Hardy-Littlewood-Sobolev inequality for $p=1$ Let $\mu $ be a positive Borel measure on $%
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^{d}$ such that $\mu \left( B\left( a,r\right) \right) \leq Cr^{n}$ for some 
$n\in (0,d]$ and for any ball $B\left( a,r\right) $ in $%
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^{d}$. Riesz potential $I_\alpha$ defined by $I_\alpha f(x)=\int_{\mathbb{R}^d} \frac{f(y)}{|x-y|^{n-\alpha}} d\mu(y)$. Could you help me to prove (disprove) that $\left\Vert I_{\alpha
}f\right\Vert _{L^{n/(n-\alpha )}\left( \mu \right) }\leq C\left\Vert
f\right\Vert _{L^{1}\left( \mu \right) }$ (Hardy-Littlewood-Sobolev inequality for $p=1$)?
In Stein's book, for lebesgue measure ($n=d$) the above inequality is not
true. By assumming that the inequality is true, one can construct a sequence
of function $\{f_{m}\}$ that implies $\int_{%
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^{d}}\frac{1}{\left\vert x\right\vert ^{d}}dx<\infty ,$ contradicting the
fact $\int_{%
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^{d}}\frac{1}{\left\vert x\right\vert ^{d}}dx=\infty $.
For (general) measure given above, I try to do the same technique and get  $
\int_{%
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^{d}}\frac{1}{\left\vert x\right\vert ^{n}}d\mu \left( x\right) <\infty ~$
(which is not always a contradiction, since there is a measure $\mu _{1}$
such that $\int_{%
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^{d}}\frac{1}{\left\vert x\right\vert ^{n}}d\mu _{1}\left( x\right) <\infty $
). Could you give a hint or reference?
Thanks a lot.
 A: One difference between the special case of the Lebesgue measure and the general case is translation-invariance. 
Since the Lebesgue measure is translation-invariant, integrals of the form $\int \phi(x-y)\,dx$ do not actually depend on $y$. As a consequence, finiteness of this integral for some value of $x$ is equivalent to it being uniformly bounded with respect to $y$. 
For a general measure $\mu$, finiteness and uniform boundedness of $\int |x-y|^{-n}\,d\mu(x)$ are different things. I claim that the crucial property is uniform boundedness. Specifically, the following are equivalent: 


*

*$\mathcal{I}_\alpha$ is a bounded operator from $L^1_\mu$ to $L^{n/(n-\alpha)}_\mu$

*There exists a constant $M$ such that for $\mu$-a.e. $y$ we have $\int |x-y|^{-n}\,d\mu(x)\le M$.


In the proofs, all Lebesgue norms are taken with respect to $\mu$ (the Lebesgue measure is never considered). 
Proof of $2\implies 1$. By duality, $\|\mathcal{I}_\alpha f\|_{n/(n-\alpha)} = \sup\{\int ( \mathcal{I}_\alpha f)\,g : \|g\|_{n/\alpha}\le 1\}$.
 For any such $g$ Hölder's inequality and assumption 2 imply
 $$\int \frac{g(x)}{|x-y|^{n-\alpha}}\,d\mu(x) \le \|g\|_{n/\alpha} \int |x-y|^{-n}\,d\mu(x) \le M$$
for $\mu$-a.e. $y$.  Hence, 
 $$ \int (\mathcal{I}_\alpha f )\,g \le \int \int \frac{|g(x)||f(y)|}{|x-y|^{n-\alpha}}\,d\mu(x)\,d\mu(y) 
 \le M\int |f(y)|\,d\mu(y) = M \|f\|_1 $$
which establishes 1. 
Proof of $1\implies 2$. Fix $y$ such that every neighborhood of $y$ has positive $\mu$-measure (this is true 
for $\mu$-a.e. point $y$). Let $f_k$ be a sequence of positive functions such that $\int f_k(x)\,d\mu(x)=1$ 
and $f_k$ is supported in the $(1/k)$-neighborhood of $y$. As in your earlier question, pointwise convergence $\mathcal{I}_\alpha f_k(x)\to |x-y|^{n-\alpha}$ together with Fatou's lemma imply that $$\int |x-y|^{-n}\,d\mu(x)\ge \liminf_{k\to\infty} \int  (\mathcal{I}_\alpha f_k)^{n/(n-\alpha)} \le C^{n/(n-\alpha)}$$
where $C$ is the operator norm of  $\mathcal{I}_\alpha$. $\Box$
