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

functional-analysisreal-analysis

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 $\blacksquare$

Best Answer

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.

In fact the main motivation (for me) to prove the equivalence of the norms is to show that any finite-dimensional normed space is a Banach space.

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