Is the 2D Ladyzhenskaya inequality true for periodic functions? I have only seen the following version of 2D Ladyzhenskaya inequality in cited references of PDE: 

Let $\Omega$ be a Lipschitz domain in ${\bf R}^2$ and let $u: \Omega → {\bf R}$ be a weakly differentiable function that vanishes on the boundary of ${\bf R}$ in the sense of trace (that is, $u$ is a limit in the Sobolev space $H^1(\Omega)$ of a sequence of smooth functions that are compactly supported in $\Omega$). Then there exists a constant $C$ depending only on $\Omega$ such that
  $$
{\displaystyle \|u\|_{L^{4}}\leq C\|u\|_{L^{2}}^{1/2}\|\nabla u\|_{L^{2}}^{1/2}}. 
$$

Is it true for periodic functions as well? More precisely, is it true that
$$
{\displaystyle \|u\|_{L^{4}(\Omega)}\leq C\|u\|_{L^{2}(\Omega)}^{1/2}\|\nabla u\|_{L^{2}(\Omega)}^{1/2}} 
$$
where $u:{\bf R}^2\to{\bf R}$ is a smooth function with the period $\Omega=[l_1,r_1]\times[l_2,r_2]$?
[Added later:] Thanks to Hannes's comment, any nonzero constant function is an easy counterexample to the statement above. I'm now looking for a proof (if it is true) of the following updated "Ladyzhenskaya inequality":
$$
{\displaystyle \|u\|_{L^{4}(\Omega)}\leq C\|u\|_{L^{2}(\Omega)}^{1/2}\|u\|_{H^{1}(\Omega)}^{1/2}} 
$$
 A: Let us start with the so-called Gagliardo-Nirenberg Inequality in $n$ dimensions,
$$
\Vert u\Vert_{L^{n/(n-1)}(\mathbb R^n)}\le c_n\Vert \nabla u\Vert_{L^{1}(\mathbb R^n)},
\tag{GN}$$
an inequality that can be applied to your function so that you get in two dimensions
$$
\Vert u\Vert_{L^{2}(\mathbb R^n)}\le c_2\Vert \nabla u\Vert_{L^{1}(\mathbb R^n)},
\tag{$\ast$}$$
and applying this to $u=v^2$, you obtain
$$
\Vert v\Vert_{L^{4}(\mathbb R^n)}^2\le 2c_2\Vert v\nabla v\Vert_{L^{1}(\mathbb R^n)}\lesssim\Vert v\Vert_{L^{2}(\mathbb R^n)}\Vert \nabla v\Vert_{L^{2}(\mathbb R^n)}.
$$
This is true for $v\in C^1_c(\Omega)$ and consequently by density in $H^1_0(\Omega)$ (which does not contain any constant non-zero function).
On periodic functions in $\mathbb R^2$: it is enough to prove $(\ast)$, but some condition must be obviously imposed. Writing for instance
$$
u(x,y)=\sum_{k,l}e^{2π i(kx+ly)}\hat u(k,l),
$$
we assume that 
$
\forall k,\ \sum_{l}\hat u(k,l)=0 ,\quad
\forall l,\ \sum_{k}\hat u(k,l)=0.
$
Then we can write
$$
u(x,y)=\int_0^x\partial_1 u(s,y) ds=\int_0^y\partial_2 u(x,t) dt,
$$
and we get $(\ast)$ by integrating wrt $x,y$ the inequality
$$
\vert u(x,y)\vert^2\le \iint_{[0,1]^2}\vert\partial_1 u(s,y)\vert\vert \partial_2 u(x,t)\vert dsdt.
$$
N.B. The proof of the Gagliardo-Nirenberg inequality (GN) in three or more dimensions is much more difficult, but mutatis mutandis, assuming as above the vanishing of some partial sums of the Fourier coefficients, we can get (GN) for periodic functions.
