Let \(p(z)\) be a polynomial, \(R>0\) any positive number, and \(m \geq 1\) an integer. Let \(M_R = \sup \{ |z^{m} p(z) - 1|: |z| = R \}\). Show that \(M_R>1\).
Let \(m \geq 1\) be an integer and \(K = \{z \in {\mathbb C}: r \leq |z| \leq R \}\) where \(r<R\). Show (i) using (1) as well as, (ii) without using (1) that there exists a positive number \(\varepsilon_0>0\) such that for each polynomial \(p(z)\), \[\sup \{|p(z) - z^{-m}|: z \in K \} \geq \varepsilon_0 \, .\]
Part (1) requires \(p\not\equiv0\): for \(p\equiv0\), \(z^mp(z)-1\equiv-1\) and \(M_R=1\). Part (2) holds for every polynomial, including \(p\equiv0\).
Solution
Throughout, \(0<r<R\), so that \(z^{-m}\) is defined on \(K\). Let \(f(z) = z^m p(z) - 1\).
Part (1).
1. \(f\) is a polynomial with \(\abs{f(0)} = 1\).
\(z^m p(z) - 1\) is a polynomial, and \(f(0) = 0^m p(0) - 1 = -1\) since \(m\ge1\).
2. If \(p \not\equiv 0\), then \(f\) is nonconstant.
The term \(z^m p(z)\) has degree \(m + \deg(p) \ge 1\).
3. Q.E.D.
Let \(p\not\equiv0\). By step 2 and the maximum modulus principle on the closed disk \(\overline{D(0,R)}\), the modulus of \(f\) at the interior point \(0\) is strictly less than its maximum on the circle \(\abs z=R\). By step 1, \(M_R > \abs{f(0)} = 1\). The remark treats \(p\equiv0\).
Part (2)(i).
4. For every polynomial \(p\), \(\sup_{z \in K} |p(z) - z^{-m}| \geq R^{-m}\).
On the circle \(\abs z=R\), which lies in \(K\), \[|p(z) - z^{-m}| = \frac{|z^m p(z) - 1|}{|z|^m} = \frac{|z^m p(z) - 1|}{R^m},\] so \(\sup_{z \in K} |p(z) - z^{-m}| \geq M_R/R^m\). If \(p\not\equiv0\), Part (1) gives \(M_R>1\). If \(p\equiv0\), then \(\sup_{z\in K}\abs{z^{-m}}=r^{-m}>R^{-m}\).
5. Q.E.D.
Step 4 gives the bound with \(\varepsilon_0 = R^{-m}\).
Part (2)(ii). Let \(\rho = \frac{r+R}{2}\) and let \(\gamma_\rho\) be the circle \(z = \rho e^{i\theta}\), \(0\le\theta\le2\pi\), which lies in \(K\). Let \(S = \sup_{z \in K} |p(z) - z^{-m}|\).
6. \(\oint_{\gamma_\rho} \big( p(z) - z^{-m} \big) z^{m-1} \, dz = -2\pi i\) for every polynomial \(p\).
By linearity the integral is \(\oint_{\gamma_\rho} p(z) z^{m-1} \, dz - \oint_{\gamma_\rho} \frac{dz}{z}\). The first integral vanishes by Cauchy’s theorem, since \(p(z)z^{m-1}\) is a polynomial, and the second equals \(2\pi i\).
7. \(S\geq\rho^{-m}\).
On \(\gamma_\rho\), \(|z^{m-1}| = \rho^{m-1}\), and \(\gamma_\rho\) has length \(2\pi \rho\). By step 6 and the \(ML\)-inequality, \[2\pi = \left| \oint_{\gamma_\rho} \big( p(z) - z^{-m} \big) z^{m-1} \, dz \right| \leq S \cdot \rho^{m-1} \cdot 2\pi \rho = 2\pi \rho^m S.\]
8. Q.E.D.
Step 7 gives the bound with \(\varepsilon_0 = \rho^{-m} = \left(\frac{2}{r+R}\right)^m\), which depends only on \(r\), \(R\) and \(m\).