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Does anyone know a simple way to derive the following inequality for smooth, compactly supported functions in $\mathbb R^3$?

$$ \max \,\{ |u(x)| \mid x \in \mathbb R^3 \} \;\leq\; K\|\, Du \|_{L^2(\mathbb R^3)}^{1/2} \| D^2 u \|_{L^2(\mathbb R^3)}^{1/2} $$

(This is sometimes called "Moser inequality")

Michael Hardy
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Paulo
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  • Do you have any reference for such an inequality? Where did you learn it? – Siminore Jul 18 '15 at 08:27
  • This inequality is part of a general family of Sobolev estimates of the form

    (1) $ ;, | , u ,|{L^{\infty}(\mathbb{R}^{n})} ,\leq; K , |, D^{m-1} u , |{L^{2}(\mathbb{R}^{n})}^{1/2} , |, D^{m+1} u ,|_{L^{2}(\mathbb{R}^{n})}^{1/2} ;;;;$ if $n = 2;!m $,

    (2) $;, | , u ,|{L^{\infty}(\mathbb{R}^{n})} ,\leq; K , |, D^{m} u , |{L^{2}(\mathbb{R}^{n})}^{1/2} , |, D^{m+1} u ,|_{L^{2}(\mathbb{R}^{n})}^{1/2} ;;;;$ if $n = 2;!m,+,1$.

     – Paulo Jul 19 '15 at 16:45
  • See e.g. Prop. 3.8, Ch. 13, in Michael Taylor's book on PDEs (M. Taylor, Partial Differential Equations, vol. III, Springer, New York, 1996). However, Taylor's proof provided there is inadequate: he simply claims the validity of an equivalent inquality, which seems just as hard to obtain. – Paulo Jul 19 '15 at 17:15
  • Estimating the supnorm in terms of derivatives of low order is very important in PDE theory; in dimension $ n = 2m $ or $ n = 2m + 1 $, any such estimate in terms of L2 norms has to involve derivatives of order at least $ m + 1 $ (see Taylor's book, Ch. 13). For example, it is easy to derive the following Nirenberg-Gagliardo estimate:

    (3) $;, |, u ,|{L^{\infty}(\mathbb{R}^{n})} ,\leq, |, u ,|{L^{2}(\mathbb{R}^{n})}^{1/(m+1)} , |, D^{m+1} u ,|_{L^{2}(\mathbb{R}^{n})}^{m/(m+1)} ;$ if $ n = 2m $,

    and

     – Paulo Jul 19 '15 at 17:26
  • (4) $;, |, u ,|{L^{\infty}(\mathbb{R}^{n})} ,\leq, |, u ,|{L^{2}(\mathbb{R}^{n})}^{1/(2m+2)} , |, D^{m+1} u ,|_{L^{2}(\mathbb{R}^{n})}^{(2m+1)/(2m+2)} ;$ if $ n = 2m + 1 $. – Paulo Jul 19 '15 at 17:28
  • In particular, the Moser estimates are very easy to show in dimension $ n = 1 $ or $ n = 2 $. The real difficulty starts in dimension $ n = 3 $: once this is obtained, it is likely that Moser's inequalities can be obtained for any $ n \geq 3 $ as well. – Paulo Jul 19 '15 at 17:30
  • Actually, Moser's inequalities seem to be not really needed anywhere in analysis, since the much easier Gagliardo-Nirenberg inequalities (3), (4) above always do the job. So, the question I raised (about showing Moser's inequality in dimension $ n = 3 $, and hopefully also for any $ n \geq 3 $) is not really vital for anything; it is simply an intelectual question. – Paulo Jul 19 '15 at 17:37
  • I looked at the papers by Moser and by Nirenberg. It seems that 1d inequalities usually suffice to cover the general case. – Siminore Jul 19 '15 at 17:37
  • One last remark: (3), (4) are direct consequences of Prop. 2.4, Ch. 13, in Taylor's book (M. E. Taylor, Partial Differential Equations, vol. III, Springer, New York, 1996). – Paulo Jul 19 '15 at 17:41
  • 1D inequalities can be very useful, true, but not in this case. – Paulo Jul 19 '15 at 17:44

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