Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Here they are: $$||f||_{\infty} \leq C ||f||_q^{\frac {qk} {n+kq}} \left( \sum_{|\mu|=k} ||D^\mu f||_{BMO} \right)^{ \frac n {n+kq}}$$ and $$||f||_{Lip_\alpha} \leq C ||f||_q^{\frac {qk} {n+kq} \frac {k-\alpha} k} \left( \sum_{|\mu|=k} ||D^\mu f||_{BMO} \right)^{ \frac n {n+kq} \frac {k-\alpha} k + \frac \alpha k};$$ also $$||f||_{\infty} \leq C ||f||_q^{\frac q r \left( \frac n {rk-n} + \frac q r \right)^{-1}} ||\nabla^k f||_r^{\frac n {rk-n} \left( \frac n {rk-n} + \frac q r \right)^{-1}} \quad (rk>n)$$ and $$||f||_{Lip_\alpha} \leq C ||f||_q^{\frac q r \left( \frac n {rk-n} + \frac q r \right)^{-1} (1 - \frac {\alpha r } {rk - n})} ||\nabla^kf||_r^{\frac n {rk-n} \left( \frac n {rk-n} + \frac q r \right)^{-1}(1 - \frac {\alpha r } {rk - n}) + \frac {\alpha r } {rk - n}},$$ here $0<\alpha \leq \frac {rk-n} r$ and $rk>n$.

share|improve this question
    
Those look like some fancy variations of Gagliardo-Nierenberg. The usual way to prove such things is combining the Littlewood-Paley decomposition with a bit of common sense. If that's what you did to obtain them, they are, indeed, members of the G-N family though your way to parameterize the powers is too exotic to figure out if they coincide with some standard versions at a glance. –  fedja Dec 23 '11 at 1:58
    
Thanks, yes, it is close to G.-N. inequality but i'm interested in exact links to these inequalities (if they exist). Also, these inequalities are obtained using special maximal functions but not Littlewood-Paley theory. –  Evgeniy Lokharu Dec 23 '11 at 4:39
    
Alas, despite the vast literature, I haven't seen any treatise where all reasonable versions of G-N would be discussed at once. I can take a look and see if the standard METHODS give your particular versions without trouble but that seems to be not quite what you are looking for. As to the literature search, here I know just as much as you and have access to more or less the same tools. –  fedja Dec 25 '11 at 3:41

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.