# Inconclusive inequality result: show $(id - f)^{-1}$ is bounded

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?

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

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