Page 273 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Measure Theory and Functional Analysis

Notes If S is countable infinite, let S be arranged in the definite order such as {e , e , …, e , …}.
1 2 n
In this case we can write

2
|(x, e )| = |(x, e )| 2 … (2)
i n
n 1
The series on the R.H.S. of (2) is absolutely convergent.
Hence every series obtained from this by rearranging the terms is also convergent and all such
series have the same sum.

2
2
Therefore, we define the sum |(x, e )| to be |(x, e )| .
i n
n 1
2
Hence the sum of |(x, e )| is an extended non-negative real number which depends only on S
i
and not on the rearrangement of vectors.
Now by Bessel’s inequality in the finite case, we have
n
|(x, e )| 2 x 2 … (3)
i
i 1
For various values of n, the sum on the L.H.S. of (3) are non-negative. So they form a monotonic
2
increasing sequence. Since this sequence is bounded above by x , it converges. Since the
sequence is the sequence of partial sums of the series on the R.H.S. of (2), it converges and we
have e S,
i

2
2
|(x, e )| = |(x, e )| x 2
i i
n 1
This completes the proof of the theorem.
2
Note: From the Bessel’s inequality, we note that the series |(x, e )| is convergent series.
n
n 1
Corollary: If e S, then (x, e ) 0 as n .
n n

2
Proof: By Bessel’s inequality, the series |(x, e )| is convergent.
n
n 1
2
Hence |(x, e )| 0 as n .
n
(x, e ) 0 as n .
n
Theorem 4: If {e } is an orthonormal set in a Hilbert space H and x is an arbitrary vector in H, then
i
{x – (x, e ) e } e for each j.
i i j
Proof: Let S = {e : (x, e ) 0}
i i
Then S is empty or countable. [See theorem (2)]
If S is empty, then (x, e ) = 0 for every i.
i
In this case, we define (x, e ) e to be a zero vector and so we get
i i
x – (x, e ) e = x – 0 = x.
i i

266 LOVELY PROFESSIONAL UNIVERSITY