Using Results in Sobolev Spaces I have two questions about using results in Sobolev Spaces and a last one on an inequality involving Holder spaces:
1.The Gagliardo-Nirenberg-Sobolev Inequality states the following:
Assume $1 \leq p < n$. Then there exists a constant $C$, depending only on $p$ and $n$, such that $||u||_{L^{p^{*}}(\mathbb{R}^{n})} \leq C||Du||_{L^{p}(\mathbb{R}^{n})}$ for all $u \in C_{c}^{1}(\mathbb{R}^{n})$ .
Can we apply this Theorem to any arbitrary $u \in W^{1,p}(\mathbb{R}^{n})$ by noting that $C_{c}^{1}(\mathbb{R}^{n})$ is dense in $W^{1,p}(\mathbb{R}^{n})$ and then taking $\{u_{m}\}_{m}^{\infty} \subset C_{c}^{1}(\mathbb{R}^{n})$ such that $u_{m} \rightarrow u$ in $W^{1,p}(\mathbb{R}^{n})$. Using the G-N-S Inequality we get $||u_{m}||_{L^{p^{*}}(\mathbb{R}^{n})} \leq C||Du_{m}||_{L^{p}(\mathbb{R}^{n})}$ and then it follows that $\text{lim}_{m \rightarrow \infty}$ $||u_{m}||_{L^{p^{*}}(\mathbb{R}^{n})} \leq \text{lim}_{m \rightarrow \infty}C||Du_{m}||_{L^{p}(\mathbb{R}^{n})}$ and therefore $||u||_{L^{p^{*}}(\mathbb{R}^{n})} \leq C||Du||_{L^{p}(\mathbb{R}^{n})}$ for $u \in W^{1,p}(\mathbb{R}^{n})$.
Is this fine?
2.Similarly does Morrey's inequality which states:
Assume $n < p \leq \infty$. Then there esists a constant $C$, depending only $p$ and $n$, such that $||u||_{C^{0,\gamma}(\mathbb{R}^{n})} \leq C||u||_{W^{1,p}(\mathbb{R}^{n})}$ for all $u \in C^{1}(\mathbb{R}^{n})$. 
Does this hold for all $u \in W^{1,p}(\mathbb{R}^{n})$ by a similar density argument as above?
3.Lastly do we have the following inequality or some variation of this:
$||u||_{L^{p}(U)} \leq ||u||_{C^{0,\gamma}(\bar{U})} \leq ||u||_{W^{1,p}(U)}$ for $n < p \leq \infty$ for $u \in W^{1,p}(U)$ where $U$ is a bounded, open subset of $\mathbb{R}^{n}$ and $\partial U$ is $C^{1}$ and where $C^{0,\gamma}(\bar{U})$ is the Holder space. 
Thanks a lot for any assistance! Let me know if something is unclear.
 A: For the first one:
you still need to prove
$$\lim\|u_m\|_{p^*} = \|u\|_{p^*}.$$
That is, you need to show that $u_m \to u$ in $L^{p^*}$. Hint: Use the GNS to show that $u_m$ is a Cauchy sequence.
The same comment applies to 2.
In order to prove 3 from 2, you still need an extension property for $U$, that is, for each $u \in W^{1,p}(U)$, there exists $v \in W^{1,p}(\mathbb R^n)$ with $\|v\| \le C \, \|u\|$.
A: I think gerw has the missing piece of your proof in the hint that you should use Cauchy sequences instead of using G-N-S and $||u_{m}-u||_{L^{p^{*}}} <= C|||Du_{m}-Du||_{L^{p}}$. Having said that I don't think his answer was written in a very informative way. 
To adapt your proof for question (1) use the following:
Since you showed that $\{Du_{m}\}_{m}$ is convergent in $L^{p}(\mathbb{R}^{n})$ it also follows that it is Cauchy in $L^{p}$. So then using the inequality $||u_{m}-u_{j}||_{L^{p^{*}}} \leq C||Du_{m}-Du_{j}||_{L^{p}}$ it is easy to see that $\{u_{m}\}_{m}^{\infty}$ is Cauchy in $L^{p^{*}}$, since $L^{p^{*}}$ is a Banach Space(complete) it follows that $u_{m} \rightarrow v$ for some $v \in L^{p^{*}}(\mathbb{R}^{n})$. You also have $u_{m} \rightarrow u$ in $W^{1,p}(\mathbb{R}^{n})$ which implies $u_{m} \rightarrow u$ in $L^{p}(\mathbb{R}^{n})$. Convergence in $L^{p}$ implies convergence in measure, so the limits are equal a.e.(so $v=u$ a.e.). So $u_{m} \rightarrow u$ in $L^{p^{*}}(\mathbb{R}^{n})$. 
It follows now from the continuity of the norm that $||u_{m}||_{L^{p^{*}}} \rightarrow ||u||_{L^{p}}$ and $||Du_{m}||_{L^{p}} \rightarrow ||Du||_{L^{p}}$, this together with the N-G-S inequality $||u_{m}||_{L^{p^{*}}} \leq C||Du_{m}||_{L^{p}}$ gives you the desired inequality $||u||_{L^{p^{*}}} \leq ||Du||_{L^{p}}$. Note that this result also give you the continuous embedding $W^{1,p}(\mathbb{R}^{n}) \subset L^{p}(\mathbb{R}^{n})$.     
The same type of density argument applies to (2). 
A good reference for the full proofs of 'Gagliardo-Nirenberg-Sobolev Inequality' and 'Morrey's Inequality' is Brezis book 'Functional Analysis, Sobolev Spaces and Partial Differential Equations'
It would be better to post (3) as a separate question if you don't necessarily want it to follow from the other two questions.
