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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, it can be shown that $x$ can take at most $2$ distinct values in its co-ordinates (or one of the co-ordinates is $0$, which just reduces the dimension of the problem).

My guess is that the $x$ which works for $q = \infty$ should also work for all $q > p$.

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With only two distinct nonzero $x$ values, you're reduced to a finite number of cases: if there are to be $k$ nonzero (and wlog positive) values of $x$ and $m$ of $y$, you want to have $k x + m y = 1$ and $k x^p + m y^p = \theta$, and there are at most two solutions of that system. – Robert Israel Dec 16 '11 at 4:35
I agree. The picture seems to be that for each $q$, one of these finite cases becomes the optimal, until such a point when $q$ is large enough and the optimal for $q = \infty$ takes over. Could it be that the optimal for $q = \infty$ is optimal throughout? – VSJ Dec 16 '11 at 5:03
I'm still stuck at the same point.. but atleast matlab calculations haven't yielded any counter-examples (i.e the optimal at $q = \infty$ is optimal throughout). It seems cute enough to permit an analytical solution. – VSJ Dec 17 '11 at 2:26
up vote 1 down vote accepted

After searching around and asking on mathoverflow, it turns out there is a theorem specifically tailored for this type of norm-constrained problems. Its called the Equal variable theorem and is proved here:

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