Let S = I + T*T: H rightarrow H, where T is linear and bounded. Show that S^-1: S(H)
rightarrow H exists.
Solution
T is a bounded linear operator. Hence, T*T is also a bounded linear operator as composition of
two bounded linear operator is a bounded linear operator.
I is the identity operator. Hence, it is bounded linear operator. So, S = I + T*T is a bounded
linear operator as sum of two bounded linear operator is a bounded linear operator.
Now, S : H --> H is a bounded linear operator.
The identity operator I : H --> H can be written as S-1S : H --> H as I = S-1S.
Let, h H. Then I(h) = h
This can be written as S-1S(h) = h
Or, S-1(S(h)) = h
As h H and S : H --> H is a bounded linear operator, S(h) S(H). Let S(h) = h1 S(H)
So, S-1(h1) = h where h H and h1 S(H)
As h is arbitrary, we can say that S-1 : S(H) --> H exists. Proved.
Transparency, Recognition and the role of eSealing - Ildiko Mazar and Koen No...
Let S = I + TT H rightarrow H, where T is linear and bounded. Show.pdf
1. Let S = I + T*T: H rightarrow H, where T is linear and bounded. Show that S^-1: S(H)
rightarrow H exists.
Solution
T is a bounded linear operator. Hence, T*T is also a bounded linear operator as composition of
two bounded linear operator is a bounded linear operator.
I is the identity operator. Hence, it is bounded linear operator. So, S = I + T*T is a bounded
linear operator as sum of two bounded linear operator is a bounded linear operator.
Now, S : H --> H is a bounded linear operator.
The identity operator I : H --> H can be written as S-1S : H --> H as I = S-1S.
Let, h H. Then I(h) = h
This can be written as S-1S(h) = h
Or, S-1(S(h)) = h
As h H and S : H --> H is a bounded linear operator, S(h) S(H). Let S(h) = h1 S(H)
So, S-1(h1) = h where h H and h1 S(H)
As h is arbitrary, we can say that S-1 : S(H) --> H exists. Proved
