Unitary norm algebra such that $u \in A$ and $1-u$ is invertible I was attempting last year exam of Analysis course( for practice before finals) and I couldn't solve this question that was asked last year:
Let $A$ be a Unitary norm algebra and let $u\in A$. Assume that $1-u$  is invertible. Show that if $\|u\| <1$ , then $\|(1-u)^{-1} \|  \leq \frac{1} {1- \|u\|}$ and if $\|u\|>1$, then $\|(1-u)^{-1} \|\leq \frac{1} {\|u\|-1}$.
I am sorry to say but I am not able to make any progress on any inequalities. The course is for masters 1st year. I have followed Rudin Functional Analysis for some parts of the course but for spectral theory I have followed only lectures notes of the course.
Please guide on how to prove these inequalities.
 A: The answer to your first question is standard in $M_n(\Bbb C)$ or more generally in a Banach algebra (in that case, the hypothesis that ${\rm id}-u$ is invertible is useless: it is a consequence of $\|u\|<1,$ the inverse being $\sum_{n=0}^\infty u^n,$ where $u^0:={\rm id}$). But since your normed algebra is not supposed to be complete, let us rewrite it carefully in your case.
If $\|u\|<1$ and $1-u$ is invertible, then
$$\begin{align}\left\|(1-u)^{-1}-\sum_{k=0}^{n-1}u^k\right\|&=\left\|(1-u)^{-1}\left(1-(1-u)\sum_{k=0}^{n-1}u^k\right)\right\|\\&=\left\|(1-u)^{-1}u^n\right\|\\&\le\left\|(1-u)^{-1}\right\|\|u\|^n\\&\to0\end{align}$$
hence
$$\begin{align}\left\|(1-u)^{-1}\right\|&=\left\|\sum_{k=0}^\infty u^k\right\|\\&\le\sum_{k=0}^\infty\left\|u\right\|^k\\&=\frac1{1-\|u\|}.
\end{align}$$
Your second statement is false. A counterexample is the operator $u$ on $\Bbb C^2$ canonically identified to the matrix $$U:=\begin{pmatrix}2&0\\0&1/2\end{pmatrix}.$$
$$(I-U)^{-1}=\begin{pmatrix}-1&0\\0&1/2\end{pmatrix}^{-1}=\begin{pmatrix} -1&0\\0&2\end{pmatrix}.$$
For any of the $p$-norms on $\Bbb C^2$ (e.g. the hermitian one), the induced operator norms of these matrices are
$\|U\|=2>1$ and $\|(I-U)^{-1}\|=2>1=\frac1{\|U\|-1}.$
A: The first inequality is an application of Neumann series (if $\|u\| < 1)$. The second inequality is wrong:
Take the algebra of $2 \times 2$-matrices endowed with the row sum norm, $I$ the identity matrix and
$$
U=\left(\begin{array}{cc}
0 & 2 \\
0 & 0
\end{array}\right).
$$
Then
$$
I-U = \left(\begin{array}{cc}
1 & -2 \\
0 & 1
\end{array}\right), \quad 
(I-U)^{-1} = \left(\begin{array}{cc}
1 & 2 \\
0 & 1
\end{array}\right).
$$
Hence $\|U\|=2$, $\|(I-U)^{-1}\|=3$ and $\frac{1}{\|U\|-1}=1$.
