Editing Inner product space (section)

==Orthonormal sequences==
{{See also|Orthogonal basis|Orthonormal basis}}

Let <math>V</math> be a finite dimensional inner product space of dimension <math>n.</math> Recall that every [[Basis (linear algebra)|basis]] of <math>V</math> consists of exactly <math>n</math> linearly independent vectors. Using the [[Gram–Schmidt process]] we may start with an arbitrary basis and transform it into an orthonormal basis. That is, into a basis in which all the elements are orthogonal and have unit norm. In symbols, a basis <math>\{e_1, \ldots, e_n\}</math> is orthonormal if <math>\langle e_i, e_j \rangle = 0</math> for every <math>i \neq j</math> and <math>\langle e_i, e_i \rangle = \|e_a\|^2 = 1</math> for each index <math>i.</math>

This definition of orthonormal basis generalizes to the case of infinite-dimensional inner product spaces in the following way. Let <math>V</math> be any inner product space. Then a collection
<math display=block>E = \left\{ e_a \right\}_{a \in A}</math>
is a {{em|basis}} for <math>V</math> if the subspace of <math>V</math> generated by finite linear combinations of elements of <math>E</math> is dense in <math>V</math> (in the norm induced by the inner product). Say that <math>E</math> is an {{em|[[orthonormal basis]]}} for <math>V</math> if it is a basis and
<math display=block>\left\langle e_{a}, e_{b} \right\rangle = 0</math>
if <math>a \neq b</math> and <math>\langle e_a, e_a \rangle = \|e_a\|^2 = 1</math> for all <math>a, b \in A.</math>

Using an infinite-dimensional analog of the Gram-Schmidt process one may show:

'''Theorem.''' Any [[Separable space|separable]] inner product space has an orthonormal basis.

Using the [[Hausdorff maximal principle]] and the fact that in a [[Hilbert space|complete inner product space]] orthogonal projection onto linear subspaces is well-defined, one may also show that

'''Theorem.''' Any [[Hilbert space|complete inner product space]] has an orthonormal basis.

The two previous theorems raise the question of whether all inner product spaces have an orthonormal basis. The answer, it turns out is negative. This is a non-trivial result, and is proved below. The following proof is taken from Halmos's ''A Hilbert Space Problem Book'' (see the references).{{citation needed|date=October 2017}}

:{| class="toccolours collapsible collapsed" width="90%" style="text-align:left"
!Proof
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| Recall that the dimension of an inner product space is the [[cardinality]] of a maximal orthonormal system that it contains (by [[Zorn's lemma]] it contains at least one, and any two have the same cardinality). An orthonormal basis is certainly a maximal orthonormal system but the converse need not hold in general. If <math>G</math> is a dense subspace of an inner product space <math>V,</math> then any orthonormal basis for <math>G</math> is automatically an orthonormal basis for <math>V.</math> Thus, it suffices to construct an inner product space <math>V</math> with a dense subspace <math>G</math> whose dimension is strictly smaller than that of <math>V.</math>

Let <math>K</math> be a [[Hilbert space]] of dimension [[Aleph-null|<math>\aleph_0.</math>]] (for instance, <math>K = \ell^2(\N)</math>). Let <math>E</math> be an orthonormal basis of <math>K,</math> so <math>|E| = \aleph_0.</math> Extend <math>E</math> to a [[Basis (linear algebra)#Related notions|Hamel basis]] <math>E \cup F</math> for <math>K,</math>where <math>E \cap F = \varnothing.</math> Since it is known that the [[Hamel dimension]] of <math>K</math> is <math>c,</math> the cardinality of the continuum, it must be that <math>|F| = c.</math>

Let <math>L</math> be a Hilbert space of dimension <math>c</math> (for instance, <math>L = \ell^2(\R)</math>). Let <math>B</math> be an orthonormal basis for <math>L</math> and let <math>\varphi : F \to B</math> be a bijection. Then there is a linear transformation <math>T : K \to L</math> such that <math>T f = \varphi(f)</math> for <math>f \in F,</math> and <math>Te = 0</math> for <math>e \in E.</math>

Let <math>V = K \oplus L</math> and let <math>G = \{ (k, T k) : k \in K \}</math> be the graph of <math>T.</math> Let <math>\overline{G}</math> be the closure of <math>G</math> in <math>V</math>; we will show <math>\overline{G} = V.</math> Since for any <math>e \in E</math> we have <math>(e, 0) \in G,</math> it follows that <math>K \oplus 0 \subseteq \overline{G}.</math>

Next, if <math>b \in B,</math> then <math>b = T f</math> for some <math>f \in F \subseteq K,</math> so <math>(f, b) \in G \subseteq \overline{G}</math>; since <math>(f, 0) \in \overline{G}</math> as well, we also have <math>(0, b) \in \overline{G}.</math> It follows that <math>0 \oplus L \subseteq \overline{G},</math> so <math>\overline{G} = V,</math> and <math>G</math> is dense in <math>V.</math>

Finally, <math>\{(e, 0) : e \in E \}</math> is a maximal orthonormal set in <math>G</math>; if
<math display=block>0 = \langle (e, 0), (k, Tk) \rangle = \langle e, k \rangle + \langle 0, Tk \rangle = \langle e, k \rangle</math>
for all <math>e \in E</math> then <math>k = 0,</math> so <math>(k, Tk) = (0, 0)</math> is the zero vector in <math>G.</math> Hence the dimension of <math>G</math> is <math>|E| = \aleph_0,</math> whereas it is clear that the dimension of <math>V</math> is <math>c.</math> This completes the proof.
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[[Parseval's identity]] leads immediately to the following theorem:

'''Theorem.''' Let <math>V</math> be a separable inner product space and <math>\left\{e_k\right\}_k</math> an orthonormal basis of <math>V.</math> Then the map
<math display=block>x \mapsto \bigl\{\langle e_k, x \rangle\bigr\}_{k \in \N}</math>
is an isometric linear map <math>V \rightarrow \ell^2</math> with a dense image.

This theorem can be regarded as an abstract form of [[Fourier series]], in which an arbitrary orthonormal basis plays the role of the sequence of [[trigonometric polynomial]]s. Note that the underlying index set can be taken to be any countable set (and in fact any set whatsoever, provided <math>\ell^2</math> is defined appropriately, as is explained in the article [[Hilbert space]]). In particular, we obtain the following result in the theory of Fourier series:

'''Theorem.''' Let <math>V</math> be the inner product space <math>C[-\pi, \pi].</math> Then the sequence (indexed on set of all integers) of continuous functions
<math display=block>e_k(t) = \frac{e^{i k t}}{\sqrt{2 \pi}}</math>
is an orthonormal basis of the space <math>C[-\pi, \pi]</math> with the <math>L^2</math> inner product. The mapping
<math display=block>f \mapsto \frac{1}{\sqrt{2 \pi}} \left\{\int_{-\pi}^\pi f(t) e^{-i k t} \, \mathrm{d}t \right\}_{k \in \Z}</math>
is an isometric linear map with dense image.

Orthogonality of the sequence <math>\{ e_k \}_k</math> follows immediately from the fact that if <math>k \neq j,</math> then
<math display=block>\int_{-\pi}^\pi e^{-i (j - k) t} \, \mathrm{d}t = 0.</math>

Normality of the sequence is by design, that is, the coefficients are so chosen so that the norm comes out to 1. Finally the fact that the sequence has a dense algebraic span, in the {{em|inner product norm}}, follows from the fact that the sequence has a dense algebraic span, this time in the space of continuous periodic functions on <math>[-\pi, \pi]</math> with the uniform norm. This is the content of the [[Weierstrass approximation theorem|Weierstrass theorem]] on the uniform density of trigonometric polynomials.