Exercise 8.28

Following the hint, for $0\le r\le 1$, $0\le t\le 2\pi$, put ${\gamma }_r(t)=g(r{e}^{\mathit{it}})$, a closed, continuous curve with values on $T$ so that we can define $({\gamma }_r)$. Note that since $|z|>1∕2$ for $z\in T$, if a trigonometric polynomial $P$ satisfies $|P-\gamma |<1∕8$ for any curve with values on $T$, then $(\gamma )=(P)$. Since $g$ is uniformly continuous on the compact set $\bar{D}$, this implies that for every $r\in [0,1]$ there is a $\delta >0$ such that if $|r_0-r|<\delta$, then $({\gamma }_{r_0})=({\gamma }_r)$, that is, $({\gamma }_r)$ is a continuous function on the connected set $[0,1]$. As shown at the end of the solution to Exercise 27, this indicates that $({\gamma }_r)$ is constant on $[0,1]$. Since ${\gamma }_0$ is the curve with the constant value $f(0)$, this constant must be $({\gamma }_0)=0$.

Put $\psi (t)=e^{\mathit{it}}{\gamma }_1(t)$, which is also a closed continuous curve with values on $T$. If $\psi (t)=-1$ for some $t$, then $g(e^{\mathit{it}})={\gamma }_1(t)=-e^{\mathit{it}}$. So if we assume that $g(z)\ne -z$ for all $z\in T$, then $\psi (z)\ne -1$, which is the intersection of $T$ with the negative real axis. Hence by the extension of Exercise 24 shown in Exercise 26, we have $(\psi )=0$. Let $P_1$ be a trigonometric polynomial such that $|P_1-\psi |<1∕8$ on $T$ so that $(P_1)=(\psi )=0$. If $P_2(t)=e^{\mathit{it}}{P}_1(t)$, then

so that

which contradicts $({\gamma }_1)=0$ shown in the first paragraph.