Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If $f\in H^2(\mathbb R^2)$, I want to show that

  1. $||f||_{L^\infty}\le c||f||_{H^2}$
  2. $||f||_{L^\infty}\le c||f||_{H^1} [1+\ln(1+||f||_{H^2})]$

For 1, I use $||f||_{L^\infty}\le \sup_{x\in \mathbb R^2} |f|\le\int_{\mathbb R^2} |\hat f(\xi)|d\xi = \int(1+|\xi|^2)^{-1}(1+|\xi|^2)|\hat f(\xi)|d\xi \le (\int (1+|\xi|^2)^{-2}d\xi)^2(\text {which is integrable in this case})||f||_{H^2}\to \text{is this correct?}$

But for 2, I am stucked. How can I get the "ln"? and how can I make it into a product of $H^1$ and $H^2$ norm?

share|improve this question
add comment

1 Answer

up vote 2 down vote accepted

Your approach for 1 is correct.

To prove 2, try this: First prove that for all $\alpha \in (0,1)$ $$ \Vert f \Vert_{L^\infty} \le C_\alpha \Vert f \Vert_{H^1}^\alpha \Vert f \Vert_{H^2}^{1- \alpha} $$ using the Fourier transform and Hoelder's inequality. Keep close track of the constant $C_\alpha$. Then minimize the right hand side with respect to $\alpha$. You should get a sharp estimate like $$ \Vert f \Vert_{L^\infty}\le c \Vert f \Vert_{H^1} \sqrt{1+\ln(1+ \Vert f \Vert_{H^2})} $$ out of this approach.

There is also a real variable approach to this, see my paper in Comm. PDE vol. 14 (1989), pp. 541-544.

share|improve this answer
    
Thank you very much, and I would like to see a real variable approach. Do you have the link of your paper? –  Zhixia Zhang Dec 15 '12 at 17:38
    
Send me an email at the canonical address and I'll send you a copy. I don't think it's on line. –  Hans Engler Dec 15 '12 at 18:33
    
Sorry, the proof is too hard for me to understand. Could you please explain where does the "ln" come from? –  Zhixia Zhang Dec 18 '12 at 16:37
    
Hi @HansEngler. How can i obtain your paper? What is this canonical address? –  Tomás Dec 22 '12 at 23:25
    
engler@georgetown.edu –  Hans Engler Dec 23 '12 at 0:47
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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