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Let $\mu $ be a positive Borel measure on $% %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion ^{d}$ 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 $% %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion ^{d}$. 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 $\int_{% %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion ^{d}}\frac{1}{\left\vert x\right\vert ^{d}}dx<\infty ,$ contradicting the fact $\int_{% %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion ^{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_{% %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion ^{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_{% %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion ^{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.

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What is $I_\alpha$? –  Eric Naslund Jan 1 '13 at 2:44
    
Also, for other users, this is related to beginner's previous question: math.stackexchange.com/questions/267990/1-xn-is-not-integrable/… –  Eric Naslund Jan 1 '13 at 2:44
    
If I remember correctly, you are reading the 1971 book by Stein, Singular integrals... It's indeed a great book for beginners, with down-to-earth selection of results and polished proofs. His later, more voluminous book Harmonic analysis develops the subject is much greater scope and generality, including introduction of weights wherever possible. It's quite possible that your question is satisfactorily explained there (unfortunately, I don't have the book). –  user53153 Jan 1 '13 at 4:32
    
References: weighted Hardy-Littlewood-Sobolev inequality –  user53153 Jan 3 '13 at 20:56

1 Answer 1

up vote 2 down vote accepted

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:

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

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