Normed and Banach Spaces - Main Theorem

One of the most fundamental results in functional analysis establishes that all finite-dimensional normed spaces share the same topological properties, regardless of the choice of norm.

TheoremEquivalence of Norms in Finite Dimensions

Let $X$ be a finite-dimensional vector space over $\mathbb{K}$ . Then all norms on $X$ are equivalent.

This remarkable theorem has far-reaching consequences. It tells us that in finite dimensions, the choice of norm is largely a matter of convenience—topological properties such as convergence, continuity, compactness, and completeness are independent of which norm we use.

Proof

It suffices to show that any norm $\|\cdot\|$ on $X$ is equivalent to a fixed reference norm. Let $\dim X = n$ and choose a basis $\{e_1, \ldots, e_n\}$ for $X$ . Define the reference norm by $\left\|\sum_{i=1}^n \alpha_i e_i\right\|_1 = \sum_{i=1}^n |\alpha_i|$

For any $x = \sum_{i=1}^n \alpha_i e_i \in X$ , by the triangle inequality: $\|x\| = \left\|\sum_{i=1}^n \alpha_i e_i\right\| \leq \sum_{i=1}^n |\alpha_i| \|e_i\| \leq C \|x\|_1$ where $C = \max_{1 \leq i \leq n} \|e_i\|$ .

For the reverse inequality, define $f: \mathbb{K}^n \to \mathbb{R}$ by $f(\alpha_1, \ldots, \alpha_n) = \|\sum_{i=1}^n \alpha_i e_i\|$ . This function is continuous with respect to the standard topology on $\mathbb{K}^n$ . The unit sphere $S = \{(\alpha_1, \ldots, \alpha_n) : \sum_i |\alpha_i| = 1\}$ is compact, so $f$ attains its minimum on $S$ . Since $f > 0$ on $S$ (by positivity of the norm), we have $c = \min_{S} f > 0$ .

For any nonzero $x = \sum_{i=1}^n \alpha_i e_i$ , we can normalize to get a point in $S$ , yielding $\|x\| \geq c \|x\|_1$ .

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ExampleConsequences of the Theorem

All Banach: Every finite-dimensional normed space is a Banach space (complete)
Compactness: A subset of $\mathbb{R}^n$ is compact if and only if it is closed and bounded, regardless of the norm used
Continuity: A linear map between finite-dimensional normed spaces is always continuous
Dimension Formula: If $X$ is a finite-dimensional normed space and $Y \subset X$ is a closed subspace, then $\dim X = \dim Y + \dim(X/Y)$

Remark

This theorem fails dramatically in infinite dimensions. Different norms on an infinite-dimensional space can induce completely different topologies, lead to different notions of convergence, and result in spaces with vastly different completeness properties. This is why the study of Banach spaces requires such careful attention to the choice of norm.

The theorem also explains why linear algebra in finite dimensions is so well-behaved compared to infinite-dimensional analysis. Many pathological phenomena in functional analysis simply cannot occur in finite dimensions.