In a previous homework, I had to show that $$c_{2}\|x\|_{\alpha}\le\|x\|_{\beta}\le c_{1}\|x\|_{\alpha}$$ where $c_{2}, c_{1}\gt0$ and $\|\cdot\|_{\alpha}$ and $\|\cdot\|_{\beta}$ are arbitrary norms on $\mathbb{R^{n}}$. This meant that all norms were equivalent on $\mathbb{R^{n}}$.
I was wondering that if I have that $\|x_{n}\|_{\alpha}$ converges to $a$, then can I say that $\|x_{n}\|_{\beta}$ converges to $a$? I said that yes it does because $$c_{2}\|x_{n}-a\|_{\alpha}\le\|x_{n}-a\|_{\beta}\le c_{1}\|x_{n}-a\|_{\alpha}$$ where the left and right side of the inequality tend to $0$. Then, we have $$0\le\lim_{n\rightarrow\infty}\|x_{n}-a\|_{\beta}\le0$$ So by the Squeeze Theorem, $\lim_{n\rightarrow\infty}\|x_{n}-a\|_{\beta}=0$, so $\|x_{n}\|_{\beta}\rightarrow a$.
Is this correct?
Only for $n=2$, to show $$ \|x\|_\beta\le c_2\|x\|_\gamma, x\in R^2 \tag1$$ is equivalent to show $$ (1+z^\beta)^\frac1{\beta}\le c_2(1+z^\gamma)^\frac1{\gamma},z\in[0,\infty) $$ or $$ (1+z^\beta)^{\gamma}\le c_2(1+z^\gamma)^{\beta}\tag2 $$ Let $$ f(z)=\frac{(1+z^\beta)^{\gamma}}{(1+z^\gamma)^{\beta}},z\in[0,\infty). $$ It is easy to see that $f(0)=f(\infty)=1$ and hence $f(z)$ reaches the maximum $C$ in $\in[0,\infty)$. So $$ f(z)\le C$$ and hence (2) is proved.