Unitary norm algebra such that $u \in A$ and $1-u$ is invertible

Question

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.

Answer 1

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 $\|(I-U)^{-1}\|=2>1=\frac1{\|U\|-1}.$

Answer 2

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 $\frac{1}{\|U\|-1}=1$.