An inequality about the $BMO$ space from Garfako's Modern Fourier Analysis

Question

In Garfako's Modern Fourier Analysis I came across the following inequality: $$ \left| \left| f \right|-\underset{Q}{\mathrm{Avg}}\left| f \right| \right|\le \left| f-\underset{Q}{\mathrm{Avg}}f \right|+\underset{Q}{\mathrm{Avg}}\left| f-\underset{Q}{\mathrm{Avg}}f \right|, $$ where $$ \underset{Q}{\mathrm{Avg}}f=\frac{1}{\left| Q \right|}\int_Q{f\mathrm{d}x}. $$ I had difficulty in showing this inequality. I attempt to do it with the inequality $|a+b|\le |a|+|b|$, but I failed. Can you kindly explain why this inequality is true? Thanks in advance.

Answer 1

$\newcommand{\d}{\,\mathrm{d}}\newcommand{\avg}{\operatorname{Avg}_Q}$A direct triangle inequality bash wasn't clear to me, but working backwards and trying to squeeze as much under the integral sign as possible does work. Very crucial is the point that $\avg a=a$ for a constant real $a$.

Observe for any fixed $x$: $$\begin{align}\avg|f|+|f(x)-\avg f|\\+\avg|f-\avg f|&=|f-\avg f|+\frac{1}{\mu(Q)}\int_Q(|f(y)|+|\avg f-f(y)|)\d\mu(y)\\&\ge|f(x)-\avg f|+\frac{1}{\mu(Q)}\int_Q|\avg f|\d\mu(y)\\&=|\avg f|+|f(x)-\avg f|\\&\ge|f(x)|\end{align}$$And: $$\begin{align}|f(x)|+|f(x)-\avg f|\\+\avg|f-\avg f|&=\frac{1}{\mu(Q)}\int_Q(|f(x)|+|\avg f-f(x)|+|f(y)-\avg f|)\d\mu(y)\\&\ge\frac{1}{\mu(Q)}\int_Q(|\avg f|+|f(y)-\avg f|)\d\mu(y)\\&\ge\frac{1}{\mu(Q)}\int_Q|f(y)|\d\mu(y)\\&=\avg|f|\end{align}$$Therefore: $$\Big||f|(x)-\avg|f|\Big|\le|f(x)-\avg f|+\avg|f-\avg f|$$Holds for any $x$, and: $$\Big||f|-\avg|f|\Big|\le|f-\avg f|+\avg|f-\avg f|$$May be written.