5
$\begingroup$

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

$\endgroup$

1 Answer 1

6
$\begingroup$

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 $\mathbb{R}^n$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.