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Let us denote the Riesz potential in $\mathbb R^d$ by $$ I_\alpha (f)(x) := c_{d, \alpha} \int_{\mathbb R^d} \frac{f(y)}{|x-y|^{d-\alpha}} \, dy.$$

By the classical Hardy-Littlewood-Sobolev theorem on fractional integration we have for $1 < p <d/\alpha$ that $$ \|I_\alpha(f)\|_{L^q} \le C_{d, \alpha, p} \|f\|_{L^p}, \quad\text{where}\ \ \tag 1 q=\frac{dp}{d-p\alpha}. $$

I am looking for a reference (with a proof) for the borderline case $p=d/\alpha$: $$I_\alpha:L^{d/\alpha} \to \rm{BMO}. \tag 2$$

I have checked Grafakos's and Stein's harmonic analysis books, but I don't think the proof is in any of them.

A related result, which would probably be enough for me, is stated as an exercise 8.11. on page 62-63 of

http://www.ms.uky.edu/~rbrown/courses/ma773/notes.pdf

but at least a quick look seems to indicate that the constant in (1) is of the form $C_{d,\alpha} q$ (for $p$ close to $d/\alpha$), so that the corresponding power series in the exercise does not converge after raising the estimate to power $q$ (the factor k^k dominates k! of the denominator). Probably I am missing something.

In any case, the best would be to find a self-contained proof of (2) directly.

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  • 1
    $\begingroup$ Hint: K^k is comparable to e^k k! by Stirling. $\endgroup$ – Fan Zheng Apr 27 '15 at 15:04
  • $\begingroup$ Oh yes, this must be the piece I was missing. Thanks! :) $\endgroup$ – Juhana Siljander Apr 27 '15 at 16:51
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Let $a \in \mathbb{R}^d$ and $r > 0$. We have $$ \frac{1}{\vert B_r \vert^2} \int_{B_r} \int_{B_r} \vert I_\alpha (f) (x) - I_\alpha (f) (y) \vert\,\mathrm{d}x\,\mathrm{d}y \le \frac{c_{d, \alpha}}{\vert B_r \vert^2} \int_{\mathbb{R^d}} \int_{B_r} \int_{B_r} \vert f (z)\vert \,\Big\vert \frac{1}{\vert z - x\vert^{d - \alpha}} - \frac{1}{\vert z - y\vert^{d - \alpha}} \Big\vert \,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z. $$ We compute $$ \int_{B_r} \int_{B_r}\Big\vert \frac{1}{\vert z - x\vert^{d - \alpha}} - \frac{1}{\vert z - y\vert^{d - \alpha}} \Big\vert \,\mathrm{d}x\,\mathrm{d}y \le \frac{C r^{2 d + 1}}{(r + \vert z \vert)^{d - \alpha + 1}}, $$ and we conclude by the classical Hölder inequality that $$ \int_{B_r} \int_{B_r} \vert I_\alpha (f) (x) - I_\alpha (f) (y) \vert\,\mathrm{d}x\,\mathrm{d}y \le C r^{2 d + 1} \Bigl(\int_{\mathbb{R}^d} \Bigl(\frac{1}{(r + \vert z \vert)^{d - \alpha + 1}}\Bigr)^\frac{d}{d - \alpha} \, \mathrm{d} z\Bigr)^{1 - \frac{\alpha}{d}} \Vert f \Vert_{L^{d/\alpha}}, $$ Since $$ \int_{\mathbb{R}^d} \Bigl(\frac{1}{(r + \vert z \vert)^{d - \alpha + 1}}\Bigr)^\frac{d}{d - \alpha} \, \mathrm{d} z = \frac{1}{r^{d + \frac{d}{d -\alpha}}}\int_{\mathbb{R}^d} \Bigl(\frac{1}{(1 + \vert z \vert/r)^{d - \alpha + 1}}\Bigr)^\frac{d}{d - \alpha} \, \mathrm{d} z = \frac{C}{r^\frac{d}{d - \alpha}}, $$ we conclude that $$\frac{1}{\vert B_r \vert^2} \int_{B_r} \int_{B_r} \vert I_\alpha (f) (x) - I_\alpha (f) (y) \vert\,\mathrm{d}x\,\mathrm{d}y\le C'' \Vert f \Vert_{L^{d/\alpha}} $$

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  • $\begingroup$ For some reason I obtain that the constant $C'$ depends on the radius $r$ (I checked the case $\alpha=1$). Are you certain about your second estimate? Maybe I did some mistake somewhere. Looks good in any case. $\endgroup$ – Juhana Siljander Apr 27 '15 at 14:33
  • $\begingroup$ I have added some details of the computation of the $L^{d/(d - \alpha)}$ norm. I hope that it helps. $\endgroup$ – Jean Van Schaftingen Apr 27 '15 at 15:25
  • $\begingroup$ Thanks, this is also what I obtained. But in order to show that $I_\alpha(f)$ is in $\rm{BMO}$, don't we want to take supremum over all balls on the left hand side. Now the bound $Cr^{-\frac d{d-\alpha}}$ blows up when $r \to 0$, so how do we know that the supremum stays bounded? $\endgroup$ – Juhana Siljander Apr 27 '15 at 16:48
  • $\begingroup$ We are taking an $L^{d/(d - \alpha)}$ norm, which blows up like $1/r$ and we divide by $r^{2 d}$ (volume of the ball squared), this compensates exactly the $r^{2 d + 1}$ coming from the second estimate. $\endgroup$ – Jean Van Schaftingen Apr 28 '15 at 6:24
  • $\begingroup$ Of course, thanks. I forgot the power outside the integral. :) $\endgroup$ – Juhana Siljander Apr 28 '15 at 10:57

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