# Equivalency of all norms on a finite dimensional vector space: compactness theorems vs. the open mapping theorem

Going through my functional analysis course notes, I feel like there are two different proofs for the following theorem.

In $$\mathbb{R}^n$$ (or $$\mathbb{C}^n$$), any two norms are equivalent.

One uses the compactness-related extreme value theorem (i.e., a continuous function on compact set must achieve its maximum and minimum values), while the other uses the open mapping theorem (i.e, for every continuous linear mapping $$T$$ from a Banach space $$X$$ onto another Banach space $$Y$$, and every $$U \in X$$ open, $$T(U)$$ is open). These two theorems use different hypothesis and are not equivalent. Therefore, I am suspicious that I am doing something wrong.

My questions is whether both these proofs are correct, or if I am doing something wrong here.

### Common steps of both proofs

• Define (recall) the $$\|.\|_\text{sup}$$ norm as $$\|x\| = \sup_{i} |x_i|$$.
• (*) Show that for any norm $$\|.\|_b$$ on $$\mathbb{R}^n$$, there exist an $$M_b > 0$$, such that for all $$x \in \mathbb{R}^n$$, $$\|x\|_b \leq M_b \|x\|_\text{sup}$$ (see for example here on how to find $$M_b$$).

### Proof with extreme value theorem

• From (*) we deduce that any norm $$\|x\|_b$$ is continuous w.r.t the $$\|x\|_\text{sup}$$ norm.
• (+) Use the fact that the unit sphere (of the sup norm) is compact in $$\mathbb R^n$$, (*), and the extreme value theorem to deduce that $$\|x\|_b$$ achieves a minimum $$m_b$$ on the unit sphere (of the sup norm). In other words, there exists $$m_b > 0$$ such that $$\|x\|_b \geq m_b \|x\|_\text{sup}$$ for all $$x \in \mathbb R^n$$.
• Combining (+) and (*), we get that any norm $$\|.\|_b$$ and $$\|.\|_\text{sup}$$ are equivalent $$\blacksquare$$

### Proof with the open mapping theorem

This is the proof that I am not sure about.

• From the open mapping theorem, one can prove that (see for example here for a proof):

Let $$\|.\|_1$$ and $$\|.\|_2$$ be norms on a Banach space $$X$$, such that $$\|x\|_1 \leq\|x\|_2$$. Then, the norms are equivalent.

• Now combine this with (*) with the fact that $$\mathbb R^n$$ is complete (Banach), and you get that any norm in $$\mathbb{R}^n$$ is equivalent to the $$\|.\|_\text{sup}$$ norm $$\blacksquare$$

• A priori you don't know that $(\mathbb{R}^n, \Vert \cdot \Vert_b)$ is a Banach space. That is one of the nice conclusions of the fact you want to prove. – Severin Schraven May 16 '19 at 8:40
• @SeverinSchraven I see now. Thank you. Would you write your comment as an answer, so that I can accept it? – Hashimoto May 16 '19 at 15:33

We cannot apply the open mapping theorem as we do not know a priori that $$(\mathbb{R}^n, \Vert \cdot \Vert_b)$$ is a Banach space.