Whether this vector norm proposition is true? $ \vert x_i \vert \ge \vert y_i \vert \Rightarrow \Vert X \Vert \ge \Vert Y \Vert$

Question

Whether this vector norm proposition is true?

$ \vert x_i \vert \ge \vert y_i \vert \Rightarrow \Vert X \Vert \ge \Vert Y \Vert$ where $X, Y \in \mathbb R ^n$

Is it true for all kinds of norms?

I've attempted some norms below and it is obviously true for them:

1. $\Vert X \Vert _p = \left ( \sum \limits_{i = 1} ^{n} \vert x_i \vert ^p \right ) ^{1/p}, p \in [1, + \infty)$

2. $\Vert X \Vert _{\infty} = \max \limits_{1 \le i \le n} \vert x_i \vert $

3. In case $n = 1,$ $\Vert X \Vert = \Vert x \Vert = \Vert 1 \Vert \vert x \vert$

Answer 1

Are you one of my students? I thought about this today in class, and snarkily remarked that if this was false I'd give back my degree. I shouldn't have, because I was wrong!

Suppose that $x$ and $y$ are linearly independent vectors, and satisfy your hypothesis. Since they span a plane, we may as well assume we are in $\mathbb{R}^2$. Now we can go ahead and make a norm that has $||x|| = 1$ and $||y|| = 2$. Since these vectors are linearly independent in $\mathbb{R}^2$, we can extend this linearly to a norm on the whole plane. It is by definition homogeneous, only $0$ for the $0$ vector, and since we have defined it to be a linear map, it satisfies the triangle inequality. However, it has $|x_i| < |y_i|$ for both $i$.

The triangle inequality may require some further comments, so let's make them. Let $p = a_1x + a_2y$ and $q = b_1x + b_2y$. This is possibly since $x$ and $y$ are a basis. Now $$||p + q|| = ||(a_1+b_1)x + (a_2+b_2)y| = (a_1+b_1) + 2(a_2+b_2)$$

But also $||p|| = a_1 + 2a_2$ and $||q|| = b_1 + 2 b_2$, so it turns out that here we have $$||p + q|| = ||p|| + ||q|| \leq ||p|| + ||q||$$ so the triangle inequality holds.

Answer 2

Define a norm on $\Bbb R^2$ by $$||(x,y)||=|x|+|x-y|.$$