Hardy Littlewood maximal function bounded in higher $p-$norm?

Question

I would like to show that for $0<s<t<\infty$ and $$L_sf(x):=\left(\sup_{r>0} \frac{1}{|B(x, r)|}\int_{B(x, r)} |f(y)|^s\, dy \right)^{\frac{1}{s}}$$ the s-th order HL-maximal function we have $||L_sf||_t \le K ||f||_t$ for some $K$ finite. Does anybody know how to do this? Looks like an application of Hoelder or anything like that.

Answer 1

If $f \in L^t$, then $|f|^s \in L^{t/s}$. But since $t/s > 1$, the usual maximal function is bounded from $L^{t/s} \to L^{t/s}$, say with norm $K$. This implies

$$ \|L_s f\|_{L^t}^s = \| (M |f|^s )^{1/s} \|_{L^{t}}^s = \| M |f|^s \|_{L^{t/s}} \leq K \cdot \| |f|^s \|_{L^{t/s}} = K \cdot \| f\|_{L^t}^s . $$ Take this estimate to the power $1/s$ to get the claim.