Here is the statement of the Hardy–Littlewood–Sobolev theorem.

Let $0< \alpha< n$, $1 < p < q < \infty$ and $\frac{1}{q}=\frac{1}{p}-\frac{\alpha}{n}$. Then: $$ \left \| \int_{\mathbb{R}^n} \frac{f(y)dy}{|x-y|^{n-\alpha} } \right\|_{L^q(\mathbb{R}^n)}\leq C\left\| f\right\| _{L^p(\mathbb{R^n})}.$$

I know two proofs of this theorem. The first one (I think the standard one) uses the Marcinckiewicz interpolation theorem.

The second one uses the Hardy–Littlewood maximal function and its boundedness from $L^p(\mathbb{R}^n)$ to $L^p(\mathbb{R}^n)$. To prove this boundedness I need the Marcinkiewicz interpolation theorem again. (Even if it's enough the "diagonal" version.)

My question is: is there a proof of the above theorem that doesn't use Marcinkiewicz? Is this interpolation theorem necessary in order to prove HLS?

  • 1
    $\begingroup$ I might be wrong, but I guess this thesis can be useful: mathematik.uni-muenchen.de/~lerdos/Stud/khotyakov.pdf $\endgroup$
    – Siminore
    Commented Dec 1, 2014 at 15:13
  • $\begingroup$ @Siminore this link is now dead; I can't find this Bsc thesis titled "Two Proofs of the Sharp Hardy-Littlewood-Sobolev Inequality" (but I found his Msc? which looks off topic) $\endgroup$ Commented Feb 20, 2021 at 6:53

2 Answers 2


There is a direct and self-contained proof of HLS inequality in Analysis by Lieb and Loss, Theorem 4.3. It uses nothing but layer cake representation, Hölder's inequality, and clever manipulation of integrals. A bit too long to reproduce here, though.

Also, the boundedness of Hardy-Littlewood maximal function is much more straightforward than the general Marcinkiewicz interpolation theorem; it is presented in the textbooks as a consequence of the latter just because the authors would like it to be one. Stein proves it as Theorem 1.1.1 in Singular integrals and differentiability properties of functions. First, the covering lemma is used to prove the weak $(1,1)$ inequality $$m(E_\alpha)\le C\alpha^{-1}\int_{\mathbb R^n} |f(x)|\,dx \tag{1}$$ where $E_\alpha = \{x:Mf(x)>\alpha\}$.

Fix $\alpha$ and let $f_1=f\chi_{|f|\ge \alpha/2}$. Since $|f|\le f_1+\alpha/2$, it follows that $$\{x:Mf(x)>\alpha\}\subset \{x:Mf_1(x)>\alpha/2\}$$ Apply $(1)$ and use the layercake representation of $\int (Mf)^p$: $$ \int_{\mathbb R^n} (Mf(x))^p\,dx = p\int_0^\infty \alpha^{p-1} m(E_\alpha)\,d\alpha \le p \int_0^\infty \alpha^{p-1} \frac{C}{\alpha}\left( \int_{|f|>\alpha/2}|f(x)|\,dx\right)\,d\alpha $$ Switch the order of integration on the right to get $$ C p \int_{\mathbb R^n}|f(x)|\,dx \int_0^{2|f(x)|} \alpha^{p-2} \,d\alpha = C'\int_{\mathbb R^n}|f(x)|^p\,dx $$ as desired.

And now that I typed all this, I see that the Wikipedia article Hardy–Littlewood maximal function also gives this proof.

  • 5
    $\begingroup$ To be fair, one has to admit that the direct proof given by you (or by Stein) is more or less the proof of the Marcinckiewicz interpolation theorem specialized to this particular case (interpolation between weak $1-1$ and $\infty-\infty$ estimate). $\endgroup$
    – PhoemueX
    Commented Dec 4, 2014 at 19:00

Are you can prove the case one dimensional without use of Marcinkiewicz interpolation theorem? If your answer is yes, then continue by induction combining Hölder, Young and Minkowski inequalities with the identity

$$\int_{\mathbb{R}^{n-1}}\frac{dy_1\cdots dy_{n-1}}{|x-y|}=\frac{c_n}{|x_n-y_n|}$$

where $c_n$ is a constant that depends only on n.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .