Surjectivity of $Id-A$ for linear operator $A$ on Banach space with $\|A\|<1$

Question

Let $X$ be Banach space and $A:X\rightarrow X$ linear opeartor such that $\|A\|<1$. It is clear that $Id-A$ is injective. Why is it also surjective?

Answer 1

Hint Prove that $Id+A+A^2+..+A^n+...$ defines an operator which is the inverse of $Id-A$.

Answer 2

You know that $B(X)$, the space of bounded linear operators on $X$ is a Banach space. Consider the series $$ \sum_{n=0}^{\infty} A^n $$ Since $\|A\| < 1$, it converges absolutely, and thus converges in $B(X)$ to an operator $B$. Now check that $B(I-A) = (I-A)B = I$ and so $(I-A)$ must be surjective.

Answer 3

That $I - A$ is both injective and surjective follow from the fact that it is invertible. Indeed, let $B:X \to X$ be any invertible operator; then we have an operator $B^{-1}$ such that

$BB^{-1} = B^{-1}B = I. \tag{1}$

To see that (1) implies $B$ is injective, suppose that

$Bx_1 = Bx_2 \tag{2}$

for some $x_1, x_2 \in X$. Then from (1)

$x_1 = Ix_1 = B^{-1}Bx_1 = B^{-1}Bx_2 = Ix_2 = x_2, \tag{3}$

so $B$ is injective. To see surjectivity, for any $y \in X$ set $x = B^{-1}y$; then again from (1)

$Bx = BB^{-1}y = Iy = y; \tag{4}$

surjectivity proved.

It remains to show $I - A$ is invertible. But it is well-known that since $\Vert A \Vert < 1$, the series

$\sum_0^\infty A^n \tag{5}$

converges (it is majorized by $\sum_0^\infty \Vert A \Vert^n$, a convergent real geometric series). Indeed we have

$(I - A)(\sum_0^\infty A^n) = (\sum_0^\infty A^n)(I - A) = I, \tag{6}$

also well-known. (6) follows from the algebraic identity

$(I - A)(\sum_0^m A^n) = I - A^{m + 1} \tag{7}$

by letting $m \to \infty$ which forces $A^{m + 1} \to 0$, since $\Vert A^{m + 1} \Vert \le \Vert A \Vert^{m + 1} \to 0$. This is a very well-known argument which has often been used here on MSE; see my answer to this question for a detailed disussion. (6) states that the inverse of $I - A$ is $\sum_0^\infty A^n$. Since $(I - A)^{-1}$ exists, the discussion of the previous paragraph shows that $I - A$ is both injective and also surjective, as sought.