Louisville’s Theorem

Reading

Section 5.2

Problems

Practice problems (page 74): 6b, 8, 9, 10, 12, 15

Challenge (optional): 13

Topics to know

1. Louisville’s Theorem: An entire bounded function is constant.
   • Fix two points \(a, b\). Consider \(R\to \infty\), and focus on circle \(C\) around \(0\) with radius \(R\).
   • \(f(b) - f(a) = \frac{1}{2\pi i}\left(\int_C \frac{f(z)}{z-a}dz - \int_C \frac{f(z)}{z-b}dz\right) = \frac{1}{2\pi i}\int_C \frac{f(z)(b-a)}{(z-a)(z-b)}dz\)
   • If \(f \ll M\), then this integral is \(\ll \frac{M(b-a) R}{(R-|a|)(R-|b|)}\).
   • As \(R\to \infty\), this goes to \(0\).
   • So \(f(b)-f(a)\) must equal \(0\).
2. Extended Louisville Theorem: If \(f\) is entire and \(|f(z)| \leq A + B|z|^k\), then \(f\) is a polynomial of degree at most \(k\).
   • Proof by induction on \(k\), base case being Louisville’s Theorem.
   • \(|g(z)| = \left|\frac{f(z) - f(0)}{z}\right| \leq C + D|z|^{k-1}\)
     • Near \(0\) it can be extended to be entire, so is bounded
     • For \(|z| \geq 1\): \(\frac{|f(z) - f(0)|}{|z|}\leq \frac{|f(z)| + |f(0)|}{|z|} \leq \frac{A + |f(0)| + B |z|^k}{|z|}\leq C + B|z|^{k-1}\)
   • By induction \(g\) must be a polynomial of degree at most \(k-1\).
   • \(f(z) = f(0) + z g(z)\) is a polynomial of degree at most \(k\).
3. Alternative proof of the above two facts (exercises 6, 7):
   • If \(f\) is bounded by \(M\) along the circle of radius \(R\) around \(0\) and \(a_k\) is the \(k\)-th coefficient in its power series expansion around \(0\), then \(|a_k|\leq \frac{M}{R^k}\). This follows from \(M-L\) formula on \(a_k = \frac{1}{2\pi i}\int_C \frac{f(z)}{z^{k+1}} dz\)
   • So if \(f\) is bounded on the entire complex plane, \(R\) can be arbitrarily large, forcing \(a_k=0\) for all \(k\geq 1\). So power series is just constant, so \(f\) is constant.
   • If \(|f(z)|\leq A + B|z|^k\), and \(j > k\), then \(|a_j| \leq \frac{A + B R^k}{R^j}\) must be valid for all \(R > 0\), so it must equal \(0\) if we consider \(R\to \infty\). Therefore the power series terminates at the \(k\)-th term, hence a polynomial.
4. Fundamental Theorem of Algebra: Any non-constant polynomial \(P\) must have a zero. And by induction, it has exactly \(\textrm{deg} P\) zeroes.
   • Suppose not. Consider \(f(z) = \frac{1}{P(z)}\).
   • \(f\) is entire.
   • \(P(z) \to \infty\) as \(z\to\infty\). So \(f\) is bounded.
   • So \(f\) must be constant. Then \(P\) is constant, a contradiction.
5. Gauss-Lucas theorem (proof on page 68): The zeroes of the derivative of a polynomial lie within the convex hull of the zeros of the polynomial.