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.



Reference request: Riesz potential $I_\alpha : L^{d/\alpha} \to \rm{BMO}$?
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 
$$\frac{1}{\vert B\_r \vert^2}
\int\_{B\_r} \int\_{B\_r} \vert I\_\alpha (f) (x) - I\_\alpha (f) (y) \vert\,\mathrm{d}x\,\mathrm{d}y\le C'' \Vert f \Vert\_{L^{d/\alpha}}
$$


