Page 274 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Unit 25: Orthonormal Sets

Hence in this case, we have to show x e for each j. Notes
j
Since S is empty, (x, e) = 0 for every j.
j
x e for every j.
j
Now let S is not empty. Then S is either finite or countably infinite. If S is finite, let

S = {e , e , …, e } and we define
1 2 n
n
(x, e ) e = (x, e ) e ,
i
i
i i
i 1
and prove that x (x, e ) e e for each j. This result follows from (2) of theorem (1).
i i j
n 1
Finally let S be countably infinite and let
S = {e , e , …, e , …}
1 2 n
Let s = (x, e ) e
n i i
i 1
2
m
For m > n, s – s 2 = (x, e )e
m n i i
i n 1
m
2
= (x, e )
i
i n 1
2
By Bessel’s inequality, the series (x, e ) is convergent.
n
n 1
2
Hence (x, e ) as m, n .
i
i n 1
s – s 0 as m, n .
2
m n
(s ) is a Cauchy sequence in H.
n
Since H is complete s s H. Now s H can be written as
n

s = (x, e ) e n
n
n 1

Now we can define (x, e ) e = (x, e ) e n .
n
i i
n 1
Before, completing the proof, we shall show that the above sum is well-defined and does not
depend upon the rearrangement of vectors.

For this, let the vector in S be arranged in a different manner as
S = {f , f , f , …, f , …}
1 2 3 n

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