Timeline for Markov-type inequalities with arbitrary exponents
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 26, 2011 at 13:36 | comment | added | fedja | If you want density, just add first $n$ functions in the standard trigonometric system to $U_n$ and reenumerate accordingly. | |
| Nov 26, 2011 at 5:21 | comment | added | Ben Adcock | Thanks for your answer. However, your construction doesn't work when you require $\cup U_n$ to be dense in $L^\infty[-1,1]$. Apologies, I should have added this to the original question - have now changed it accordingly. | |
| Nov 26, 2011 at 5:18 | history | edited | Ben Adcock | CC BY-SA 3.0 |
added 60 characters in body
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| Nov 26, 2011 at 3:36 | comment | added | fedja | Then $\sin [k^\alpha] x$ is such a system ($U_n$ is the span of the first $n$ functions). | |
| Nov 26, 2011 at 2:06 | comment | added | Ben Adcock | Thanks. However, I should have said that it's the case $\alpha \geq 1$ that's of interest. | |
| Nov 26, 2011 at 1:39 | comment | added | fedja | Sorry, the last inequality should read $\lambda\ge n$. | |
| Nov 26, 2011 at 1:38 | comment | added | fedja | No. We must have $\alpha\ge 1$ at the very least. Indeed, assume that $\|f'\|\le \lambda \|f\|$ for every $f\in U_{n+1}$. Choose a non-zero $f$ orthogonal to all characteristic functions of intervals $I_k=[\frac kn,\frac{k+1}n]$. Then $\|f\|\le \max_k\operatorname{osc_{I_k}}f\le \frac 1n\|f'\|\le \frac{\lambda}n\|f\|$, so $\lambda n\ge 1$. | |
| Nov 26, 2011 at 0:47 | history | asked | Ben Adcock | CC BY-SA 3.0 |