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

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    $\begingroup$ 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. $\endgroup$ Sep 7, 2021 at 20:32
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    $\begingroup$ Put $u_i= |a_i|$, $v_i = \|e_i\|$, then apply the CS inequality as you quote it and take square roots. $\endgroup$
    – Rob Arthan
    Sep 7, 2021 at 20:33

1 Answer 1


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

  • $\begingroup$ 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 $X$. $\endgroup$
    – Hermi
    Sep 7, 2021 at 21:34
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    $\begingroup$ 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 $\endgroup$
    – Physor
    Sep 8, 2021 at 17:06

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