# Sobolev-type inequality.

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

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Nice statement. Thank you for sharing it. What would you like to know about it? –  András Bátkai Oct 19 '12 at 9:36

This is the standard Hardy-Littlewood-Sobolev inequality(or the theorem of fractional integration).A more direct approach is write $$\int{f(x-y)|y|^{\alpha-n}dy}=\int_{|y|<R}+\int_{|y|\ge R}$$ For the second term on the RHS,using Holder inequality,and easy to see that it's dominated by $\|f\|_{L^p}R^{-\frac{q}{n}}$. For the first term,one can use the majorizationgiven by the maximal function M,and to see that $$|f\ast |y|^{\alpha-n}|(x)\leq C(M(f)(x)\cdot R^{\alpha}+\|f\|_{L^p}\cdot R^{-\frac{q}{n}})$$ Choosing a proper constant R to make the two terms above be equal,and then the desired inequality hold by intergration(note that the maximal operator is bounded on $L^p$ for $1<p<\infty$).

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Thank you, that is what I was looking for! –  Felice Oct 20 '12 at 9:44
Can you prove it without using maximal function? –  timur Oct 20 '12 at 13:29
There is a proof of this inequality by the layer cake representation of functions and the Hölder inequality in Lieb and Loss, Analysis, 2001, section 4.3. –  Jean Van Schaftingen Dec 4 '13 at 15:28
The function $\vert x\vert^{\alpha-n}$ is radial homogeneous of degree $\alpha-n$, so its Fourier transform is radial homogeneous of degree $-(\alpha-n)-n=-\alpha$ (both locally integrable since $\alpha >0$ and $-\alpha>-n$ so both are distributions which are easily seen as temperate: Fourier transforms make sense), so your convolution operator is in fact the Fourier multiplier $\vert D_x\vert^{-\alpha}$. The question at hand is thus (with homogeneous spaces) $$\Vert u\Vert_{W^{-\alpha,q}}\lesssim \Vert u\Vert_{W^{0,p}},\quad \text{i.e. }W^{0,p}\subset W^{-\alpha,q},$$ which is a particular case of Sobolev injection since $$0>-\alpha,\quad p < q,\quad \frac{1}{p}-\frac{1}{q}=\frac{\alpha}{n}.$$