- Can every continuous function on \(\bar \DD\) be uniformly approximated by polynomials in the variable \(z\)?
Hint: compare to Weierstrass for the real interval.
- Suppose \(f\) is analytic, defined on all of \(\CC\), and for each \(z_0 \in \CC\) there is at least one coefficient in the expansion \(f(z) = \sum_{n=0}^\infty c_n(z-z_0)^n\) is zero. Prove that \(f\) is a polynomial.
Hint: use the fact that \(c_n n! = f^{(n)}(z_0)\) and use a countability argument.
Let \(|\alpha|<r<|\beta|\), and let \(\gamma\) be the positively oriented circle centered at \(0\) of radius \(r\). Show that \[ \int_\gamma \frac{dz}{(z-\alpha)(z-\beta)} =\frac{2\pi i}{\alpha-\beta}. \]
Assume \(f\) is continuous on \[ \{x+iy:x\ge x_0,\ 0\le y\le b\} \] and that \(f(x+iy)\to A\) as \(x\to+\infty\) uniformly with respect to \(y\in[0,b]\). If \[ \gamma_x=\{x+it:0\le t\le b\}, \] show that \[ \lim_{x\to+\infty}\int_{\gamma_x}f(z)\,dz=iAb. \]
If \(f\) is analytic on a region \(D\), show that \[ g(z)=\overline{f(\bar z)} \] is analytic on the reflected region \(\bar D=\{z:\bar z\in D\}\).
Solution
Part 1.
1. No, not every continuous function on \(\bar\DD\) is uniformly approximable by polynomials in \(z\).
1.1. A uniform limit of holomorphic polynomials on \(\bar\DD\) is holomorphic on the interior \(\DD\).
a uniform limit of holomorphic functions is holomorphic (Morera’s theorem).
1.2. But there are continuous functions on \(\bar\DD\) that are not holomorphic on \(\DD\).
e.g. \(f(z) = \bar z\) is continuous on \(\bar\DD\) but not holomorphic.
1.3. Hence \(\bar z\) (or any such function) cannot be uniformly approximated by polynomials in \(z\).
1 and 2.
2. Q.E.D. (part 1).
1.
Part 2.
1. For each \(n \ge 0\), let \(E_n = \{z \in \CC : f^{(n)}(z) = 0\}\).
define the zero sets of the derivatives.
2. Each \(E_n\) is closed, and \(\CC = \bigcup_{n=0}^{\infty} E_n\).
\(f^{(n)}\) is continuous so \(E_n\) is closed; the hypothesis says that for each \(z_0\) some coefficient \(c_n = f^{(n)}(z_0)/n!\) is zero, i.e. \(f^{(n)}(z_0) = 0\) for some \(n\), so \(z_0 \in E_n\).
3. By the Baire category theorem, some \(E_n\) has nonempty interior.
\(\CC\) is a complete metric space and is the countable union of the closed sets \(E_n\), so one of them has nonempty interior.
4. Hence \(f^{(n)} \equiv 0\) on \(\CC\).
\(f^{(n)}\) is entire and vanishes on a set with nonempty interior (an open disk), so by the identity theorem it vanishes identically.
5. Therefore \(f\) is a polynomial of degree at most \(n-1\).
\(f^{(n)} \equiv 0\) implies \(f\) is a polynomial of degree \(< n\).
6. Q.E.D. (part 2).
5.
Part 3. Since \(|\alpha|<r<|\beta|\), the integrand has exactly one pole inside \(\gamma\), namely the simple pole at \(\alpha\). Its residue is \[ \operatorname{Res}_{z=\alpha} \frac1{(z-\alpha)(z-\beta)} =\frac1{\alpha-\beta}. \] Therefore the residue theorem gives \[ \boxed{\int_\gamma\frac{dz}{(z-\alpha)(z-\beta)} =\frac{2\pi i}{\alpha-\beta}}. \]
Part 4. Parametrize \(\gamma_x\) by \(z=x+it\), \(0\le t\le b\). Then \[ \int_{\gamma_x}f(z)\,dz=i\int_0^b f(x+it)\,dt. \] Hence \[ \left|\int_{\gamma_x}f(z)\,dz-iAb\right| \le b\sup_{0\le t\le b}|f(x+it)-A|. \] The right-hand side tends to \(0\) by the assumed uniform convergence, so \[ \boxed{\lim_{x\to\infty}\int_{\gamma_x}f(z)\,dz=iAb}. \]
Part 5. Let \(z\in\bar D\). For small \(h\ne0\) with \(z+h\in\bar D\), \[ \frac{g(z+h)-g(z)}h =\frac{\overline{f(\bar z+\bar h)-f(\bar z)}}h =\overline{ \frac{f(\bar z+\bar h)-f(\bar z)}{\bar h} }. \] As \(h\to0\), also \(\bar h\to0\), so the last expression tends to \(\overline{f'(\bar z)}\). Thus \(g\) is complex differentiable at every point of \(\bar D\), with \[ \boxed{g'(z)=\overline{f'(\bar z)}}. \]