Proving that $||v||=\sqrt{2v_1^2+3v_2^2}$ defines a norm on $\mathbb{R}^2$

Question

To prove a norm, I have to show 3 things:

for all $v,w\in V$ and $c\in \mathbb{R}$

(1) $||v||\ge 0$ along with $||v||=0$ iff $v=\textbf{0}$

(2) $||cv||=|c|*||v||$

(3) $||v+w||\le ||v||+||w||$

I have questions about (2) and (3).

(2) I believe this can be proven by: $||cv||=\sqrt{2(cv_1)^2+3(cv_2)^2}=\color{red}{|c|}\sqrt{2v_1^2+3v_2^2}$ (I'm only questioning if I should have $c_1,c_2$ instead of just $c$)?

(3) I'm wondering if there's a quick way to prove the triangle inequality because I ended up doing a fair amount of algebra to end up with $48v_1w_1v_2w_2\le 24v_2^2w_1^2+24v_1^2w_2^2$ which may be true? The RHS is always $>0$?

Answer 1

(2) I believe this can be proven by: $||cv||=\sqrt{2(cv_1)^2+3(cv_2)^2}=c\sqrt{2v_1^2+3v_2^2}$ (I'm only questioning if I should have $c_1,c_2$ instead of just $c$)?

No, a single $c$ is what you need. The axiom $(2)$ is about scalar multiplication, which involves a single scalar multiplying all components of the vector $v$.

(3) I'm wondering if there's a quick way to prove the triangle inequality because I ended up doing a fair amount of algebra to end up with $48v_1w_1v_2w_2\le 24v_2^2w_1^2+24v_1^2w_2^2$ which may be true? The RHS is always $>0$?

It is true. Your inequality is equivalent to: $$24(v_1^2w_1^2 - 2v_1w_1v_2w_2 + v_2^2w_2^2) \ge 0,$$ which is precisely, $$24(v_1w_1 - v_2w_2)^2 \ge 0.$$

Alternatively, if you're happy to accept that the Euclidean norm $\|v\|_2 = \sqrt{v_1^2 + v_2^2}$ satisfies $(3)$, you can write $\|v\| = \|Tv\|_2$, where $T(v_1, v_2) = (\sqrt{2}v_1, \sqrt{3}v_2)$. Then $T$ is linear, so, $$\|v + w\| = \|T(v + w)\|_2 = \|Tv + Tw\|_2 \le \|Tv\|_2 + \|Tw\|_2 = \|v\| + \|w\|.$$