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FALSE: all norms on a real vector space are equivalent

Statement

False claim: any two norms on a real vector space are equivalent (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms, Equivalent norms, and the dictionary with equivalent metrics).

What is true is the same statement for $R^{n}$ with $n$ a natural number, which is For $n \geq 1$ all norms on $R^{n}$ are equivalent. Dropping the hypothesis that the space is one of the $R^{n}$ makes the claim false, and the witness below is built from published material only.

The witness. Let $R^{N}$ be the function space of all functions $N \to R$ with pointwise operations (The vector space $F^{X}$ of all functions $X \to F$ with pointwise operations, and $F^{n}$ as the case $X = n = {0, 1, \dots, n - 1}$ ), and let

$V : = {v \in R^{N} : there is K \in N with v_{j} = 0 for every j \geq K}$

be the set of finitely supported sequences. On $V$ define

$N_{1} (v) : = \sum_{j < K} ∣ v_{j} ∣, N_{\infty} (v) : = \max {∣ v_{j} ∣ : j < K},$

for any $K \geq 1$ with $v_{j} = 0$ for $j \geq K$ . Both are norms on $V$ , both values are independent of the admissible $K$ chosen, and no real $C$ satisfies $N_{1} \leq C N_{\infty}$ on $V$ .

Facts & Assumptions

Given: The space $R^{N}$ , the subset $V$ , the functions $N_{1}$ and $N_{\infty}$ above, and, for $m \geq 1$ , the vector $u^{(m)} \in V$ with $u_{j}^{(m)} = 1$ for $j < m$ and $u_{j}^{(m)} = 0$ for $j \geq m$ . For $i \in N$ , $e_{i} \in R^{N}$ is the function with $e_{i} (i) = 1$ and $e_{i} (j) = 0$ for $j \neq i$ .

[A1]

The refuted claim: any two norms on a real vector space are equivalent.

[L1]

$R^{N}$ is a vector space over $R$ with pointwise operations, and a nonempty $W$ with $λ u + w \in W$ for all $λ$ and all $u, w \in W$ is a linear subspace, hence itself a vector space (The vector space $F^{X}$ of all functions $X \to F$ with pointwise operations, and $F^{n}$ as the case $X = n = {0, 1, \dots, n - 1}$ , Vector space over a field, Linear subspace of a vector space, One-step subspace test: a nonempty $W \subseteq V$ is a linear subspace if and only if $λ u + v \in W$ for all $λ \in F$ and $u, v \in W$ ).

[L2]

Finite sums in a function space are pointwise, for an arbitrary index set: $(\sum_{l < q} y_{l}) (j) = \sum_{l < q} y_{l} (j)$ (The standard list $e : n \to F^{n}$ with $e_{i} (i) = 1_{F}$ and $e_{i} (j) = 0_{F}$ for $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ and dimension $0$ clause 1, stated there for an arbitrary $X$ ; Linear combination of a finite list, and the span $span (S)$ as the smallest linear subspace containing $S$ , Finite sums and finite products, by recursion).

[L3]

Laws of finite sums (Laws of finite sums and finite products, Finite sums and finite products, by recursion): additivity, scaling, splitting, monotonicity, a sum of nonnegative terms is nonnegative, a vanishing sum of nonnegative terms has all terms $0$ , and $\sum_{j < m} 1 = ι (m)$ .

[L4]

Maxima (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set): a nonempty finite set of reals has a maximum, which belongs to it and bounds it above.

[L5]

Absolute value (Absolute value in an ordered field, Basic properties of the absolute value, The triangle inequality): $∣ t ∣ \geq 0$ ; $∣ t ∣ = 0$ exactly when $t = 0$ ; $∣ s t ∣ = ∣ s ∣ ∣ t ∣$ ; $∣ s + t ∣ \leq ∣ s ∣ + ∣ t ∣$ .

[L6]

The Archimedean property: for every real $x$ there is a natural $m \geq 1$ with $x < ι (m)$ (Every complete ordered field is Archimedean, The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, Canonical naturals are positive and strictly increasing).

[L7]

Dimension: if $V$ has a spanning set with $q$ elements then every linearly independent subset of $V$ is finite with at most $q$ elements; a finite-dimensional space is one with a finite basis (If $V$ has a spanning set with $n$ elements, then every linearly independent subset of $V$ is finite with at most $n$ elements; in particular $V$ has no linearly independent subset equinumerous with $N$ , Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis, Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis, Linear independence: a finite list $v : n \to V$ is independent when $\sum_{i < n} λ_{i} v_{i} = 0_{V}$ forces every $λ_{i} = 0_{F}$ , and a subset $S \subseteq V$ is independent when every injective finite list into $S$ is independent, Finite, countably infinite, countable, uncountable, Equinumerous sets, $A \approx B$ and $A ⪯ B$ , The pigeonhole principle on $N$ ).

[L8]

Norm equivalence: $M$ and $N$ are equivalent when $c M \leq N \leq C M$ for some reals $c, C > 0$ (Equivalent norms, and the dictionary with equivalent metrics); the norm axioms are (N1), (N2), (N3) (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms); and induction (The principle of mathematical induction).

Refutation

technique · direct

1.1

$V$ is a linear subspace of $R^{N}$ , hence a real vector space: it contains $0$ , and if $v_{j} = 0$ for $j \geq K$ and $w_{j} = 0$ for $j \geq K^{'}$ then $(λ v + w)_{j} = 0$ for $j \geq \max {K, K^{'}}$ .

L1L4

1.2

The values $N_{1} (v)$ and $N_{\infty} (v)$ do not depend on the admissible $K$ . If $K \leq K^{'}$ are both admissible, then splitting the sum gives $\sum_{j < K^{'}} ∣ v_{j} ∣ = \sum_{j < K} ∣ v_{j} ∣ + \sum_{j = K}^{K^{'} - 1} ∣ v_{j} ∣$ , and the second part is a sum of zeros; and $\max {∣ v_{j} ∣ : j < K^{'}} = \max {∣ v_{j} ∣ : j < K}$ because the extra entries are $0$ and the maximum over $j < K$ is $\geq ∣ v_{0} ∣ \geq 0$ .

L3L4L5

1.3

The hypothesis that fails is finite-dimensionality. For every $p \in N$ the set ${e_{i} : i < p}$ is a subset of $V$ with $p$ elements, the map $i \mapsto e_{i}$ being injective because $e_{i} (i) = 1 \neq 0 = e_{i^{'}} (i)$ for $i \neq i^{'}$ ; and it is linearly independent, since for an injective list $l \mapsto e_{i_{l}}$ into it and scalars $λ$ , evaluating $\sum_{l < q} λ_{l} e_{i_{l}} = 0$ at the point $i_{l_{0}}$ gives $λ_{l_{0}} = 0$ , the list $l \mapsto λ_{l} e_{i_{l}} (i_{l_{0}})$ vanishing off the single index $l_{0}$ .

L2L3L5

2.1

$N_{1}$ is a norm on $V$ . (N1): $N_{1} (v) = 0$ forces every $∣ v_{j} ∣ = 0$ for $j < K$ , hence $v = 0$ ; and $N_{1} (0) = 0$ . (N2): $λ v$ is admissible with the same $K$ and $\sum_{j < K} ∣ λ v_{j} ∣ = ∣ λ ∣ \sum_{j < K} ∣ v_{j} ∣$ . (N3): with $K$ admissible for both $v$ and $w$ , $\sum_{j < K} ∣ v_{j} + w_{j} ∣ \leq \sum_{j < K} ∣ v_{j} ∣ + \sum_{j < K} ∣ w_{j} ∣$ termwise.

step 1.2L3L5L8

2.2

$N_{\infty}$ is a norm on $V$ . (N1): $N_{\infty} (v) = 0$ forces $∣ v_{j} ∣ \leq 0$ and $\geq 0$ for every $j < K$ , hence $v = 0$ . (N2): $\max {∣ λ v_{j} ∣} = ∣ λ ∣ \max {∣ v_{j} ∣}$ , since $∣ λ ∣ ∣ v_{j} ∣ \leq ∣ λ ∣ N_{\infty} (v)$ for every $j$ with equality at an index attaining the maximum. (N3): $∣ v_{j} + w_{j} ∣ \leq ∣ v_{j} ∣ + ∣ w_{j} ∣ \leq N_{\infty} (v) + N_{\infty} (w)$ for every $j < K$ , and the maximum on the left is one of those numbers.

step 1.2L4L5L8

2.3

For $m \geq 1$ the vector $u^{(m)}$ lies in $V$ , and $K = m$ is admissible for it; so $N_{1} (u^{(m)}) = \sum_{j < m} 1 = ι (m)$ and $N_{\infty} (u^{(m)}) = \max {1, \dots, 1} = 1$ .

step 1.2L3L4L5

2.4

So $V$ has no finite basis: a basis $B$ with $q$ elements would span $V$ , forcing every linearly independent subset to have at most $q$ elements, while step 1.3 produces one with $q + 1$ . Hence $V$ is infinite-dimensional, and For $n \geq 1$ all norms on $R^{n}$ are equivalent, which is a statement about $R^{n}$ for a natural $n$ , does not apply to it.

step 1.3L7

3.1

Suppose $N_{1}$ and $N_{\infty}$ were equivalent, so that in particular $N_{1} (v) \leq C N_{\infty} (v)$ for every $v \in V$ and some real $C > 0$ . Then $ι (m) \leq C$ for every $m \geq 1$ , by step 2.3.

step 2.3L8

4.1

That contradicts the Archimedean property, which supplies a natural $m \geq 1$ with $C < ι (m)$ . So $N_{1}$ and $N_{\infty}$ are not equivalent, and [A1] is false.

step 3.1A1L6

5.1

The claim [A1] is therefore false, and the true statement in its neighbourhood is For $n \geq 1$ all norms on $R^{n}$ are equivalent, whose proof spends compactness of the Euclidean unit sphere, a property step 2.4 shows $V$ has no analogue of.

step 4.1step 2.4A1∎

Remarks

No classification of infinite-dimensional normed spaces is claimed here. What is exhibited is one real vector space carrying two inequivalent norms, which is all that is needed to refute the claim.
Where the proof of For $n \geq 1$ all norms on $R^{n}$ are equivalent breaks on $V$ . That proof takes the unit sphere of $∥ \cdot ∥_{2}$ , which is closed and bounded, and concludes compactness from Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line — a theorem about $R^{n}$ for a natural $n$ , proved by bisecting finitely many coordinates. On $V$ there is no such theorem, and indeed FALSE: in every normed space a closed bounded set is compact refutes the corresponding claim on the same space.
The two norms are the honest analogues of $∥ \cdot ∥_{1}$ and $∥ \cdot ∥_{\infty}$ , and the ratio $N_{1} / N_{\infty}$ at $u^{(m)}$ is exactly $ι (m)$ , the same constant that appears in the comparison chain of The finite and reverse triangle inequalities for a norm; and for $n \geq 1$ every norm $N$ on $R^{n}$ satisfies $N (x) \leq C ∥ x ∥_{1}$ and is Lipschitz, hence continuous, for $d_{2}$ on $R^{m}$ . In finite dimensions that constant is a bound; on $V$ it grows without bound, and the Archimedean property is what turns that into a refutation.

Depends on

Used by

FALSE: in every normed space a closed bounded set is compact False statement

Dependency tree · two levels

91 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Norm (mathematics) (Wikipedia) (standard reference, not scraped)
Archimedean property (Wikipedia) (standard reference, not scraped)
J. Demmel, MA221 Lecture 3: Vector Norms (standard reference, not scraped)
G. Zitelli, Math 641 Functional Analysis, Part I (standard reference, not scraped)