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Let $T:X \rightarrow X$ be a bounded linear operator ($X$ is a Banach space) and $ ||T|| < 1 $. Show that the operator $$I-T$$ is invertible, where $I$ denotes the identity operator.

Hint. How can you represent $\frac{1}{1-z}$ when $z$ is a complex number such that $|z|<1$?

Invertibility requires us to show that there exists an $(I-T)^{-1}$ operator such that composing it with $(I-T)$ yields $I$.

I think the hint I'm provided with suggests that $\frac{1}{1-z}$ is a sum of an infinite series $\sum_{n=0}^{\infty}z^{n}$ but how am I supposed to use this fact here?

I'd be grateful for a detailed explanation as funcional analysis is a hard nut to crack for me.

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    $\begingroup$ Prove that the series $\sum_{n=0}^{\infty}T^n$ converges, and show that it is the inverse of $(I-T)$. $\endgroup$ Commented Jun 28, 2017 at 14:19
  • $\begingroup$ @uniquesolution Provided the space of bounded linear operators is Banach (I'm not sure, could you clarify?), then the series $\sum_{n=0}^{\infty}||T||^{n}$ converges given $||T||<1$, hence $\sum_{n=0}^{\infty}T^{n}$ converges. Correct? $\endgroup$
    – Theta
    Commented Jun 28, 2017 at 14:28
  • $\begingroup$ Yes,correct. If X is Banach, then the space of bounded linear operators is Banach $\endgroup$
    – Fred
    Commented Jun 28, 2017 at 14:32
  • $\begingroup$ Nevertheless, the space of bounded linear operators seems quite a vague notion to me. If someone would be so kind as to elaborate on the topic, I'd appreciate that. $\endgroup$
    – Theta
    Commented Jun 28, 2017 at 14:36
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    $\begingroup$ Operators on complex vector spaces are analogous to complex numbers. For instance the complex number analogy for an operator and its adjoint are a complex number and its conjugate. So by suggesting that $\frac{1}{1-z} = \sum_0^{\infty} z^n$ for $|z|<1$, the question hints that the inverse of $I - T$ is $\sum_0^{\infty} T^n$. (since $1 - z$ corresponds to $I - T$ and also since $\frac{1}{1-z}$ is the multiplicative inverse of $1-z$). $\endgroup$ Commented Jun 28, 2017 at 14:58

2 Answers 2

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The space of bounded linear operators on $X$, denoted $\mathcal{L}(X)$, is a Banach space, since $X$ is complete. The norm associated with $\mathcal{L}(X)$ is the operator norm, i.e. \begin{align*} \|T\| = \sup_{\|x\| = 1} \|Tx\|. \end{align*}

Recall that a normed vector space is complete if and only if every absolutely convergent series converges.

So consider the series \begin{align*} \sum_{n=0}^{\infty} \|T\|^n \end{align*} Since $\|T\| < 1$, this series converges, hence \begin{align*} \sum_{n=0}^{\infty} T^n \end{align*} converges in $\mathcal{L}(X)$ to some element, call it $S$. To show that $S = (I-T)^{-1}$, it suffices to show that $S(I-T) = I$. Observe, \begin{align*} S(I - T) &= \sum_{n=0}^{\infty} T^n (I - T) \\ &= \sum_{n=0}^{\infty} T^n - \sum_{n=0}^{\infty} T^{n+1} \\ &= \sum_{n=0}^{\infty} T^n - \sum_{n=1}^{\infty} T^{n} \\ &= T^0 \\ &= I \end{align*} Thus, $S = (I-T)^{-1}$.

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    $\begingroup$ Thank you very much for your extensive answer, it is of great help to me. :) $\endgroup$
    – Theta
    Commented Jun 28, 2017 at 15:21
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    $\begingroup$ Very carefully tailor-made for the customer, indeed. $\endgroup$ Commented Jun 28, 2017 at 18:27
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When the operator $a*I_d - b ad(Y)$ is invertible, $a$, $b$ - complex numbers, $I_d$ - identity operator, $ad(Y)$ - adjoint operator generated by matrix $Y$. All acts in some linear finite dimensional vector space. By "when" I mean necessary and sufficient conditions on $a$, $b$, $Y$.

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