Sobolev inequality as a consequence of the Hardy-Littlewood-Sobolev inequality

Question

In my introductory graduate analysis class, we learned the Hardy-Littlewood Sobolev inequality (in dimension $d$): if $p, q > 1$, $0 < \lambda < d$ are such that $$ \frac{1}{p} + \frac{1}{q} + \frac{\lambda}{d} = 2, $$ then for all compactly supported smooth functions on $\mathbb{R}^d$, $$ \int_{\mathbb{R}^d}\int_{\mathbb{R}^d} f(x) |x - y|^{-\lambda}g(y)\, dx\, dy \leq C\|{f}\|_p \|{g}\|_q $$ for some constant $C$. We were then told that Sobolev's inequality, $$ \|u\|_{L^{p^*}} \leq C\|Du\|_{L^p} $$ for some (different than above, perhaps) constant $C$, was a consequence of Hardy-Littlewood-Sobolev. However, I had no idea how to show this. Firstly, the $L^{p^*}$ norm of a function $u$ (here, $p^{*}$ is the Sobolev conjugate of $p$) is a single integral, and I don't understand how we can recover this from the double integral in H-L-S.

My only idea was to somehow relate the $|x - y|$ to the difference quotient of $u$, but I don't see how to do this (or if this is even right). Even with this, I have no idea how to pick the right $f$ and $g$.

Any help in the right direction would be appreciated.

Answer 1

From double to single integral

Let $I_{n-\lambda} f(y) = \int_{\mathbb{R}^d} f(x) |x - y|^{-\lambda} \, dx $, this is the Riesz potential of $f$ of order $n-\lambda$. The HLS inequality states that $$ \int I_{n-\lambda }f g \le C\|f\|_p\|g\|_q \tag1 $$ From the discussion of the equality case in Hölder's inequality it follows that there is $g$ such that $\|g\|_q = 1$ and $$\int I_{n-\lambda }f g = \| I_{n-\lambda }f \|_{q'} \tag2$$ where as usual, $q'=q/(q-1)$. Thus, the HLS inequality is simply the boundedness of $I_{n-\lambda}$ from $L^p$ to $L^{q'}$.

Riesz potential and the gradient

Suppose $u$ is smooth and has compact support. Then $u(x)$ can be estimated by integrating $\nabla u$ along any half-line emanating from $x$; namely, $$ |u(x)|\le \int_0^\infty |\nabla u(x+r\xi)|\,dr \tag3 $$ for any unit vector $\xi$. We don't know which direction is best, so just average over all: $$ |u(x)|\le C\int_{\|\xi\|=1}\int_0^\infty |\nabla u(x+r\xi)|\,dr\,d\xi \tag4 $$ This looks a lot like integrating $|\nabla u|$ in polar coordinates, except the factor $r^{d-1}$ is missing. Well, this factor is exactly what the Riesz potential supplies with $\lambda = d-1$. Hence, $$ |u(x)|\le C I_1(|\nabla u|) \tag5 $$

Conclusion

Now that we know that we'll use $\lambda = d-1$, the relation $$ \frac{1}{p} + \frac{1}{q} + \frac{\lambda}{d} = 2, \tag6 $$ can be rearranged as $$ \frac{1}{p} - \frac{1}{d} = 1 - \frac{1}{q} \tag7 $$ which tells us that $q' = p^*$. From Part 1 we know that $$ \|I_1 |\nabla u| \|_{p^*} \le C\|\nabla u\|_p \tag8 $$ And Part 2 implies $$ \|u \|_{p^*} \le C\|\nabla u\|_p \tag9 $$ As usual, the inequality extends to general Sobolev functions by density.