# A bound for the product of two functions in BMO

The question: Let's consider $f\in L^\infty(\mathbb{T})$ and $g\in BMO(\mathbb{T})$. I'm trying to figure out if the following inequality is true $$\|fg\|_{BMO}\leq C\|f\|_{L^\infty}\|g\|_{BMO}.$$

My approach: I tried using the duality with $H^1$. I get $$\|fg\|_{BMO}\leq \sup_{\|h\|_{H^1}\leq 1} 2\|f\|_{L^\infty}\|g\|_{BMO}\||h|\|_{H^1}.$$ So, if in the periodic case we have $h\in H^1\rightarrow |h|\in H^1$, I'm done.

In the real line, in order $H h\in L^1$ (H is the Hilbert transform) one needs $h$ with zero mean (this implies that $h\in H^1\rightarrow |h|\in H^1$ isn't true). However, this restriction seems like it might be bypassed in the periodic setting. For instance let's take $h(x)=c$ with $c$ a constant. Then $h\in H^1$ and $|h|\in H^1$.

I'm not very familiar with harmonic analysis, so, this question could be quite trivial for an expert.

PD: I ask this question in mathoverflow too.

-

The answer is no. The multipliers of BMO are bounded functions that are in LMO, meaning that $$\log (\frac 1{|I|})\frac 1{|I|}\int_I|f-f_{I}|dx\leq C.$$ This is a well-known theorem, due to S. Janson. See S. JANSON, On functions with conditions on the mean oscillation, Ark. Mat., 14 (1976), 189-196 and also D. A. STEGENGA, Bounded Toeplitz operators on Hl and applications of dualit between H1 and the functions of bounded mean oscillation, Amer. J. Math., 98 (1976), 573-589.

Also, it is well-known that a positive function is in $H^1$ if and only if it is in $L Log L$. This is due to E. Stein, see his books.

-

I believe a function like $\log|x|$ is in BMO$([-1,1])$, whereas $I_{x>0}\log x$ is not.

You can do this directly from the definition for $g(x) = I_{x>0} \log x$. Just consider $Q = [-\epsilon,\epsilon]$, and compute $\left(\frac1{|Q|} \int_Q |g - g_Q|^2 \, dx\right)^{1/2}$ and see that it blows up like $\log(1/\epsilon)$ as $\epsilon \to 0$.

To show that $\log|x|$ is in BMO, you can work directly from the definition. But you could also use the $H^1$-BMO duality. See that for any function $f \in H^1$ we have $$\left|\int_{-1}^1 \log|x| f(x) \, dx \right| = \left|\int_{-1}^1 \int_{|x|}^1 \frac{dy}y f(x) \, dx \right| = \int_0^1 \frac1y \int_{-|y|}^{|y|} f(x) \, dx \, dy \\ \le \int_0^1 Mf(y) \, dy \le \|Mf\|_1 = \|f\|_{H^1}$$ where $M$ denotes the maximal function.

You can also go from the $H^1$ side. Consider the functions $$f_\epsilon(x) = \frac1\epsilon \text{sign}(x) I_{|x|<\epsilon} .$$ Then show that $f_\epsilon$ is uniformly bounded in $H^1$, whereas $\||f_\epsilon|\|_{H^1}$ blows up like $\log(1/\epsilon)$ as $\epsilon\to 0$.

-
Thank you for your answer. However, this means that it's not true that $h\in H^1\implies |h|\in H^1$. But it doesn't say anything on the bound for the product, right? – guacho Dec 18 '13 at 5:45
So consider the operator $T_f:g\mapsto fg$, Its adjoint is also $T_f$. So $T_f$ is a bounded operator on $H^1$ if and only if it is a bounded operator on BMO, – Stephen Montgomery-Smith Dec 18 '13 at 12:46
My first example shows that multiplication by $I_{x>0}$ is not a bounded operator on BMO. My second example shows that multiplication by $\text{sign}(x)$ is not a bounded operator on $H^1$, – Stephen Montgomery-Smith Dec 18 '13 at 12:47
Are any of you aware of any proofs on this site that $\log|x|$ is a BMO using directly the definition? – Jessy Cat Apr 29 '14 at 2:23