Inequalities Concerning Norm While I was going through a paper, I came across an inequality and I got stuck.
Here is the question:
Let $\alpha_{ij}:\mathbb{R}^N\rightarrow [0,1]$ be functions. Let $\bar{u}_i,\bar{u}_j$ be such that $\bar{u}_i=\bar{u}_j$ for some $\bar{U}\in \mathbb{R}^N$. Then following hold:
$$|\alpha_{ij}(U)(u_j-u_i)|\leq|u_j-u_i|\leq \sqrt{2}\|U-\bar{U}\|$$
for any $U=(u_1,u_2,...,u_N)\in \mathbb{R}^N$.
Now while proving this here is my approach:
$$|\alpha_{ij}(U)(u_j-u_i)|\leq|u_j-u_i|=|u_j+\bar{u}_j-\bar{u}_j-u_i|\leq |u_j-\bar{u}_j|+|\bar{u}_j-u_i|$$
Now as $\bar{u}_i=\bar{u}_j$ we get
$$|\alpha_{ij}(U)(u_j-u_i)|\leq|u_j-\bar{u}_j|+|\bar{u}_i-u_i|$$
Now by definition of norm (Euclidean Norm) we have $|u_j-\bar{u}_j|\leq \|U-\bar{U}\|$, so we get 
$$|\alpha_{ij}(U)(u_j-u_i)|\leq|u_j-u_i|\leq 2\|U-\bar{U}\|$$
I think I am making mistake in this step $|u_j-\bar{u}_j|\leq \|U-\bar{U}\|$, so if anyone can help me with this it would be nice.
Reason for the last statement:
$$\|U-\bar{U}\|=\sqrt{\sum_{i=1}^N|u_i-\bar{u}_i|^2}$$
So, we have $|u_j-\bar{u}_j|^2\leq\sum_{i=1}^N|u_i-\bar{u}_i|^2$ for any $j\in{1,...,N}$. So by taking the square roots we get the result.
 A: Since $\alpha_{ij}:\mathbb{R}^N\rightarrow [0,1]$ thati is 
$$0\le \alpha_{ij}(U)\le 1\implies |\alpha_{ij}(U)(u_j-u_i)| =|u_j-u_i|\alpha_{ij}(U)\le |u_j-u_i|~~$$
Hence the fact that $$\color{blue}{|\alpha_{ij}(U)(u_j-u_i)|\le |u_j-u_i|~~~\text{always true!!!}}$$ is independent neither on $i,j$ nor on  $(u_j-u_i)$.
$\color{red}{\text{This question as nothing to do with the function}}$ $\color{blue}{\alpha_{i,j}}$. 

Your Problem:
  But rather 
  You would like to know if the following is true!!!
  $$ |u_j-u_i|\leq \sqrt{2}\|U-\bar{U}\|$$
  For very $U=(u_1,u_2,...,u_N)\in \mathbb{R}^N$ and $\bar{U}=(\bar{u}_1 ,\bar{u}_2,...,\bar{u}_N)\in \mathbb{R}^N$ whenever $\bar{u}_i=\bar{u}_j$.

THE ANSWER IS YES!!!
Since  $\bar{u}_i=\bar{u}_j$
$$|u_j-u_i|=|u_j+\bar{u}_j-\bar{u}_j-u_i|\leq |u_j-\bar{u}_j|+|\bar{u}_j-u_i| = |u_j-\bar{u}_j|+|\bar{u}_i-u_i| $$
that is $$ |u_j-u_i|^2 \le  (|u_j-\bar{u}_j|+|\bar{u}_i-u_i|)^2  = |u_j-\bar{u}_j|^2+|\bar{u}_i-u_i|^2+2|u_j-\bar{u}_j||\bar{u}_i-u_i|$$
But we know that $$(a-b)^2\ge 0\Longleftrightarrow 2ab \le a^2+b^2$$
ie 
$$2|u_j-\bar{u}_j||\bar{u}_i-u_i|  \le |u_j-\bar{u}_j|^2+|\bar{u}_i-u_i|^2$$
Hence 
$$ |u_j-u_i|^2 \le  |u_j-\bar{u}_j|^2+|\bar{u}_i-u_i|^2+2|u_j-\bar{u}_j||\bar{u}_i-u_i| \\ \le 2(|u_j-\bar{u}_j|^2+|\bar{u}_i-u_i|^2 )\leq 2\sum_{i=1}^N|u_i-\bar{u}_i|^2\\= 2\|U-\bar{U}\|^2.$$
i.e 
$$ |u_j-u_i| \le \sqrt{2}\|U-\bar{U}\|.$$
