**Ladyzhenskaya inequality** For a smooth, compactly supported scalar function $u$ in $\mathbb{R}^2$, we have: $$(1)\hspace{25pt}\int_{\mathbb{R}^2} \abs{u(x)}^4 dx \leq \int_{\mathbb{R}^2} \abs{u(x)}^2 dx \int_{\mathbb{R}^2} \abs{\grad u(x)}^2 dx,$$ $$(1')\hspace{25pt}\int_{\mathbb{R}^2} \abs{u(x)}^8 dx \leq \int_{\mathbb{R}^2} \abs{u(x)}^6 dx \int_{\mathbb{R}^2} \abs{\grad u(x)}^2 dx.$$ In $\mathbb{R}^3$, we have $$(2)\hspace{25pt}\int_{\mathbb{R}^2} \abs{u(x)}^4 dx \leq \big(\int_{\mathbb{R}^2} \abs{u(x)}^2 dx\big)^{1/2}\big( \int_{\mathbb{R}^2} \abs{\grad u(x)}^2 dx\big)^{3/2}.$$ **Remark** The fact that no dimensional constant appears in the right hand side of Ladyzhenskaya inequality is due to the fact that these inequality are invariant by dilation, or homothety ($x\rightarrow \lambda x$), or, in physical terms, that both sides of these inequality have the same dimension. The proof of (1) is based on the following Agmon's type inequality in space dimension one: $$(3)\hspace{25pt}\abs{g(x)}^2 \leq \big(\int_{\mathbb{R}} \abs{g(\xi)}^2 d\xi\big)^{1/2}\big(\int_{\mathbb{R}} \abs{g'(\xi)}^2 d\xi\big)^{1/2},\quad \forall\,x\in\mathbb{R},$$ where $g$ is a smooth function with compact support in $\mathbb{R}$. We are going to prove (1). In dimension $2$, we use Agmon's inequality (3) twice, once in each space direction. We have $$\begin{split} u(x_1,x_2)^4&=u(x_1,x_2)^2u(x_1,x_2)^2~\\ &\leq \bigg[\int_\mathbb{R} \abs{u(\xi_1,x_2}^2d\xi_1\int_\mathbb{R} \abs{u_{x_1}(\xi_1,x_2)}^2 d\xi_1 ~\\ &\quad \int_\mathbb{R} \abs{u(x_1,\xi_2}^2d\xi_2\int_\mathbb{R} \abs{u_{x_2}(x_1,\xi_2)}^2 d\xi_2\bigg]^{1/2}. \end{split}$$ Then integrating in $\mathbb{R}^2$ ($\int_{\mathbb{R}^2}dx_1dx_2$), and using Cauchy-Schwarz inequality implies the result (1). In order to obtain $(1')$, we use the following Agmon's type inequality: $$(3')\hspace{25pt}\abs{g(x)}^4 \leq \big(\int_{\mathbb{R}} \abs{g(\xi)}^6 d\xi\big)^{1/2}\big(\int_{\mathbb{R}} \abs{g'(\xi)}^2 d\xi\big)^{1/2},\quad \forall\,x\in\mathbb{R},$$ and apply $$\begin{split} u(x_1,x_2)^8&=u(x_1,x_2)^4u(x_1,x_2)^4~\\ &\leq \bigg[\int_\mathbb{R} \abs{u(\xi_1,x_2}^6d\xi_1\int_\mathbb{R} \abs{u_{x_1}(\xi_1,x_2)}^2 d\xi_1 ~\\ &\quad \int_\mathbb{R} \abs{u(x_1,\xi_2}^6d\xi_2\int_\mathbb{R} \abs{u_{x_2}(x_1,\xi_2)}^2 d\xi_2\bigg]^{1/2}, \end{split}$$ and finally use Cauchy-Schwarz inequality; we then obtain $(1')$. **Agmon's inequality** Let $u\in H^2(\Omega)\cap H^1_0(\Omega)$ with $\Omega\subset \mathbb{R}^3$, then $$\norm{u}_{L^\infty} \leq \norm{u}_{L^2}^{1/4} \norm{u}_{H^2}^{3/4},\quad\text{ and },\quad\norm{u}_{L^\infty} \leq \norm{u}_{H^1}^{1/2} \norm{u}_{H^2}^{1/2}.$$ **Remark** We know by Sobolev embedding $\norm{u}_{L^\infty}$ is bounded by $\norm{u}_{H^2}$, hence the Agmon's inequality is somehow sharper. === Reference [FMRT01] Foias, Ciprian; Manley, O.; Rosa, R.; Temam, R. Navier-Stokes Equations and Turbulence. Cambridge: Cambridge University Press, 2001