Suppose that $B$ be a ball in $\mathbb{R}^n$ and $f\in L^1_{\rm loc}(\mathbb{R}^n)$, let \begin{equation*} \|f\|_{BMO}=\sup_{B}\frac{1}{|B|} \int_{B}|f(x)-f_{B}|dx, \end{equation*} where $$ f_{B}=\frac{1}{|B|} \int_{B} f(x)dx. $$ Define $$ BMO(\mathbb{R}^n)=\{ f\in L^1_{\rm loc}(\mathbb{R}^n) ~ : ~ \| f \|_{BMO} < \infty \}. $$
Claim: For all $r>0$, there exists $x_0\in \mathbb{R}^n$ such that $$|B(x_0,r)|^{-1} \,\|f(\cdot)-f_{B(x_0,r)}\|_{L_{1}(B(x_0,r))} > \frac{\|f\|_{BMO}}{2}.$$
I need such a property. But i can not show that if it is valid or not?
My attempt: I use the property $\sup A=a\iff \forall \epsilon>0~~ \exists x_0\in A~~ x_0>a-\epsilon.$
Suppose $\|f\|_{BMO}>0$ and choose $\epsilon=\frac{\|f\|_{BMO}}{2}$ then we have there exist a ball $B_0$ such that $$ |B_0|^{-1} \,\|f(\cdot)-f_{B_0}\|_{L_{1}(B_0)} > \frac{\|f\|_{BMO}}{2}. $$ After that how can i continue?
The claim is not true. Take $n=1$, $f= \chi_{(0,1)}$. Then certainly $\|f\|_{BMO}>0$.
Now let $r>1$, let $B$ be a ball with radius $r$ such that $|B\cap (0,1)| = \delta\in [0,1]$. Then $$ f_B = \frac \delta {2r} $$ and $$ \frac1{|B|}\int_\mathbb R| f-f_B| =\frac1{2r}\left ((2r-\delta) \frac \delta {2r} + \delta (1- \frac \delta {2r}) \right) \le \frac1{2r}\left ( {2r} \frac \delta {2r} + \delta ) \right) \le \frac1r $$ which tends to zero for $r\to\infty$.
So the claim is false for this particular function and sufficiently large $r$.