Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 3 / 2F (Topics In Analysis)

The polynomial form of Runge theorem says that if $K\subset \mathbb{C}$ is compact, $\mathbb{C}\setminus K$ is connected, and $f$ is holomorphic on a neighbourhood of $K$, then for every $\epsilon >0$ there is a polynomial $P$ such that

Statement (i) is false. Take the entire function $f(z)=e^{-z}$. It is bounded on the closed first quadrant $E$. If polynomials $P_n$ converged uniformly to $f$ on $E$, then each sufficiently late $P_n$ would be bounded on the positive real axis. A polynomial bounded on that ray must be constant. No sequence of constants converges uniformly to $e^{-x}$ on $[0,\infty )$, since its values at zero and near infinity differ by one.

Statement (ii) is true. LetEach $K_n$ is compact and has connected complement. By the polynomial Runge theorem, choose a polynomial $P_n$ withEvery fixed $z\in E$ belongs to $K_n$ for all sufficiently large $n$, so $P_n(z)\to f(z)$. Thus the convergence is pointwise on $E$.

Statement (iii) is false. Let $\Omega$ be a thin open annulus about the unit circle, let $f(z)=1/z$, and split the circle into two proper closed arcs $K_1,K_2$ whose union is the whole circle. Each arc has connected complement, so polynomial approximation of the reciprocal on a proper circular arc gives uniform polynomial approximations to $f$ on each $K_j$ separately.

If polynomials $P_n$ converged uniformly to $1/z$ on $K_1\cup K_2$, then uniform convergence on the unit circle and the Cauchy integral theorem would givea contradiction.

Statement (iv) is true. Apply the Weierstrass approximation theorem separately to the real and imaginary parts of the continuous function $f$ on the compact interval $[0,1]$. Combining the two real polynomial sequences gives complex polynomials converging uniformly to $f$ on $I$.