Equivalence of norms in finite-dimensional spaces

Question

Assume we have a separable, reflexive Banach space $X$ such that $\{e_{i}\}$ is a basis.

Let $X_{n} = \text{span}\{e_{1},\ldots,e_{n}\}$ be finite-dimensional subspaces where we define members $u_{n} \in X_{n}$ by $u_{n} = \sum_{k=1}^{n}c_{k}^{n}e_{k}$ where $c^{n} = (c^{n}_{1},...,c^{n}_{n}) \in \mathbb{R}^{n}$. What I want to know is if it is true that for all $R_{o} > 0$ there exists a $S \in \mathbb{R}$ such that for all $c^{n} \in \mathbb{R}^{n}$ where $|c^{n}|=S$(using the usual norm in $\mathbb{R}^{n}$) it follows that $\Vert u_{n} \Vert_{X} = R_{o}$(using the norm of $X_{n}$)? (the reals are denoted as $\mathbb{R}$)

I am trying to use this idea in a larger proof. The proof that I am referring to is the Browder and Minty Theorem. It uses the following Lemma:

The First step of the proof of the Theorem is as follows:

The part of the proof of the Theorem I don't follow is how you can apply Lemma 1.11 directly by considering '$c^{n} \in \mathbb{R}^{n}$ with $\Vert u_{n} \Vert_{X} = R_{o}$' instead of $\forall x \in \mathbb{R}^{n}$ with $|x| = R$ as is the requirement of the Lemma 1.11. I thought my idea above would resolve this.

Answer 1

You should reconsider the idea. Nothing like this is true in general. A few remarks:

1. The complicated condition $\forall R_0\ \exists S$ is redundant. Norms respect the scaling of vectors. (If you know which vectors have norm $1$, you know everything about the norm.) So, if the condition $\forall R_0\ \exists S$ holds, you can state is more simply: there is $k>0$ such that $\|u_n\|_X = k\, |c^n|$.

2. The equivalence of two norms (denoted $\|\cdot\|$ and $\|\cdot\|_*$) is usually understood as the existence of $C$ such that for every vector $v$, $\|x\|\le C\|x\|_*$ and $\|x\|_*\le C\|x\|$. This is substantially weaker than 1.

3. But even the condition stated in item 2 will fail, unless your $X$ is a Hilbert space in disguise. Take $\ell_p$ for $p\ne 2$, with the standard basis. The $\ell_p$ norm of $(1,1,\dots,1)\in X^n$ is $n^{1/p}$, while the Euclidean norm is $n^{1/2}$. Since $n$ can be arbitrarily large, the ratio of two norms is not controlled as in item 2.