How to check if a given piecewise defined function on $\mathbb R^2$ is a norm?

Question

I want to check if the function $\parallel (x,y)\parallel = \left\{ \begin{array}{cc} \sqrt{x^2+y^2} & \mbox{if } xy \geq 0 \\ \max\{\vert x\vert, \vert y\vert\} & \mbox{if } xy < 0 \end{array} \right.$

is a norm on $\mathbb R^2$. I could not prove it to be a norm, also I could not disprove by any counterexample. Please help!

Answer 1

Some things to check:

1. Is it zero at the origin?
2. Is it positive everywhere else?
3. Is it homogeneous, meaning that $\|(tx,ty)\| = t\|(x,y)\|$ for all $t>0$ and all $x,y\in\mathbb R$?

If the answers to those questions are yes (and they are) we have to deal with the hard part: the triangle inequality. Luckily, the homogeneity makes it easier:

The triangle inequality holds provided that the set $\{(x,y):\|(x,y)\|\le 1\}$ is convex.

Draw a picture of this set: its boundary consists of two quarter-circles and two line segments. Yes, it is convex.

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To prove the highlighted statement, take two points $A$, $B$. Since $A/\|A\|$ and $B/\|B\|$ are in the unit ball, so is their convex combination $$\frac{\|A\|}{\|A\|+\|B\|} \frac{A}{\|A\|} +\frac{\|B\|}{\|A\|+\|B\|}\frac{B}{\|B\|}$$ this means exactly that $\|A+B\|\le \|A\|+\|B\|$.