Let $\Omega \subset \mathbb R^d, d\ge 2$, be bounded and denote a ball in $\mathbb R^d$ by $B$. Denote also $$ f_B:= \frac1{|B|}\int_B f \, dx. $$ Suppose $f \in L_{\rm loc}^p(\Omega)$ for all $1<p<\infty$ and $$\tag 1 \left|\{x \in B : |f(x)-f_{2B}| > \lambda\}\right| \le c_1 \exp\left(-c_2\frac\lambda{\left[\frac1{|2B|}\int_{2B} |f(x)-f_{2B}|^\delta \, dx\right]^\frac1\delta}\right)|B| $$ for all $\delta>0$ and for all balls $2B \subset \Omega$, with some constants $c_1, c_2=c_2(\delta) >0$. My question is whether this - or some suitable modification of this - would be enough to guarantee that $$ f \in BMO(\Omega), $$ where $$ BMO(\Omega) := \left\{f \in L^1(\Omega): \sup_{B\subset \Omega} \frac1{|B|}\int_B |f(x)-f_B| \, dx < \infty\right\}. $$

Any references which might be useful are greatly appreciated.

  • $\begingroup$ Unfortunately, nothing short of (one of many equivalent forms of) the definition itself works. So, in this generality the question is unanswerable. Try to ask exactly what you need and then you may have a chance. $\endgroup$ – fedja Jun 9 '15 at 13:24
  • $\begingroup$ Thanks for the feedback; I added some details for what exactly I am after. $\endgroup$ – Juhana Siljander Jun 10 '15 at 15:20

In the simpler dyadic setting the answer to your question is no. Consider independent Bernoulli ($\pm 1$) random variables, aka Rademacher functions, $X_1,X_2,\dots$. Take any function that is quadratic in them:

$$f := \sum_{i < j} c_{ij} X_i X_j,$$ $$\sum_{i<j} |c_{ij}|^2 < \infty$$

(it can include diagonal terms too, but they are not important).

Take the dyadic partition, $\mathcal F_n := \sigma(X_{\le n})$ and decompose $f$ conditionally on $\mathcal{F}_n$ into the "conditionally linear" part and the "conditionally quadratic" part:

$$f - \mathsf{E}\left[f \middle| \mathcal F_n\right] = \sum_{i \le n} X_i \sum_{j>n} c_{ij} X_j + \sum_{n<i<j} c_{ij} X_i X_j$$

By our construction, conditionally on $\mathcal{F}_n$, $f$ is still a quadratic polynomial in $X$, and it is known that on those the $L^1$ norm is equivalent to the exponential Orlicz norm (this follows from hypercontractivity, see e.g. Theorem 6.7 in Janson "Gaussian Hilbert spaces"), so the John-Nirenberg-type bound does hold. On the other hand, the conditional variance of $f$ given $\mathcal{F}_n$ is bounded from below by the conditional variance of the linear part, which is:

$$\sum_{j > n} \left(\sum_{i \le n} c_{ij} X_i \right)^2$$

This has no reason to be bounded, so $f$ is generally not in $\mathrm{BMO}$.

  • $\begingroup$ Can you remind me what is the exponential Orlicz norm of a function $f$? In particular, does the condition (1) induce some known function space with such a norm? $\endgroup$ – Juhana Siljander Jun 12 '15 at 7:00
  • $\begingroup$ I meant any of the equivalent norms in the Orlicz space with a function that grows exponentially - e.g. $\Phi(x) := \cosh x - 1$. For the definition of Orlicz space see en.wikipedia.org/wiki/Birnbaum%E2%80%93Orlicz_space. $\endgroup$ – Alexander Shamov Jun 12 '15 at 13:27
  • $\begingroup$ Thanks! It took me some time to figure out what you mean, but I think you are correct. $\endgroup$ – Juhana Siljander Jun 17 '15 at 7:46

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