A well-known result from functional analysis is that all norms on a finite-dimensional space $X$ are equivalent. The source 1 proves it by showing that every norm on a finite-dimensional space is equivalent to the Euclidean norm $\vert\vert\cdot\vert\vert_{2}$. The gist goes as follows:
Let $\dim(X) = n$ and let $\{e_1, \dots, e_n\}$ be a basis of $X$. Then it follows from the triangle-inequality for $\vert\vert\cdot\vert\vert$ and the Cauchy-Schwartz inequality that $$\vert\vert x\vert\vert = \vert\vert \sum_{i= 1}^{n}\alpha_i e_i \vert\vert \leq \sum_{i = 1}^{n}\vert\alpha_i\vert \ \vert\vert e_i\vert\vert \leq \sqrt{ \sum_{i=1}^{n}\vert\alpha_i\vert^2} \sqrt{\sum_{i=1}^{n}\vert\vert e_i\vert\vert^2}.$$
I am afraid I do not understand the second inequality and how the Cauchy-Schwartz inequality comes into play here. According to Wikipedia, we can write the CS inequality as $$\vert\langle u, v\rangle\vert^2 \leq \vert\vert u\vert\vert^2 \cdot \vert\vert v\vert\vert^2.$$ I simply do not see how we can apply this to $\vert\alpha_i\vert \ \vert\vert e_i\vert\vert$.
1 Dirk Werner. Funktionalanalysis. Springer. $8$th edition
Best Answer
Consider two $n$-tuple $(|\alpha_1|,...,|\alpha_n|)$ and $n$-tuple $(\|e_1\|,...,\|e_n\|)$ as vectors in $\mathbb R^n$ and their (standard) inner product $$ \langle (|\alpha_1|,...,|\alpha_n|), (\|e_1\|,...,\|e_n\|) \rangle = \sum_{k=1}^n |\alpha_k| \|e_k \| $$