Inner Product Space - Orthonormal Sequences

Orthonormal Sequences

Let V be a finite dimensional inner product space of dimension n. Recall that every basis of V consists of exactly n 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 is orthonormal if if and for each i.

This definition of orthonormal basis generalizes to the case of infinite-dimensional inner product spaces in the following way. Let V be any inner product space. Then a collection is a basis for V if the subspace of V generated by finite linear combinations of elements of E is dense in V (in the norm induced by the inner product). We say that E is an orthonormal basis for V if it is a basis and if and for all .

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

Theorem. Any separable inner product space V has an orthonormal basis.

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

Theorem. Any complete inner product space V 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).

Proof
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 as we shall see, the converse need not hold. Observe that if G is a dense subspace of an inner product space H, then any orthonormal basis for G is automatically an orthonormal basis for H. Thus, it suffices to construct an inner product space space H with a dense subspace G whose dimension is strictly smaller than that of H. Let K be a Hilbert space of dimension (for instance, ). Let E be an orthonormal basis of K, so . Extend E to a Hamel basis for K, where . Since it is known that the Hamel dimension of K is c, the cardinality of the continuum, it must be that . Let L be a Hilbert space of dimension c (for instance, ). Let B be an orthonormal basis for L, and let be a bijection. Then there is a linear transformation such that for, and for . Let and let be the graph of T. Let be the closure of G in H; we will show . Since for any we have, it follows that . Next, if, then for some, so ; since as well, we also have . It follows that, so, and G is dense in H. Finally, is a maximal orthonormal set in G; if for all then certainly, so is the zero vector in G. Hence the dimension of G is, whereas it is clear that the dimension of H is c. This completes the proof.

Proof

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 as we shall see, the converse need not hold. Observe that if G is a dense subspace of an inner product space H, then any orthonormal basis for G is automatically an orthonormal basis for H. Thus, it suffices to construct an inner product space space H with a dense subspace G whose dimension is strictly smaller than that of H.

Let K be a Hilbert space of dimension (for instance, ). Let E be an orthonormal basis of K, so . Extend E to a Hamel basis for K, where . Since it is known that the Hamel dimension of K is c, the cardinality of the continuum, it must be that .

Let L be a Hilbert space of dimension c (for instance, ). Let B be an orthonormal basis for L, and let be a bijection. Then there is a linear transformation such that for, and for .

Let and let be the graph of T. Let be the closure of G in H; we will show . Since for any we have, it follows that .

Next, if, then for some, so ; since as well, we also have . It follows that, so, and G is dense in H.

Finally, is a maximal orthonormal set in G; if

for all then certainly, so is the zero vector in G. Hence the dimension of G is, whereas it is clear that the dimension of H is c. This completes the proof.

Parseval's identity leads immediately to the following theorem:

Theorem. Let V be a separable inner product space and {e_k}_k an orthonormal basis of V. Then the map

is an isometric linear map V → ℓ 2 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 polynomials. Note that the underlying index set can be taken to be any countable set (and in fact any set whatsoever, provided ℓ 2 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 V be the inner product space . Then the sequence (indexed on set of all integers) of continuous functions

is an orthonormal basis of the space with the L2 inner product. The mapping

is an isometric linear map with dense image.

Orthogonality of the sequence {e_k}_k follows immediately from the fact that if k ≠ j, then

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 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 with the uniform norm. This is the content of the Weierstrass theorem on the uniform density of trigonometric polynomials.

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