A problem about Hardy-Littlewood maximal function from Folland real analysis book

Question

For $x \in \Bbb R^n$, define $H^* f(x)=\sup{\frac{1}{m(B)}\int_B|f(y)|\,dy}$, where $B$ is a ball and $x \in B$ and $$H f(x)=\sup_{r>0} \frac{1}{m(B(x,r))} \int_{B(x,r)}|f(y)| \, dy.$$ Show that $Hf \leq H^* f \leq 2^n Hf.$

Answer 1

We show that $Hf \leq H^* f$ but first we write the definition of your function $$ H^* f(x)=\sup_{B \textrm{ ball such that }x\in B}{\frac{1}{m(B)}\int_B|f(y)|\,dy} $$ and $$H f(x)=\sup_{r>0} \frac{1}{m(B(x,r))} \int_{B(x,r)}|f(y)| \, dy.$$ So this inequality is obvious because the sup on a large set is almost bigger than the sup on a small one.

*$H^* f\leq 2^n Hf.$

Let $B=B(y,r)$ a ball containing $x$ so $B\subset B(x,2r)$ in fact : let a $z\in B$ so $\|z-y\|<r$ then $$ \|z-x\|=\|z-y+y-x\|\leq \|z-y\|+\|y-x\|<2r $$ So $z\in B(x,2r)$ and then $$ \int_B |f(t)|dt\leq \int_{B(x,2r)} |f(t)|dt $$ To complete it's enough to proof that $$ m(B(x,2r))= 2^n m(B) $$ First since Lebesgue measure is invariant by translation we have $m(B)=m(B(x,r))$. and it's clear that $$ m(B(x,2r))= 2^n m(B(x,r))=2^n m(B) $$ (because $m(\alpha A +w)=\alpha^n m(A)$)

So \begin{eqnarray} \frac{1}{2^n m(B)} \int_B |f(t)|dt &\leq &\frac{1}{m(B(x,2r))} \int_{B(x,2r)} |f(t)|dt\\ \frac{1}{ m(B)} \int_B |f(t)|dt &\leq &\frac{2^n}{m(B(x,2r))} \int_{B(x,2r)} |f(t)|dt\\ H^*f(x) &\leq &2^n H f(x)\\ \end{eqnarray}