Let S = I + TT H rightarrow H, where T is linear and bounded. Show.pdf
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The document demonstrates that the operator s defined as s = i + t*t, where t is a bounded linear operator, is also a bounded linear operator. It proves that the inverse operator s^-1 exists, establishing that s: h --> h is invertible by showing that s(h) can recover the original element h in the Hilbert space. The argument relies on the properties of bounded linear operators and the identity operator.
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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