[Math] Proof of Sobolev Inequality Theroem

functional-analysislp-spacessobolev-spaces

I have a short question about the proof of Theorem 2 below. I have included Theorem 1's statement since it is used in the proof of Theorem 2.

Definition: If $1 \leq p < n$, the Sobolev Conjugate of $p$ is $p* := \frac{np}{n-p}$, note that p* > p.

Consider the following two Theorems related to Sobolev Inequalites:

Theorem 1(Gagliardo-Nirenberg-Sobolev inequality)
Assume $1\leq p<n$. There exists a constant $C$, depeding only on $p$ and $n$, such that $||u||_{L^{p*}(\mathbb{R}^{n})} \leq C||Du||_{L^{p}(\mathbb{R}^{n})}$ for all $u \in C_{c}^{1}(\mathbb{R}^{n})$.

Where in the proof of Theorem 1 it is determined that $C = \frac{p(n-1)}{n-p} = p* – \frac{p}{n-p}$.

Theorem 2
Assume $U$ is a bounded open subset of $\mathbb{R}^{n}$. Suppose $u \in W_{0}^{1,p}(U)$ for some $1 \leq p < n$. Then we have the estimate $||u||_{L^{q}(U)} \leq C||Du||_{L^{P}(U)}$ for each $q \in [1,p*]$, the constant $C$ depending only on $p,q,n$ and $U$.

Proof:
Since $u \in W_{0}^{1,p}(U)$, there exist functions $u_{m} \in C_{c}^{\infty}(U)$ ($m=1,2…$) converging to $u$ in $W^{1,p}(U)$. We extend each function $u_{m}$ to be 0 on $\mathbb{R}^{n}-\bar{U}$ and apply Theorem 1 to discover $||u||_{L^{p*}(U)} \leq C||Du||_{L^{p}(U)}$. As $|U| < \infty$, we furthermore have $||u||_{L^{q}(U)} \leq C||u||_{L^{p*}(U)}$ if $1 \leq q \leq p*$. $\square$

How does the final part of the proof follow? The part which states "…As $|U| < \infty$, we furthermore have $||u||_{L^{q}(U)} \leq C||u||_{L^{p*}(U)}$ if $1 \leq q \leq p*$."

Thanks for any assistance. This proof is from book 'Partial Differential Equations' by Lawrence Evans.

Best Answer

Generalized Holder will do the trick, taking the $L^r$ norm of the indicator on $U$ and the $L^{p\ast}$ norm of $u$. Note the inequality needs $1/q=1/p\ast + 1/r$ which shows what range of $q$ this applies to.

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