For a probability vector $(p_1,p_2,\ldots,p_n)$, where $p_1 \geqslant p_2 \geqslant \cdots\geqslant p_n\geqslant 0$ and $\sum_{i=1}^n p_i = 1$, define the "inverse" of it as $(q_1,q_2,\ldots,q_n)$, where $$ q_i=\frac{1-p_i^r}{\sum_{k=1}^n(1-p_k^r)}$$ and $r$ is a positive integer. The term $\sum_{k=1}^n(1-p_k^r)$ in the denominator is for normalization. When $r = 1$, it is not hard to prove $$\sum_{i=1}^{m}p_i \geqslant \sum_{i=n-m+1}^{n}q_i$$ therefore $(p_1,p_2,\ldots,p_n)$ majorizes $(q_1,q_2,\ldots,q_n)$ (just use the monotonicity of $p_i$s). But for $r > 1$, I failed to work out the proof. I found a similar problem in this link and it provides a following lemma:
Lemma Let $\boldsymbol{u} \ge 0$ and $f: \mathbb R_{\ge 0} \to \mathbb R_{\ge 0}$ be a function for which $\frac{f(x)}{x}$ is decreasing and $f$ is increasing on the set $\{u_1,\dots,u_n \}$ where $ \sum_{i=1}^n u_i>0$ and $ \sum_{i=1}^n f(u_i)>0$, $\frac{\boldsymbol{u}}{\sum_{i=1}^n u_i}$ majorizes vector $\frac{(f (u_1),\dots,f (u_n))}{\sum_{i=1}^n f(u_i)}$.
However, the lemma requires such $f$ is increasing on the domain, but here in my problem , $f(x) = 1-x^r$ is decreasing on $[0,1]$.
I also tried to prove the difference $q_{n-i+1} - q_{n-i}$ is not bigger than $p_i - p_{i+1}$ to show the majorization, but it's not true. What's more, $q_{i+1} - q_{i} \leqslant p_i-p_{i+1}$ holds but it can't prove the majorization.
Does such majorization holds, or are there some counterexamples?
Edit: To make things clear, here I post some examples:
r = 2
P = [0.9, 0.1, 0.0], Q = [0.05, 0.45, 0.5]
Note: the "inverse" operation turns over the monotonicity of the
probabilities
------------------------------------------------------------------
r = 100
P = [0.9, 0.1, 0.0], Q = [0.333, 0.333, 0.333]
Note: when r approaches +inf, Q approaches uniform distribution
------------------------------------------------------------------
r = 3
P = [0.4, 0.25, 0.25, 0.1], Q = [0.227, 0.253, 0.253, 0.267]
Note: even if certain p_i is exactly 1/n, the corresponding item
in Q still can be bigger
------------------------------------------------------------------
r = 2
P = [0.7, 0.3], P_inv = [0.3, 0.7],
P_inv_inv = [0.7, 0.3], P_inv_inv_inv = [0.3, 0.7]
Note: in most cases, P_inv_..._inv approaches uniform distribution
, but this is not true for n = r = 2
------------------------------------------------------------------
r = 4
P = [0.8, 0.2], Q = [0.3297, 0.6703],
Note: when r > n = 2, the case is not like above
------------------------------------------------------------------
r = 8
P = [0.2, 0.2, 0.2, 0.2, 0.2], Q = [0.2, 0.2, 0.2, 0.2, 0.2]
Note: if P is uniform distribution, then Q = P
------------------------------------------------------------------
r = 2
P = [0.5, 0.5, 0.0], Q = [0.25, 0.25, 0.5]
Note: a counterexample of $q_{n-i+1} - q_{n-i}$ is not bigger than
$p_i - p_{i+1}$
------------------------------------------------------------------
Last but not least: $q_{i+1} - q_{i}$ <= $p_i-p_{i+1}$ holds but
it can't prove the majorization. Let
P1 = [0.45, 0.25, 0.2, 0.1], P2 = [0.05, 0.25, 0.3, 0.4]
since 0.1 > 0.05, it's clear that P1 doesn't majorize P2.
