When does one vector has bigger norm than the other? Let $u,v\in\mathbb{R}^n$ be two probability vectors (i.e., with all components $\geq 0$ and summing up to 1). Then I was wondering what are necessary and sufficient conditions for $$\|u\|_p\leq \|v\|_p \quad \forall p\in[1,\infty).$$
I could check that $u_k\leq v_k$ is a sufficient condition, but it need not be necessary.
 A: we put $f(x)=\|x\|_p^p$ , $C=\{ x=(x_i)_{i=1}^n\in R^n: x\geq0, \sum_{i=1}^nx_i=1\}$ and $\partial C=\{x\in C : x_i=0 ,\quad for\quad some \quad i=1,...,n\}$. $f(x)$ is strictly convex function on convex set $C$, and it has unique point that is called by $x_*$ which $x_{*i}$ are $\frac{1}{n}$ for each $i=1,...,n$.  for all $x\in \partial C$, $g_x(t)=tx_*+(1-t)x \quad t\in [0,1]$. If both $f(x)>f(y)$ and $\|x-x_*\|_1\leq \|y-x_*\|_1$ are true then At least two points exist such as $x_1,\quad y_1$ such that $\|x_1-x_*\|_1\leq \|y_1-x_*\|_1$ and $f(x_1)=f(y_1)$ . this is Contrary to the assumption with strictly convex function  $f(x)$, because $\nabla f(tx_1+(1-t)y_1).(x_1-y_1)=0, \quad t\in(0,1)$.
However, we show $\|x\|_p\leq\|y\|_p$ if and only if $ \|x-x_*\|_1\leq \|y-x_*\|_1 $. thus $max\{ x_{i} \}_{ i = 1 } ^n \leq max\{ y_{i}\}_{i=1} ^n
$ if and only if $\|x\|_p\leq\|y\|_p$ for every $p\in [1,\infty]$.
A: It can be $ max\{ u_{i} \}_{ i = 1 } ^n \leq max\{ v_{i}\}_{i=1} ^n$.
Proof: 
it is clear for $p=1$.
We prove for each $1<p\leq \infty $. We know that $\|.\|_p$ is (strictly for all $p<\infty$) convex function and has an unique minimize on feasible solution ($\{x\in R^n : \Sigma_{i=1}^n x_i=1\} $) which it is common for all $p$. We denote this point with $u^*$. It is clear that if $u$ approch from $v$ via $\|.\|_p  \quad \forall p\in(1,\infty]$ to $u^*$. However, it can be investigated by one of $1<p\leq \infty$.
