The uncentered Hardy-Littlewood maximal function of $f$ is $$\tilde{M}(f)=\sup\limits_{|y-x|<\delta}\frac{1}{|B(y,\delta)|}\int_{B(y,\delta)}|f|$$ If we denote the original Hardy-Littlewood maximal function as $M(f)$. In Grafakos's book, he states that $\tilde{M}(f)\leq 2^nM(f)$. I don't know how to prove this.
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If $|y-x|<\delta$ and $|z-y|<\delta$ then $|z-x|<2\delta$. Therefore $B(y,\delta)\subset B(x,2\delta)$, hence $$ \frac{1}{|B(y,\delta)|}\int_{B(y,\delta)}|f|\leq \frac{|B(x,2\delta)|}{|B(y,\delta)|}\frac{1}{|B(x,2\delta)|}\int_{B(x,2\delta)}|f|$$ $$ =2^n\frac{1}{|B(x,2\delta)|}\int_{B(x,2\delta)}|f|\leq 2^nMf(x)$$ and the result follows by taking the supremum.
carmichael561
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How is $\frac{|B(x,2\delta)|}{|B(y,\delta)|} = 2^n$ ? – gaufler Jun 10 '21 at 15:33
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1@gaufler The volume of an $n$-dim ball of radius $r$ is proportional to $r^n$, so doubling $r$ ... – r.e.s. Jun 10 '21 at 15:54