simple inequality for vector norm

Question

For vectors x,y $\in \mathbb{R}^n_+$ with entries ordered decreasingly: if $$ \sum_{i=1}^kx_i \leq \sum_{i=1}^ky_i \ \forall 1\leq k\leq n $$ then $\|x\|_p\leq\|y\|_p$.

From https://math.stackexchange.com/a/2819688/780264

Answer 1

This result is a consequence of Corollary II.3.4 of the same reference in the linked (Bhatia's Matrix Analysis). Here is an adaptation of the proof given in the text.

I will use $x \prec_w y$ to mean that when $x,y$ are reindexed in decreasing order, $\sum_{i=1}^kx_i \leq \sum_{i=1}^ky_i$ for $1\leq k\leq n$. $x \prec y$ denotes the same with the additional condition that $\sum_{i=1}^kx_i = \sum_{i=1}^ky_i$. I will use $x \leq y$ to mean that $x_i \leq y_i$ for all $i$. We will first prove the following:

Claim: If $x,y \in \Bbb R_+^n$ satisfy $x\prec y$, then $x^p \prec_w y^p$ (where $(x_1,\dots,x_n)^p = (x_1^p,\dots,x_n^p)$).

We need the following facts:

1. The function $f(x) = x^p$ is convex over $\Bbb R_+$ for $p \geq 1$
2. If $x \leq y$, then $\|x\|_p \leq \|y\|_p$
3. If $x \prec y$, then there exists a doubly stochastic matrix $A$ (i.e. the row and column sums of $A$ are all $1$) for which $x = Ay$ (cf. Bhatia Theorem II.1.10)
4. The Birkhoff von-Neumann theorem: such a matrix $A$ can be written in the form $\sum_j P_j$ for permutation matrices $P_j$ and positive $t_j$ with $\sum_j t_j = 1$.

With that, the proof proceeds as follows. We have $$ x^p = [Ay]^p = \left[\sum_j t_j P_j y \right]^p \leq \sum_j t_j (P_j y)^p = \sum_j t_j P_j(y)^p = A y^p $$ So, we have $x^p \leq Ay^p$ and $Ay^p \prec y^p$. It follows that $x^p \prec_w y^p$.

From the fact that $x^p \prec_w y^p$, it follows that the sum of the entries of $x^p$ is less than that of $y^p$, from which it follows that $\|x\|_p \leq \|y\|_p$.

It now remains to be shown that $x \prec_w y \implies \|x\|_p \leq \|y\|_p$. To that end, it suffices to note that if $x \prec_w y$, there exists a vector $u$ (ordered decreasingly) for which $$ x \leq u \prec y, $$ (cf. Bhatia Theorem II.2.8, a consequence of fact 3). It follows that $\|x\|_p \leq \|u\|_p \leq \|y\|_p$ .