Spring 2020 - Problem 11

Solution.

1. This follows immediately from Runge's theorem since $\C \setminus K$ has one unbounded connected component. Explicitly, this is because $\conj{z} = \frac{1}{z}$ on $\T$, so it suffices to show that we can approximate $\frac{1}{z}$ on $K$ by polynomials.

   First, observe that we may "push" poles as follows:

   $$\frac{1}{z - z_0} = \frac{1}{\p{z - z_1} - \p{z_0 - z_1}} = \frac{1}{z - z_1} \sum_{n=0}^\infty \p{\frac{z_0 - z_1}{z - z_1}}^n,$$

   and this will converge uniformly as long as $z_1$ is chosen such that $\abs{z_0 - z_1} \leq \frac{1}{2} \abs{z - z_1}$, e.g., with $\abs{z_0 - z_1} \leq \frac{1}{3}d\p{z_0, K} = r$, since

   $$\abs{z - z_1} \geq \abs{z - z_0} - \abs{z_0 - z_1} \geq 2r \implies \abs{\frac{z_0 - z_1}{z - z_1}} \leq \frac{1}{2}.$$

   In other words, given a rational function with a pole only at $z_0$, we can approximate it uniformly by rational functions with a single pole in $B\p{z_0, r}$, so by iterating this construction, we can move the poles as far away as we want. In particular, we can approximate $\frac{1}{z}$ uniformly on $K$ by a function with a single pole arbitrarily far away from $K$. In particular, we can prescribe the pole to have $\abs{z_0} \geq 2 \sup_{z \in K}\,\abs{z}$. Then

   $$\frac{1}{z - z_0} = -\frac{1}{z_0} \frac{1}{1 - \frac{z}{z_0}} = -\frac{1}{z_0} \sum_{n=0}^\infty \p{\frac{z}{z_0}}^n$$

   converges uniformly on $K$, which completes the proof.

2. Observe that for any polynomial $p\p{z} = \sum_{k=1}^n a_k z_k$, we have

   $$\begin{aligned} 2\pi &= \abs{\int_\T \abs{z}^2 \,\diff{z}} \\ &= \abs{\int_\T z\conj{z} \,\diff{z}} \\ &= \abs{\int_\T z\p{\conj{z} - p\p{z}} \,\diff{z} + \int_\T zp\p{z} \,\diff{z}} \\ &= \abs{\int_\T z\p{\conj{z} - p\p{z}} \,\diff{z}} && \p{\text{Cauchy's theorem}} \\ &\leq \int_\T \abs{\conj{z} - p\p{z}} \,\diff{z} \\ &\leq 2\pi \norm{\conj{z} - p}_{L^\infty}. \end{aligned}$$

   In particular, $\norm{\conj{z} - p}_{L^\infty} \geq 1$ for any polynomial, so $\conj{z}$ cannot be approximated by a polynomial uniformly on all of $\T$.