Agmon's inequality on an open subset $\Omega$ of $\mathbb{R}^3$ I'm looking for a reference for what we call the Agmon's inequality on a regular open bounded subset $\Omega$ of $\mathbb{R}^3$ :
$$\|u\|_{L^\infty(\Omega)}\leq C \|u\|_{H^1(\Omega)}^{1/2} \|u\|_{H^2(\Omega)}^{1/2}.$$
This limit case for Sobolev embeddings can be found for example in the book of JC Robinson et al. "The Three-Dimensional Navier-Stokes Equations: Classical Theory" in Thm. 1.20, or in the book of Foias et al. " Navier-Stokes equations and turbulence" but I didn't manage to find a precise proof (the lecture notes of Agmon only provide the exponants $1/4, 3/4$ on the norms in the inequality).
Thank you !
 A: I think I have actually managed to get a direct proof, with the good exponants, which I write down here.
First, we extend our fixed function $u \in \mathrm{H}^2(\Omega)$ on the whole space thank to a continuous extension operator $\mathcal{E} : \mathrm{H}^s(\Omega) \rightarrow \mathrm{H}^s(\mathbb{R}^3)$ (note that this one is the same for all values of $s$, see for example the book of Stein , Singular integrals and differentiability properties of functions. Princeton Mathematical Series, No. 30. Princeton University Press, Princeton, N.J., 1970. )
So, if we set $v:=\mathcal{E} u$, we have $\Vert u \Vert_{\mathrm{L}^{\infty}(\Omega)} \leq \Vert v \Vert_{\mathrm{L}^{\infty}(\mathbb{R}^3)} $ and since we have $$\Vert v \Vert_{\mathrm{H}^1(\mathbb{R}^3)}^{1/2} \Vert v \Vert_{\mathrm{H}^2(\mathbb{R}^3)}^{1/2} \leq c \Vert u \Vert_{\mathrm{H}^1(\Omega)}^{1/2} \Vert u \Vert_{\mathrm{H}^2(\Omega)}^{1/2},$$
it remains to prove that the following estimate holds :
$$ \Vert v \Vert_{\mathrm{L}^{\infty}(\mathbb{R}^3)} \lesssim\Vert v \Vert_{\mathrm{H}^1(\mathbb{R}^3)}^{1/2} \Vert v \Vert_{\mathrm{H}^2(\mathbb{R}^3)}^{1/2}.$$
In order to do so, we use Sobolev embedding for $v \in \mathrm{H}^2(\mathbb{R}^3) \hookrightarrow \mathscr{C}(\mathbb{R}^3)$ with the estimate
$$\Vert v \Vert_{\mathrm{L}^{\infty}(\mathbb{R}^3)} \lesssim \Vert \widehat{v} \Vert_{\mathrm{L}^{1}(\mathbb{R}^3)}.$$
Then, we write
$$\Vert \widehat{v} \Vert_{\mathrm{L}^{1}(\mathbb{R}^3)}=\int_{B(0,A)}\widehat{v}(\xi) \, \mathrm{d}\xi \ + \int_{B(0,A)^c}\widehat{v}(\xi) \, \mathrm{d}\xi,  $$ with $A>0$ which will be chosen later. We use Cauchy-Schwarz inequality on each term to get
$$\int_{B(0,A)}\widehat{v}(\xi) \, \mathrm{d}\xi \leq \left(\int_{B(0,A)}\dfrac{\mathrm{d}\xi}{1+\vert \xi \vert^2} \right)^{1/2} \left( \int_{B(0,A)} (1+\vert \xi \vert^2) \vert \widehat{v}(\xi) \vert^2 \, \mathrm{d}\xi \right)^{1/2},$$
$$\int_{B(0,A)^c}\widehat{v}(\xi) \, \mathrm{d}\xi \leq \left(\int_{B(0,A)^c}\dfrac{\mathrm{d}\xi}{1+\vert \xi \vert^4} \right)^{1/2} \left( \int_{B(0,A)^c} (1+\vert \xi \vert^4) \vert \widehat{v}(\xi) \vert^2 \, \mathrm{d}\xi \right)^{1/2}. $$
Thus, we have
$$\Vert v \Vert_{\mathrm{L}^{\infty}(\mathbb{R}^3)} \lesssim \left(\int_0^A \dfrac{r^2}{1+r^2} \mathrm{d}r \right)^{1/2}\Vert v \Vert_{\mathrm{H}^1(\mathbb{R}^3)}+\left(\int_A^{+\infty} \dfrac{r^2}{1+r^4} \mathrm{d}r \right)^{1/2}\Vert v \Vert_{\mathrm{H}^2(\mathbb{R}^3)}.$$
We can easily compute the first integral and bound it exactly by $A^{1/2}$. For the other one, it's a convergent integral whose integrand is equivalent to $1/r^2$ when $r \rightarrow +\infty$ so that it is equivalent to the rest of the integral of $1/r^2$ on [A,+infty[, which can be compared to $A^{-1/2}$ (equivalent of the rest of a Riemann integral).
We end up with
$$\Vert v \Vert_{\mathrm{L}^{\infty}(\mathbb{R}^3)} \lesssim A^{1/2}\Vert v \Vert_{\mathrm{H}^1(\mathbb{R}^3)} + A^{-1/2}\Vert v \Vert_{\mathrm{H}^2(\mathbb{R}^3)},$$ for $A$ large enough. We finally choose $A$ of order $\Vert v \Vert_{\mathrm{H}^2(\mathbb{R}^3)} \, / \, \Vert v \Vert_{\mathrm{H}^1(\mathbb{R}^3)}$ to complete the proof.
