2
$\begingroup$

I've read on another topic that general interpolation result from Gagliardo-Nirenberg inequality can be read as follow :

\begin{equation} \|D^ju\|^1_{L^p} \leq C \|D^mu\|^a_{L^r} \|u\|^{1-a}_{L^q} \end{equation}

with some relations between $a$, $r$, $q$ and $p$, $j$ and $m$.

Does this inequality stands in $\mathbb{R}^n$ ? More precisely, I would make sure that $C$ only depends on $f$, and not of its support.

Thanks for any help!

Improve this question
$\endgroup$
3
  • $\begingroup$ What's $f$? For functions on $R^n$, the constant $C$ depends only on the dimension $n$ and the numbers $p$, $q$, $r$, $j$, and $m$. It does not depend on the function $u$ at all. $\endgroup$
    – Deane Yang
    Commented Aug 20, 2012 at 14:19
  • $\begingroup$ There is a very nice account of general GN inequalities in the first chapter of Volume III of Taylor's PDE. I think it is not worth to repeat the whole argument here $\endgroup$
    – Piero D'Ancona
    Commented Aug 20, 2012 at 18:42
  • $\begingroup$ Oh sorry, by "f" I meant "u" of course. Thanks a lot for the reference Piero, I'll take a look at it. $\endgroup$
    – Welfar
    Commented Aug 21, 2012 at 10:03

1 Answer 1

Reset to default
8
$\begingroup$

The mother of all Gagliardo-Nirenberg inequalities is $$ \Vert u\Vert_{L^{\frac{n}{n-1}}(\mathbb R^n)}\le c_n \Vert \nabla u\Vert_{L^{1}(\mathbb R^n)}, \tag {GN} $$ where $c_n$ depends only on $n$ and $u$ runs say in $C^1_c(\mathbb R^n)$. Applying this to $u=v^2$, you get with $p=\frac{2n}{n-1}$ $$\Vert v\Vert_{L^p}^2= \left(\int\vert v\vert^{\frac{2n}{n-1}} dx\right)^{\frac{n-1}{n}}\le c_n \int2 \vert v\vert\vert\nabla v\vert dx\le 2c_n\Vert v\Vert_{L^p}\Vert \nabla v\Vert_{L^{p'}}, $$ which is $ \Vert v\Vert_{L^{\frac{2n}{n-1}}}\le 2c_n\Vert \nabla v\Vert_{L^{\frac{2n}{n+1}}}. $ Manipulations of the same type (apply (GN) to $u=v^\alpha$) induce the Sobolev inclusions $$ \dot W^{s,p}\subset \dot W^{t,q},\quad s>t,\ p< q,\quad (s-t)/n=1/p-1/q $$ with $p,q\in(1,+\infty)$. Derivatives are somehow a convertible currency that you can exchange against a fixed amount of $L^p$ regularity according to the exchange rate displayed above, but $L^p$ regularity is a non-convertible currency which cannot buy derivatives.

Improve this answer
$\endgroup$
3
  • 5
    $\begingroup$ But you should also explain how the first inequality is proved using only the 1-dimensional fundamental theorem of calculus. And how the general inequality is proved using induction, integration by parts, and the Holder inequality. This is one of the easiest proofs of a deep fact that I've ever seen. I learned it from Nirenberg himself in a class he taught on PDE's. I can still remember it quite clearly, because it was so mind-boggling to me how easy it was. $\endgroup$
    – Deane Yang
    Commented Aug 20, 2012 at 14:21
  • 1
    $\begingroup$ And you can figure out the value of $a$ by demanding that both sides of the inequality scale the same under rescaling the co-ordinate variable $x$ by a constant factor. This can also be said by assuming that $x$ has units (or, as you say, a "currency") associated to it and demanding that the induced units (which are powers of the units for $x$) corresponding to the two sides of the inequality must agree. $\endgroup$
    – Deane Yang
    Commented Aug 20, 2012 at 14:33
  • $\begingroup$ Well, thank you for your answers Deane, that was very helpful! Actually I already known the easy proof for the basic inequality, I just didn't know too much about the interpolation one. $\endgroup$
    – Welfar
    Commented Aug 21, 2012 at 10:04

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .