# Which domains make Sobolev inequality weaker?

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 $$\tag{2} \left\lVert u\right\rVert_{p^\star, \mathbb{R}^n_+}\le C\left\lVert \nabla u \right\rVert_{p, \mathbb{R}^n_+}.$$

Questions.

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 $(1)$?

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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: $$\tag{EE} \begin{split} \lVert \overline{u}_i\rVert_{p, \mathbb{R}^n} &\le C_0\lVert u_i\rVert_{p, U_i} \\ \lVert \nabla \overline{u}_i\rVert_{p, \mathbb{R}^n} &\le C_1\lVert \nabla u_i \rVert_{p, U_i}. \end{split}$$ 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: $$\tag{**} \begin{split} \lVert \nabla \overline{u}_i\rVert_{p, \mathbb{R}^n}&\le C_1 \lVert \nabla u_i\rVert_{p, D} \\ &\le C_1\lVert (\nabla u)\rho_i\rVert_{p, D} + \lVert u \nabla\rho_i\rVert_{p, D} \\ &\le C_2\left( \lVert \nabla u\rVert_{p, D} + \lVert u \rVert_{p, D}\right). \end{split}$$ 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.

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