Q:

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. 

A:

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
$$
\lambda\in\left(-\tfrac14,\tfrac14\right)\cup\left(\tfrac34,\tfrac54\right)
$$

