Can someone verify that everything I did in this proof is correct?

Every two norms on a finite dimensional (real or complex) vector space $V$ are equivalent.

Proof: We assume that $\dim(V)=n$ and that $V$ is a real vector space. If $V$ is complex, the proof is completely analogue.

Let $E:= \{e_1, \dots, e_n\}$ be a basis of $V$. Write $v = \sum_i v_ie_i$ for every vector $v \in V$.

Then, define $\Vert\cdot\Vert_2: V \to \mathbb{R}^+: v \mapsto \left(\sum_{i}|v_i|^2\right)^{1/2}$. This is a norm on $V$. We prove that every norm is equivalent with this norm, and then the result will follow because the relation "two norms are equivalent" is an equivalence relation.

Let $\Vert \cdot\Vert$ be an arbitrary norm on $V$. Let $M:= (\sum\Vert e_i\Vert^2)^{1/2}$. Then, we have, for $v \in V:$

$$\Vert v\Vert = \Vert \sum v_ie_1 \Vert \leq \sum|v_i|\Vert e_i\Vert \leq (\sum|v_i|^2)^{1/2}(\sum\Vert e_i\Vert^2)^{1/2} = M\Vert v\Vert_2$$ where we used Cauchy's inequality for the last inequality.

It also follows that $\Vert\cdot \Vert: (V, \Vert \cdot \Vert_2) \to \mathbb{R}^+$ is a continuous function. Indeed, let $w \in V, \epsilon > 0$

Then, for $v \in V$ satisfying $\Vert v-w\Vert _2 < \delta$, it follows that:

$$|\Vert v\Vert - \Vert w\Vert| \leq \Vert v-w\Vert \leq M\Vert v - w \Vert_2 < M \delta = \epsilon$$

if we take $\delta :=\epsilon/M$.

Now, we prove that $S^{n-1}:= \{v \in V \mid \Vert v \Vert_2 =1\}$ is compact in $(V,\Vert \cdot \Vert_2)$. Because $V$ is finite dimensional, it suffices to prove that $S^{n-1}$ is closed and bounded. Boundedness is trivial.

For closedness, it suffices to notice that $S^{n-1}= \Vert \cdot \Vert_2^{-1}(\{1\})$ and because $\Vert \cdot \Vert_2: (V, \Vert \cdot\Vert_2) \to \mathbb{R}^+$ is continuous and $\{1\}$ is closed, it follows that $S^{n-1}$ is closed in $(V, \Vert \cdot \Vert_2)$

Hence, the function $\Vert \cdot \Vert: S^{n-1} \to \mathbb{R}^+$ attains a minimum $m > 0$ ($m=0$ is not possible because the domain are vectors with norm 1)

So, for every $v \in S^{n-1}$, we have:

$$0 < m \leq \Vert v \Vert$$

and hence, for $v \neq 0$, it follows that:

$$m \leq \Vert \frac{1}{\Vert v \Vert_2} v \Vert$$

and for $v=0$ the inequality is obvious.

From this, it follows that $m \Vert v\Vert _2 \leq \Vert v \Vert$ such that the two norms are equivalent, as desired.


1 Answer 1


I found nothing wrong with your proof, but perhaps that you could justify the assertion “Boundedness is trivial”.

  • $\begingroup$ For example, $S^{n-1} \subseteq B_{\Vert \cdot \Vert_2}(0, 2)$? $\endgroup$
    – user370967
    Feb 18, 2018 at 15:47
  • $\begingroup$ @Math_QED Why is that true? $\endgroup$ Feb 18, 2018 at 15:49
  • $\begingroup$ Let $x \in S^{n-1}$. Then, $\Vert x \Vert_2 = 1 < 2$, so $\Vert x - 0 \Vert_2 < 2$, meaning that $x$ is in the given ball? $\endgroup$
    – user370967
    Feb 18, 2018 at 15:51
  • $\begingroup$ @Math_QED Forget my doubts. I thought that $S^{n-1}$ was the unit sphere with respect to the other norm. Sorry about that. $\endgroup$ Feb 18, 2018 at 15:53
  • $\begingroup$ Ok. I thought I was overlooking something. Thanks for the verification! If nobody else answers soon, I'll accept this answer. $\endgroup$
    – user370967
    Feb 18, 2018 at 15:54

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