Why $\|u\|_{L^r}\leq C\|\nabla u\|_{L^p}^{a}\|u\|_{L^q}^{1-a}$? Let $u\in \mathcal C_c^1(\mathbb R^n)$ and $p&gt;n$. Let $r&gt;0$ s.t. $$\frac{1}{r}=a\left(\frac{1}{p}-\frac{1}{n}\right)+\frac{1-a}{q},$$
with $a\in [0,1]$, $p&gt;1$ and $q\geq 1$. I'm trying to understand why $$\|u\|_{L^r(\mathbb R^n)}\leq C\|u\|_{W^{1,p}(\mathbb R^n)}^a\|u\|_{L^q(\mathbb R^n)}^{1-a},$$
with $C$ independent of $u$. 


I thought first using the fact that $W^{1,p}(\mathbb R^n) \subset L^\infty (\mathbb R^n)$, but it's not conclusive. 
But I know using Morrey's theorem that there is $C$ independent of $u$ s.t. $$\frac{|u(x)-u(y)|}{|x-y|^{1-\frac{n}{p}}}\leq C\|\nabla u\|_{L^p(\mathbb R^n)},$$
but even with this information, I can't conclude. 
By a scaling argument, I can consider $\|u\|_{L^q}=\|u\|_{W^{1,p}}=1$, and thus, I have to prove that $u$ is uniformly bounded for the $r$ norm. I know that the injection $W^{1,p}(\mathbb R^n)\subset L^\infty (\mathbb R^n)$ is continuous and thus there is $C$ independent of $u$ s.t. $$\|u\|_{L^\infty }\leq C.$$
Now, I know that on a compact $K$, $L^{\infty }(K)$ is continuously injected in $L^{r}(K)$. So if I denote $K$ the support of $u$, I indeed have $\|u\|_{L^r(\mathbb R^n)}=\|u\|_{L^r}(K)\leq C$, and thus I can conclude, but since $K$ depend on $u$, the constant will depend on $u$, and thus, it doesn't solve my problem.


Any help is welcome :-)
$$\int_\Omega |u|^r=\int_\Omega |u|^{r-q}|u|^q\leq \|u^{r-q}\|_{L^\infty }\|u\|_{L^q}^q\leq C,$$
et voilà !
Please see the proof below, which is from Adams' book "Sobolev space".  Also, in the Theorem 5.9 in Adams' book, the restriction $p\leq q$ is removed.