TITLE: Function and its Gradient with Prescribed Norms
QUESTION [5 upvotes]: I'm not sure if the following question is too elementary for Mathoverflow. I'm sorry if it is the case. 
Question: 
Let $n\in\mathbb{N}$ and let $1\leqslant p<\infty$. Let $\alpha,\beta>0$. What is the necessary and sufficient condition $\alpha,\beta$ for which there exists a $u\in C^\infty_{0}(\mathbb{R}^n)$ such that 
$$
\|u\|_p=\alpha\text{ and }\|Du\|_p=\beta?
$$
What happens if we replace $\mathbb{R}^n$ with an open (not necessarily bounded) set $U\subseteq\mathbb{R}^n$? 
Probably, it has something to do with the Poincare Inequality or the first eigenvalue of the domain. But I'm unable to make it precise. 
Thank you. 
Stefan

REPLY [6 votes]: Regarding $\mathbb{R}^n$, that's a simple matter of scaling. Assume w.l.o.g. that $\alpha = 1$ (the quantity that does matter in your problem is the ratio $\frac{\beta}{\alpha}$). Let $u$ be your favorite smooth cut-off function and assume w.l.o.g. that its $L^p$ norm is equal to $1$. Denote by $N$ the quantity $\|Du\|_p$.
Define $u_{\lambda}(x) := \lambda^{-\frac{n}{p}} u(\frac{x}{\lambda})$. Then, notice that $u_{\lambda}$ has still an $L^p$ norm equal to $1$, while that of its derivative is $\lambda^{-1}N$. Choosing $\lambda$ such that $\lambda^{-1}N = \frac{\beta}{\alpha}$, you are done.
Regarding the case of a general open subset $U \subset \mathbb{R}^n$, if you choose $\|u\|_p = 1$ (say), then the possible choices for $\frac{\beta}{\alpha}$ should lie in something like $\left[\sqrt{\mu_1(p)}, + \infty \right[$, where $\mu_1(p)$ is the first eigenvalue of the Dirichlet p-Laplacian on $U$. I'm no expert on this subject, but if things go like in the $p=2$ case, then having $U$ bounded in one direction gives a nontrivial range, contrary to what happens with $\mathbb{R}^n$.