MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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. – Deane Yang Aug 20 '12 at 14:19
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 – Piero D'Ancona Aug 20 '12 at 18:42
Oh sorry, by "f" I meant "u" of course. Thanks a lot for the reference Piero, I'll take a look at it. – Welfar Aug 21 '12 at 10:03
up vote 7 down vote accepted

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.

share|cite|improve this answer
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. – Deane Yang Aug 20 '12 at 14:21
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. – Deane Yang Aug 20 '12 at 14:33
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. – Welfar Aug 21 '12 at 10:04

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.