Gagliardo-Nirenberg inequality?

For $1\leq p<n$, we define the Sobolev conjugate of$p$ as $p^{\ast}=\frac{np}{n-p}$.

Recall the Gagliardo-Nirenberg inequality $$||u||_{L^{p^{\ast}}(\mathbb{R}^n)}\leq C ||\nabla u||_{L^p(\mathbb{R}^n)}$$ for some constant $C$ depending only on $n$ and $p$ for any function $u\in C_c^1(\mathbb{R}^n)$.

From this, one derives easily the following estimate for $W^{1,p}$: let $U$ be a bounded open subset of $\mathbb{r}^n$, an open disk say. Assume that $1\leq p<n$ and let $u\in W^{1,p}(U)$. Then $u\in L^{p^{\ast}}(U)$ with $$||u||_{L^{p^{\ast}}(U)}\leq C||u||_{W^{1,p}(U)}$$

I came accross the following inequality $$||u||_{L^2(\partial U)}\leq C||u||_{L^p(U)}^{1-1/p} ||u||_{W^{1,p}(U)}^{1/p}, \quad u\in W^{1,p}(U)$$

Is this a consequence of the previous inequalities?

I am even confused about the appropriateness of the 'u' in the left hand side: I would expect the restriction to $\partial U$ of a function $v\in W^{1,p}(U)$. Since the restriction operator $\tau:W^{1,p}(U)\rightarrow L^2(\partial U)$ is continuous, then $$||v_{|\partial U}||_{L^2(\partial U)}=||\tau(u)||_{L^2(\partial U)}\leq C||u||_{W^{1,2}(U)}$$ and from here I am wondering if the inequalities above shed some light.

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If I may ask, where did this inequality come from? – Jose27 Apr 4 '12 at 6:09
Your statement of Gagliardo-Nirenberg is wrong: the right hand side should have $L^p$ and note $L^{p^*}$. – Willie Wong Apr 4 '12 at 8:23
Apologies, I fixed it. I appreciate a lot your observations below. – Luc Apr 4 '12 at 14:55

If $p \geq 2$, then an appropriate version of the Hölder inequality shows that for $U$ bounded, $L^p(U) \hookrightarrow L^2(U)$, hence $W^{1,p}\hookrightarrow W^{1,2} \hookrightarrow L^2$ using the trace inequality. So you have

$$\|u\|_2 \leq \|u\|_2^{1-1/p}\|u\|_2^{1/p} \lesssim \|u\|_p^{1-1/p} \|u\|^{1/p}_{W^{1,2}} \lesssim \|u\|^{1-1/p}_p \|u\|^{1/p}_{W^{1,p}}$$

which does not require passing through the Gagliardo-Nirenberg inequality. (Though one can argue that the proof of the GN inequality can be recycled to prove the trace theorem, so the two aren't completely unrelated.)

The interesting case is when $p < 2$, in which your estimate follows partially from the generalized Sobolev imbedding theorems (which allows also for trace estimates). See, for example, Theorem 4.12 Case C of Robert Adams' Sobolev Spaces. In particular the numerology requires $p \leq 2 \leq (n-1)p / (n-p)$ which implies

$$\frac{2n}{n+1} \leq p \leq 2$$

must hold for the classical trace theorem $W^{1,p}(\Omega) \hookrightarrow L^2(\partial\Omega)$. (If you believe in fractional Sobolev spaces, then you can also "obtain" the above by combining a fractional version of Gagliardo-Niremberg from $W^{1,p}\hookrightarrow H^s$ for some $s > 1/2$ and using the fractional trace theorem $H^s(\Omega)\hookrightarrow L^2(\partial\Omega)$; note that this route naively will require the lower bound on $p$ to be a strict inequality due to the failure of $H^{1/2}(\Omega)\not\hookrightarrow L^2(\partial\Omega)$.) In particular, for dimension $n > 1$ the $L^1$ endpoint is not covered by this theorem.

This failure at $L^1$ is not a problem of the method: it is a genuine failure of the inequality. Consider the function in two dimension $u(x,y) = \left[ (x-1)^2 + y^2\right]^{1/4}$, which is just a translated version of the function $\frac{1}{\sqrt{r}}$. It is not $L^2$ integrable when restricted to the circle, since it has a Logarithmic singularity at $(1,0)$. But $\frac{1}{\sqrt{r}}$ is certainly $L^1_{loc}(\mathbb{R}^2)$, and its derivative which has a $\frac{1}{r^{3/2}}$ singularity is also $L^1_{loc}(\mathbb{R}^2)$. (The same example also works in arbitrary $n \geq 2$; you can either use the same principle via functions like translates of $\frac{1}{r^{(n-1)/2}}$, or just argue through the method of descent.)

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