Why is $\tau_x(\lambda) - \lambda \in \sigma(-x)$ for every $\lambda \in \mathbb{C}$? Spectral Translation Explanation Needed.

Question

Let $A$ be a Banach algebra and $\phi$ a multiplicative functional on $A$ such that $\phi(x)\in \sigma(x)$ for each $x \in A$.

Now for each $x \in A$, define $\tau_x(\lambda):=\phi(\lambda\textbf{1}-x)$. Assume that $\tau_x(\lambda)$ is entire.

The paper then mentions the following:

Observe that, by spectral translation, we have that $$\tau_x(\lambda) - \lambda \in \sigma(-x)$$ for each $\lambda \in \mathbb{C}.$

I am trying to convince myself why this is true.

How does this follow from the fact that $\phi(x) \in \sigma(x)$?

I know that this means that $\tau_x(\lambda) \in \sigma(\lambda\textbf{1}-x)$, but how do I use this to obtain that $\tau_x(\lambda) - \lambda \in \sigma(-x)$?

Answer 1

Presumably $A$ is an algebra with unit. A nonzero multiplicative linear functional $\phi$ satisfies $\phi(x)\in\sigma(x)$ for all $x\in A$ and $\lambda\in \sigma(x)$ iff there is such a multiplicative functional such that $\lambda=\phi(x)$. In your notation, $\tau_x(\lambda):=\phi(\lambda \mathbf{1} -x)\in \sigma(\lambda\mathbf{1}-x)$. This means that $$\tau_x(\lambda)\mathbf{1}-(\lambda\mathbf{1}-x)=(\tau_x(\lambda)-\lambda)\mathbf{1}+x=\phi(-x)\mathbf{1}-(-x)$$ is not invertible; hence $\tau_x(\lambda)-\lambda\in\sigma(-x)$.

Answer 2

From your notation, I will be assuming that $A$ is unital. Moreover, note that $\phi(1) = \phi(1^2) = \phi(1)^2$ so $\phi(1)=1$ since $\phi$ is not the zero map (indeed, if $\phi=0$, then your assumption gives $0 \in \sigma(x)$ for all $x \in A$, in particular $0 \in \sigma(1)$, an absurdity).

Assume to the contrary that $\tau_x(\lambda)-\lambda\notin \sigma(-x)$. Then $$(\phi(\lambda 1-x) -\lambda)1+x= \tau_x(\lambda)-\lambda +x$$ is invertible, so there is $y\in A$ such that $$(x-\phi(x)1)y=((\phi(\lambda 1-x)-\lambda)1 + x)y = 1.$$ Applying $\phi$ to both sides yields $$0 = 1$$ a contradiction.