Re: inverse opeerator

From: David C. Ullrich (ullrich_at_math.okstate.edu)
Date: 01/26/05 

Date: Wed, 26 Jan 2005 05:44:23 -0600

On 25 Jan 2005 22:37:58 -0800, saemtuarg@yahoo.it (Saem T.) wrote:

>>Saying "||A-B||_{L(X,Y)} -> 0" doesn't really make sense.
>
>why??
>
>>Yes, it's true that if ||A-B||_{L(X,Y)} is small enough then
>>I-(A-B)A^{-1} is an invertible operator - this has nothing
>>to do with anything except the fact that if ||T|| < 1 then
>>I - T is invertible.
>
>May you explain me the reason for wich ||T|| < 1 then
>I - T is invertible, please ?

Because then the series I + T + T^2 + T^3 + ... converges
in norm.

>>>(note that I = I_Y : Y --> Y denotes the identity operator on Y.)
>>>
>>>May you help me please ?
>
>>With what?
>
>?
>
>>David C. Ullrich
>
>ST

************************

David C. Ullrich

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