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Hardy-Littlewood-Sobolev inequality states that if $1<p<q<\infty$, $1/r=1-1/p+1/q$, then we have $$\left\|\frac{1}{|x|^{n/r}}\ast f\right\|_{L^q(\mathbb R^n)}\le\|f\|_{L^p(\mathbb R^n).}$$

Note that here $q=\infty$ is not allowed. My question is, is it possible to get some bounds for $q=\infty$, if we weaken the RHS norm to be in Lorentz space $L^{p,1}$:

$$\left\|\frac{1}{|x|^{n/r}}\ast f\right\|_{L^q(\mathbb R^n)}\le\|f\|_{L^{p,1}(\mathbb R^n).}$$

Any comments/references are welcome. Thanks.

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  • $\begingroup$ In the Further Results section of Chapter V, Singular Integrals and Differentiability Properties of Functions, Stein refers to a paper of R. O'Neil for a treatment of fractional integration in the setting of Lorentz spaces. You might look there. $\endgroup$ Commented Mar 27, 2016 at 1:19
  • $\begingroup$ @Matt Thanks! It seems to me that R. O'Neil's paper also excludes the possibility for $q$ to be $\infty$, maybe it's not possible to have $q=\infty$ in this kind of bounds. $\endgroup$
    – forevenone
    Commented Mar 27, 2016 at 2:16
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    $\begingroup$ The $q=\infty$ case follows from the duality of $L^{p,1}$ and $L^{p',\infty}$. This is not difficult to prove, but I don't know of a precise reference (you could try Grafakos's text, perhaps). $\endgroup$
    – Terry Tao
    Commented Mar 28, 2016 at 17:23

2 Answers 2

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The Young inequality in Lorentz spaces covers these cases: if $p_1,p_2,p\in]1,\infty[$, $q_1,q_2,q\in[1,\infty]$, \begin{equation} \|f\ast g\|_{L^{p,q}}\leq C\|f\|_{L^{p_1,q_1}} \|g\|_{L^{p_2,q_2}},\qquad p_1^{-1}+p_2^{-1} =1+p^{-1},\ q_1^{-1}+q_2^{-1}\geq q^{-1}; \end{equation} and if $p_{1},p_2\in ]1,\infty[$, $q_1,q_2\in[1,\infty]$, then \begin{equation} \|f\ast g\|_{L^\infty}\leq C\|f\|_{L^{p_1,q_1}} \|g\|_{L^{p_2,q_2}},\qquad p_1^{-1}+p_2^{-1} =1,\ q_1^{-1}+q_2^{-1}\geq1. \end{equation} Note that the second inequality contains your estimate.

The oldest reference I know for this is a paper by R.O'Neil, Convolution operators and $L(p,q)$ spaces, Duke Math. J. 30 (1963), 129-142.

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The right substitute for $L^{\infty}$ is the space $BMO$, in the following sense: Singular integral operators in $L^1$ is not bounded from $L^1$ to itself, but is bounded from $H^1$ to $L^1$. Hence, by duality, these operators are bounded from $L^\infty$ to $BMO$.

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  • $\begingroup$ I note that Wikipedia says: "The space of functions of bounded mean oscillation (BMO), is a function space that, in some precise sense, plays the same role in the theory of Hardy spaces $H^p$ that the space $L^\infty$ of essentially bounded functions plays in the theory of $L^p$-spaces:". Thanks for pointing this out, but a little more detail would have been more helpful to justify this claim. $\endgroup$
    – David Roberts
    Commented Jan 16, 2021 at 4:39
  • $\begingroup$ The right substitute for $L^{\infty}$ is the BMO space in the following sense: Singular integral operators in $L^1$ is not bounded from $L^1$ to itself, but is bounded from $H^1$ to $L^1$. Hence, by duality, these operators are bounded from $L^\infty$ to $BMO$. $\endgroup$ Commented Jan 17, 2021 at 9:42
  • $\begingroup$ Thanks, I have edited this into the answer. $\endgroup$
    – David Roberts
    Commented Jan 17, 2021 at 11:15

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