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I'm trying to solve Exercise 7.1.4 of Grafakos' book $\textit{Modern Fourier Analysis}$: consider two real numbers $K<L$ and a function $f \in BMO(\mathbb{R^n})$. Define the truncation of $f$ $$f_{KL}(x)= \begin{cases} K &\text{ if } f(x)<K\\ f(x) &\text{ if } K\le f(x) \le L \\ L & \text{ if } L < f(x) \end{cases}.$$ $f_{KL}$ is in $L^{\infty}(\mathbb{R^n})$, then it is in $BMO(\mathbb{R}^n)$. The goal is to prove that $$\|f_{KL}\|_{BMO} \le \frac{9}{4}\|f\|_{BMO}.$$

I've tried to solve it bluntly. Consider a cube $Q$ of $\mathbb{R}^n$. If $Q$ is totally contained in $\{f<K\}$ or in $\{f>L\}$, then $\int_{Q} |f_{KL}(x)-(f_{KL})_{Q}|dx = 0$. If $Q$ is totally contained in $\{K\le f \le L\}$, then $$\frac{1}{|Q|}\int_{Q} |f_{KL}(x)-(f_{KL})_{Q}|dx = \frac{1}{|Q|}\int_{Q} |f(x)-(f)_{Q}|dx \le \|f\|_{BMO}.$$ Then the real difficulties appear when $Q$ is not totally contained in one of these three sets. I first assume that $Q \cap \{f<K\} = \emptyset$ and $Q \cap \{L<f\}\neq \emptyset \neq Q \cap \{K \le f \le L\}$. My first idea was to introduce $f$ using the triangle inequality: \begin{eqnarray} \frac{1}{|Q|}\int_{Q}|f_{KL}(x)-(f_{KL})_{Q}|dx & \le & \frac{1}{|Q|}\left( \int_{Q}|f_{KL}(x)-f(x)|+|f(x)-f_{Q}|dx \right) + |f_{Q}-(f_{KL})_{Q}|\\ & = & \frac{1}{|Q|}\int_{Q\cap\{L<f\}} |L-f(x)| dx + \frac{1}{Q}\int_{Q}|f(x)-f_{Q}|dx+ \left|\frac{1}{|Q|}\int_{Q} f_{Q}-f_{KL}(x) dx\right|\\ & \le & \frac{1}{|Q|}\int_{Q\cap\{L<f\}} (f(x)-L) dx + \|f\|_{BMO} + \frac{1}{|Q|}\int_{Q} |f_{Q}-f_{KL}(x)| dx \\ & \le & \int_{Q\cap\{L<f\}} (f(x)-L) dx + \|f\|_{BMO} + \frac{1}{|Q|} \int_{Q\cap \{L<f\}} |f_{Q}-L|dx + \frac{1}{|Q|} \int_{Q\cap \{ L \ge f \}} |f_{Q}-f(x)| dx \\ & \le & \int_{Q\cap\{L<f\}} (f(x)-L) dx + \|f\|_{BMO} + |Q \cap \{L<f\}| \frac{|f_{Q}-L|}{|Q|} + \frac{1}{|Q|} \int_{Q} |f_{Q}-f(x)| dx\\ & \le & 2\|f\|_{BMO} + \int_{Q\cap\{L<f\}} (f(x)-L) dx + |Q \cap \{L<f\}| \frac{|f_{Q}-L|}{|Q|} \end{eqnarray} but I don't know how to go on next... Does anybody know how to procced, or if there is a totally different way to handle the problem?

David Tewodrose
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2 Answers2

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Forget about $(f_{KL})_Q$! You don't need to worry about that to prove what you want. An often useful lemma:

Lemma For any $\alpha\in\Bbb C$ we have $\frac{1}{|Q|}\int_Q|f-f_Q| \le2\frac{1}{|Q|}\int_Q|f-\alpha|$.

Proof: $$\begin{align}\frac{1}{|Q|}\int_Q|f-f_Q|&\le\frac{1}{|Q|}\int_Q|f_Q-\alpha| +\frac{1}{|Q|}\int_Q|f-\alpha| \\&=|\alpha-f_Q|+\frac{1}{|Q|}\int_Q|f-\alpha| \\&=|(\alpha-f)_Q|+\frac{1}{|Q|}\int_Q|f-\alpha| \\&\le 2\frac{1}{|Q|}\int_Q|f-\alpha|.\end{align}$$

Cor If $|\phi(t)-\phi(s)|\le|t-s|$ then $||\phi\circ f||_{BMO}\le 2||f||_{BMO}$.

Proof $$\frac{1}{|Q|}\int_Q|\phi\circ f-(\phi\circ f)_Q| \le 2\frac{1}{|Q|}\int_Q|\phi\circ f-\phi(f_Q)| \le2\frac{1}{|Q|}\int_Q|f-f_Q|.$$

I don't know what $c=9/4$ is doing there, since getting $c=2$ is so simple...

David C. Ullrich
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This is late to the game, but someone else may be wondering the same thing.

As explained in an answer above (setting $\phi(t)=|t|$), $\lVert |f| \rVert_{BMO} \leq 2 \lVert f \rVert_{BMO}$ for $f\in BMO(\mathbb{R}^n)$. Using this, the triangle inequality, and writing $\max(f,g)=\frac{1}{2}[f+g+|f-g|]$ we have that \begin{align} \lVert \max(f,g) \rVert_{BMO} &\leq \frac{\lVert f \rVert_{BMO}+\lVert g \rVert_{BMO} + \lVert |f-g| \rVert_{BMO}}{2}\\&\leq \frac{\lVert f \rVert_{BMO}+\lVert g \rVert_{BMO} + 2\lVert f-g \rVert_{BMO}}{2}\\&\leq \frac{\lVert f \rVert_{BMO}+\lVert g \rVert_{BMO} + 2(\lVert f \rVert_{BMO}+\lVert g \rVert_{BMO})}{2}\\&=\frac{3}{2}(\lVert f \rVert_{BMO}+\lVert g \rVert_{BMO}) \end{align} if both $f,g\in BMO(\mathbb{R}^n)$.

Now we come to it. Let $f\in BMO(\mathbb{R}^n)$ and consider the truncations $f_{KL}$ as defined above. One can rewrite this in the following way: $$ f_{KL}(x)=\max(\min(f(x),K), L). $$ Since constant functions have zero $BMO$ norm, the previous result implies that \begin{align} \lVert f_{KL} \rVert_{BMO} =\lVert \max(\min(f,K), L) \rVert_{BMO} \leq \frac{3}{2}\lVert \min(f,K) \rVert_{BMO}\leq \frac{9}{4}\lVert f\rVert_{BMO}. \end{align}

Of course, as explained in an above answer, one can more easily show that $\lVert f_{KL} \rVert_{BMO}\leq 2\lVert f\rVert_{BMO}$.