For a Banach space $X$ and a bounded linear operator $T\in \mathscr{L}(X)$ with $\|T\|< 1$ the operator $I - T$ is bijective

Question

Let $X$ be a Banach space and let $T\colon X\to X$ be a bounded linear operator with $\|T\|<1$. Why is $I-T$ one-to-one and surjective? Why does it have a continuous inverse?

Can anyone please explain?

Answer 1

That's not necessarily the case. The most obvious situation occurs when $T=I$, and then $I-T=0$.

Answer 2

You need norm strictly less than one, since $I$ has norm one, and $I - I = 0$ is not onto.

The spectral radius formula tells us that if $T$ is a bounded linear operator, and $\lambda \in \mathbb{C}$ with $\lambda I - T$ not invertible, then $|\lambda| \leq ||T^k||^{1/k}$ for all $k \in \mathbb{N}$. Thus if $||T|| < 1$, no element outside the open unit disc can be in the spectrum of $T$, and so $I - T$ must be invertible, i.e. bijective with bounded inverse.

Answer 3

Suppose $\|T\|<1$, consider the sequence $U_n=I+T+...+T^n$, for every $x\in X$, $\|U_{n+m}(x)-U_n(x)\|\leq \|T\|^{n+1}\|(\|x\|+\|T\|\|x\|+..\|T\|^{m-1}\|x\|)\leq \|T\|^n({1\over{1-\|T\|}})$. This implies that $U_n(x)$ is a Cauchy sequence. Since $X$ is a Banach space there $U(x)=lim_nU(x)$. $U$ is linear and bounded: $\|U(x)\|\leq \sum_{n\geq 0}\|T\|^n\|x\|\leq {1\over{1-\|T\|}}\|x\|$.

$U$ is the inverse of $I-T$, $((I-T)U-I)(x)=lim_n(I-T)\sum_{n\geq 0}T^n(x)-x=lim_{n}-T^{n+1}(x)=0$ since $\|T^n(x)\|\leq \|T^n\|x\|$ and $\|T\|<1$. A similar argument shows that $U(I-T)=I$.

Answer 4

You can use the Contraction Mapping Principle: $$|| Tx - Ty|| = || T(x-y)|| \leq ||T|| ||x-y||$$

Since $||T||<1$, this mapping is a contraction and therefore there exists a unique fixed point. Also, notice that $0$ is a fixed point since $T$ is linear.

To prove that $I-T$ is onto, for every $y\in X$ you can define $Lx = Tx + y$ and use the same idea.