Prove that $0 < x < y$ implies $\|x\| < \|y\|$ for any norm.

Question

All vectors are real. Prove that $0 < x < y$ (element-wise) implies $\|x\| < \|y\|$ for any norm. This is probably very basic, but I don't seem to get the hang of it.

Edit: it turns out this is not true.

This is the first quadrant view of the unit ball of the norm suggested by @Daniel Fischer, which violates this.

Answer 1

It's not true. Consider $\mathbb{R}^2$ with the norm

$$\lVert (x,y)\rVert = \sqrt{\tfrac{1}{2}(x+y)^2 + 100(x-y)^2},$$

and look at $\left(\frac{3}{4},\frac{1}{2}\right)$ and $(1,1)$. You can generalise this example to arbitrary dimensions.

Answer 2

In the setting of Banach lattices this is not true:

take $x=(1-\tfrac{1}{n})_{n=1}^\infty, y = (1)_{n=1}^\infty\in \ell_\infty$. Then $x<y$ but $\|x\|=\|y\|=1$.

Answer 3

This holds in $\mathbb R^n$ equipped with the $L^p$ norm if $1 \leq p < \infty$, while this fails if $p = \infty$.