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I have the following statement to prove:

Let $V$ be a normed vector space (so not necessarily complete or finite dimensional) over $\mathbb{R}$. Take any norm $\Vert \cdot \Vert_V$ on $V$ and let $\Vert \cdot \Vert_{V\rightarrow V}$ be the induced norm. Suppose $f : V \rightarrow V$ is bounded and $\Vert f \Vert_{V\rightarrow V} \leq \beta < 1$. If $(id - f)^{-1} $ exists, show that $\Vert (id - f)^{-1} \Vert_{V\rightarrow V} \leq \frac{1}{1-\beta}$, where $id :V \rightarrow V$ is the identity operator.

My attempt:

First by reverse triangle inequality, we get $$ \Vert id - f \Vert_{V \rightarrow V} \geq \Vert id \Vert_{V \rightarrow V} - \Vert f\Vert_{V \rightarrow V} \geq 1 - \beta \implies \frac{1}{\Vert id - f \Vert_{V \rightarrow V}} \leq \frac{1}{1 - \beta}. $$

Then by submultiplicity of the induced norm, we get $$ 1 = \Vert(id - f)(id - f)^{-1} \Vert_{V \rightarrow V} \leq \Vert id - f \Vert_{V \rightarrow V}\Vert (id - f)^{-1} \Vert_{V \rightarrow V}, $$ which implies $$ \Vert (id - f)^{-1} \Vert_{V \rightarrow V} \geq \frac{1}{\Vert id - f \Vert_{V \rightarrow V}} \leq \frac{1}{1-\beta}. $$

So this attempt ended up with an inconclusive result. I'm guessing I made a silly mistake somewhere in the inequalities... Any hint on what I did wrong here?

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    $\begingroup$ Show that if the inverse exists then $(I-f)^{-1} = I +f +f^2+ \cdots$. Then $\| (I-f)^{-1}\| \le 1+ \beta + \beta^2+\cdots$. $\endgroup$
    – copper.hat
    Commented Feb 2, 2021 at 0:15
  • $\begingroup$ Ah yes I actually attempted that way as well but didn't finish to show the bound is $\frac{1}{1-\beta}$, continuing now! $\endgroup$
    – user594147
    Commented Feb 2, 2021 at 0:17
  • $\begingroup$ LOL I missed the geometric series at the very end when I tried this way yesterday... I was one step away hahaha $\endgroup$
    – user594147
    Commented Feb 2, 2021 at 0:37

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More simply, if $y = (I-f)^{-1}(x)$, then \begin{equation} (I - f)(y) = x \Rightarrow y = x + f(y)\Rightarrow \|y\|\le\|x\| + \beta \|y\| \end{equation} Hence \begin{equation} (1-\beta)\|y\|\le\|x\|\quad\Rightarrow\quad \|(I-f)^{-1}(x)\| = \|y\|\le\frac{1}{1-\beta}\|x\| \end{equation}

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  • $\begingroup$ +1 Nicer than my suggestion. $\endgroup$
    – copper.hat
    Commented Feb 2, 2021 at 1:46

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