Find a norm on $\Bbb{R}^2$ for which $\|(0, 1)\| = 1 = \|(1, 0)\|$ and for which $\|(1, 1)\| = 0.0001$. Find a norm on $\Bbb{R}^2$ for which $\|(0, 1)\| = 1 = \|(1, 0)\|$ and for which $\|(1, 1)\| = 0.0001$.  Show that your answer works.
I know obviously that the $\Bbb{R}^2$ norm is the Euclidean norm, but I am unsure on how to do the actual question. Any help much appreciated!
 A: I'd try a norm based on a positive definite symmetric matrix:
$$\|(x,y)\|^2=\pmatrix{x&y}A\pmatrix{x\\y}$$
defines a norm if the matrix $A$ is positive definite. To get $\|(1,0)\|=\|(0,1)\|=1$ one needs
$$A=\pmatrix{1&b\\b&1}$$
and this is positive definite if $b^2<1$. You want $\|(1,1)\|^2=10^{-8}$. But
$$\|(1,1)\|^2=2+2b.$$
Does the solution of $2+2b=10^{-8}$ for $b$ also satisfy $b^2<1$?
A: Consider the norm
$$\|(x,y)\| := 0.00005|x|+0.00005|y| + 0.99995 |x-y|$$
Then $$\|(1,0)\| = \|(0,1)\| = 0.00005+0.99995 = 1$$
and $$\|(1,1)\| = 0.00005 + 0.00005 = 0.0001.$$
A: Any convex set $D$ with an open interior symmetrical about the origin defines a norm for which it is the unit disk. The norm of any vector is the ratio of the (Euclidean) distance to the origin over the distance along that ray to the boundary of $D$.
Now draw a long skinny ellipse with major axis the line $y=x$ passing through the points $(1,0)$ and $(0,1)$. If it's long enough the new distance to $(1,1)$ can be as small as you like.
[maybe I will come back later and insert a picture, or someone can edit to do that]
