MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If $P$ is an elliptic pseudodifferential operator of order 1 in the sense that its principal symbol is invertible, then we have the a priori estimate

$\|u\|_{H^1(U)} \le C (\|Pu\|_{L^2(W)} + \|u\|_{L^2(W)}), \quad (C>0 \text{ a constant})$

for any $\overline{U}\subset W\subset \mathbb{R}^n$ with $W$ bounded. Say I'm only interested in an upper bound of $\|u\|_{L^2(U)}$. Then the term $\|u\|_{L^2(W)}$ on the right hand side above feels a bit redundant. Can I obtain

$\|u\|_{L^2(U)} \le C\|Pu\|_{L^2(W)} \quad \text{?}$

share|cite|improve this question

closed as no longer relevant by Ryan Budney, Willie Wong, Andy Putman, Deane Yang, Bill Johnson Dec 2 '11 at 17:32

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

The answer to your question is no. Take any non-injective operator $P.$

share|cite|improve this answer
I forgot to mention I work semi-classically, so for a small parameter elliptic operators are injective. Anyway, I think I solved the problem. Thank you. – flavio Dec 2 '11 at 12:41
If your operator is injective then you can even estimate the H^1 -norm of u in terms of the L^2-norm of u. This is true because any injective operator with closed image satisfies an injectivity estimate. – Orbicular Dec 2 '11 at 12:50

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