# Equivalence of Norms on Finite-Dimensional Spaces: Questions

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

• Here is the important point: the proof is applying the CS inequality to the numbers $|\alpha_i|$ and $||e_i||$, NOT to elements of X. Sep 7, 2021 at 20:32
• Put $u_i= |a_i|$, $v_i = \|e_i\|$, then apply the CS inequality as you quote it and take square roots. Sep 7, 2021 at 20:33

## 1 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 $$\langle (|\alpha_1|,...,|\alpha_n|), (\|e_1\|,...,\|e_n\|) \rangle = \sum_{k=1}^n |\alpha_k| \|e_k \|$$

• You wrote that $(|\alpha_1|,...,|\alpha_n|)$, $(\|e_1\|,...,\|e_n\|)\in\mathbb R^n$. But wouldn't this then mean that on the right side of the CS-inequality, it would have to read $\sum_{k=1}^n |\alpha_k| \|e_k \| \leq \sqrt{\sum_{i = 1}^{n}|\alpha_i|^2} \sqrt{\sum_{i = 1}^{n}\| e_i\|_{2}^{2}}$? However, in the textbook it reads $\sum_{k=1}^n |\alpha_k| \|e_k \| \leq \sqrt{\sum_{i = 1}^{n}|\alpha_i|^2} \sqrt{\sum_{i = 1}^{n}\| e_i\|^{2}}$, which somehow seems to me as if we were mixing up the Euclidean norm with the other norm $\|\|$ we are considering on $X$. Sep 7, 2021 at 21:34
• I am not mixing any norms, take the $|\alpha_j|$ and $\| e_j\|$ merely as positive numbers and use the proven Cauchy-Schwarz inequality for Euclidean (in fact any) inner product on $\mathbb R^n$. It is just an inequality of numbers, that's it Sep 8, 2021 at 17:06