Sobolev inequality as a consequence of the Hardy-Littlewood-Sobolev inequality 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.
 A: 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.
