Page 327 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Measure Theory and Functional Analysis

Notes 2 2
x Px
Px x .

(iv) We have P sup Px : x 1

But Px x x H (by(iii))

sup Px : x 1 1
Hence P 1

This completes the proof of the theorem.

Example: If P and Q are the projections on closed linear subspaces M and N of H. Show
that PQ is a projection PQ = QP. In this case, show that PQ is the projection on M N.

Solution: Since P and Q are projections on H, therefore P = P, P* = P, Q = Q, Q* = Q. Also it is given
2
2
that M is range of P and N is the range of Q.
Now suppose PQ is projection on H. Then to prove that PQ = QP.
Since PQ is a projection on H.
(PQ)* = PQ
Q* P* = PQ

QP = PQ ( Q* = Q, P* = P)
Conversely, let PQ = QP. We shall show that PQ is a projection on H.
We have (PQ)* = Q*P* = QP = PQ.
2
Also (PQ) = (PQ) (PQ) = (PQ) (QP)
2
= PQ P = PQP
= QPP = QP 2
= QP = PQ

Thus (PQ)* = PQ and (PQ) = PQ.
2
PQ is a projection on H.
Finally we are to show that PQ is the projection on M N, i.e. we are to show that range of PQ
is M N.
Let R (PQ) = range of PQ.
Let x M N x M, x N we have
(PQ) (x) = P (Qx) = Px [ N is range of Q and x N Qx = x]

= x [ M is range of P and x P]
(PQ)x = x
x R (PQ)

x M N x R (PQ)

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