Prove whether norm is or is not metric Let $(X,d)$ be a metric space and $\left\lVert\cdot\right\rVert$ be a norm on $\mathbb{R}^2$. Define $\tilde{d}$ by $\tilde{d}((x_1,x_2),(y_1,y_2)) = \left\lVert(d(x_1,y_1),d(x_2,y_2))\right\rVert$. Is this $\tilde{d}$ a metric on $X \times X$? It quite clearly is when e.g. $\left\lVert\cdot\right\rVert$ is an $l^p$ norm, but how to prove or disprove it in the general case?
 A: Since $\|\cdot\|$ is a norm, clearly $\tilde{d}:X\times X\to [0,+\infty)$.
$\tilde{d}((x_1,x_2),(y_1,y_2)) = \left\lVert(d(x_1,y_1),d(x_2,y_2))\right\rVert=0\Leftrightarrow(d(x_1,y_1),d(x_2,y_2))=(0,0)\Leftrightarrow d(x_1,y_1)=0\land d(x_2,y_2)=0\Leftrightarrow x_1=y_1\land x_2=y_2\Leftrightarrow(x_1,x_2)=(y_1,y_2)$.
$\tilde{d}((x_1,x_2),(y_1,y_2)) = \left\lVert(d(x_1,y_1),d(x_2,y_2))\right\rVert=\|(d(y_1,x_1),d(y_2,x_2))\|=\tilde{d}((y_1,y_2),(x_1,x_2))$.
All the previous properties are valid in general. Unfortunately the last property we need, the triangle inequality, can fail:
Take $X=\Bbb R$ with $d(a,b)=|a-b|$ and consider the norm $\|(x,y)\|=|x|+|x-y|$ on $\Bbb R^2$. Then $\tilde{d}((2,0),(0,0))=\|(d(2,0),d(0,0))\|=\|(2,0)\|=4\nleq2=\|(1,1)\|+\|(1,1)\|=\|(d(2,1),d(0,1))\|+\|d(1,0),d(1,0))\|=\tilde{d}((2,0),(1,1))+\tilde{d}((1,1),(0,0))$.
However, we can require $\|\cdot\|$ to be monotone respect the following order: $(x_1,y_1)\le(x_2,y_2)$ iff $|x_1|\le |x_2|\land |y_1|\le |y_2|$, meaning that whenever $(x_1,y_1)\le(x_2,y_2)$ we have $\|(x_1,y_1)\|\le\|(x_2,y_2)\|$. The previous partial order can be interpreted as ordering by the distance to the origin, so we are demanding the norm to respect the "farness" of the points of $\Bbb R^2$. Here's a link mentioning this monotone norms and a characterisation.
In our case we don't need the absolute values in that partial order of $\Bbb R^2$, since $d$ is nonnegative, so if your norm $\|\cdot\|$ has that property then $\tilde{d}((x_1,x_2),(y_1,y_2))=\|(d(x_1,y_1),d(x_2,y_2))\|\le\|(d(x_1,z_1)+d(z_1,y_1),d(x_2,z_2)+d(z_2,y_2))\|=\|(d(x_1,z_1),d(x_2,z_2))+(d(z_1,y_1),d(z_2,y_2))\|\le\|(d(x_1,z_1),d(x_2,z_2))\|+\|(d(z_1,y_1),d(z_2,y_2))\|=\tilde{d}((x_1,x_2),(z_1,z_2))+\tilde{d}((z_1,z_2),(y_1,y_2))$.
So with that additional requirement $\tilde{d}$ would be a metric on $X\times X$.
