Page 326 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Unit 30: Projections

From (1) and (2), we have Notes

x 2 Px 2 I P x 2

2
I P x 0  by hypothesis Px x
I P x 0

x Px 0

Px 0
This completes the proof of the theorem.
Theorem 3: If P is a projection on a Hilbert space H, then
(i) P is a positive operator i.e. P 0

(ii) 0 P 1

(iii) Px x x H.

(iv) P 1.

Proof: P, projection on H P 2 P,P * P.
Let M = range of P.

(i) Let x H. Then
Px,x PPx,x

*
Px,P x Px,Px Px 2 0
Px,x 0 x H.

P is a positive operator i.e. P 0.
(ii) P is a projection on H I – P is also a projection on H.
I – P 0. (by (i))

P I
But P 0, consequently 0 P 1.

(iii) Let x H. If M is the range of P, then M is the range of (I – P).

Now Px is in M and (I – P)x is in M .
Therefore Px and (I – P)x are orthogonal vector. So by Pythagorean theorem, we have

2 2 2
Px I P x Px I P x

2 2 2
x Px I P x Px+ I–P x x

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