2
$\begingroup$

Let $B_R$ be an open bounded ball in $\mathbb{R}^n$. I am trying to show that if $u\in BMO(B_R)$ then $u\in \mathcal{L}^{1, n}(B_R)$ and that \begin{equation} \|u\|_{\mathcal{L}^{1, n}(B_R)}\leq C\|u\|_{BMO(B_R)}. \end{equation} I would like to prove this without using the John-Nirenberg theorem, since most books say it follows from the definition of the two spaces.

So let $K_{\rho}(x)$ be a ball with $x\in B_R$ and $0<\rho<2R$. It suffices to show that \begin{equation*}\tag{1} \rho^{-n}\int_{B_R\cap K_{\rho}(x)}|u-u_{B_R\cap K_{\rho}}|\ \mathrm{d}y\leq C\|u\|_{BMO(B_R)}\quad\forall\ x\in B_R,\ \rho\in(0, 2R). \end{equation*} So I thought this is what I should try to show but I got stuck. Below is my working.

Let $r_x$ denote the distance from $x\in B_R$ to $\partial B_R$. We will denote $B_R$ by $B$ and $K_{\rho}(x)$ by $K$ in the following. Then \begin{align} \rho^{-n}\int_{B\cap K}|u-u_{B\cap K}|\ \mathrm{d}y&\leq\rho^{-n}\int_{B\cap K}|u-u_{B_{r_x}}|\ \mathrm{d}y+\rho^{-n}\int_{B\cap K}|u_{B_{r_x}}-u_{B\cap K}|\ \mathrm{d}y\\ &\leq\rho^{-n}\int_{B\cap K}|u-u_{B_{r_x}}|\ \mathrm{d}y+\frac{\alpha(n)}{|B\cap K|}\int_{B\cap K}|u-u_{B_{r_x}}|\ \mathrm{d}y\\ &\leq \rho^{-n}\int_{B\cap K}|u-u_{B_{r_x}}|\ \mathrm{d}y+\frac{\alpha(n)}{A\rho^n}\int_{B\cap K}|u-u_{B_{r_x}}|\ \mathrm{d}y\\ &\leq C_1\rho^{-n}\int_{B\cap K}|u-u_{B_{r_x}}|\ \mathrm{d}y, \end{align} where the second last inequality used the fact that balls are domains of type $A$ so $|B\cap K|\geq A\rho^n$. With the last estimate, keeping in mind that $B_{r_x}\subset B\cap K$ and that $\rho>r_x$, I could only get as far as

\begin{align} C_1\rho^{-n}\int_{B\cap K}|u-u_{B_{r_x}}|\ \mathrm{d}y &\leq \frac{\alpha(n)C_1}{|B_{r_x}|} \int_{B_{r_x}}|u-u_{B_{r_x}}|\ \mathrm{d}y+C_1\rho^{-n}\int_{B\cap K\setminus B_{r_x}}|u-u_{B_{r_x}}|\ \mathrm{d}y\\ &\leq C_2|u|_{\ast}+C_1\rho^{-n}\int_{B\cap K\setminus B_{r_x}}|u-u_{B_{r_x}}|\ \mathrm{d}y. \end{align}

I don't know how to get the second term either bounded above by the $L^1$ norm of $u$ or the seminorm $|u|_{\ast}$ to conclude (1).

$\endgroup$
  • $\begingroup$ When I look at the definition of Campanato space and put $p=1$, $\lambda=n$ there, I see the definition of BMO. Which raises the question: what are the definitions that you are working with? You stated neither. $\endgroup$ – user129630 Feb 19 '14 at 5:19
  • $\begingroup$ The definition of $BMO(B_R)$ I am using is $u\in BMO(B_R)$ if $u\in L^1(B_R)$ and \begin{equation}\|u\|_{BMO(B_R)}=\|u\|_{1}+|u|_{\ast, B_R}<\infty\end{equation} where\begin{equation}|u|_{\ast, B_R}=\sup_{B\subset B_R}\frac{1}{|B|}\int_{B}|u-u_B|\ \mathrm{d}x.\end{equation} I can see how for balls that are wholly contained in $B_R$ the definition of Campanato spaces implies the definition of BMO with perhaps a constant, but I'm not sure about balls that are not wholly contained in $B_R$. $\endgroup$ – Nirav Feb 19 '14 at 7:46
  • $\begingroup$ I am using the same definition of Campanato space as in your link. $\endgroup$ – Nirav Feb 19 '14 at 7:49
  • $\begingroup$ The books saying that the equivalence is by definition are probably concerned with spaces of functions on $\mathbb R^n$. For spaces of functions on $\mathbb R^n$ this is a tricky business; essentially the matter amounts to showing that extending a BMO function on a ball by $0$ preserves the BMO property. This (if true; I'd have to check) is not an easy consequence of definition. $\endgroup$ – user129630 Feb 19 '14 at 14:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.