Page 314 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Unit 29: Normal and Unitary Operators

Let T be any limited point of M. Then to show that T M i.e. to show that T is a normal operator Notes
on H.
Since T is a limited point of M, therefore a sequence T of distinct point of M such that
n
T T. We have
n
T * n T * T n T * T n T 0.

T * n T * 0 T * n T *.

Now, TT * T * T TT * T T * T T * n T * T TT * T T * n T T * n T * T
n
n
n
n
TT * T T * T T * n T * T n T * T n T * T
n
n
n
n

TT * T T * n T T * n T * T n T * T n T * T
n
n
n
n
TT * T T * n 0 T * T n T * T
n
n
 T n M T is a normal operator on H i.e. T T* T* T and 0 0
n
n
n
n
n
TT * T T * T * T T * T 0 since T T and T * T *
n n n n n n
Thus, TT * T * T 0 TT * T * T 0
TT* T * T T is normal operator on H.
T M and so M is closed.
Now every self adjoint operator is normal. Therefore the set M contains the set of all self-adjoint
operators on H.
Finally, we show that M is closed with respect to scalar multiplication i.e. T M
T M, is any scalar.
In other words, we are to show that if T is a normal operator on H and is any scalar, then T
is normal operator on H. Since T is normal, therefore TT* =T*T.

We have T * T *.

Now T T * T T * TT * .

Also T * T T T T * T ( )(TT*)

T T * T * T

T is normal.
This completes the proof of the theorem.
Theorem 3: If N , N are normal operators on a Hilbert space H with the property that either
1 2
commutes with the adjoint of the other, then N + N and N N are also normal operators.
1 2 1 2
Proof: Since N , N are normal operators, therefore
1 2
N N* = N* N and N N* 2 N* N 2 ...(1)
1
1
1
2
2
1
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