Completeness

Theorem 1. If is any finite orthonormal set in an inner product space, if is any vector, and if , then (Bessel’s inequality) The vector is orthogonal to each and, consequently, to the subspace spanned by .

Proof. For the first assertion: for the second assertion: ◻

Theorem 2. If is any finite orthonormal set in an inner product space , the following six conditions on are equivalent to each other.

The orthonormal set is complete.
If for , then .
The subspace spanned by is the whole space .
If is in , then .
If and are in , then (Parseval’s identity)
If is in , then

Proof. We shall establish the implications

(1) (2) (3) (4) (5) (6) (1).

Thus we first assume (1) and prove (2), then assume (2) to prove (3), and so on till we finally prove (1) assuming (6).

(1) (2). If for all and , then we may adjoin to and thus obtain an orthonormal set larger than .

(2) (3). If there is an that is not a linear combination of the , then, by the second part of Theorem 1, is different from and is orthogonal to each .

(3) (4). If every has the form , then

(4) (5). If and , with and , then

(5) (6). Set .

(6) (1). If were contained in a larger orthogonal set, say if is orthogonal to each , then so that . ◻

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