Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 4 / 12H (Topics in Analysis)

Letand choose $0<\eta <\delta$. Write $C_{\eta }$ for the positively oriented boundary of the square with $|x|,|y|\le 1+\eta$. The Cauchy integral formula gives, for every $z\in S$,Because $C_{\eta }$ and $S$ are compact and separated by a positive distance, Riemann sums for this contour integral converge uniformly in $z\in S$. Thus, for any $\epsilon >0$, there are points ${\zeta }_j\in C_{\eta }$ and constants $c_j$ such that

It remains to approximate each resolvent by polynomials on $S$. Let $\Lambda$ be the set of $\lambda \notin S$ for which $(z-\lambda {)}^{-1}$ is polynomially uniformly approximable on $S$. If $|\lambda |>{\sup }_{z\in S}|z|$, the geometric seriesshows that $\lambda \in \Lambda$. Moreover, the resolvent propagation across a complementary component argument shows that $\Lambda$ is both open and closed relative to $\mathbb{C}\setminus S$. The complement of a square is connected, and $\Lambda$ is nonempty, so $\Lambda =\mathbb{C}\setminus S$.

In particular, every ${\zeta }_j$ lies in $\Lambda$. Approximate the finitely many resolvents accurately enough that their weighted sum differs by less than $\epsilon /2$. Combining this with the Riemann-sum estimate produces a polynomial $p$ satisfyingThis proves the claimed elementary polynomial approximation on a square.

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