3.2 Analyticity and Holomorphy
The Cauchy–Goursat Formula (Theorem 3.1.8) can also be generalized into a result that equates complex integration and differentiation:
Theorem 3.2.1 (Cauchy–Goursat).
Let be an open region bounded by a simple closed boundary , and let be holomorphic and continuous over . Then , , exists, and
Additionally, since is open, , such that the closed disk , has the uniformly and absolutely convergent Taylor expansion
where
for .
Proof.
, , by Theorem 3.1.8,
and dividing by , the above is equal to
Since
Let be the distance from to ; then . Then since and , . Then the absolute value of the integrand of (3.2.4) is bounded above by , where is the maximum of , which exists by Theorem 1.2.13. Then,
As , the difference vanishes, and therefore,
Now inductively assume that (3.2.1) is true for a given , or
Notice the expansion of the kernel, convergent since :
Then,
where the remainder terms resemble
The difference quotient is equal to
As , the remainder terms vanish, and
By induction, (3.2.1) is valid. By substituting (3.2.5) into Theorem 3.1.8, we obtain
Because is continuous over , it is bounded by a constant . Additionally, since and consequently , the sum is termwise uniformly bounded by the convergent series
By the Weierstrass –Test (Theorem 2.3.2), the integrand uniformly converges, and we can justify
which verifies (3.2.2) and (3.2.3).
Remark.
By induction, we have shown that assuming the existence of the first order derivative of a holomorphic function , the -th order derivative of exists and is holomorphic over the same region as . Furthermore, if is holomorphic, then , there exists an open disk enclosing such that has a convergent Taylor series expansion. This property is known as analyticity, and Theorem 3.2.1 tells us that all holomorphic functions are analytic. Analytic functions can be expanded into power series, which are termwise differentiable, and therefore complex differentiable. Thus, analyticity and holomorphy are logically equivalent, which is a fundamental difference between real and complex functions.
The differentiation formula above can be thought of as a generalization of Theorem 3.1.8, and provides similar utility in the evaluation of integrals:
Example 3.2.1.
A Legendre polynomial is a polynomial whose explicit equation is given by
Prove the integral form
where is a simple closed curve enclosing .
Theorem 3.2.2 (Cauchy's Estimate).
For a function holomorphic over and and such that , ,
where
Proof.
By the Differentiation Formula (Theorem 3.2.1), ,
Because is continuous over the boundary , it is bounded by . Thus,
as desired.
Theorem 3.2.5 will profoundly generalize this statement significantly. The relationship between the derivatives of a holomorphic function and the function itself is an important property of holomorphic functions.
Example 3.2.2.
Let be entire and , . Prove that , and
Proof.
is obviously true by letting . Then , by Cauchy’s Estimate (Theorem 3.2.2),
By letting , the conclusion follows. In fact, this is the tightest possible inequality. Consider to be a function of . It attains its minimum as its derivative vanishes:
To confirm it as a minimum, we calculate the second order derivative:
which is positive and convex.
The following theorem, albeit originally proven by Cauchy in 1844, shows a fundamental difference between holomorphic functions on proper subsets of and entire functions.
Theorem 3.2.3 (Liouville).
Any bounded entire function is constant.
Proof.
Let be entire. Then, , , is holomorphic over . By Theorem 3.2.2,
where . By letting , where is any arbitrary value in . Therefore, is constant.
Proof.
Let be distinct and arbitrarily chosen. Let be entire and bounded such that for some . Let . Since , such that . By the Cauchy–Goursat Theorem (Theorem 3.1.7), we have
Since is holomorphic on the disk centered at and is holomorphic on the disk centered at , by the Cauchy–Goursat Formula (Theorem 3.1.8), we have
On the contrary, we also have
We conclude that
for all distinct complex and . Hence, is a constant function.
Theorem 3.2.4 (Morera).
Let and be continuous over . If for any closed triangular contour ,
then is holomorphic over .
Proof.
Let be arbitrary. Since is open, such that . Define
where the path is a straight line segment, and is well-defined for . Now
Note that the first three integrals sum to form a closed triangular curve and hence vanish by assumption. Therefore,
By the continuity of at , for any , such that . Then, for ,
Thus, for all . Since was arbitrary, is holomorphic over .
Theorem 3.2.5.
Let be open, let be compact and be open such that is compact ( is relatively compact in ). Let be holomorphic in . Then there exists a sequence dependent only on and (independent of and ) such that ,
where denotes
Proof.
Let satisfy and be identically equal to over some open neighborhood of relatively compact in . Since , by the Cauchy–Pompeiu Theorem (Theorem 3.1.3) on ,
By the product rule,
and since , the first term vanishes, resulting in
Let denote , and , . Therefore,
We can differentiate within the integral as is and bounded over , and thus the integrand is uniformly bounded by an integrable function independent of :
and by the triangle inequality,
Notice that over , , , and is disjoint from (or that ). Then, the distance between and is positive and the two are disjoint. Therefore, such that
and thus,
can be bounded by a sequence , independent of and dependent only on and the sets and . Then,
Because is compact, it has a finite area , and we can define a new sequence to find that
The problem now stands to prove that exists in the first place, which requires a topological argument to be later discussed in Theorem 3.2.12.
Corollary 3.2.1.
Let be open, let be compact and be open such that . For any holomorphic function in , there exist constants (independent of and ) such that
Proof.
Starting from (3.2.7), observe that
and we can define a new set of constants equal to , which are still independent of .
For the next theorem we will briefly introduce the concept of analytic continuation.
Definition 3.2.1 (Analytic Continuation).
Let be open, and let be holomorphic. Let be open with . A function
is an analytic continuation of to if:
- is holomorphic on , and
- on .
The concept of analytic continuation and its consequent problems and properties will be discussed in detail in a later chapter. For now, we will prove a theorem that is a direct consequence of the Cauchy–Goursat Differentiation Formula (Theorem 3.2.1) and the existence of holomorphic functions with removable singularities.
Theorem 3.2.6 (Riemann).
Let (known as a punctured disk), and be holomorphic and bounded. Then can be analytically continued to .
Proof.
Define the auxiliary function
is bounded and continuously differentiable on and satisfies the Cauchy–Riemann Equations since
meaning that . For ,
As , , meaning that is holomorphic over . By Theorem 3.2.1,
which is convergent over . Then we can define
over the same disk of convergence. Over the punctured disk, , and therefore is an analytic continuation of .