Let $K$ be a compact subset of $\Bbb C$. Let $A_K$ denote the set of all normal elements $x$ with $\sigma_A(x)\subset K$. If $f$ is a continuous function on $K$, then the functional calculus :$x\in A_K\to f(x)\in A$ is continuous.
To prove it, the author says, for every $\epsilon>0$, there is a polynomial $p$of $\lambda$ , $\bar\lambda$ such that $\sup|p(\lambda,\bar\lambda) - f(\lambda)|<\epsilon.$ Also there is a $\delta>0$ such that $||p(x,x^*) - p(y,y^*)||<\epsilon$ if $||x-y||<\delta$. Then show that $||f(x) - f(y)||<3\epsilon$
My question is that how does he say there is a $\delta>0$ such that $||p(x,x^*) - p(y,y^*)||<\epsilon$ if $||x-y||<\delta$. I can accept this inequality for functions, but now $p(x,x^*)$ is just an element of $A$.
Please help me to understand it.