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Suppose Id denotes the identity operator defined by $Id(v)=v$ for all $v \in V$ where V is the normed vector space $V$. Let also $M$ denote the operator $M:V \to V$ satisfying $\left\| M \right\| \le \gamma $ for some $\gamma$ less than 1 and greater than zero. I want to show that if $(Id-M):V\to V $ is invertible, then the inverse operator $(Id-M)^{-1}$ is bounded and $\left\| {\left( {Id - M} \right)} \right\| \le \frac{1}{{1 - \gamma }}$.

My answer: The first part follows from the open mapping theorem. Given a bijection, it maps open sets to open sets. Using this I could prove the inverse does same and hence bounded. However, I could not prove the second part (the bound on its norm).

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Hint: $Id+M+M^{2}+\cdots$ converges in operator norm. If we denote this sum by $T$ then $T(Id-M)=(Id-M)T=I$. So $\|(Id-M)^{-1}|| =\|T\| \leq \sum_n \|M^{n}\| \leq \sum_n \gamma ^{n}=\frac 1 {1-\gamma}$.

Kavi Rama Murthy
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  • But my question was not for the infinite sum, but only for the operator Id-M. I can not see the relevance of your answer to my question. –  Dec 01 '21 at 00:34
  • @FethiBencherki this answer is giving the inverse to $Id-M$ explicitly which then helps to solve the problem. The infinite sum gives the inverse. – Randall Dec 01 '21 at 00:36
  • ohh, my bad, I got it now. Thanks. –  Dec 01 '21 at 00:38
  • but I do appreciate if the argument could be developed using just the fact that $Id-M$ is bounded, and using that somehow bounding the norm to $(Id-M)^2$ –  Dec 01 '21 at 00:43
  • I don't know what you mean but $|(Id-M)^{2}|\leq |Id -M|^{2}$. – Kavi Rama Murthy Dec 01 '21 at 04:57