Example 3

Tag 0795, permanent. Cite it as [Bourbaki, Tag 0795].

Topological Vector Spaces, chapter I, § 2. Linear varieties in a topological vector space

In the normed space $\mathscr{C}(I; \mathbf{C})$ of continuous functions defined over $I = \{0, 1\}$ , the restrictions to I of the functions $e^{2n\pi i x}$ ( $n \in \mathbf{Z}$ ) form a topologically independent family. If $f(x)$ is the linear combination $\sum_{k \neq n} c_k e^{2k\pi i x}$ (where all but finitely many of the $c_k$ are zero) then

\int_0^1 |e^{2n\pi i x} - f(x)|^2 \, dx = 1 + \sum_{k \neq n} |c_k|^2 \geq 1

and, a fortiori, by the mean value theorem

\sup_{x \in I} |e^{2n\pi i x} - f(x)| \geq 1

which shows that

e^{2\pi i n x}

does not belong to the closed vector subspace of

\mathscr{C}(I; \mathbf{C})

generated by

e^{2k\pi i x}, \ k \neq n

The set of elements of a topologically independent family is called a topologically independent set of E. Every subset of a topologically independent subset is topologically independent; every subset consisting of a single point $x \neq 0$ is topologically independent if E is a Hausdorff space.

A topologically independent family is independent (in the algebraic sense; cf. A, II, § 7.1, Remark), but the converse is incorrect.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

Nothing.

Cites

II, § 7