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

I would like to ask for some help (hints, ideas) in solving the following problem:

Given integer $n>0$ and real $\alpha>0,\beta>1$ we want to show, that

if we define for any $x\in\mathbb{R}^{n}$ the function $$ g(\alpha,x) = \prod\limits_{i=0}^{n-1}(1+\alpha i)\cdot \left(\sum\limits_{i=1}^{n}x_i^{\alpha}\right)^{-\frac{1}{\alpha}-n} \cdot \left(\prod\limits_{i=1}^{n}x_i \right)^{\alpha-1} $$ and the domain $$ A = \{x>{0}:\ \sum\limits_{i} x_i^{\frac{1}{\beta}}>1 \} $$ then the integral given by $$ \int\limits_{A} g(\alpha,x) \mbox{d}x $$ is strictly monotonic with respect to $\alpha$. It could be shown that limits at $0$ and $\infty$ are equal to $n$ and $n^{\beta}$, respectively. It has been proved, that this expression is monotonic. Argument uses representation of $g$ as a limit of certain sequence of functions and cannot prove strict monotonicity.

I tried the following approach which fails: if we show that $g(\alpha,x)$ is analytic with respect to $\alpha$ then, according to the standard procedure (Leibnitz Integral Rule), we prove that integral is analytic. In fact, we easily get for $\alpha>1$ $$|g(\alpha,x)| \leqslant n^{n} \alpha^{n} \\|x\||_{\alpha}^{-n-1} $$ hence $g(\alpha,x)$ is locally uniformly integrable on $A$ with respect to $\alpha$.

However, the expression $ \left(\sum\limits_{i=1}^{n}x_i^{\alpha}\right)^{-\frac{1}{\alpha}-n}$ seems to be non-analytic (with respect to any domain in $\mathbb{C}$ extending positive real line). I wish I knew what other methods could be applied to such kind of problems.

share|improve this question
    
The obvious thing is to try $\partial / \partial \alpha$ and see if what you get is "obviously" positive. Since the domain $A$ doesn't depend on $\alpha$, it's enough to show (if true) that $\partial g/ \partial \alpha > 0$. Sorry, but I am too busy (i.e., lazy) to work the details out myself and see if it works or not, especially since I haven't got a handy Computer Algebra package. I'm a bit confused why you talk about analyticity at the end, though; how is it related to monotonicity? –  Zen Harper Sep 14 '11 at 8:03
    
Thank you for your response. I have missed one important fact: monotonicity (no strict) is proved by other method. The question stated in paper from which the problem comes, is to check whether monotonicity is strict (authors suspect that there is). The obvious way you have described above, involves quite complicated integral (differentiation under the integral sign) and is rather hard to deal with it. Analyticity proves strict monotonicity, if only we know that weak monotonicity holds. –  Maciej S. Sep 14 '11 at 9:06

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.