$\alpha N_1-\beta N_2$ is a norm where $N_2<N_1$

Question

Given $N_2(x)\le N_1(x)$ norms on some vector space over $F$ with equality only holding iff $x=0$ and $N_1\ne\delta N_2$, is it possible that $\alpha N_1-\beta N_2$ is a norm ($\alpha>\beta>0$)
I'm having diffcult proving that trianlge ineqaulity would be violated so I was wondering if counterexamples exist

Answer 1

It is possible. Let $N_0, N_2$ be norms such that $N_2 \le N_0$ and $N_0$ is not a multiple of $N_2$. Then $N_1 = N_0 + \frac{1}{2}N_2$ is a norm as in your question. We have $$1 N_1 - \frac{1}{2}N_2 = N_0 .$$