Uniqueness, Mean Value, Maximum Modulus Theorems

Reading

Section 6.3

Problems

Practice problems (page 90): 4, 5, 6, 7, 8, 10

Challenge: 12, 13

Topics to know

1. Uniqueness Theorem: On a domain \(U\), if there is a sequence \(z_n\subset U\) and \(z\in U\) such that \(z_n\to z\), and \(f(z_n) = 0\), then \(f\) is identically zero on \(U\).
   • Note that power series uniqueness says: If \(y_n\to y \in U\) and \(f(y_n) = 0\) and \(D(y, r)\subset U\), then \(f(z) = 0\) for all \(z\in D(y, r)\).
   • Define \(A = \{a \in U \mid a \textrm{ is the limit point of a sequence of zeros } \}\) (in particular \(f(z) = 0\) for all \(z\in A\)).
   • Define \(B\) be the rest.
   • We will show they are both open sets. Because U is connected one of them has to be empty. Since \(A\) is nonempty, \(B\) must be empty.
   • \(A\) is open because of the power series uniqueness: If \(z\in A\) then we can find a sequence of zeros going to \(z\). Since \(U\) is open there is an \(D(z, r)\subset U\), and \(f\) is identically \(0\) there. All these points are limit points of zeros of \(f\), so \(D(z, r)\subset A\).
   • \(B\) is open: If \(z\in B\), then there must be an \(r>0\) such that \(D(z, r)\subset B\). If not, we will build a sequence of zeros going to \(z\):
     • For each \(\epsilon > 0\) there is a \(w\in D(z, \epsilon/2)\) with \(w\not\in B\).
     • This \(w\) must be the limit point of a sequence of zeros, so there is a zero \(y\) such that \(y\in D(w, \epsilon/2)\).
     • Then \(y\) is in \(D(z, \epsilon)\).
     • So for each \(\epsilon > 0\) there is a zero of \(f\) within \(\epsilon\) of \(z\).
     • So \(z\) would be a limit point of zeros. Contradiction.
   • NOTE: A function can have infinitely many zeros. But their limit point must be outside the domain of analyticity \(U\).
2. Corollary: If two functions agree on a converging sequence in a domain \(U\), then they must be equal throughout the domain \(U\).
   • Use case: An identity between analytic functions, that holds for real numbers, must also hold for complex numbers.
3. Corollary: If \(f\) is an entire function and \(f(z)\to\infty\) as \(z\to \infty\), then \(f\) must be a polynomial.
   • There is an \(M\) so that \(|f(z) \geq 1|\) for all \(|z|\geq M\).
   • All zeros are restricted to \(D(0, M)\). So there must be finitely many of them.
   • \(g(z) = \frac{f(z)}{(z-a_1)(z-a_2)\cdots(z-a_n)}\) is entire, and has no zeros.
   • \(h(z) = 1/g(z)\) satisfies \(|h(z)|\leq A + |z|^n\). So is polynomial.
   • But \(h(z)\neq 0\), so it must be constant. So \(f(z) = C (z-a_1)(z-a_2)\cdots(z-a_n)\).
4. Mean Value Theorem: \(f(a)\) equals the mean value taken around the boundary of a disc: \(f(a) = \frac{1}{2\pi}\int_0^{2\pi} f\left(a + re^{i\theta}\right)d\theta\)
   • Follows directly by Cauchy Integral Formula.
5. Maximum Modulus Theorem: A non-constant analytic function has no interior maximum points: For each \(z\in D\) and \(\delta > 0\) there is a \(w\in D_z(\delta)\) such that \(|f(w)| > |f(z)|\).
   • From mean value theorem: \(|f(z)|\leq \max_\theta|f\left(z + re^{i\theta}\right)\)
   • Since \(f\) cannot be constant on any such circle, there must be a point such that \(|f(z)| < |f(w)|\).
6. Minimum Modulus Theorem: An interior point for an analytic function can only be a relative minimum for the modulus if it is actually \(0\).
   • Apply Maximum Modulus to \(1/f\).