For which coefficients $a,b$, the expression: $$\|\vec x\|=a(|x_1|+|x_2|)+b\max(|x_1|,|x_2|)$$ is a norm in $\mathbb R^2$?

My attempt: I need to verify the properties of the norm:

  1. Triangle inequality: $\|\vec x+\vec y\|\le\|\vec x\|+\|\vec y\|$

$$\|\vec x+\vec y\|=a(|x_1+y_1|+|x_2+y_2|)+b\max(|x_1+y_1|,|x_2+y_2|)$$

I know that $|x_1+y_1|+|x_2+y_2|\le (|x_1|+|x_2|)+(|y_1|+|y_2|)$ then in order to preserve the inequality we need to have that $a\ge 0$

on the other hand $$\max(|x_1+y_1|,|x_2+y_2|)\le \max(|x_1|,|x_2|)+\max(|y_1|,|y_2|)$$ so if we want to preserve the inequality $b\ge 0$

But we can do this in another way:

First we note that $$a(|x_1|+|x_2|)+b\max(|x_1|,|x_2|)=(a+b)\max(|x_1|,|x_1|)+a min(|x_1|,|x_2|)$$

Then we know that $$\max(|x_1+y_1|,|x_2+y_2|)\le \max(|x_1|,|x_2|)+\max(|y_1|,|y_2|)$$

So in order to preserve the inequality we need to have that $a+b\ge 0$ similarly we have that $$\min(|x_1+y_1|,|x_2+y_2|)\le \min(|x_1|,|x_2|)+\min(|y_1|,|y_2|)$$ so we need to have that $a\ge 0$

In both cases we have that $a\ge 0$ but in the first one we have that $b\ge 0$ while in the second one we have that $a+b\ge 0$ so which one is the correct approach to solve this problem?

I would really appreciate your help.


Let $p(t) = t(1,0)+(1-t)(0,1) = (t,1-t)$.

$\|p(0)\| = \|p(1)\| = a+b$ and $\|p({1 \over 2})\| = a+{1 \over 2} b$.

Convexity requires $\|p({1 \over 2})\| \le {1 \over 2} (\|p(0)\| + \|p(1)\|)$, or ${1 \over 2} b \le b$. In particular, this implies $b \ge 0$.

Now let $q(t) = t (-1,1)+ (1-t)(1,1) = (1-2t,1)$.

$\|q(0)\| = \|q(1)\| = 2a+b$ and $\|q({1 \over 2} ) \| = a+b$.

Similarly, convexity implies $a \le 2a$, hence $a \ge 0$.

It should be clear that if $a=b = 0$, then the expression is not a norm.

It is straightforward to show that if $n_1, n_2$ are norms, then $a n_1+bn_1$ is a norm whenever $a,b \ge 0$ and $(a,b) \neq (0,0)$.

Hence the above expression is a norm iff $a,b \ge 0$ and $(a,b) \neq (0,0)$.


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