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pxchg1200
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how to use the sobolev inequality to obtain the embedding theorem

I am reading Luca Capogna's article An Embedding theorem and the Harnack inequalitiy for nonlinear subelliptic equations. In this article, the authors proved the following theorem

(Theorem 2.3) Let $U\subset \mathbb{R}^n$ be a bounded open set and denote by $Q$ the homogeneous dimension relative to $U$. Let $1<p<Q$. Then there exist $C>0$ and $R_{0}>0$ such that for any $x\in U$, $B_{R}=B(x,R)$ ($B_{R}$ is the subunit ball) with $R\leq R_{0}$, we have $$ \left(\frac{1}{|B_{R}|}\int_{B_{R}}|u|^{sp}dx \right)^{\frac{1}{sp}}\leq CR\left(\frac{1}{|B_{R}|}\int_{B_{R}}|D_{L}u|^{p}dx\right)^{\frac{1}{p}}$$ for any $u\in S_{0}^{1,p}(B_{R})$, Here, $1\leq s\leq \frac{Q}{Q-p}$.

The author says that a standard partition of the unity argument implies $$ S_{0}^{1,p}(U)\hookrightarrow L^{q}(U)$$ for any $U\subset\subset \mathbb{R}^n$. I don't know how to use the partition of unity to obtain this claim. Can someone show it in detail?. Furthermore, can we deduce the following fact ? $$ \left(\int_{U}|u|^{q}dx\right)^{\frac{1}{q}}\leq C\left(\int_{U}|D_{L}u|^{p}dx \right)^{\frac{1}{p}},$$ for $U\subset\subset \mathbb{R}^n$ instead the subunit ball $B_{R}$?

My approach: since $\overline{U}$ is a compact set, then there exist $n$ subunit ball $B_{i}(x_{i},r_{i}) (i=1,\ldots,n)$ which cover $\overline{U}$ (We can assume that each $r_{i}\leq R_{0}$). Then there exists a partition of unity of $B_{i}(x_{i},r_{i}) (i=1,\ldots,n)$ satisfy

(1)$0\leq \phi_{i}\leq 1, \text{supp}\phi_{i}\subset B_{i}(x_{i},r_{i}) $ and $\phi_{i}\in C_{0}^{\infty}(\mathbb{R}^n)$.

(2) $$ \sum_{i=1}^{n}\phi_{i}=1 \qquad \forall x\in U $$

Then for a function $f\in S_{0}^{1,p}(U)$, we have $$ f=\sum_{i=1}^{n}f\phi_{i} $$ \begin{align*} \left(\int_{U}|f|^{q}dx \right)^{\frac{1}{q}}&=\left(\int_{U}|\sum_{i=1}^{n}f\phi_{i}|^{q}dx \right)^{\frac{1}{q}} \end{align*} Then I stuck here and don't know how to continue, it seems far away to the right side. Can some one help me? thank you very much!

pxchg1200
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