from generalized inequality, we now that for $p>q$, we have $M_p(\mathbf{x})\ge M_q(\mathbf{x})$. now I am curious to know if we can find a constant $\alpha(p,q)$ which is only function of $p,q$ such that $M_p(\mathbf{x})\le \alpha(p,q) M_q(\mathbf{x})$. if these exists such constant, how can we find it preferably in closed form?
$\begingroup$
$\endgroup$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
7
Well, if you accept dependence on the the maximum and minimum components (which are the mean values for plus and minus infinity), Mitrinovic's "Analytic Inequalities" book has such bounds on ratios and differences of the two means, even their weighted versions. E.g., $$ \alpha(p,q)\leq \left(\frac{q(C^p-C^q) }{(p-q)(C^q-1) }\right)^{1/p} \left(\frac{p(C^q-C^p) }{(q-p)(C^p-1) }\right)^{-1/q} $$ where $C=M_{\infty}(\mathbf{x})/M_{-\infty}(\mathbf{x}),$ thus those max and min values must be bounded away from zero and infinity.
-
$\begingroup$ In fact power means are norms on finite dimensional spaces. All such norms are equivalent, so the inequalities you asked for exist and constants in them are embedding space theorems constants. $\endgroup$– SergeiCommented Apr 6, 2015 at 4:21
-
$\begingroup$ in general not all power means are norm, only for p>1. the provided bound looks very fascinating, however the dependency of $C$ on $\mathbf{x}$ violates the requirement of the question. but because it is yet very useful and application, I would accept that. $\endgroup$ Commented Apr 6, 2015 at 8:06
-
$\begingroup$ @sergei , would please give us a link or reference on embedding space theorems constants? i have looked over net and couldn't find anything yet. $\endgroup$ Commented Apr 6, 2015 at 10:50
-
$\begingroup$ In Wiki we find exact inequalities: for $p>r>0$ it follows $\endgroup$– SergeiCommented Apr 6, 2015 at 12:37
-
1$\begingroup$ $||x||_p \le ||x||_r \le n^{1/r-1/p} ||x||_p$, $\endgroup$– SergeiCommented Apr 6, 2015 at 12:40