I just want my proof checked that all norms on $\mathbb{R}$ are equivalent to the Euclidean norm. It's as follows:
Let $||\cdot || $be a norm on $\mathbb{R}^n .\ $Define $f:S^{n-1}\rightarrow \mathbb{R}^n, x\mapsto ||x||.$ Let $ \{ e_j \}_{j=1}^{n} $ be the standard basis of $\mathbb{R}^n $ Let $S$ denote $S^{n-1}$ for ease. Now, $f \ $ is continuous since $$||f(x)-f(y)||\leq ||\sum_{j=1}^{n} (x_j-y_j)e_j ||\leq n ||x-y||_{\infty}\leq n||x-y||_2 \ .$$ Now $f$ is continuous and $S$ is compact so by the extreme value theorem there are $C_1, C_2 \geq 0 $ such that $$C_1\leq ||x|| \leq C_2 \ .$$ We must have $C_1>0 $ as otherwise $f(x)=||x||=0 $ for some $x\in S$ but this means $x=0$ which is impossible. Now for any $x\in S$ we have $x=y/||y||_2 $ for some non zero $y\in \mathbb{R}^n .$ So $$C_1\leq \frac{||y||}{||y||_2} \leq C_2 $$ for all non zero $y$ and so $$C_1||y||_2\leq ||y|| \leq ||C_2 ||y||_2 $$ holds for all $y \in \mathbb{R}^n $.
Is this ok?