# How to prove the surjectivity of an operator

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.

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 $$I-A=I-(I-T)=T$$.

• 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 $TS$= $ST$ = $I$? – Abhimanyu Swami Jul 31 at 0:48
• 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. – Mark Jul 31 at 8:38

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.

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