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Consider a vector $x \in \mathbb R_+^n$ and $p,q \in \mathbb R$ such that $1 < p < q$

We fix $\sum \limits_{i=1}^{n}|x_i| = 1$ and $ \left(\sum \limits_{i=1}^{n}|x_i|^p \right)^\frac{1}{p} = \theta < 1$.

In other words, $x$ is a probability mass function, with a fixed $p$-norm.

The question is to find that $x$ which maximises the $q$-norm i.e. maximises $\left( \sum \limits_{i=1}^{n}|x_i|^q \right)^\frac{1}{q}$.

Some observations:

  1. If $q = \infty$, the problem reduces to finding that $x$ which has the largest value for $\sup \limits_i \; x(i)$ leading to the distribution which looks like $\left( \alpha, \frac{1-\alpha}{n-1}, \frac{1-\alpha}{n-1}, \ldots, \frac{1-\alpha}{n-1} \right)$.

  2. Using perturbation arguments (Lagrangian multipliers), it can be shown that $x$ can take at most $2$ distinct non-zero values in its co-ordinates. This reduces the problem to a finite number of cases.

My guess is that the distribution $\left( \alpha, \frac{1-\alpha}{n-1}, \frac{1-\alpha}{n-1}, \ldots, \frac{1-\alpha}{n-1} \right)$ which works for $q = \infty$ should also work for all $q > p$. It seems intuitively true and matlab calculations also support this guess. Any pointers/related problems/inequalities are very much welcome.

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closed as no longer relevant by Bill Johnson, Mark Meckes, Gjergji Zaimi, Andrés E. Caicedo, Ryan Budney Dec 20 '11 at 21:11

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You will find an answer to a more general question here… – Gjergji Zaimi Dec 19 '11 at 10:49
A very similar question is considered in: – Igor Rivin Dec 19 '11 at 12:36
This is a nice question but rather elementary for MO. Since the OP has been pointed in the right direction, I vote to close. – Bill Johnson Dec 19 '11 at 16:07
This is exactly what I had been searching for. Thanks! Just for the sake of completion, link to the proof of Equal variable theorem: – VSJ Dec 20 '11 at 9:20