91Ó°ÊÓ

Since T is invertible, we have T^(-1)T = I and TT^(-1) = I. Let S = T + D, where D = S - T. In this case, we have to show that there exists a δ > 0 such that if ‖D‖ < δ, then ‖S^(-1) - T^(-1)‖ < ε for any ε > 0.

We know T is invertible, and we want to show that for ‖D‖ sufficiently small, S is also invertible. In order to find the required δ, consider the following formula, which is derived from the Neumann series of invertible operators: (1 - D)T^(-1) = Sum(from k=0 to infinity) ((-DT^(-1))^k) We want to find a bound for ‖D‖, so that this series converges. By using the properties of the norm, we can bound this series by: Sum(from k=0 to infinity) (‖D‖^k)‖T^(-1)‖^k < Infinity Using the geometric series, this will converge if ‖D‖ < 1/‖T^(-1)‖ and will also guarantee that S = T + D is invertible. Let δ = 1/‖T^(-1)‖.

Now, we can use the bound on ‖D‖ and write S^(-1) in terms of T^(-1): S^(-1) = (1 - D)T^(-1)Sum(from k=0 to infinity) ((-DT^(-1))^k) Now, we need to show that for ‖D‖ < δ, ‖S^(-1) - T^(-1)‖ < ε. To do this, consider the following expression: ‖S^(-1) - T^(-1)‖ = ‖(1 - D)T^(-1)Sum(from k=0 to infinity) ((-DT^(-1))^k) - T^(-1)‖ Using the properties of norms, we can bound this by: ‖S^(-1) - T^(-1)‖ ≤ ‖(1 - D)T^(-1)‖ * Sum(from k=1 to infinity) ‖(-D)T^(-1)‖^k We know that ‖D‖ < δ and δ = 1/‖T^(-1)‖, so: ‖S^(-1) - T^(-1)‖ ≤ ‖T^(-1)‖ * Sum(from k=1 to infinity) (‖D‖^k)‖T^(-1)‖^k Now, we want this to be less than ε for some ε > 0. We have a geometric series, so choosing ε = ‖T^(-1)‖(2δ) suffices: ‖S^(-1) - T^(-1)‖ ≤ ‖T^(-1)‖ * Sum(from k=1 to infinity) (‖D‖^k)‖T^(-1)‖^k = ‖T^(-1)‖ * (2δ) Thus, we have shown that if ‖D‖ < δ, then the distance between S^(-1) and T^(-1) is less than ε. Therefore, the function T ↦ T^(-1) is continuous on Inv(V).