This post is written in spirit of Terrance Tao’s post on problem solving strategies in real analysis. Our emphasis will be on concrete techniques with plenty of examples (many are taken from past prelim questions in UW).

**1. Quantitative estimates using Cauchy’s integral formula**

The starting point is ** Cauchy’s integral formula**. If is analytic in a domain , then

,

where is a positively oriented contour enclosing . (We need that the winding number is .) In many applications, will be a circle. The basic idea is to use bounds for on (the “boundary”) to bound . Besides , we can also bound the higher order derivatives, or equivalently the coefficients of the power series centered at . Explicitly, if , then

.

To estimate the integral, we always use the following basic inequality (which is essential when applying the * residue theorem*):

.

Here is the basic example:

**Example 1.1**. **(Liouville’s Theorem)** If is a bounded entire function, then is constant.

*Proof. *Let and suppose is bounded by everywhere. We claim that for . For any , applying Cauchy’s integral formula for the circle gives

.

Hence

,

which goes to as . (The last inequality is called * Cauchy’s estimate.*)

The proof of Liouville’s Theorem can be modified to yield a more quantitative statement.

**Example 1.2.** Let be entire. If there exists such that , then is a polynomial with degree at most .

*Proof*. Again we write . By Cauchy’s integral formula,

which goes to as for . Hence , where is the integer part of .

In the above examples we let to obtain the desired result. Sometimes, after apply Cauchy’s integral formula, we need to choose the best to **optimize the estimate**.

**Example 1.3.** **(Bloch functions) **A holomorphic function on is called a * Bloch function* if

.

Prove that there is a positive constant , independent of , so that

.

*Proof*. Since we have a condition on , it is natural to express in terms of . Since , for we have

,

which gives

.

We need to prove that

.

For each , the infimum is attained (by calculus) at . Since

,

the sequence is bounded above. We are done.

For functions defined in a disc , the circle is the most natural contour. Sometimes it is better to use other contours.

**Example 1.4.** (Problem 5.4.22 in Berkeley Problems in Mathematics) Let be analytic on such that for all . Prove that .

*Proof*. Fix . Instead of using for , we use the contour . Applying Cauchy’s integral formula, we have

.

Now on the circle is at most . Plugging this in and after some algebra, we get .

We give an example were the appropriate contour is not a circle.

**Example 1.5.** (Problem 5.5.13 of Berkeley Problems in Mathematics) Let the function be entire and satisfy the inequality (with )

off the imaginary axis. Prove that is constant.

*Proof*. By Liouville’s theorem, it suffices to prove that is bounded. From the hypothesis, is bounded by for . Fix with . Consider the the square contour with vertices at (the reason will become clear later). By Cauchy’s integral formula,

.

Since for , it suffices to bound . Since is bounded for , the integral on each vertical segment is bounded by . For each horizontal segment, the integral is bounded above by

.

Hence is bounded.

Finally we give a simple but more involved example.

**Example 1.6.** Let be holomorphic on a neighborhood of , and let be holomorphic on . Prove that

exists. Show that, with fixed, if is a sequence of functions holomorphic on and if uniformly on compacts – whence is holomorphic on – then .

*Proof*. Write and . By assumption and Hadamard’s formula, we have

, .

For , by uniform convergence, we have

.

(Here we use the standard fact that .) Consider the last power series. Since

,

the power series convergence for . Hence the limit as exists, and equals

.

We note that the series converges absolutely.

Now suppose that locally uniformly. Write , and we want to show that

.

Choose with . Using Cauchy’s integral formula and integrating along the circle , we have the estimate

,

where . Next consider estimating the difference

.

Let be given. Since and uniformly on , we may choose an integer so that for and . Now we split the sum into two parts:

.

Now since , the first sum (having finitely many terms) tends to as . The second sum is bounded above by

.

Letting , we have

.

Since is arbitrary, we are done.

Quantitative estimates can be derived in other ways, say using * Schwarz’s lemma* (related to

*).*

**conformal mappings****2. Functions in the disc**

The unit disc is one of the most fundamental domains. For example, the * Riemann mapping theorem* states that any simply connected domain not equal to the whole plane is conformally equivalent to . Many things for the unit disc are known explicitly (e.g. the

*) . Moreover, every automorphism of is a*

**Poisson kernel***:*

**Mobius map**.

Using a Mobius map, we can move any point to the origin, where the analysis is usually simpler. We illustrate this with the proof of Pick’s lemma, which is the “invariant form” of * Schwarz’s lemma*.

**Example 2.1.** **(Pick’s lemma)** Let be analytic. Then, for (with ), we have

.

Moreover, equality holds for some (and hence every) pair if and only if is a Mobius map.

*Proof*. Fix in . Let and be Mobius maps defined by and . Then is analytic and sends to . By Schwarz’s lemma, for any we have

.

Now letting gives the result. As in Schwarz’s lemma, equality holds if and only if is a Mobius map.

Pick’s lemma can be interpreted in terms of the pseudohyperbolic distance on . In this terminology, Pick’s lemma says that any analytic is Lipschitz in , and it is an isometry if and only if is a Mobius map. From this viewpoint the following result is “obvious”.

**Example 2.2.** Let be analytic. If has two fixed points, then is the identity.

*Proof*. Since , is a Mobius map, and a simple calculation shows that is the identity.

Here is another example where it is useful to move a point to the origin.

**Example 2.3.** Let be a bounded domain, and let be analytic with for . Show that if is non-constant, then .

*Proof*. Suppose is non-constant. By the * maximum principle*, . Suppose on the contrary that omits the value in . Let . Then omits .

Now is analytic on and is unimodular on . By the maximum principle, on . Now on by the hypothesis. It follows that , and hence , is constant. This contradiction proves that .

**Remark 2.4.** If does not vanish on a domain , there are a few things to try. First, is analytic (as in the above proof). Second, if is simply connected we may write for some analytic .

For bounded analytic functions on a unit disc, a very useful technique is the * Blaschke product*.

**Theorem 2.5.** Let be analytic on , continuous on , such that for . Then has the form

,

where are the zeros of on (duplicated according to multiplicities), and is a unimodular constant.

The product above is called a * Blaskche product*.

*Proof*. Let be the product on the right (without the constant). It is easy to verify that is analytic on , continuous on , and on . Then is analytic on , continuous on , and on . By the maximum principle, on . But satisfies the same conditions, and so on . Hence ; so for some .

The proof of the following result is similar.

**Theorem 2.6.** Let be analytic on , , and are zeros of (maybe repeated not more than multiplicities). Then

,

where is analytic in and on .

**Example 2.8.** Let be analytic in the closed unit disc, with and for all with . Find the best possible upper bound of .

*Proof*. The function is analytic and bounded by one (maximum principle). By Theorem 2.6, write

,

where is analytic and bounded by . It follows that

and so . Equality is attained, for example, when .

**Example 2.7. (Blaschke condition)** Let be analytic and bounded on , with zeros . Prove that

.

*Proof. *Without loss of generality we may assume is bounded by . First, we consider the case where . For each , by Theorem 2.6 there is an analytic function with such that

.

Putting , we have

.

Taking logarithm, we have

.

For , we have the inequality . It follows that

.

Letting , we have .

Now if , then we may write , where is the order of the zero at , and is analytic and . Now we may apply the proved result to .

**3. Local behaviors – zeros, poles, and singularities**

First we try to understand * zeros*. If is holomorphic, non-constant, the zeros are isolated. A rewording is the very important

*: if vanishes on a set with an accumulation point in the domain, then is identically zero.*

**identity principle****Example 3.1.** Suppose is entire and for all . Prove that where is entire.

*Proof*. Recall that if is holomorhpic on a * simply connected domain*, then does not vanish if and only if for some holomorphic . Hence, it suffices to prove that does not vanish. Consider the function

which is entire (note that this is the formula for the * Schwarz reflection*). Then is entire and, for $x \in {\Bbb R}$,

.

Hence is identically by the identity principle, and so never vanishes.

If , then near we may write

for some integer (called the ** order**), where is analytic near and . For generally, for any , the function has a zero at . And to say that is equivalent to saying that has a zero at . This viewpoint is quite useful. For example, to prove that is

**is equivalent to proving that has no other zeros other than .**

*univalent*The local behavior of near can be decribed by the **local mapping theorem**. If and the order of the zero at is , then on a neighborhood of we can write

,

where is conformal and . (Actually for an appropriate branch.) A usual way to detect zeros is the ** argument principle**. If is holomorphic and does not vanish on a contour , the number of zeros it encloses is (counting multiplicities)

.

(We only consider simple domains such as discs and rectangles, and topological considerations are not needed.) It can also be interpreted as

,

which is the * winding number *of around . An application of this idea is the proof of

*, which states that if on , then and has the same number of zeros inside .*

**Rouche’s theorem****Example 3.2.** Suppose is analytic on and one-to-one on . Prove that is one-to-one on .

*Proof*. Let be fixed. Since is non-constant, there exists $latex \frac{1}{2} < r < 1$, aribrarily close to , such that for . We want to prove that has at most one zero in . By the argument principle, the number of zeros of in (written ) is

.

which is the winding number of around . Since is univalent for , the curve is a Jordan curve. Hence the winding number around any point is either or . Letting , we get the desired result.

**Example 3.3.** Let be a real-valued non-constant harmonic function on a neighborhood of . Describe the structure of the level set near .

*Proof*. In a neighorhood of , we have where is some non-constant analytic function. By translation, we may assume . Let be the order of the zero (which is unknown based on the given information). By the local mapping theorem, we may write

where is conformal near . Now that is the preimage of the imaginary axis under . The preimage of under the map is equally spaced straight lines passing through the origin. Hence is the conformaly image of equally spaced straight lines passing through .

Next we consider the following problem.

**Example 3.4.** Suppose that is entire, non-constant and

for all .

For , let equal the number of zeros of having modulus less than or equal to . Prove that there exist non-negative constants and such that for all ,

.

(An example of such a function is .)

Here the argument principle is not useful because what we have is an* upper bound* of . Here the appropriate tool is **Jensen’s formula.**

**Jensen’s formula.** If is analytic on , and are the zeros of in (counted according to multiplicities), and . Then

.

The idea to prove Jensen’s formula is to observe that is harmonic and then apply the mean value property. If , we may write , and then we may apply the formula to .

*Proof of Example 3.4*. Write , where . Fix and suppose are the zeros of in . By Jensen’s formula,

.

Now and

.

Combining, we have, for any ,

.

For we can drop the term, and for we simply use a large enough constant term. Explicitly, if we pick and , then for all .