Finite Dimensional Normed Spaces and Riesz Lemma — Examples

Let $\mathbb{K}\in \{\mathbb{R},\mathbb{C}\}$ and let $x=(x_0,\dots ,x_{n-1})\in {\mathbb{K}}^n$ with $n\ge 1$. Define

$\mathrm{\parallel }x{\mathrm{\parallel }}_{\mathrm{\infty }}:={\max }_{j<n}\mathrm{\mid }{x}_j\mathrm{\mid },\mathrm{\parallel }x{\mathrm{\parallel }}_2:={\left({\sum }_{j<n}\mathrm{\mid }{x}_j{\mathrm{\mid }}^2\right)}^{1\mathrm{/}2},\mathrm{\parallel }x{\mathrm{\parallel }}_1:={\sum }_{j<n}\mathrm{\mid }{x}_j\mathrm{\mid }\mathrm{.}$

Then

$\mathrm{\parallel }x{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }x{\mathrm{\parallel }}_2\le \mathrm{\parallel }x{\mathrm{\parallel }}_1\le \sqrt{n}\mathrm{\parallel }x{\mathrm{\parallel }}_2\le n\mathrm{\parallel }x{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{.}$

In particular, the abstract norm-equivalence theorems on the A page can be read with explicit constants on these three standard coordinate norms.

• The constants are sharp in the standard basis: $x=(1,\dots ,1)$ makes $\mathrm{\parallel }x{\mathrm{\parallel }}_1=n\mathrm{\parallel }x{\mathrm{\parallel }}_{\mathrm{\infty }}$ and $\mathrm{\parallel }x{\mathrm{\parallel }}_1=\sqrt{n}\mathrm{\parallel }x{\mathrm{\parallel }}_2$.

Assume Dependent Choice. Let $X$ be a normed space that is not the span of any finite list. Then the unit sphere of $X$ contains an infinite subset whose distinct points are all more than $1\mathrm{/}2$ apart.

Refuted claim: every closed and bounded subset of an infinite-dimensional Banach space is compact.

Take the Hilbert space

${\mathrm{\ell }}^2:=\{x=(x_n{)}_{n\ge 0}:{\sum }_{n=0}^{\mathrm{\infty }}\mathrm{\mid }{x}_n{\mathrm{\mid }}^2<\mathrm{\infty }\}\mathrm{.}$

Its closed unit ball is closed and bounded, but it is not compact.

• The Banach-space adjective in the refuted claim is there because the failure is not an incompleteness issue. The obstruction is infinite dimension.

Refuted claim: the polynomial space over a scalar field can be made into a Banach space by some norm.

Let $\mathbb{K}\in \{\mathbb{R},\mathbb{C}\}$ and let $\mathbb{K}[x]$ be the polynomial ring of The polynomial ring over a commutative ring as finitely supported coefficient sequences with convolution, regarded as a vector space over $\mathbb{K}$. Then no norm on $\mathbb{K}[x]$ is complete.

Let $M=\{o,a,b\}$ with metric determined by

$d(o,a)=1,d(a,b)=1,d(o,b)=2.$

Using $o$ as basepoint, the based Kuratowski map $K_o:M\to C_b(M)$ is given by

$K_o(o)=(0,0,0),K_o(a)=(1,-1,-1),K_o(b)=(2,0,-2),$

where each function is recorded by its values on $(o,a,b)$. These three points in ${\mathbb{R}}^3$ are at the same pairwise sup distances as the original metric space.

On an infinite-dimensional Banach space, the existence of a discontinuous linear functional is a choice-sensitive statement. Full Choice gives one immediately from an infinite Hamel basis: pick the basis, prescribe an unbounded coefficient functional on it, and extend linearly. The opposite direction is not a theorem of ZF + DC. Howard and Tachtsis record the consistency boundary, and Heil's discussion of Hamel bases spells out the elementary "basis gives an unbounded functional" half.

The point for this page is negative rather than positive: the explicit discontinuous functional on the companion example below lives on the incomplete space $c_{00}$, not on a Banach space. That distinction is exactly why this remark is kept separate from the example and why it never serves as a dependency.

Let

$c_{00}:=\{x=(x_n{)}_{n\ge 0}:x_n=0\text{ for all but finitely many }n\}$

with the supremum norm $\mathrm{\parallel }x{\mathrm{\parallel }}_{\mathrm{\infty }}:={\sup }_n\mathrm{\mid }{x}_n\mathrm{\mid }$, a norm in the sense of A norm on a real vector space, the induced metric, and the dictionary with the metric axioms. Define

$f:c_{00}\to \mathbb{R},f(x):={\sum }_{n=0}^{\mathrm{\infty }}(n+1)x_n\mathrm{.}$

Because every $x\in c_{00}$ has finite support, the displayed sum is really finite. The map $f$ is linear but unbounded, hence discontinuous.

• This is the explicit incomplete-space witness promised by the companion remark. No choice principle is used anywhere in the construction.