Any two norms on a finite dimensional linear space are equivalent.

Suppose not, and that $||\cdot||$ is a norm such that for any other norm $||\cdot||'$ and any constant $C$, $C||x||'<||x||$ for all $x$. Define $||\cdot||''=\sum |x_i|\cdot||e_i||$ (*). This is a norm and attains at least as large values as $||\cdot||$ for all $x$.

Could this be used as part of a proof? That two norms $||\cdot||_1,||\cdot||_2$ are equivalent means that there are $m,M$ such that $m||x||_1 \leq ||x||_2 \leq M||x||_1$ for all $x$. In the above I only say that there cannot be a norm such that there does NOT exist an $M$ such that $||x||\leq M||x||'$ for any other norm $||\cdot||'$, but I'm really not sure that proves the entire assertion.

If this is utter nonsense, some hints would be appreciated, thank you.

(*The $e_i$ form a base and sloppily assumed the space to be real.)

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    $\begingroup$ Normally you prove this directly by proving everything is equivalent to the Euclidean norm. $\endgroup$ Commented Dec 9, 2013 at 14:09
  • $\begingroup$ The negation of two norms being equivalent in the above is wrong. You can not claim to have a constant C... $\endgroup$ Commented Dec 9, 2013 at 14:20

3 Answers 3


Note that it is sufficient to prove that any norm $\|\cdot\|$ is equivalent to the sup norm $\|\cdot\|_\infty$.

First write $a=\sum_{i=1}^n a_ie_i$ as a linear combination of basis elements of the $n$-dimensional vector space, say $K$. Then $\| a\|=\left\|\sum_{i=1}^n a_i e_i\right\|$ and use triangle inequality to get $c$ such that $\| a\|\leq c\|a\|_\infty.$

For the other direction consider the identity map $$id:(K,\|\cdot\|_\infty)\longrightarrow(K,\|\cdot\|)$$

Let $S$ be the unit sphere in $(K,\|\cdot\|_\infty )$. Then

  • Show that $S$ is compact.
  • Show that $id$ is continuous.

Then $S$ is compact and hence bounded in $\|\cdot\|$. Now take any $a\in K$ and let $b=a/\|a\|_\infty$. Then $\| b\|\geq c^\prime$ for some positive constant $c^\prime$ and you are done.

  • $\begingroup$ same directions? $\endgroup$
    – Lookout
    Commented Apr 7, 2016 at 1:19
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    $\begingroup$ @user166445: ??? Could you be more explicit? $\endgroup$
    – Alex M.
    Commented Sep 6, 2016 at 19:41
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    $\begingroup$ @pritam, couldn't we use the compacity of $S$ and continuity of $id$ to prove also that $||b||\leq c$? $\endgroup$
    – rmdmc89
    Commented Jun 9, 2018 at 21:34
  • $\begingroup$ At the end, you really want to use that $\{\|x\| : x \in S\}$ is compact, hence closed, so there exists $c>0$ such that $\|x\| > c$ for $x \in S$. $\endgroup$ Commented May 15, 2019 at 18:04

This lecture note answers this question quite well. https://math.mit.edu/~stevenj/18.335/norm-equivalence.pdf


The negation of the statement that all norms on a finite dimensional space are equivalent is different than what you claim. All you can say is that at least one norm exists for which the inequality doesn't hold, not that it doesn't hold for all norms.

So no, your attempt doesn't work.


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