Reference request: Riesz potential $I_\alpha : L^{d/\alpha} \to \rm{BMO}$? 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.
 A: 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 
$$\frac{1}{\vert B_r \vert^2}
\int_{B_r} \int_{B_r} \vert I_\alpha (f) (x) - I_\alpha (f) (y) \vert\,\mathrm{d}x\,\mathrm{d}y\le C'' \Vert f \Vert_{L^{d/\alpha}}
$$
