An imbedding inequality in PDE. u is a function of 3-dimension, I'm trying to prove this:
$\|u\|_4^4 \leq C \|u\|^2_{H^1} \|u\|_{L^2}^2$
Anyone can shed light on this? Thanks very much.
 A: I don't believe the inequality is correct, an application of the Gigliardo-Nirenberg interpolation inequality http://en.wikipedia.org/wiki/Gagliardo%E2%80%93Nirenberg_interpolation_inequality, for $p,q,r\in[1,\infty]$, $j,m\in\Bbb N_0$ where 
$$\frac{1}{p} = \frac{j}{n} + \left( \frac{1}{r} - \frac{m}{n} \right) \alpha + \frac{1 - \alpha}{q},$$
then $$
\| \mathrm{D}^{j} u \|_{L^{p}} \leq C \| \mathrm{D}^{m} u \|_{L^{r}}^{\alpha} \| u \|_{L^{q}}^{1 - \alpha}.$$
Here $q=r=2$, $p=4$, $n=3$, $j=0$ $m=1$, so 
$$\frac{1}{4}=0+(\frac{1}{2}-\frac{1}{3})\alpha+\frac{1}{2}-\frac{\alpha}{2},$$
$$\alpha = \frac{3}{4}.$$
This yields $\|u\|_4^4\le C\|\nabla u\|_2^3\|u\|_2$. This can be proven directly assuming $u\in H^1_0$. 
Firstly $$\|u\|^4_4=\|u^4\|_1=\||u||u|^3\|_1\le\|u\|_2\|u\|_6^3,$$ the latter following from holders inequality, now my assumption is that $u\in H^1_0$ which is compactly embedded in $L^{2*}$, where $2^*=\frac{2n}{n-2}=6$ (for $n=3$), so $H^1\subset\subset L^6$, i.e. $\|u\|_6\le C\|u\|_{H^1_0}$, and so 
$$\|u\|^4_4\le \|u\|_2\|u\|_6\le C\|u\|_2\|\nabla u\|_{2}^3.$$ 
