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By a Markov-type inequality I mean an inequality of the form $$ \| p^{(k)} \| \leq \lambda_{k,n} \| p \|,\quad \forall p \in U_n, $$ for some $\lambda_{k,n} > 0$, where $U_n \subset L^\infty[-1,1]$ is a subspace of dimension $n$ and $$ \| g \| = \sup_{x \in [-1,1]} | g(x) |. $$ If $U_n$ is the space of algebraic polynomials of degree $n$, then it is well known that $$ \lambda_{k,n} = O(n^{2k}),\quad n \rightarrow \infty, $$ for each fixed $k$, whereas for trigonometric polynomials, $$ \lambda_{k,n} = O(n^{k}),\quad n \rightarrow \infty. $$ My question is as follows. Is it possible to find subspaces $U_1 \subset U_2 \subset U_3,\ldots$ with $\bigcup^{\infty}_{n=1} U_n$ dense in $L^\infty[-1,1]$ possessing Markov-type inequalities with arbitrary exponents, i.e. $$ \lambda_{k,n} = O(n^{\alpha k}),\quad n \rightarrow \infty, $$ for some given $\alpha > 0$? Moreover can one find such subspaces in a constructive manner?

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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 $\lambda n\ge 1$. –  fedja Nov 26 '11 at 1:38
    
Sorry, the last inequality should read $\lambda\ge n$. –  fedja Nov 26 '11 at 1:39
    
Thanks. However, I should have said that it's the case $\alpha \geq 1$ that's of interest. –  Ben Adcock Nov 26 '11 at 2:06
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Then $\sin [k^\alpha] x$ is such a system ($U_n$ is the span of the first $n$ functions). –  fedja Nov 26 '11 at 3:36
    
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. –  Ben Adcock Nov 26 '11 at 5:21

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