Cauchy Integral Formula for Derivatives
Formula for the Derivative:
Let U C be an open set and let f be holomorphic on U. Then f C(U). Moreover, if D(P, r) U and z D(P, r), then
This formula is obtained simply by differentiating the standard Cauchy formula under the integral sign.
The Cauchy Estimates:
If f is a holomorphic on a region containing the closed disc and if | f | ≤ M on then,
This is proved by direct estimation of the Cauchy Formula.
Entire Functions and Liouville’s Theorem:
A function f is said to be entire if it is defined and holomorphic on all of C, i.e., f : C → C is holomorphic. For instance, any holomorphic polynomial is entire, ez is entire, and sin z, cos z are entire. The function f(z) = 1/z is not entire because it is undefined at z = 0. [In a sense that we shall make precise later, this last function has a “singularity” at 0.] The question we wish to consider is: “Which entire functions are bounded?” This question has a very elegant and complete answer as follows:
Liouville’s Theorem A bounded entire function is constant.
Proof: Let f be entire and assume that | f(z) | ≤ M for all z C. Fix a P C and let r > 0. We apply the Cauchy estimate for k = 1 on . So
Since this inequality is true for every r > 0, we conclude that
Since P was arbitrary, we conclude that
The end of the last proof bears some commentary. We prove that df/dz ≡ 0. But we know, since f is holomorphic, that df/dz ≡ 0. It follows from linear algebra that df/dx ≡ 0 and df/dy ≡ 0. Then calculus tells us that f is constant. The reasoning that establishes Liouville’s theorem can also be used to prove this more general fact: If f : C → C is an entire function and if for some real number C and some positive integer k, it holds that,
for all z, then f is a polynomial in z of degree at most k.