# Do the pth powers of $p$-norms define the same partial ordering on the set of all probability distributions for all $p>1$?

Consider the $p$-th power of the Schatten $p$-norm $||q||_p$ of a probability distribution $q$ , ie, the function $\sum_j q_j^p$, where $\sum_j q_j = 1$ and $q_j \geq 0$. For fixed $q$ and $p>1$ this is a nonincreasing function of $p$.

The question is: given $p_1, p_2$ both $>1$, is it true that:

whenever $\left(||q||_{p_1} \right)^{p_1} \geq \left(||s||_{p_1} \right)^{p_1}$ , then also $\left(||q||_{p_2} \right)^{p_2} \geq \left(||s||_{p_2} \right)^{p_2}$?

In other words, do all these functions define the same partial order on the set of probability distributions?

Note: in information-theoretic terms, the question can also be equivalently stated in terms of the Tsallis entropies $T_p = \frac{1-\sum_j q_j^p}{p-1}$),

and also the closely related Rényi entropies $R_p ( q)= \frac{1}{1-p} \ln \sum_j q_j^p$

It is false. For example, let $q=(\frac{1}{2},\frac{1}{2})$ and let $s=(\frac{3}{5},\frac{1}{5},\frac{1}{5})$. Then $\|q\|_2^2=\frac{1}{2}> \frac{11}{25}=\|s\|_2^2$. However, when $p>4$, $\|q\|_p^p=2^{1-p}<(\frac{3}{5})^p<\|s\|_p^p$.

• So it is. Thanks, I had tried some examples around that one but hadn't found it. In other words, it seems that deciding which distribution is "more disordered" depends on the measure you choose... – David Apr 12 '13 at 6:53
• @David: You are welcome. – 23rd Apr 12 '13 at 6:58