Two Functional Analysis Questions

[Combinatorics]

Homework Statement

1. Given an operator [tex]H[/tex] , and a sequence [tex]\{ H_n \} _{n\geq 1 } [/tex] in an arbitrary Hilbert Space , such that both [tex]H[/tex] and [tex] H_n [/tex] are self-adjoint .

How can I prove that if [tex]||(H_n+i)^{-1} - (H+i) ^ {-1} || \to 0 [/tex] and if [tex]H[/tex] has an isolated eigenvalue [tex]\lambda[/tex] of multiplicity one, then for large enough [tex]n[/tex], [tex]H_n[/tex] also have isolated eigenvalues [tex]\lambda _ n[/tex] of multiplicty one that converge to [tex]\lambda[/tex].

2. Given an operator [tex]H[/tex] , and a sequence [tex]\{ H_n \} _{n\geq 1 } [/tex] in an arbitrary Hilbert Space , such that both [tex]H[/tex] and [tex] H_n [/tex] are self-adjoint and non-negative.

How can I prove that [tex]||(H_n+1)^{-1} - (H+1) ^ {-1} || \to 0 [/tex] is equivalent to
[tex]||(H_n+i)^{-1} - (H+i) ^ {-1} || \to 0 [/tex] ?

Homework Equations

The Attempt at a Solution

I really have no idea about it... I assume it has something to do with the self-adjointness...maybe some estimations on the inner-product...but can't figure out what

Thanks in advance

 

[mighty2000]

for any help!

Thank you for your interesting question. I am always happy to help others understand complex concepts and solve challenging problems. Here is my attempt at a solution to your questions:

1. To prove that H_n has isolated eigenvalues \lambda_n of multiplicity one that converge to \lambda, we can use the following steps:

Step 1: First, let us consider the resolvent operator R_n = (H_n + i)^{-1}. As both H_n and H are self-adjoint, their resolvent operators are also self-adjoint. Therefore, we can write R_n = (H_n + i)^{-1} = (H_n - i)^{*} = (H_n^{*} - i)^{-1}.

Step 2: Now, let us consider the difference between the resolvent operators of H and H_n, i.e., ||R_n - R|| = ||(H_n + i)^{-1} - (H + i)^{-1}||.

Step 3: Using the triangle inequality, we can write ||R_n - R|| \leq ||R_n - R + \lambda_n R_n - \lambda_n R|| + ||\lambda_n R_n - \lambda_n R||.

Step 4: Since ||R_n - R|| \to 0, we can choose a large enough n such that ||R_n - R|| < \frac{\epsilon}{2}, where \epsilon is a small positive number.

Step 5: Now, using the fact that H has an isolated eigenvalue \lambda of multiplicity one, we can write ||\lambda_n R_n - \lambda_n R|| < \frac{\epsilon}{2} for large enough n.

Step 6: Combining steps 4 and 5, we get ||R_n - R + \lambda_n R_n - \lambda_n R|| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.

Step 7: Finally, using the fact that R_n and R are self-adjoint, we can write ||R_n - R + \lambda_n R_n - \lambda_n R|| = ||(R_n - R)(I + \lambda_n)|| = ||R_n - R|| ||I + \lambda_n||, where I is the identity operator.