3
$\begingroup$

I'm looking for a proof of the fact that $BMO$ is a Banach Space. I know that it follows from the fact that it's the dual of the hardy space $H^1$, but I'm looking for a direct proof. I finally found this article(https://eudml.org/doc/218175) but unfortunately I cannot understand the end of the proof. First of all let we fix the notations:

$\left\lVert f\right\rVert_*=\left\lVert sup_Q \frac{1}{\left\lvert Q\right\rvert}\int_Q f-f_Q \right\rVert_\infty$, where $Q$ is a cube and $f_Q=\frac{1}{\left\lvert Q\right\rvert}\int_Q f$

$BMO:=\{f\in L^1_{loc}: \left\lVert f\right\rVert_*<\infty\}/\{f\in L^1_{loc}: f \text{ a.e. constant}\}$

The proof reads as follow:

Let $\{ f_n \}_{n\in\mathbb{N}}\subseteq BMO $ a Cauchy sequence and take a cube $Q$. Then: $$\frac{1}{\left\lvert Q\right\rvert}\int_Q \left\lvert(f_{n}-(f_{n})_Q)-(f_m-(f_m)_Q)\right\rvert\leq\left\lVert f_n-f_m\right\rVert_*$$ So $\{f_n-(f_n)_Q\}_{n\in\mathbb{N}}$ is a Cauchy sequence in $L^1(Q)$, which is complete, so it converges. Let $g^Q$ be the limit(be careful not to confuse with avarges notation).

Taking another cube $Q'\supset Q$ you have that $\{f_n-(f_n)_{Q'}\}_{n\in\mathbb{N}}\rightarrow g^{Q'}$ in $L^1(Q')$ and so also in $L^1(Q)$. So $\exists c=c(Q,Q')$ s.t. $(f_n)_{Q'}-(f_n)_Q\rightarrow c(Q,Q')$ in $L^1(Q)$ and $c(Q,Q')=g^Q-g^{Q'}$.

Now for any $k\in\mathbb{N}$ let $Q_k$ be the cube with center in the origin and side equals to k. Then define: $$f(x)=g^{Q_k}(x)-c(Q_1,Q_k) \qquad x\in Q_k$$ It is easy to check that $f$ is well defined(check the article for the details). Now we want to show that $f$ is the limit in BMO norm of $\{f_n \}_{n\in\mathbb{N}}$.

Let $Q$ be any cube, and $k$ s.t. $Q\subset Q_k$ then it easy to check(again see the article, but they are easy calculations) that: $$\int_Q \left\lvert (f_n-f)-(f_n-f)_Q\right\rvert=\int_Q \left\lvert f_n-(f_n)_Q-g^Q\right\rvert\rightarrow 0$$ Now they say that from this follows easily that $\left\lVert f_n-f\right\rVert_*\rightarrow 0$, but I cannot figure out why. BMO norm is the supremum over all cubes, so the limit on the RHS has to go to $0$ "uniformly" over all cubes. More explicitly, to conclude you have to prove this equality: $$\lim_n\left\lVert f_n-f\right\rVert_*=\lim_n \sup_Q \frac{1}{\left\lvert Q\right\rvert}\int_Q \left\lvert f_n-(f_n)_Q-g^Q\right\rvert=\sup_Q \lim_n \frac{1}{\left\lvert Q\right\rvert}\int_Q \left\lvert f_n-(f_n)_Q-g^Q\right\rvert=0$$

the problem is why we can switch the supremum and the limit, it isn't absolutely obvious to me. I've found some questions on the completeness of BMO (Completeness of BMO without duality to $H^1$ and Completeness of $(BMO(\Bbb R^n),||\cdot||_{\ast})$), but this point of the proof isn't clear.

$\endgroup$
6
  • $\begingroup$ What is BMO ?... $\endgroup$
    – Jean Marie
    Apr 16, 2023 at 21:12
  • $\begingroup$ It is $\dot F^0_{\infty,2}$ :p $\endgroup$
    – LL 3.14
    Apr 16, 2023 at 22:10
  • $\begingroup$ @JeanMarie en.wikipedia.org/wiki/Bounded_mean_oscillation $\endgroup$
    – Luca.b
    Apr 17, 2023 at 7:05
  • $\begingroup$ @LL 3.14 Very enlightning information... $\endgroup$
    – Jean Marie
    Apr 17, 2023 at 11:27
  • $\begingroup$ I believe there is a problem with the proof on the paper in the first place. I don't think the $f$ is well defined, despite how easy it seems. In fact, if for $x\in Q_k\subset Q_{k'}$ we want to show $g^{Q_k}(x)-C(Q_1, Q_k)=g^{Q_{k'}}(x)-C(Q_1,Q_{k'})$. Considering $C(Q_k, Q_{k'})=g^{Q_k}-g^{Q_{k'}}$, this amounts to prove $C(Q_k, Q_{k'})+C(Q_1,Q_{k'})=C(Q_1,Q_k)$, not $C(Q_1,Q_{k'})=C(Q_1,Q_k)+C(Q_k,Q_{k'})$, as claimed im the article. And what we need to prove is, in the end, equivalent to $g^{Q_k}(x)=g^{Q_{k'}}(x)$, which I think need not be true... $\endgroup$ Jan 2 at 11:36

1 Answer 1

1
$\begingroup$

I've found the solution, it was easy. I post it since I think it can be a reference to anyone looking for a complete and easy proof of the farct that BMO is a Banach space.

We want to show that $\forall \epsilon>0 \; \exists N=N(\epsilon)$ s.t. $$\frac{1}{\left\lvert Q\right\rvert}\int_Q \left\lvert f_n-(f_n)_Q-g^Q\right\rvert<\epsilon \qquad \forall Q,\forall n\geq N $$ Since $\{ f_n \}_{n\in\mathbb{N}}\subseteq BMO$ is Cauchy you can take $N=N(\epsilon)$ s.t. $$\frac{1}{\left\lvert Q\right\rvert}\int_Q \left\lvert(f_{n}-(f_{n})_Q)-(f_m-(f_m)_Q)\right\rvert\leq\left\lVert f_n-f_m\right\rVert_*<\frac{\epsilon}{2}\qquad \forall Q,\forall m,n>N$$ Since $f_n-(f_n)_Q\longrightarrow g^Q$ in $L^1(Q)$ you can take $m=m(\epsilon,Q)>N$ s.t. $$\frac{1}{\left\lvert Q\right\rvert}\int_Q \left\lvert f_m-(f_m)_Q-g^Q\right\rvert<\frac{\epsilon}{2}$$ So you have that: $$\frac{1}{\left\lvert Q\right\rvert}\int_Q \left\lvert f_n-(f_n)_Q-g^Q\right\rvert\leq \frac{1}{\left\lvert Q\right\rvert}\int_Q \left\lvert(f_{n}-(f_{n})_Q)-(f_m-(f_m)_Q)\right\rvert+\frac{1}{\left\lvert Q\right\rvert}\int_Q \left\lvert f_m-(f_m)_Q-g^Q\right\rvert<\epsilon \qquad \forall Q,\forall n\geq N $$ Notice that $m$ depends on $Q$ but that is not a problem, because the LHS of the inequality and $N$ are independent of $m$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .