Prove that operator is surjective.

Question

Take a sequence of bounded operators $S_n \in \mathcal{B}(X,X),$ where $X$ is a Banach space. Suppose that $S_n \rightarrow I,$ in the operator norm, for $n \to \infty.$ Then It´s easy to check that $S_n$ is eventually injective. I would like to prove that it´s also surjective, because I would like the inverse to be in $\mathcal{B}(X,X)$. Is it true? In particular I need this to prove that every uniformly continuous operator semigroup can be written as exponential of some operator (which i need to be defined on all of $X$).

Answer 1

If $S_n \to I$, then $\|I-S_n\| <1$ when $n$ large. So

$$S_n = I - (I-S_n)$$

is invertible. Indeed,

$$ S_n^{-1} = (I- (I -S_n))^{-1} = I + (I-S_n) + (I-S_n)^2 + \cdots + (I-S_n)^k + \cdots $$

Answer 2

The invertible elements form an open set of $\mathcal{B}(X,X)$.