1
$\begingroup$

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

$\endgroup$
2
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

3
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.