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Suppose $T$ is a bounded linear operator on a banach space $X$, such that $\|I-T\| < 1$. Show that $T$ has an inverse and it is bounded.

To show that an inverse exists, we have to show that $T$ is injectice and surjective. Now, I can prove tha injective part, but cannot prove the surjectivity, and how is the inverse bounded.

I know a similar question was asked here but the answer there did not help at all. I also cannot ask for an explanation in the comments either because my reputation is not yet 50.

Edit: $I$ is the identity operator.

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Lemma: If $A:X\to X$ is a bounded operator defined on a Banach space and $||A||<1$ then $I-A$ is invertible and the inverse is bounded.

Proof: For all $k\in\mathbb{N}$ we have $||A^k||\leq ||A||^k$. Since $||A||<1$ the series $\sum_{k=0}^\infty ||A^k||$ converges. Since $X$ is a Banach space we know that the space of bounded linear operators $L(X,X)$ is also a Banach space. And it is well known that in a Banach space an absolutely convergent series is convergent. Hence $\sum_{k=0}^\infty A^k$ converges to an element $S\in L(X,X)$. Now we can define the sequence of partial sums $S_n=\sum_{k=0}^n A^k$. We have:

$(I-A)S_n=S_n(I-A)=I-A^{n+1}$

By taking $n\to\infty$ we get $(I-A)S=S(I-A)=I$. So $S$ is an inverse of $I-A$, hence the inverse exists and it is bounded.

Well, now in your case just take $A=I-T$. Then $I-A=I-(I-T)=T$.

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  • $\begingroup$ So, we do not need to show the bijectivity of the operator. In the future, if I have to prove that an inverse of an operator $T$ exists, is it sufficient to show that there exists an operator $S$ such that $TS$= $ST$ = $I$? $\endgroup$ – Abhimanyu Swami Jul 31 at 0:48
  • $\begingroup$ If you want to show that an inverse exists then yes, a function having an inverse is equivalent to it being bijective-this is just set theory. But note that proving the inverse is also a bounded operator is a different exercise. In my solution we got that $S$ is bounded without much work, it followed from $S$ being a limit in the space $L(X,X)$ which is a Banach space. $\endgroup$ – Mark Jul 31 at 8:38
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An addition to the other answers: we can prove that $T$ is surjective directly. Let $T^*$ denote the adjoint of $T$, noting that $\|T\| = \|T^*\|$. Moreover, we have $\|T^* - I\| = \|(T-I)^*\| = \|T-I\|$. $T^*$ is injective for the same reason that $T$ is injective.

Let $\alpha = \|T-I\| = \|T^* - I\| < 1$. We have $$ \|(T^*-I)x\| \leq \alpha\|x\| \quad \text{for all }x \in X \implies\\ \|T^*x - x\| \leq \alpha\|x\| \quad \text{for all }x \in X \implies\\ \|x\| - \|T^*x\| \leq \alpha\|x\| \quad \text{for all }x \in X \implies\\ \|T^*x\| \geq (1 - \alpha)\|x\| \quad \text{for all }x \in X. $$ So, there exists a $c>0$ such that $\|T^*x\| \geq c\|x\|$. In other words, $T^*$ is injective and is "bounded from below", which means that $T$ is surjective.

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Since $\|I-T\|<1$, $\sum_n I+(I-T)^n$ is invertible and its inverse is $(I-(I-T)=T$

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