# Equivalent norms without Cauchy-Schwarz inequality

Let $X$ be a finite-dimensional vector space over $\mathbb{F}$. ($\mathbb{R}$ or $\mathbb{C}$)

Theorem: All norms on $X$ are equivalent.

Proof: $a_k$s and $c_k$s will refer to elements of $\mathbb{F}$. Let $(e_k)_{k=0}^{n-1}$ be a basis of $X$. Define $\lVert \sum_{k=0}^{n-1} c_k e_k \rVert_1 = \sum_{k=0}^{n-1} |c_k|$. Evidently $\lVert \cdot \rVert_1$ is a norm. Let $\lVert \cdot \rVert$ be another norm on $X$. Then if we define $c_0 = 1/\max(\lVert e_k \rVert)_{k=0}^{n-1}$, for $x = \sum_{k=0}^{n-1} a_k e_k$, $$c_0\lVert x \rVert = c_0\left\lVert \sum_{k=0}^{n-1} a_k e_k \right\rVert \leq c_0 \sum_{k=0}^{n-1} |a_k| \lVert e_k \rVert \leq \sum_{k=0}^{n-1} |a_k| = \lVert x \rVert_1$$ Conversely, observe that if $c_1 > 0$ so that $\lVert x \rVert_1 \leq c_1 \lVert x \rVert$, (We may assume that $\lVert x \rVert_1 \neq 0$ by the identity of indiscernibles) $$1 = \frac{\lVert x \rVert_1}{\lVert x \rVert_1} \leq \frac{c}{\lVert x \rVert_1} \lVert x \rVert = c_1\left\lVert \frac{x}{\lVert x \rVert_1} \right\rVert$$ so it suffices to show that $\lVert\cdot\rVert$ is bounded below on the unit sphere $S =\{x \in X\mid \lVert x \rVert_1 = 1\}$.

However, this is the part I'm having trouble with. Obviously if $\mathbb{F} = \mathbb{R}$, I can use the fact that every bounded set in $\mathbb{R}^n$ is totally bounded, and thus show the compactness of $S$.(which implies every sequence of norms on $S$ has a convergent subsequence) But what if $\mathbb{F} = \mathbb{C}$? How does the argument work here? Also, the sequence of norms mentioned earlier converges iff $\lVert\cdot\rVert$ is continuous, which is what I'm trying to show.

• You have shown that the natural map $(X,\|\cdot\|_1)\rightarrow(X,\|\cdot\|)$ is continuous. The unit sphere $S_1 = \{ x\in X : \|x\|_1 = 1 \}$ is compact. Therefore $S_1$ is compact in $(X,\|\cdot\|)$. Because $0 \notin S_1$, this gives the existence of $\alpha$ such that $0 < \alpha \le \inf_{x\in S}\|x\|$. Hence, $\alpha \le \|\frac{1}{\|x\|_1}x\|$ for $x \ne 0$, or $\|x\|_1 \le \frac{1}{\alpha}\|x\|$. – DisintegratingByParts Feb 24 '16 at 5:19
• I realised that the $1$-norm is not the usual Euclidean norm on $X$, so how is $S_1$ compact? – Henricus V. Mar 1 '16 at 17:21
• Every coordinate is bounded, and there you can find a convergent subsequence. Take the new subsequence and do the same in the next coordinate. Etc. until you have a convergent subsequence. Show that the limit is on unit sphere in the $1$ norm. – DisintegratingByParts Mar 1 '16 at 18:01

The same argument work, because compact set in $\mathbb{C}^n$ are exactly closed and bounded set (think about $\mathbb{C}^n$ as $\mathbb{R}^{2n}$), but you can also avoid all this argument by using Open mapping theorem as above :
the identity maps : $$\begin{array}{c} I : (E,\|.\|_1) \rightarrow (E, \|.\|) \end{array}$$
Is surjective and continuous (that what you have proved) and the space $E$ is a Banach space for $\|.\|_1$ so it inverse will be continuous.
• I think the continuity of identity map only implies that the norms induce the same topology, but not equivalent ($\exists c_0,c_1 > 0\mid \forall x \in X, c_0 \lVert x \rVert \leq \lVert x \rVert_1 \leq c_1 \lVert x \rVert$). I'm using the "strong equivalence" definition of wikipedia. – Henricus V. Feb 24 '16 at 2:14
• The are equivalent in the case of normed space because saying that the identity is an homeomorphism mean that (using the fact that $I$ is linear) the norms will be equivalent – Hamza Feb 24 '16 at 2:17