A special interpolation inequality I am reading a paper, but getting stuck in the following inequality. I want to know whether it is correct or not.
$\Omega=(-\infty,\infty)\times [0,1]$ and for any smooth function $u=0$ only on the lower boundary $(-\infty,\infty)\times\{0\}$ then
$$||u||_{4}\leq C||\nabla u||_{2}^{\frac{3}{4}}||\partial_{1} u||_{2}^{\frac{1}{4}}$$
Is this true? How to prove it? Thanks!
 A: First we copy the proof of the Gagliardo-Nirenberg inequality:
Notice that $C^\infty_c(\Omega)$ is dense in $L^4(\Omega)$, and so we can assume the functions we are working with have compact support. For convenience I will denote $x\in \mathbb{R}$ and $y\in [0,1]$, then by fundamental theorem of calculus
$$ |u(x,y)| \leq \int_{\mathbb{R}} |\partial_x u(s,y)| ~ds, \qquad |u(x,y)| \leq \int_0^1 |\partial_y u(x,t)| ~dt $$
where the first expression we used decay at infinity for $x$, and the second we used $u(x,0) = 0$. 
Thus 
$$ |u(x,y)|^4 \leq 4 \left(\int_{\mathbb{R}}|u(s,y)| |\partial_x u(s,y)| ~ds \right)\cdot\left( \int_0^1 |u(x,t)| |\partial_y u(x,t)| ~dt \right) $$
and hence
$$ \|u\|^4_4 \leq 4 \| u \partial_x u\|_1 \cdot \|u \partial_y u\|_1 \leq 4 \|u\|_2^2 \|\partial_x u\|_2 \|\partial_y u\|_2 $$
Next we integrate in $y$ to control $\|u\|_2$:
Since 
$$ |u(x,y)|^2 = \int_0^y u(x,t) \partial_y u(x,t) ~dt $$
after integrating in $x$ and applying Fubini we have 
$$ \int_{\mathbb{R}} |u(x,y)|^2 ~dx \leq \int_0^y \|u(\cdot,t)\|_{L^2(\mathbb{R})} \|\partial_y u(\cdot,t) \|_{L^2(\mathbb{R})} $$
Or that 
$$ \|u\|^2_{L^\infty_y L^2_x} \leq \|u\|_{L^\infty_y L^2_x} \|\partial_y u\|_{L^1_y L^2_x} $$
This implies
$$ \|u\|_2 \leq \|u\|^2_{L^\infty_y L^2_x} \leq \|\partial_y u\|_{L^1_y L^2_x} \leq \|\partial_y u\|_{2} $$
So putting everything together we have
$$ \|u\|_4^4 \leq 4 \|\partial_y u\|_2^3 \|\partial_x u\|_2^1 $$
which implies the desired claim. 
