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": $$ {\displaystyle \|u\|_{L^{4}(\Omega)}\leq C\|u\|_{L^{2}(\Omega)}^{1/2}\|u\|_{H^{1}(\Omega)}^{1/2}} $$