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For instance, given a finite dimensional vector space $V$ we want to show that any norms in this $V$ is equivalent. However, all proofs online consider only two norms, mainly $|| . ||$ and $|| . ||_{p}$ where $1<p< \infty $ when trying to show this statement and they say this is due to "transitivity".

Why is this the case? I think we can define infinitely many norms on a finite dimensional vector space, not just those of the form $||.||_{p}$ for all $1<p< \infty$.

  • https://math.stackexchange.com/questions/57686/understanding-of-the-theorem-that-all-norms-are-equivalent-in-finite-dimensional – Shweta Aggrawal Oct 20 '18 at 08:12
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    You may want to quote exactly the passage because different people may read different books, and different books may be written differently. For instance, mine did something like fixing two values of $p$ (I think they were $p_1=\infty$ and $p_2=1$, but I don't remember exactly), then it did something like proving that every norm satisfies $c\lVert\bullet\rVert_{p_1}\le\lVert \bullet\rVert\le c'\lVert\bullet\rVert_{p_2}$, then it proved that $\lVert\bullet\rVert_{p_2}\sim\lVert\bullet\rVert_{p_1}$. At some point, there was some topology involved, probably for $p_1$. –  Oct 20 '18 at 08:16

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If you know there are constants $c>0, C>0$ such that $$c||.|| \le ||.||_p \le C||.||$$ where $||.||$ is an arbitrarily chosen norm, and if you then have to norms $|.|_a, |.|_b$ you then know there are corresponding constants $c_a, C_a, c_b, C_b$, such that $$c_a |.|_a \le ||.||_p \le C_b |.|_b$$ and $$c_b|.|_b \le ||.||_p \le C_a |.|_a$$

which implies that $|.|_a$ and $|.|_b$ are also equivalent.

(Of course, in the initial statement, $||.||$ must not be some fixed norm, but any norm, with $c,C$ then depending on this norm and on $p$. Also the indices $a$, $b$ are not meant to imply a certain definition of the norm like the $p$-norms, but are just meant to give a name to the two norms).

Thomas
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