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According to Wikipedia, Agmon's inequality provides a bound on the $L^\infty$ norm of a $H^2$ function on a (regular) subset of $\mathbb{R}^3$. In the book of JC Robinson et al. "The Three-Dimensional Navier-Stokes Equations: Classical Theory" in Thm. 1.20 the same statement holds on the whole of $\mathbb{R}^3$, i.e.

$$\|u\|_{L^\infty(\mathbb{R}^3)}\leq C \|u\|_{H^1(\mathbb{R}^3)}^{1/2} \|u\|_{H^2(\mathbb{R}^3)}^{1/2}.$$

My question is:

  • where can I find a proof of this inequality?
  • are there any other useful inequalities which allow to bound the $L^\infty$ norm on $\mathbb{R}^3$.
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  • $\begingroup$ I don't know how to prove the inequality, but if you just want to see that $H^2$ functions are bounded in $d=3$, then you can get this easily by estimating $\|\widehat{u}\|_1$ with Cauchy-Schwarz. $\endgroup$ Commented Aug 3, 2017 at 18:20
  • $\begingroup$ In other words, this will show that $\|u\|_{\infty}\lesssim \|u\|_{H^2}$, but your inequality is stronger because a large second derivative has less of an effect on the RHS. $\endgroup$ Commented Aug 3, 2017 at 18:24

1 Answer 1

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We note that $u$ is Hölder continuous, and we may write for any $M>0$ (to be chosen later), $$ u(x)=\int_{\vert \xi\vert\le M} e^{2π i x\xi}\hat u(\xi)(1+\vert \xi\vert) (1+\vert \xi\vert)^{-1}d\xi +\int_{\vert \xi\vert\ge M} e^{2π i x\xi}\hat u(\xi)\vert \xi\vert^2 \vert \xi\vert^{-2}d\xi, $$ so that using Cauchy-Schwarz inequality, we get \begin{multline} \Vert u\Vert_{L^\infty}\le c_n\Vert u\Vert_{H^1}\left(\int_0^M\frac{r^2}{1+r^2} dr\right)^{1/2}+ c_n\Vert u\Vert_{H^2}\left(\int_M^{+\infty}{r^2}r^{-4} dr\right)^{1/2} \\\le c_n\Vert u\Vert_{H^1} M^{1/2} +c_n\Vert u\Vert_{H^2}M^{-1/2}. \end{multline} Choosing now $M=\Vert u\Vert_{H^2}\Vert u\Vert_{H^1}^{-1}$ gives the sought answer. It is easy to get generalizations of this inequality in $n$ dimensions following the same method with given $s_0, s_1$ such that $s_0<\frac n2< s_1$, yielding $$ \Vert u\Vert_{L^\infty}\le C_{n, s_0,s_1}\Vert u\Vert_{H^{s_0}}^{1-\theta}\Vert u\Vert_{H^{s_1}}^\theta, \quad \frac n2 =(1-\theta) s_0+\theta s_1.$$ Needless to say the above inequality is not true for $s_0=n/2=s_1$.

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  • $\begingroup$ How to prove it in 2D? This proof doesn’t work. $\endgroup$ Commented Jul 9, 2020 at 5:14

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