Here is my question -
Let $\|\cdot\|_s:\mathbb{R}^2\to\mathbb{R}^2$ be defined by:
$$\|(x_1,x_2)\|_s=\left\{ \begin{array}{l l} \|(x_1,x_2)\|_2 & \quad \text{$x_1x_2\geq 0$}\\ \|(x_1,x_2)\|_1 & \quad \text{$x_1x_2\leq 0$} \end{array} \right.$$
Is this a norm? Find $m,M\gt 0$ such that for every $\mathbb{R}^2$ $$m\|x\|_\infty\leq\|x\|_s\leq M\|x\|_\infty$$
Here is what I have (I was hoping for feedback on it):
$\|\cdot\|_s$ is a norm
1) We know as they are both norms, $\|x\|_1\geq 0$ and $\|x\|_2\geq 0$, and that $\|x\|_1= 0$ if and only if $x= 0$, and $\|x\|_2= 0$ if and only if $x= 0$. Therefore we have three cases:
a) $x_1x_2\lt 0$, then $\|(x_1,x_2)\|_s=\|(x_1,x_2)\|_1\gt 0$
b) $x_1x_2\gt 0$, then $\|(x_1,x_2)\|_s=\|(x_1,x_2)\|_2\gt 0$
c) $x_1x_2= 0$, then either $x_1=0,x_2=0$ or both equal $0$. From this we know that if both are not equal to $0$ (say $x_1=0$ but $x_2\neq 0$), $\|(0,x_2)\|_s=\|(0,x_2)\|_1\gt 0$ or $\|(0,x_2)\|_s=\|(0,x_2)\|_2\gt 0$. But if $x_1=0$ and $x_2=0$, $\|(0,0)\|_s=\|(0,0)\|_1=0$ or $\|(0,0)\|_s=\|(0,0)\|_2=0$, but either way $\|(0,0)\|_s=0$ therefore $\|(x_1,x_2)\|=0$ if and only if $x_1=0$ and $x_2=0$.
2) We know as they are both norms, $\|\alpha x\|_1=|\alpha|\|x\|_1$ and $\|\alpha x\|_2=|\alpha|\|x\|_2$. Therefore we have two cases:
a) $x_1x_2\lt 0$, then $\|(\alpha x_1,\alpha x_2)\|_s=\|(\alpha x_1,\alpha x_2)\|_1=|\alpha |\|(x_1,x_2)\|_1=|\alpha |\|(x_1,x_2)\|_s$
b) $x_1x_2\gt 0$, then $\|(\alpha x_1,\alpha x_2)\|_s=\|(\alpha x_1,\alpha x_2)\|_2=|\alpha |\|(x_1,x_2)\|_2=|\alpha |\|(x_1,x_2)\|_s$
3) We know as they are both norms, $\|(x_1+y_1,x_2+y_2)\|_1\leq\|(x_1,x_2)\|_1+\|(y_1,y_2)\|_1$ and $\| (x_1+y_1,x_2+y_2)\|_2\leq\|(x_1,x_2)\|_2+\|(y_1,y_2)\|_2$. Therefore we have two cases:
a) $x_1x_2\lt 0$, then $\|( x_1+y_1, x_2+y_2)\|_s=\|(x_1+y_1,x_2+y_2)\|_1\leq\|(x_1,x_2)\|_1+\|(y_1,y_2)\|_1=\|(x_1,x_2)\|_s+\|(y_1,y_2)\|_s$
b) $x_1x_2\gt 0$, then $\|( x_1+y_1, x_2+y_2)\|_s=\|(x_1+y_1,x_2+y_2)\|_2\leq\|(x_1,x_2)\|_2+\|(y_1,y_2)\|_2=\|(x_1,x_2)\|_s+\|(y_1,y_2)\|_s$
$m,M>0$
First, let us note a few things:
$$\|(x_1,x_2)\|_\infty=\sup\{|x_1|,|x_2|\}$$ $$\|(x_1,x_2)\|_1=|x_1|+|x_2|$$ $$\|(x_1,x_2)\|_2=\sqrt{|x_1|^2+|x_2|^2}$$
First, let us assume that $|x_1|>|x_2|$. Then we have the following:
$$(1/2)\sup\{|x_1|,|x_2|\}=(1/2)|x_1|\leq |x_1|\leq |x_1|+|x_2|$$
$$(1/2)\sup\{|x_1|,|x_2|\}=(1/2)|x_1|\leq |x_1|=\sqrt{|x_1|^2}\leq \sqrt{|x_1|^2+|x_2|^2}$$
So in either case, if $m=1/2$ then $m\|x\|_\infty\leq \|x\|_s$. Next we have the following:
$$2\sup\{|x_1|,|x_2|\}=2|x_1|=|x_1|+|x_1|\geq |x_1|+|x_2|$$
$$2\sup\{|x_1|,|x_2|\}=2|x_1|=\sqrt{2|x_1|^2}\geq \sqrt{|x_1|^2+|x_2|^2}$$
So in either case, if $M=2$ then $m\|x\|_\infty\geq \|x\|_s$.
Here is an approach to showing subadditivity:
Let $B = \{x | \|x\|_s < 1\}$ and note that $B$ is convex (this is obvious if you draw a picture, also you could note that $B = B_2(0,1) \cap \{x | |x_1-x_2| < 1\|$).
Let $\mu_B(x) = \inf\{ t > 0 | x \in t B \}$ (the Minkowski functional). Since $x \in (\|x\|_s+\epsilon)B$ for all $\epsilon$, we have $\mu_B(x) \le \|x\|_s$, and since $x \notin \|x\|_sB$, we have $\mu_B(x) \ge \|x\|_s$, hence $\mu_B(x) = \|x\|_s$.
Now suppose $x \in s B, y \in u B$. In particular, we have $x=sb_1, y= u b_2$ for some $b_1,b_2 \in B$. Then $x+y = (s+u)({ s \over s+u} b_1 + { u \over s+u} b_2) \in (s+u)B$ (since $B$ is convex) and hence $\mu_B(x+y) \le s+u$. Taking the $\inf$ over allowable $s,u$ gives $\mu_B(x+y) \le \mu_B(x)+ \mu_B(y)$, hence $\|\cdot\|_s$ is subadditive.
For the bounds, note that we have $\|x\|_\infty \le \|x\|_2 \le \sqrt{2} \|x\|_\infty$ and $\|x\|_\infty \le \|x\|_1 \le 2 \|x\|_\infty$, hence we have $\|x\|_\infty \le \|x\|_s \le 2 \|x\|_\infty$.