Page 276 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Unit 25: Orthonormal Sets

Hence (x – (x, e ) e , e) = (x, e) = 0 since e S. Notes
i i j j j
n
If e S, then (s , e) = (x, e ) e , e … (5)
j n j i i j
i 1
n
If n > j, we get (x, e ) e , e j = (x, e) j … (6)
i
i
i 1
From (5) & (6), we get

lim (s , e ) = (x, e).
n n j j
So, in this case
(x – (x, e ) e , e) = (x, e) – (x, e ) = 0
i i j j j
Thus (x – (x, e ) e , e) = 0 for each j.
i i j
Hence x – (x, e ) e e for each j.
i i j
This completes the proof of the theorem.
Theorem 5: A Hilbert space H is separable every orthonormal set in H is countable.

Proof: Let H be separable with a countable dense subset D so that H = D .
Let B be an orthonormal basis for H.
We show that B is countable.

For x, y B, x y, we have
x – y 2 = x + y = 2
2
2
Hence the open sphere
1 1
S x; z : z x = as x B are all disjoint.
2 2
1
Since D is dense, D must contain a point in each S x, .
2
Hence if B is uncountable, then B must also be uncountable and H cannot be separable contradicting
the hypothesis. Therefore B must be countable.

Conversely, let B be countable and let B = {x , x , …}. Then H is equal to the closure of all finite
1 2
linear combinations of element of B. That is H = L(B) . Let G be a non-empty open subset of H.
n
Then G contains an element of the form a x with a C. We can take a C. We can take a
i i i i i
i 1
to be complex number with real and imaginary parts as rational numbers. Then the set

n
D = a x , n 1, 2, , a rational
i
i
i
i 1
is a countable dense set in H and so H is separable.
This completes the proof of the theorem.

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