\ &=\inf_{c\in\R}\sup_B\fint_B\left|f(x)+c-\fint_B(f(y)+c)dy\right|\,dx\\ &=\inf_{c\in\R}\sup_B\fint_B|f(x)-f(B)|\,dx\\ &=\sup_B\fint_B|f(x)-f(B)|\,dx\\ &=\|f\|_q \end{align*} Take now a sequence of classes of functions $\{[f_m]\}_m\subseteq\text{BMO}$ such that $\sum_m\|[f_m]\|_*<+\infty$.

So we have a sequence of functions $\{f_m\}_m\subseteq\text{quasi-BMO}$ such that $\sum_m\|f_m\|_q<+\infty$.

Now let $B$ be a closed ball with radius $R$ centered in $x_0\in\Bbb R^n$. Then \begin{align*} \frac{R^{\epsilon}}{(2R)^{n+\epsilon}}\|f_m\|_{L^1(B)} &=R^{\epsilon}\int_B\frac{|f_m(x)|}{(2R)^{n+\epsilon}}\,dx\\ &\le R^{\epsilon}\int_B\frac{|f_m(x)-f_m(B)|}{(2R)^{n+\epsilon}}\,dx+\frac{R^{\epsilon}}{|B|(2R)^{n+\epsilon}}\overbrace{\int_B|f_m(x)|\,dx}^{\|f_m\|_{L^1(B)}}\\ \end{align*} from which (we take wlog $|B|>1$) \begin{align*} \left(1-\frac1{|B|}\right)\left(\frac{R^{\epsilon}}{(2R)^{n+\epsilon}}\right)\|f_m\|_{L^1(B)} &\le R^{\epsilon}\int_B\frac{|f_m(x)-f_m(B)|}{(2R)^{n+\epsilon}}\,dx\\ &\le R^\epsilon \int_{B} \frac{|f_m(x)-f_m(B)|}{(R + |x-x_0|)^{n + \epsilon}} \, dx\\ %&=R^\epsilon \int_{B} \frac{|f_m(x)-f_m(B)|}{(R + |x |)^{n + \epsilon}} \, dx\\ &\le R^\epsilon \int_{\Bbb R^n} \frac{|f_m(x)-f_m(B)|}{(R + |x-x_0|)^{n + \epsilon}} \, dx\;\tag{see (\ref3)}\\ &\le C_{n,\epsilon} \|f_m\|_q\;\;.\\ %&\le C_{n,\epsilon} \|[f_m]\|_*\\ \end{align*} Take now a compact subset $K\Subset\Bbb R^n$; so there exist $R>0$, $x_0$ such that $K\subseteq B$ and $|B|>1$; from this we deduce that $$\sum_m\|f_m\|_{L^1(K)}\le \sum_m\|f_m\|_{L^1(B)}\le C\sum_m\|f_m\|_q<+\infty$$ so being $L^1(K)$ complete, by the Theorem stated above we have that $\sum_m f_m$ converges in $L^1(K)$, to a function we call $F_K$, i.e. $\|\sum_m f_m-F_K\|_{L^1(K)}\to0$ as $m\to+\infty$.\ Define $F:\Bbb R^n\to\Bbb R$ to be $F_K$ on every compact subset $K$. Let's see $F$ is well defined.\ If $K_1,K_2\Subset\Bbb R^n$ are such that $K_1\cap K_2\neq\emptyset$, let's show that $${{F_{K_1}}_|}_{K_1\cap K_2}\equiv {{F_{K_2}}_|}_{K_1\cap K_2}.$$ Clearly $\|\sum_m f_m-{F_K}_1\|_{L^1(K_1\cap K_2)}\le\|\sum _m f_m-{F_K}_1\|_{L^1(K_1)}\to0$; similarly $\|\sum_m f_m-{F_K}_2\|_{L^1(K_1\cap K_2)}\to0$ thus $\sum_m f_m$ converges to both $F_{K_1}$ and $F_{K_2}$ in $L^1(K_1\cap K_2)$, so (possibly passing to a subsequence) $\sum_m f_m$ converges pointwise, a.e. on $K_1\cap K_2$, to both $F_{K_1}$ and $F_{K_2}$ which hence must here coincide, as wanted.\

So we have our $F\in L_{loc}^1$ which is well defined and $\sum_m f_m$ converges to $F$ in $L_{loc}^1(\Bbb R^n)$.\ Let's prove now that $\sum_m f_m$ converges to $F$ in $\text{quasi-BMO}$: \begin{align*} \fint_B \left|\sum_{m=m_0}^N f_m(x)-F(x)-\left(\sum_{m=m_0}^N f_m-F\right)(B)\right|\,dx &\le\fint_B \left|\sum_{m=m_0}^N f_m(x)-F(x)\right|\,dx+\left|\left(\sum_{m=m_0}^N f_m-F\right)(B)\right|\\ &\le\frac2{|B|}\int_B \left|\sum_{m=m_0}^N f_m(x)-F(x)\right|\,dx\\ &=\frac2{|B|}\left\|\sum_{m=m_0}^N f_m-F_B\right\|_{L^1(B)} \end{align*} from which we get \begin{align*} \left\|\sum_{m=m_0}^N f_m-F\right\|_q \le\sup_B\frac2{|B|}\left\|\sum_{m=m_0}^N f_m-F_B\right\|_{L^1(B)} \end{align*} thus \begin{align*} \lim_N\left\|\sum_{m=m_0}^N f_m-F\right\|_q &\le\lim_N\sup_B\frac2{|B|}\left\|\sum_{m=m_0}^N f_m-F_B\right\|_{L^1(B)}\\ &=\sup_B\frac2{|B|}\lim_N\left\|\sum_{m=m_0}^N f_m-F_B\right\|_{L^1(B)}=0 \end{align*} I know that the interchange between $\lim$ and $\sup$ should be checked, but it seems reasonable to accept it.\ Thus $\sum_m f_m$ converges to $F$ in $\text{quasi-BMO}$ and since \begin{align*} \left\|\sum_{m=m_0}^N f_m-F\right\|_q &=\left\|\pi\left(\sum_{m=m_0}^N f_m-F\right)\right\|_* \tag{$\pi$ is linear}\\ &=\left\|\sum_{m=m_0}^N [f_m]-[F]\right\|_* \end{align*} we can conclude that the given sequence $\{[f_m]\}_m\subset \text{BMO}$ taken such that $\sum_m\|[f_m]\|_m<+\infty$, is then such that $\sum_m[f_m]$ converges in $\text{BMO}$, which is thus complete by the Theorem above.