For vectors $\vec v = (v_1,v_2,v_3) \in \mathbb R^3$, does $||\vec v|| = |v_1| + \max\{ |v_2|,|v_3|\}$ define a norm on $\mathbb{R}^3$?
I know I need to show positivity, homogeneity and the triangle inequality.
I can do positivity, I'm lost for the other two conditions though.
Let $c \in \mathbb R$ and $v,w \in \mathbb R^3$.
\begin{equation}\begin{split} ||c\vec v|| &= |cv_1| + \max\{|cv_2|,|cv_3|\} \\ &= |c||v_1| + \max\{|c||v_2|,|c||v_3|\} \\ &= |c||v_1| + |c|\max\{|v_2|,|v_3|\} \\ &= |c|(|v_1| + \max\{|v_2|,|v_3|\}) \\ &= |c|||\vec v|| \end{split}\end{equation}
And
\begin{equation}\begin{split} ||\vec v+ \vec w|| &= |v_1 + w_1| + \max\{|v_2+w_2|,|v_3+w_3|\} \\ &\leq |v_1| + |w_1| + \max\{|v_2|+|w_2|,|v_3|+|w_3|\} \\ &= (|v_1| + \max\{|v_2|,|v_3|\}) + (|w_1| + \max\{|w_2|,|w_3|\}) \\ &= ||\vec v||+||\vec w|| \end{split}\end{equation} Thus $||\vec v+ \vec w|| ≤ ||\vec v|| + ||\vec w|| $
That's what I've tried so far.
For the Triangle Inequality, we need to show the following
$|v_1+u_1|+max{\{(v_2+u_2),(v_3+u_3)\}}\leq (|v_1|+max{\{v_2,v_3}\})+(|u_1|+max{\{u_2,u_3\}})$.
note you can use the Triangle inequality for each component to arrive at the Triangle Inequality for the norm.
we have $|v_1+u_1|<|v_1|+|u_1|$ for the first component, and also we have $max{\{|v_2+u_2|,|v_3+u_3|\}}$. Note that you can split this into two cases where if $|v_2+u_2|>|v_3+u_3|$ and $|v_2+u_2|<|v_3+u_3|$ the Triangle Inequality (for the norm) naturally follows with a little bit of computation. So the proposed norm is indeed a norm.