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

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    $\begingroup$ I think you need to exclude nonzero constant functions for such an inequality to work, or you need to replace $\|\nabla u\|_{L^2(\Omega)}$ by $\|u\|_{H^1(\Omega)}$. $\endgroup$
    – Hannes
    Commented Feb 19, 2018 at 16:33
  • $\begingroup$ @Hannes: Thanks. Sounds very similar to the situation in the Poincare inequality. Do you have a sketch or a reference for the proof if one replaces the gradient norm by the $H^1$ norm you mentioned? $\endgroup$
    – user14319
    Commented Feb 19, 2018 at 16:51
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    $\begingroup$ Does the updated version not follow essentially from Sobolev? You have $\|u\|_{L^4(\Omega)} \lesssim \|u\|_{H^{1/2}(\Omega)} \lesssim \|u\|_{L^2(\Omega)}^{1/2} \|u\|_{H^1(\Omega)}^{1/2}$ the first inequality is the sharp Sobolev embedding in dimension 2, the second interpolation follows by taking the Fourier transform (since you are on a periodic domain). $\endgroup$
    – Willie Wong
    Commented Feb 19, 2018 at 17:36
  • $\begingroup$ @WillieWong: Thanks for the comment! I was looking at a proof of the Ladyzhenskakaya (for $C_c^\infty({\bf R}^2)$ functions) which uses some rather elementary observations and the Cauchy-Schwarz inequality and tried to adapt it to the periodic case (which did not work). I didn't know that one may use the Sobolev embedding. $\endgroup$
    – user14319
    Commented Feb 19, 2018 at 18:09
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    $\begingroup$ If by the proof using "elementary observations" you mean the one where you start by using the fundamental theorem of calculus, you can write down the analogous proof by integrating, instead of from $-\infty$, from a point $\bar{x}$ where $u(\bar{x}) = \frac{1}{|\Omega|} \int_\Omega u(y) ~\mathrm{d}y$. (The point where $u$ attains its mean value.) This would give you the following inequality: $$ \| u(x) - u(\bar{x})\|_{L^4} \lesssim \|u(x) - u(\bar{x})\|_{L^2}^{1/2} \|\nabla u\|_{L^2}^{1/2} $$ $\endgroup$
    – Willie Wong
    Commented Feb 19, 2018 at 18:42

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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.

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  • $\begingroup$ Thanks. How does this relate to the case of periodic functions? $\endgroup$
    – user14319
    Commented Feb 19, 2018 at 23:01
  • $\begingroup$ I have added a comment on periodic functions in my answer. $\endgroup$
    – Bazin
    Commented Feb 20, 2018 at 11:00
  • $\begingroup$ Thanks for the update! One more question: your argument for the periodic functions can be used to show that $$ {\displaystyle \|u\|_{L^{4}(\Omega)}\leq C\|u\|_{L^{2}(\Omega)}^{1/2}\|u\|_{H^{1}(\Omega)}^{1/2}} $$ is true without the "we assume that" assumption (if one uses the $H^1$ norm instead of the gradient norm in the inequality $ \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)} $ , then no further conditions are needed), correct? $\endgroup$
    – user14319
    Commented Feb 21, 2018 at 0:42
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    $\begingroup$ @Jack Yes, but it is not completely obvious, you have to write$$u(x,y)=u(a,y)+\int_a^x(\partial_1u)(s,y) ds$$ and the similar formula for $u(x,y)=u(x,b)+\dots$. Then you have an expression for $\vert u(x,y)\vert^2$ that you can apply to $u=v^2$. $\endgroup$
    – Bazin
    Commented Feb 22, 2018 at 12:50

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