Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 4 / 12H / b / Solution

For $\lambda \notin K$, write $g_{\lambda }(z)=(z-\lambda {)}^{-1}$. If $\lambda \in \Lambda$ andthenuniformly on $K$. Uniformly polynomially approximable functions form an algebra closed under uniform limits, so $\mu \in \Lambda$. Hence $\Lambda$ is open.

The set $\Lambda$ is also closed relative to $\mathbb{C}\setminus K$. Indeed, if ${\lambda }_j\in \Lambda$ and ${\lambda }_j\to \lambda \notin K$, thenso $g_{\lambda }$ is again uniformly approximable by polynomials. Since $K$ is closed, every point of $\Gamma$ has a neighbourhood in $\mathbb{C}\setminus K$; relative closedness of $\Lambda$ then gives a smaller neighbourhood disjoint from $\Lambda$. Thus both $\Lambda$ and $\Gamma$ are open.

It is not always true that $\Lambda$ is nonempty. Take $K=\mathbb{R}$. Any polynomial uniformly close on $\mathbb{R}$ to a bounded resolvent $g_{\lambda }$, with $\lambda \notin \mathbb{R}$, must itself be bounded on $\mathbb{R}$, and hence must be constant. Uniform limits of constants are constant, whereas $g_{\lambda }$ is not. Therefore

Boundedness of $K$ does not force $\Gamma$ to be empty either. Take $K=\{z:|z|=1\}$ and $\lambda =0$. If polynomials $p_n$ converged uniformly to $1/z$ on $K$, then uniform convergence and contour integration would givea contradiction. Hence