Sobolev Embeddings (reference request) The following two results are often used in research papers:

$$W^{2,1}_p(J\times \Omega)\hookrightarrow L^\infty(J\times\Omega)\qquad\text{and}\qquad \|u^2\|_{L^p(J\times\Omega)}\le \|u\|^2_{W^{2,1}_p(J\times \Omega)},$$
  where $$W^{2,1}_p(J\times \Omega):=L^p(J,W^2_p(\Omega))\cap W^1_p(J,L^p(\Omega)),\qquad J\subset\mathbb{R},\quad \Omega\subset\mathbb{R}^n\quad\text{bounded, smooth}$$

However, I could not find a book/article/proof I could use as a reference for these embeddings. It is not clear to me why these results hold and for which $p$ it holds. I suppose that there is a lower bound for $p$? 
 A: Adams and Fournier does not cover Sobolev spaces with values in a Banach spaces. For those your best bet is some book on evolution equations, for example Franck Boyer • Pierre Fabrie Mathematical Tools for the Study of the Incompressible Navier-Stokes Equations and Related Models  Boyer-Fabrie
or some book on semigroups, for example Barbu, Viorel, Nonlinear semigroups and differential equations in Banach spaces
Barbou. 
Update To get the results you need you can use the theory of anisotropic Sobolev spaces. The book you want is Besov. Unfortunately, it is not easy to read. 
Consider the anisotropic Sobolev space $W_{p}^{\overrightarrow{m}}%
(J\times\Omega)$, where $\overrightarrow{m}=(m_{1},m_{2},\ldots,m_{n+1})$ is a
vector of positive integers (so you are taking $m_{1}$ derivatives with
respect to $t$, $m_{2}$ with respect to $x_{1}, \ldots, m_{n+1}$ derivatives
with respect to $x_{n}$. If $p<q\leq\infty$ and
$$
\left(  \frac{1}{p}-\frac{1}{q}\right)  \sum_{k=1}^{n+1}\frac{1}{m_{k}}<1
$$
then
$$
\Vert u\Vert_{L^{q}(J\times\Omega)}\leq c\sum_{k=1}^{n+1}\Vert\partial^{m_{k}%
}u\Vert_{L^{p}(J\times\Omega)}+c\Vert u\Vert_{L^{p}(J\times\Omega)}.
$$
In your case $\overrightarrow{m}=(1,2,\ldots,2)$ (so you have the time
derivative of order one in $L^{p}$ and space derivatives of order two in
$L^{p}$). If you take $q=\infty$ you get
$$
\frac{1}{p}\sum_{k=1}^{n+1}\frac{1}{m_{k}}=\frac{1}{p}\left(  1+\frac{n}
{2}\right)  <1
$$
that is $p>1+\frac{n}{2}$. 
On the other hand, if you take $q=2p$ you need
$$
\left(  \frac{1}{p}-\frac{1}{2p}\right)  \left(  1+\frac{n}{2}\right)  <1,
$$
that is, $p>\frac{1}{2}+\frac{n}{4}$.
**1: https://link.springer.com/book/10.1007%2F978-1-4614-5975-0
