# Gagliardo-Niremberg inequality on annuli (proof)

I already ask the question on Math Stack Exchange (here) but without any answer. I hope I'll have more chance here.

Let $$\frac{1}{\tau}=a\left(\frac{1}{p}-\frac{1}{d}\right)+\frac{1-a}{q},$$ $\tau>0, p\geq 1, a\in [0,1]$, $q\geq 1$ and $d\geq 1$. I know that $$\|u\|_{L^\tau(\mathbb R^d)}\leq C\|\nabla u\|_{L^p(\mathbb R^d)}^a\|u\|_{L^q(\mathbb R^d)}^{1-a}.\tag{E}$$

I want to prove that

$$\|u-\bar u\|_{L^\tau(\mathcal D)}\leq C\|\nabla u\|_{L^p(\mathcal D)}^a\|u-\bar u\|_{L^q(\mathcal D)}^{1-a},$$ where $$\mathcal D=\{x\in\mathbb R^d\mid r<|x|<R\},$$ and $$\bar u=\frac{1}{|\mathcal D|}\int_{\mathcal D}u.$$

Q1) If $\frac{1}{p}-\frac{1}{d}>0$ it's a consequence of Sobolev inequality. But how can I conclude when $\frac{1}{p}-\frac{1}{d}\leq 0$ ? It's probably using a prolongement extension of $u-\bar u$ on $\mathbb R^d$ and using (E) but I don't see how to make it.

Q2) Why do we need $u-\bar u$ instead of $u$ in $$\|u-\bar u\|_{L^\tau(\mathcal D)}\leq C\|\nabla u\|_{L^p(\mathcal D)}^a\|u-\bar u\|_{L^q(\mathcal D)}^{1-a} \ \ ?$$ If $u$ doesn't work, I don't really see why.

I'm sure the answer is in the paper "ulteriori proprietà di alcune classi di funzioni in più variabili" of Gagliardo, but I try to find it on the web without any success.

Can we do as following ? Let $P:W^{1,p}(\mathcal D)\longrightarrow W^{1,p}(\mathbb R^d)$ a prolongement of $u-\bar u$. Then, by (E) $$\|P(u-\bar u)\|_{L^\tau(\mathbb R^d)}\leq \|\nabla Pu\|_{L^p(\mathbb R^d}^a\|P(u-\bar u)\|_{L^q(\mathbb R^d)}^{1-a}.$$ We know that $$\|\nabla Pu\|_{L^p(\mathbb R^d)}\leq C\|\nabla u\|_{L^p(\mathcal D)},$$ and thus $$\|u-\bar u\|_{L^\tau(\mathcal D)}\leq C\|\nabla u\|_{L^p(\mathcal D)}^a\|P(u-\bar u)\|_{L^q(\mathbb R^d)}^{1-a}.$$ Now, can I say that $$\|P(u-\bar u)\|_{L^q(\mathbb R^d)}^{1-a}\leq C\|u-\bar u\|_{L^q(\mathcal D)}^{1-a},$$ or not really ?

• Regarding Q2: If $u$ is a constant function then $\nabla u = 0$ and the inquality will be false. In infinite-measure domains that is not required since constans are not in $L^p$. – Adrián González-Pérez Feb 8 '18 at 14:50
• @AdriánGonzález-Pérez: Thanks a lot for your remark, that's help a lot to understand why takin $u-\bar u$ instead of $u$. – idm Feb 8 '18 at 15:03
• Do you require your constants to be independent of $R$ and $r$? – Adrián González-Pérez Feb 8 '18 at 16:17
• @AdriánGonzález-Pérez: No, it can be dependent of $r$ and $R$. – idm Feb 8 '18 at 16:27
• @AdriánGonzález-Pérez: I added a proof in my post. Do you think it's correct ? Thanks. – idm Feb 9 '18 at 8:17

Let me give a solution of Q1 in the case in which $r^{-1} = p^{-1} - d^{-1}$ satisfies that $r = \infty$. The case of $r < 0$ is a litle bit more involved.
Case of $r = \infty$:
Use first that: $$\|u - \bar{u} \|_{\tau} \lesssim \| u - \bar{u} \|_\infty^{a} \, \| u - \bar{u} \|_{q}^{1-a}.$$ Now, $\| u - \bar{u} \|_\infty \leq \| u \|_{BMO}$ and by the Sobolev embedding theorem in the critical range $p = d$ we have that $\| u \|_{BMO} \leq \| \nabla u \|_{p}$. You can prove that inequality by hand using the mean value theorem or look in [S:Theorem 1.4.4], where something stronger is proven.
• Thank you for your answer. How do you get $\|u-bar u\|_\tau \leq \|u-\bar u\|_\infty ^a\|u-\bar u\|_{q}^{1-a}$ ? (I guess the norm are on $\mathcal D$, no ?). And for $r<0$ it goes the same ? – idm Feb 8 '18 at 16:09
• ok, it's fine. Nice see for the BMO norm. For $r<0$ you do have an idea ? (thanks a lot for your help). By the way, do you think that the inequality $$\|u\|_{L^\tau(\mathbb R^d)}\leq C\|\nabla u\|_{L^p(\mathbb R^d)}^a\|u-u\|_{L^q(\mathbb R^d)}^{1-a}$$ can be usefull ? (because in a paper I'm reading, the inequality I'm asking for it's a consequence of it's inequality). – idm Feb 8 '18 at 16:30
• I didn't have time to work out the details of the other case but i will proceed as follows: First try to prove that $$\| u - \bar{u} \|_\tau \lesssim \| u \|_{C^{1-\frac{d}{p}}}^{a} \, \| u - \bar{u} \|_{q}^{1-a}$$ and then apply Morrey's inequality. For the first step, express $u -\bar{u}$ in as an integral of $|u(x) - u(y)|$ (this is a standard trick). Then, use that a priori $u \in C^{1-\frac{d}{p}}$ to extract a term depending of $|x-y|$ and try to integrate that extra term (probably using that you are working on an annulus to avoid the $x = y = 0$ singularity for integrability purposes). – Adrián González-Pérez Feb 8 '18 at 16:35