Perturbation of Identity is isomorphism Let $X$ be a Banach space and $K:X\to X$ be a linear, bounded operator with $\Vert K\Vert<1$. Show that $(I-K)^{-1}$ exists and is bounded.
Let $\Vert K\Vert = 1-\varepsilon$. Then we get $$\varepsilon\Vert x\Vert \leq \Vert (I-K)x\Vert \leq (2-\varepsilon)\Vert x\Vert,$$ so $I-K$ is bounded and injective. If we have surjectivity, it follows that $I-K$ is an isomorphism since $X$ is Banach. While it seems clear that it should be surjective, I have so far failed to prove it.
 A: Consider the sequence
$S_n = \displaystyle \sum_0^n K^i = I + K + K^2 + \ldots + K^n \in \mathcal B(X), \tag 1$
where $\mathcal B(X)$ is the Banach space of bounded linear operators $X \to X$, with the operator norm.  I claim $S_n$ is Cauchy in $\mathcal B(X)$; for with $m > n$ we have
$S_m - S_n = \displaystyle \sum_0^m K^i - \sum_0^n K^i = \sum_{n + 1}^m K^i = K^{n + 1} \sum_0^{m - n - 1} K^i, \tag 2$
whence
$\Vert S_m - S_n \Vert = \left \Vert  K^{n + 1} \displaystyle \sum_0^{m - n - 1} K^i \right \Vert \le \Vert K \Vert^{n + 1} \left \Vert \displaystyle \sum_0^{m - n - 1} K^i \right \Vert \le \Vert K \Vert^{n + 1} \displaystyle \sum_0^{m - n - 1} \Vert K \Vert^i; \tag 3$
now since
$\Vert K \Vert < 1, \tag 4$
it follows that the strictly increasing geometric series
$L = \displaystyle \sum_0^\infty \Vert K \Vert ^i \tag 5$
converges; furthermore,  we have
$L - 1 = \displaystyle \sum_1^\infty \Vert K \Vert^i = \Vert K \Vert \sum_0^\infty \Vert K \Vert^i = \Vert K \Vert L, \tag 6$
whence
$L(1 - \Vert K \Vert) = 1, \tag 7$
or
$L = \dfrac{1}{1 - \Vert K \Vert}; \tag 8$
incorporating formulas (5)-(8) into (3) yields
$\Vert S_m - S_n \Vert \le \Vert K \Vert^{n + 1} \displaystyle \sum_0^{m - n - 1} \Vert K \Vert^i \le  \Vert K \Vert^{n + 1} \sum_0^\infty \Vert K \Vert^i = \Vert K \Vert^{n + 1} L = \dfrac{\Vert K \Vert^{n + 1}}{1 - \Vert K \Vert}; \tag 9$
since $\Vert K \Vert < 1$, the right-hand side of this equation may be made arbitrarily small by taking $n$ sufficiently large; which shows that $S_n$ is indeed Cauchy in $\mathcal B(X)$; since $X$ is Banach, so is $\mathcal B(X)$ in the operator norm, hence $S_n$ is possessed of a limit $S \in \mathcal B(X)$; and indeed, said limit is in fact 
$S = \displaystyle \sum_0^\infty K^i, \tag{10}$
since, in similarity to (9),
$\Vert S - S_n \Vert = \left \Vert \displaystyle \sum_{n + 1}^\infty K^i \right \Vert \le  \displaystyle \sum_{n + 1}^\infty \Vert K \Vert^i = \Vert K \Vert^{n + 1} \sum_0^\infty \Vert K \Vert^i = \dfrac{\Vert K \Vert^{n + 1}}{1 - \Vert K \Vert} \to 0 \; \text{as} \; n \to \infty. \tag{11}$
Finally, we have
$(I - K)S_n = (I - K) \displaystyle \sum_0^n K^i = \sum_0^n K^i - \sum_1^{n + 1}K^i = I - K^{n + 1}; \tag{12}$
thus,
$\Vert (I - K)S - I \Vert = \displaystyle \lim_{n \to \infty} \Vert (I - K)S_n - I \Vert = \lim_{n \to \infty} \Vert K^{n + 1} \Vert \le \lim_{n \to \infty} \Vert K \Vert^{n + 1} = 0, \tag{13}$
and we see that, by virtue of the fact that $KS = SK$ (cf. (10)),
$S(I - K) = (I - K)S = I, \tag{14}$
or
$(I - K)^{-1} = S = \displaystyle \sum_0^\infty K^i. \tag{15}$
It is easy to see that $S$ is bounded, since
$\Vert S \Vert = \left \Vert \displaystyle \sum_0^\infty K^i \right \Vert \le \displaystyle \sum_0^\infty \Vert K \Vert^i = L. \tag{16}$
This whole spiel is of course a re-iteration of the standard theory.
In closing, I have a brief query of my own about our OP Analysis801's question; I would like to know how the left-most inequality in
$\epsilon \Vert x \Vert \le \Vert (I - K)x \Vert \le (2 - \epsilon) \Vert x \Vert \tag{17}$
comes about.  I can derive the right-hand inequality myself; from
$\Vert K \Vert = 1 - \epsilon, \tag{18}$
we have
$\Vert I - K \Vert \le \Vert I \Vert + \Vert K \Vert = 1 + 1 - \epsilon = 2 - \epsilon; \tag{19}$
thus,
$\Vert (I - K) x \Vert \le \Vert I - K \Vert \Vert x \Vert \le (2 - \epsilon) \Vert x \Vert; \tag{21}$
what I haven't been able to figure out is why
$\epsilon \Vert x \Vert \le \Vert (I - K)x \Vert; \tag{22}$
the best I have been able to manage is, 
$\epsilon \Vert x \Vert = (1 - (1 - \epsilon)) \Vert x \Vert = \vert \Vert I \Vert - \Vert K \Vert \vert \Vert x \Vert \le \Vert I - K \Vert \Vert x \Vert, \tag{23}$
but I haven't been able to show that
$\epsilon \Vert x \Vert \le \Vert (I - K)x \Vert; \tag{24}$
so if anyone wants to leave remarks, proofs or counterexamples in the comments, I would be most appreciative.
