Claim: $\Vert (\lambda_n 1-T)^{-1}\Vert \rightarrow \infty$

Question

Let $\mathcal{A}$ denote a unital Banach algebra, $T \in \mathcal{A}$ and $\sigma(T)$ then spectrum of $T$. Suppose that $(\lambda_n)_n \subseteq \mathbb{C}\setminus\sigma(T), \ \lambda \in \sigma(T)$ and $\lambda_n \rightarrow \lambda$ as $n \to \infty$.

Then an exercise in my textbook claims: $\Vert (\lambda_n 1-T)^{-1}\Vert \rightarrow \infty$. (Here $\Vert \cdot\Vert$ is the usual operator norm).

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I have trouble proving this and even finding a good place to start. I have tried using the Banach inequality and other inequalities I have at my disposal but all of them yields an upper bound, not a lower bound which is what we need for convergence to $\infty$. Say something on the form of $$ \ldots \geq \frac{1}{\vert \lambda_n -\lambda\vert} $$ so we can use the assumed convergence in $\mathbb{C}$. However the fact that $\lambda \in \sigma(T)$ (meaning $\lambda1-T$ is not invertible) also has to come into play somehow.

The exercise belongs to a section where the main result is a formula for the spectral radius (see first equation here https://encyclopediaofmath.org/wiki/Spectral_radius). Maybe this is relevant but I can't see the connection.

Can anyone help me?

Answer 1

Here is one possible way to approach the problem.

Let $G(A)$ denote the elements of the Banach-algebra $A$ that are invertible. This set is known to be an open subset of $A$. This can be seen by noticing that if $x\in G(A)$ then $$ x+h =x(e+x^{-1}h)$$ Taking $h$ small enough so that $\|h\|<\frac{1}{2\|x^{-1}\|}$ we see that $B(x;\frac{1}{2\|x^{-1}\|})\subset G(A)$.

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Now, suppose $x_n\in G(A)$ approaches a point $x$ in $\partial (G(A))$. We show that $\|x^{-1}_n\|\xrightarrow{n\rightarrow\infty}\infty$. Suppose that is not the case ans that there is a sequence $(x_n:n\in\mathbb{N})\subset G(A)$ and a point $x\in \partial G(A)$ such that $\|x_n-x\|\xrightarrow{n\rightarrow\infty}0$ and $\sup_n\|x^{-1}_n\|:M<\infty$. Let $N\in\mathbb{N}$ be such that $\|x_n-x\|<\frac1M$ for all $n\geq N$. As $$e-x^{-1}_Nx=x^{-1}_N(x_N-x),$$ we have that $$\|e-x^{-1}_Nx\|<1$$ and so, $e-(e-x^{-1}_nx)=x^{-1}_Nx\in G(A)$. Consequently, $x=x_N(x^{-1}_Nx))\in G(A)$. This contradicts the assumption that $x\in\partial G(A)$, for this means that $x\notin G(A)$.

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For you particular problem, notice that if $\lambda_n\notin\sigma(x)$, and $\lambda_n\xrightarrow{n\rightarrow\infty}\lambda\in\sigma(A)$, then $y_n:=\lambda_ne-x\in G(A)$, and $y_n\xrightarrow{n\rightarrow\infty}y=\lambda e -x\in\partial G(A)$.