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Show that if $f\in L^1(\mathbb{R}^d)$ and $E\subset \mathbb{R}^d$ has finite measure, then for any $0<q<1$, $$\int_E |f^{*}(x)|^q dx\leq C_q|E|^{1-q}||f||_{L^1(\mathbb{R}^d)}^{q}$$ where $C_q$ is a positive constant depending only on $q$ and $d$.

Here the function $f^*(x)=\sup_{x\in B}\frac{1}{|B|}\int_B |f(y)|dy$ is the Hardy-Littlewood maximal function.

Notes
It seems to me the weak type estimate $\forall \alpha>0,\enspace |\{x: f^*(x)>\alpha\}|\leq \frac{3^d}{\alpha}||f||_{L^1(\mathbb{R}^d)}$ is of great use but I am having trouble putting this to any use. Any help is appreciated.

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Indeed, the weak type estimate is useful. Using Fubini's theorem, we have $$\int_E|f^{*}(x)|\mathrm dx=q\int_0^\infty t^{q-1}\lambda\{|f^*(x)|\chi_E\geqslant t\}\mathrm dt.$$ Notice that $$\lambda\{|f^*(x)|\chi_E\geqslant t\}\leqslant \min\left\{|E|;\frac{3^d}t\lVert f\rVert_{\mathbb L^1}\right\},$$ hence cut the integrals and conclude.

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  • $\begingroup$ I understand What you have done and it makes sense, but I don't understand how you went straight to the conclusion from that, I'm just not seeing it. Maybe I'll just have to think about it some more. $\endgroup$ – CWsl2 Jan 13 '14 at 22:16
  • $\begingroup$ Did you compute the integrals which come from my last bound? $\endgroup$ – Davide Giraudo Jan 13 '14 at 22:22
  • $\begingroup$ This is what I am having trouble doing - Computing the integral from the last bound, specifically with the minimum involved. I need to review some of these concepts $\endgroup$ – CWsl2 Jan 13 '14 at 22:51
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    $\begingroup$ I was not too sure what you meant by "cut" the integrals at first, but now it is fairly clear. One may split the integral at the beginning of your statement into two integrals $\int_{0}^A+\int_A^{\infty}$, use your bound to estimate each integral independently and then one can optimize A, and the result follows. Thanks $\endgroup$ – CWsl2 Jan 15 '14 at 21:16

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