Let $0<\alpha<n$. The fractional maximal operator $M_{\alpha}$ is given by \begin{equation*} M_{\alpha} f(x)=\sup_{B\ni x}|B|^{-1+ \frac{\alpha}{n}}\int _{B} |f(y)|dy, \end{equation*} where the supremum is taken over all balls $B\subset\mathbb{R}^n$ containing $x$.
I want to prove the following equality: For any fixed ball $B_0 \subset \mathbb{R}^n$ and all $x \in B_0$
$$ M_{\alpha}(\chi_{B_0})(x)=|B_0|^{\alpha/n}, $$ where $\chi_{B_0}$ is the characteristic function of $B_0$.
My attempt:
$$ M_{\alpha}(\chi_{B_0})(x)=\sup_{B\ni x}|B|^{-1+ \frac{\alpha}{n}}\int_{B} \chi_{B_0}(y)dy\geq |B_0|^{\alpha/n} $$
Now i have to show $M_{\alpha}(\chi_{B_0})(x)\le|B_0|^{\alpha/n}$. Let $B\subset B_0$ and $x\in B$, then
$$ |B|^{-1+ \frac{\alpha}{n}}\int_{B} \chi_{B_0}(y)dy=|B|^{\frac{\alpha}{n}}\le |B_0|^{\frac{\alpha}{n}}. $$ In this case i can only take supremum over balls $B\subset B_0$ and $x\in B$, therefore i can not obtain fractional maximal function in the left-hand side of inequality. I need some help.
When $|B_0|\leq |B|$, observe that $$|B|^{-1+\frac{\alpha}{n}}\int_{B} \chi_{B_0}(y)dy=|B|^{-1+\frac{\alpha}{n}} |B_0\cap B|\leq |B_0|^{-1+\frac{\alpha}{n}}|B_0|=|B_0|^{\frac{\alpha}{n}} $$ because $-1+\frac{\alpha}{n}< 0$ and $B_0\cap B\subseteq B_0$.
In the case when $|B|\leq |B_0|$, observe that $$ |B|^{-1+\frac{\alpha}{n}}\int_{B} \chi_{B_0}(y)dy = |B|^{-1+\frac{\alpha}{n}} |B_0\cap B|\leq |B|^{-1+\frac{\alpha}{n}}|B|=|B|^{\frac{\alpha}{n}}\leq |B_0|^{\frac{\alpha}{n}} $$ because $|B|\leq |B_0|$ and $\frac{\alpha}{n}> 0$
Which implies that $$|B|^{-1+\frac{\alpha}{n}}\int_{B} \chi_{B_0}(y)dy \leq |B_0|^{\frac{\alpha}{n}}$$ regardless of size of the ball.
Therefore, we get $$M_\alpha\chi_{B_0}(x)\leq |B_0|^{\frac{\alpha}{n}}.$$