Page 272 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Unit 25: Orthonormal Sets

Theorem 2: If {e } is an orthonormal set in a Hilbert space H and if x is any vector in H, then the Notes
i
set S = {e : (x, e ) 0} is either empty or countable.
i i
Proof: For each positive integer n, consider the set
2
x
2
S = e : (x, e ) .
n i i n
If the set S contains n or more than n’ vectors, then we must have
n
2
2 x
(x, e ) > n x … (1)
i
n
e i S n
By theorem (1), we have
2
(x, e ) x 2 … (2)
i
e i S n
which contradicts (1).

Hence if (2) were to be valid, S should have at most (n – 1) elements. Hence for each positive n,
n
the set S is finite.
n
2
Now let e S. Then (x, e ) 0. However small may be the value of |(x, e )| , we can take n so
i i i
large that
2
x
|(x, e )| > .
2
i
n
Therefore if e S, then e must belong to some S . So, we can write S =  S .
i i n n
n 1
S can be expressed as a countable union of finite sets.
S is itself a countable set.
If (x, e ) = 0 for each i, then S is empty. Otherwise S is either a finite set or countable set.
i
This completes the proof of the theorem.

2
Theorem 3: Bessel’s Inequality: If {e } is an orthonormal set in a Hilbert space H, then |(x, e )|
i i
x for every vector x in H.
2
Proof: Let S = {e : (x, e ) 0}.
i i
By theorem (2), S is either empty or countable.
If S is empty, then (x, e ) = 0 i.
i
2
So if we define |(x, e )| = 0, then
i
|(x, e )| = 0 x .
2
2
i
Now let S is not empty, then S is finite or it is countably infinite.
If S is finite, then we can write S = {e , e , …, e } for some positive integer n.
1 2 n
In this case, we have
n
2
|(x, e )| = |(x, e )| x 2 … (1)
2
i i
i 1
which represents Bessel’s inequality in the finite case.

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