Show that $\log(x)$ is a Bounded Mean Oscillation (BMO) As an extension of our class notes, we were asked to show that the function $w =\log(x)$ is a Bounded Mean Oscillation (BMO).
First off, I believe our professor made a mistake, and really wanted us to show that $w = \log|x|$ was a BMO, just to keep everything real.
That being said, by the John-Nirenburg Lemma, a function $w$ is a bounded mean oscillation (BMO) if $\displaystyle \frac{1}{|B_{r}|}\int_{B_{r}}|w-\overline{w}_{B_{r}}|\leq C$, where $B_{r}$ is a ball of radius $r$, $\displaystyle \overline{w}_{B_{r}}$ is the average value of $w$ over the ball of radius $r$ (so $\displaystyle \overline{w}_{B_{r}}$ can also be written as $\displaystyle \frac{1}{|B_{r}|}\int_{B_{r}}|w|$), and $C$ is a constant.
I would like to do and understand this exercise, but so far, all I've been able to do is set it up.  I figured that since $\displaystyle w = \log|x|$, then $\displaystyle \frac{1}{|B_{r}|}\int_{B_{r}}\left|\,\log|x|-\frac{1}{|B_{r}|}\int_{B_{r}}\log|x|\, \right|$.
I'm not sure how to manipulate this expression in order to show that it is $\leq C$.  Should I add and subtract some quantities? Should I try to introduce some exponentials?  I think the two integrals are tripping me up a bit, as well.  So, any suggestions and/or assistance you could offer would be much appreciated :)
Thank you in advance! 
 A: The nice thing about the interaction between $\log x$ and that expression is that they are both scale-invariant, which allows us to eliminate one of the degrees of freedom. To be precise, suppose the boundaries of the ball are $(xy,x)$ where $x>0$ and $0<y<1$. Then:
\begin{align}
\bar w(x,y)&=\frac1{x(1-y)}\int_{xy}^x\log t\, dt\\
&=\frac1{1-y}\int_y^1(\log u+\log x)\, du=\bar w(1,y)+\log x\\
MO(x,y)&=\frac1{x(1-y)}\int_{xy}^x\left|\log t-\bar w(x,y)\right|\,dt\\
&=\frac1{1-y}\int_y^1\left|\log u+\log x-\bar w(x,y)\right|\,du\\
&=\frac1{1-y}\int_y^1\left|\log u-\bar w(1,y)\right|\,du=MO(1,y)
\end{align}
Now that everything is only in terms of $y$, let's just do the integral:
$$\bar w(1,y)=\frac1{1-y}\int_y^1\log t\, dt=\frac{y\log y-y+1}{y-1}$$
$$MO(1,y)=\frac1{1-y}\int_y^1\left|\log t-\frac{y\log y-y+1}{y-1}\right|\,dt$$
Since $\log x$ is an increasing function, we can expect this integrand to have two regions. Left of the critical point $t_0$ such that $\log t=\frac{y\log y-y+1}{y-1}$, the function will look like $-\log t$, and to the right of $t_0$, it will look like $\log t$ (but shifted vertically). The right side is easy to bound: the integrand is bounded by $1$:
$$\log t-\frac{y\log y-y+1}{y-1}\le\frac{y\log y}{1-y}+1\le1,$$
because $x\log x<0$ for $x\in(0,1)$. The left side is unbounded, but we can use a simple trick to bound the value of the integral. By definition, the mean value $\bar w$ is chosen so that the areas under the left part and the right part are equal; but we have just shown that the area of the right part is at most $1\cdot(1-t_0)\le1-y$, so the total area under the function must be at most $2(1-y)$, and so $MO(1,y)=MO(x,y)\le2$.
