Page 343 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Measure Theory and Functional Analysis

Notes (a) The M ’s are pairwise orthogonal and span H:
i
(b) The P ’s are pairwise orthogonal and P + P + … + P = I and T = P + P + … + P .
i 1 2 m 1 1 2 2 m m
(c) T is normal operator on H.
Proof: We shall show that

(i) (ii) (iii) (i)
(i) (ii)
Assume that M ’s are pairwise orthogonal and span H. Hence every x H can be represented
i
uniquely as
x = x + x + … + x … (1)
1 2 m
where x M for i = 1, 2, …, m
i i
by hypothesis M ’s are pairwise orthogonal. Since P ’s are projections in M ’s P ’s are pairwise
i i i i
orthogonal, i.e. i j P P = 0.
i j
If x is any vector in H, then from (1) for each i,
P (x) = P (x + x + … + x )
i i 1 2 m
= P x + P x + … + Px … (2)
i 1 i 2 i m
Since P is the range of M , P x = x .
i i i i i
For i j M M since x M for each j we have
j i j j
x M for j i.
j i
Hence x M (null space of P )
j i i
x M P x = x
j i i j i
from (2) we get
Px = x … (3)
i i
Since I is the identity mapping on H, we get
Ix = x + x + … + x … (by (1))
1 2 m
= P x + P x + … + P x … (by (3))
1 2 m
= (P + P + … + P ) x x H.
1 2 m
This show that I = P + P + …… + P .
1 2 m
For every x H, we have from (1)
T (x) = T (x + x + … + x )
1 2 m
= Tx + Tx + … + Tx
1 2 m
Since x M Tx = x
i i i i
T = x + x + …… + x … (4)
x 1 1 2 2 m m
= P x + P x + …… + P x … (5)
1 1 2 2 m m
T = P + P + …… + P
1 1 2 2 m m
(ii) (iii)

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