# Talk:Sobolev inequality

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Field: Analysis

## Removing ${\displaystyle q=\infty }$

The current statement of the Sobolev embedding reads:

"Let W k,p(Rn) denote the Sobolev space consisting of all real-valued functions on Rn whose first k weak derivatives are functions in Lp. Here k is a non-negative integer and 1 ≤ p ≤ ∞. The first part of the Sobolev embedding theorem states that if k > and 1 ≤ p < q ≤ ∞ are two extended real numbers such that (k)p < n and:

${\displaystyle {\frac {1}{q}}={\frac {1}{p}}-{\frac {k-\ell }{n}},}$

then

${\displaystyle W^{k,p}(\mathbf {R} ^{n})\subseteq W^{\ell ,q}(\mathbf {R} ^{n})}$

and the embedding is continuous."

I don't see why ${\displaystyle q=\infty }$ is allowed. In standard measure theory convention, ${\displaystyle q=\infty }$ would mean ${\displaystyle 1/p-(k-\ell )/n=1/q=0\Leftrightarrow 1/p=(k-\ell )/n\Leftrightarrow (k-\ell )p=n}$, which is directly excluded by the additional requirement ${\displaystyle (k-\ell )p. There is very good reason for not being able to improve this strict inequality to allow for equality. For example, it is well known that ${\displaystyle W^{1,n}(\mathbb {R} ^{n})}$ does not embed continuously into ${\displaystyle L^{\infty }(\mathbb {R} ^{n})}$ (which corresponds to ${\displaystyle p=n,q=\infty ,k=1,\ell =0}$ in the notation of the article).

I'm therefore removing the possibility that ${\displaystyle q=\infty }$ from the statement.

--68.5.85.99 (talk) 06:08, 26 January 2015 (UTC)

## Gagliardo--Nirenberg--Sobolev

Hi Igny. I would like your help if possible on the GNS inequality. I am onaraighl and I am the one who edited your article, but I am an applied mathematician and not an expert here.

It's just I have a problem with the GNS inequality if you don't subtract off a constant term: taking u=constant\neq0 on the open unit disc would satisfy the conditions of the theorem, yet gives constant\leq0, which seems to be a contradiction. I have spoken to some of my professors about the GNS inequality and consulted the functional analysis book by Kantorovich, and they all say that you need to subtract off the mean of the function in the GNS inequality.

I would dearly like it if it were the case that you did not have to subtract off the mean, since then some estimates that I am working on for the Cahn--Hilliard equation would work out a lot nicer, yet I just don't see how you can do that.

Also, perhaps the article would benefit from a discussion on the optimal constants for the bound, which clearly depends on the domain in question.

Thanks.Onaraighl 17:57, 13 May 2007 (UTC)

Actually, GNS is formulated for functions which are defined in whole R^n but have a compact support. By a standard 'lifting' argument GNS can be extended to functions from W^1_p(R^n). Next, if you have function defined in whole R^n, you'll have a lot of trouble defining its mean. You can substract the mean if you want GNS in bounded domain (equivalently, you get GNS for functions with zero mean), but this isn't the only case, for example, you can have GNS for functions with zero trace on the boundary (Dirichlet conditions).Endeavor 00 (talk) 20:53, 20 August 2009 (UTC)

## L2 estimates

There doesn't seem to be much explanation of how the various inequalities stated relate to each other or how they can be used to prove the general Sobolev imbedding theorem. Failing that I'll like to add the proof of the easy form of Sobolev imbedding only for ${\displaystyle L^{2}}$ estimates, that ${\displaystyle W^{k,2}}$ consists of strongly differentiable functions for ${\displaystyle k>n/2}$, but that would be informative if someone could add more about the general theorem or a proof sketch based on the inequalities stated here. Compsonheir (talk) 20:58, 1 July 2009 (UTC)

Well, general Sobolev embeddings are actually composed of several inequalities, Gagliardo-Nirenberg-Sobolev and Morrey being the main. There are several ways to prove them, but they all are quite involved (say, proof of GNS inequality takes about two pages in Evans' book). Furthermore, proofs are essentially different for different inequalities. Assuming ${\displaystyle p=2}$ doesn't simplify matters as far as I know. Next, ${\displaystyle W^{k,2}}$ consists not of strongly differentiable, but of Holder continuous functions - check the Morrey inequality.Endeavor 00 (talk) 20:24, 20 August 2009 (UTC)