Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. 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

Let $p<n$ and consider the Sobolev inequality on $W^{1,p}(\mathbb{R}^n)$ space: $$\tag{1} \left\lVert u \right\rVert_{p^\star, \mathbb{R}^n} \le C\left\lVert \nabla u \right\rVert_{p,\mathbb{R}^n}.$$ Inequality $(1)$ cannot hold verbatim for a bounded domain $D$: indeed, testing it against the constant function $1$ we get the contradiction $$\tag{!!} 0<\left\lVert1\right\rVert_{p^\star, D}\le C \left\lVert \nabla 1\right\rVert_{p, D}=0.$$ What we do have on $D$, provided that its boundary is not too wild, is the following weaker version of $(1)$: $$\tag{1weak} \left\lVert u \right\rVert_{p^\star, D}\le C\left\lVert u\right\rVert_{1, p, D}.$$

Now let us consider the unbounded open set $\mathbb{R}^n_+=\left\{(x', x_n)\in \mathbb{R}^{n-1}\times \mathbb{R}\ :\ x_n>0\right\}$. Here inequality $(1)$ does hold (if I am not mistaken), because if $u\in C^1\left(\overline{\mathbb{R}^n_+ }\right)\cap W^{1, p}(\mathbb{R}^n_+)$ then we can extend it by reflection: $$\overline{u}(x', x_n)=\begin{cases} u(x', x_n)& x_n \ge 0 \\ -3u(x', -x_n)+4\left(x', -\frac{x_n}{2}\right) & x_n <0 \end{cases}$$ obtaining a function $\overline{u}\in C^1(\mathbb{R}^n)\cap W^{1, p}(\mathbb{R}^n)$ satisfying the following pair of estimates: \begin{align*} \left\lVert \overline{u}\right\rVert_{p, \mathbb{R}^n}&\le C_0 \left\lVert u\right\rVert_{p, \mathbb{R}^n_+}\\ \left\lVert \nabla\overline{u}\right\rVert_{p, \mathbb{R}^n}&\le C_1 \left\lVert \nabla u\right\rVert_{p, \mathbb{R}^n_+}. \end{align*} Thus applying inequality $(1)$ to the extended function $\overline{u}$ we immediately get $$\tag{2} \left\lVert u\right\rVert_{p^\star, \mathbb{R}^n_+}\le C\left\lVert \nabla u \right\rVert_{p, \mathbb{R}^n_+}.$$


  1. Where does this procedure fail on a bounded domain?
  2. Which (unbounded) domains are such that the Sobolev inequality holds in its stronger form $(1)$?

Thank you for reading.

share|cite|improve this question
I'm starting to think that the question is connected to the geometry of $\partial D$. In the case of $\mathbb{R}^n_+$, the reflection method outlined above gives an extension operator $E\colon W^{1,p}(\mathbb{R}^n_+)\to W^{1,p}(\mathbb{R}^n)$ which is continuous with respect to each one of the seminorms $\lVert u\rVert_p$ and $\lVert \nabla u \rVert_p$. [...] – Giuseppe Negro Jan 3 '13 at 14:39
If the boundary $\partial D$ is compact, one might still be able to define an extension operator by means of the reflection method, but in doing so one should summon a partition of unity. This disrupts continuity with respect to the seminorm $\lVert \nabla u \rVert_p$. – Giuseppe Negro Jan 3 '13 at 14:41

I will answer question 1 above. Let $D$ be a bounded domain with a $C^1$ boundary (*). Our goal is to prove inequality (1weak) above by means of an extension technique, mimicking what we did for $\mathbb{R}^n_+$. In doing so we will highlight the reason why we lose the stronger inequality (1).

Let $(U_0, U_i\ :\ i=1\ldots m)$ be open subsets of $\mathbb{R}^n$ which cover $D$ and are such that $U_0\subset D$, $U_i\cap \partial D \ne \varnothing$ and for each $i$ there exists a coordinate system on $U_i$ which straightens $\partial D \cap U_i$. Take a partition of unity $\rho_0, \rho_i$ on $D$ subordinate to this covering. Fix $u\in W^{1,p}(D)$ and write $$ u= (u\rho_0)+\sum_{i=1}^m (u\rho_i)=u_0+\sum_i u_i.$$
Since each function $u_i$ is supported in $U_i$, switching to local coordinates and applying the reflection technique outlined above we can define functions $\overline{u}_i\in W^{1,p}(\mathbb{R}^n)$ which extend $u_i$ and satisfy the following pair of estimates: \begin{equation}\tag{EE} \begin{split} \lVert \overline{u}_i\rVert_{p, \mathbb{R}^n} &\le C_0\lVert u_i\rVert_{p, U_i} \\ \lVert \nabla \overline{u}_i\rVert_{p, \mathbb{R}^n} &\le C_1\lVert \nabla u_i \rVert_{p, U_i}. \end{split} \end{equation} Set $\overline{u}_0=u_0 \in W^{1, p}(\mathbb{R}^n)$ (this is no problem since $u_0 \in W^{1,p}_0(D)$). Then the function $$ \overline{u}=\overline{u}_0+\sum_i \overline{u}_i \in W^{1, p}(\mathbb{R}^n)$$ is an extension of $u$.

We can estimate $\lVert \nabla \overline{u}\rVert_{p, \mathbb{R}^n}$ in terms of $\lVert u\rVert_{1, p, D}$ as follows: \begin{equation}\tag{**} \begin{split} \lVert \nabla \overline{u}_i\rVert_{p, \mathbb{R}^n}&\le C_1 \lVert \nabla u_i\rVert_{p, D} \\ &\le C_1\lVert (\nabla u)\rho_i\rVert_{p, D} + \lVert u \nabla\rho_i\rVert_{p, D} \\ &\le C_2\left( \lVert \nabla u\rVert_{p, D} + \lVert u \rVert_{p, D}\right). \end{split} \end{equation} Summing up we finally arrive at the inequality $$ \lVert \nabla\overline{u}\rVert_{p, \mathbb{R}^n} \le K \lVert u \rVert_{1, p, D}.$$

To conclude we apply the Sobolev inequality in $\mathbb{R}^n$ to the extended function $\overline{u}$. That is, $$ \lVert u\rVert_{p^\star, D}\le \lVert \overline{u}\rVert_{p^\star, \mathbb{R}^n} \le C\lVert \nabla \overline{u}\rVert_{p, \mathbb{R}^n} \le CK\lVert u\rVert_{1, p, D}. $$ We have thus proven the sought inequality (1weak). $\square$

The gray box marks the spot in which we need to sacrifice an estimate on the seminorm $\lVert \nabla u_i\rVert_{p, D}$, getting in exchange a weaker estimate on the norm $\lVert u \rVert_{1, p, D}$. The point is that the partition of unity $\{\rho_0, \rho_i\ :\ i=1\ldots m\}$ is entering the estimate here: to estimate $\lVert \nabla (\rho_i u)\rVert_{p, D}$ you need both $\lVert \nabla u\rVert_{p, D}$ and $\lVert u\rVert_{p, D}$, the seminorm $\lVert \nabla u \rVert_{p, D}$ alone won't do.

(*) Both assumptions on $D$ and on $\partial D$ can be substantially weakened but I don't think there is the need to go into this.

share|cite|improve this answer

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.