Does the spectrum at a point vary continuously in this case?

Let $$A$$ be a C$$^{*}$$-algebra. Let $$\hat{A}$$ denote the set of all irreducible representations of $$A$$. Suppose $$\pi\in\hat{A}$$ has the following property: for all $$a\in A$$, the map from $$\hat{A}\to\mathbb{R}$$ given by $$\tau\mapsto\|\tau(a)\|$$ is continuous at $$\pi$$.

Let $$e\in \pi(A)$$ be a projection. Then there is a positive element $$y\in A$$ such that $$\pi(y)=e$$. We have that $$\operatorname{sp}(\pi(y))\subseteq\{0,1\}$$.

How can we show that there is a neighbourhood $$V\subseteq\hat{A}$$ of, $$\pi$$, such that for all $$\tau\in V$$, $$\operatorname{sp}(\tau(y))\subseteq (-\frac{1}{4},\frac{1}{4})\cup(\frac{3}{4},\frac{5}{4})$$?

This is part of a proof in a paper by Dixmier. He says it is a well known argument, but I don't believe I've seen it before. Any incite is appreciated. Thanks.

Choose $$V$$ such that $$|\|\tau(y)\|-\|\pi(y)\|\,|<\varepsilon$$ for all $$\tau\in V$$. Then $$\|\tau(y)^2-\tau(y)\|=\|\tau(y^2-y)\|<\varepsilon+\|\pi(y^2-y)\|=\varepsilon+\|e^2-e\|=\varepsilon.$$ The element $$\tau(y)^2-\tau(y)$$ is selfadjoint. For any $$\lambda\in\sigma(\tau(y)^2-\tau(y))$$, we have $$|\lambda|<\varepsilon$$. Also, $$\sigma(\tau(y)^2-\tau(y))=\{\lambda^2-\lambda:\ \lambda\in\tau(y)\}.$$ Thus we have $$|\lambda^2-\lambda|<\varepsilon$$ for all $$\lambda\in\sigma(\tau(y))\subset\mathbb R$$. This forces $$\tag1 \frac{1-\sqrt{1+4\varepsilon}}2<\lambda<\frac{1-\sqrt{1-4\varepsilon}}2,$$ or $$\tag2 \frac{1+\sqrt{1-4\varepsilon}}2<\lambda<\frac{1+\sqrt{1+4\varepsilon}}2.$$ Now choose $$\varepsilon$$ small enough such that $$(1)$$ and $$(2)$$ imply $$\lambda\in\left(-\tfrac14,\tfrac14\right)\cup\left(\tfrac34,\tfrac54\right)$$
• Thanks, Martin. I understand the spectrum calculation you are making, but I'm a bit confused as to how to apply it. $a=e=\pi(y)$ and $b=\tau(y)$ are elements in distinct C$^{*}$-algebras ($\pi(A)$ and $\tau(A)$) aren't they? So how are we to make sense of $e-\tau(y)$? – ervx May 20 at 14:19