Horman Der Companion 1

8/2/2019 Horman Der Companion 1

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THE H ORMANDER COMPANION

JENNIFER HALFPAP

As the title suggests, these notes are designed as a companion to H ormandersnow-classic book An Introduction to Complex Analysis in Several Variables [Hor90].This book has survived for many decades because it is concise and efficient. On theip side, these characteristics can make it hard for the student to read. It assumesconsiderable background and tends to leave details to the reader.

Thus, in addition to being a graduate course in the analysis of functions of one

and several complex variables, this is a course on learning how to learn mathematics.Sometimes it is tempting to abandon a hard book or paper because one doesnthave all the background. This book, for example, uses the language of differentialforms and the wedge product on the second page. If you havent met these ideasbefore and hear that they are taught in a course on manifolds, you might thinkyou need to wait until after such a course to read this book. Similarly, this bookuses theorems commonly rst encountered in a course on measure and integrationtheory. You might think that therefore such a course is also a prerequisite.

My perspective is that if you wait to read a book or paper until you have mas-tered all the prerequisites, you wont progress. Instead, when you come up againstmaterial assumed as background with which you are unfamiliar, aim to developenough of a working knowledge to be able to proceed with your main goal. As yourinterests develop, it will become clear which auxiliary areas require more attentionon your part and which are peripheral.

These notes are designed to help you learn to do this. They include some expos-itory material that may make reading H ormanders exposition easier, they includeexercises that allow you to practice using denitions and results, and they help youll in some of the details left to the reader.

1. Analytic Functions of One Complex Variable

Let be an open subset of C and consider f : C . Write f in terms of itsreal and imaginary parts, i.e., if z = x + iy, f (z) = u(x, y ) + iv(x, y ) for real-valuedfunctions u and v of the two real variables x and y. Hormander denes f to beanalytic on if f C

1() and f z = 0, where

z :=12 (

x + i

y ). It then follows

that for f analytic, df = f z

dz for z

:= 12

( x

i y

).

Exercise 1. Using this denition of analyticity, determine the system of (real)partial differential equations satised by u and v.

In most undergraduate texts on complex analysis, analyticity is dened as fol-lows:

Denition 1. f : C is analytic at z0 if (1) lim

zz 0f (z) f (z0)

z z01


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2 JENNIFER HALFPAP

exists, in which case the value is denoted by f (z0). f is said to be analytic on theopen set if it is analytic at every point of .

Exercise 2. Consider a function f = u + iv for which the above limit exists for some z0 = x0 + iy0 . Show that at this point, u and v satisfy the same system of partial differential equations identied in the previous exercise. (Hint: Consider restricted approaches to z0 .) Then show that this denition of f agrees with what H ormander would write as f z .

The point of H ormanders approach is that it reects the intuition that analyticfunctions are those functions on C which are independent of z. Such intuition isreinforced by the following elementary exercises:

Exercise 3. Dene ez := ex cos y + iex sin y. Show that this denes an analytic function on all of C . (Such functions are called entire .)

Exercise 4. The modulus of a complex number z = x + iy is its usual Euclidean distance from the origin and is denoted by |z|. Show that |z|2 = zz, and determineall points of C at which f (z) = |z|2 is analytic.Exercise 5. Show that f (z) = zn is entire for n

N . This can be done in a number of ways. In particular, either denition of analyticity can be used. Try it both ways.

2. Stokes/Greens Theorem, the Cauchy Integral Formula, andConsequences

Most undergraduate books in complex analysis take a long time working up tothe Cauchy Integral Formula, and then usually only give partial proofs for ratherspecial domains. Few of these presentations do what H ormander does and obtainthe result from a version of Stokes/Greens Theorem using the language of differ-ential forms and the exterior derivative. Furthermore, these latter topics are oftentreated in a similarly incomplete manner in the undergraduate curriculum. Suchpedagogical choices are appropriate and understandable; a thorough treatment of Stokes theorem is usually left to a course in differential topology or smooth mani-folds.

We will commit similar sins here and neither prove Stokes theorem nor developthe underlying ideas fully. Such a development would be inappropriate here becausewhat is really being used is the more elementary special case of Stokes Theoremknown as Greens Theorem. On the other hand, consistent with the spirit of thisexposition, we aim to move the reader one step closer to an understanding of thisdeep result by introducing, albeit informally, notions of manifold, tangent space,and differential form, followed by two statements of Greens theorem and the proof

of the Cauchy Integral Formula given in H ormanders Section 1.2.2.1. What is a manifold? Roughly speaking, a k-dimensional manifold is a setwhich near every point resembles R k . For example, consider the curve in the planewith equation y = x2 . This set is most certainly not a vector subspace of R 2 .However, this curve is locally linear; at every point along the curve there is awell-dened tangent line. We mention two other (trivial) classes of examples of manifolds: open subsets U of R 2 , and any nite set of points in R 2 .

For our purposes in this section, the three examples above are about all we need;we will consider curves in the plane which have a well-dened tangent line at all


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THE H ORMANDER COMPANION 3

but nitely many points (e.g., the unit circle or the boundary of some polygonalregion in the plane), open sets in the plane whose boundaries are these sorts of

curves, and nite collections of points in the plane.For any manifold, there is an associated notion of dimension , corresponding tothe dimension of the Euclidean space it resembles locally. Thus an open set U in R 2is a manifold of dimension 2, the unit circle in R 2 is a manifold of dimension 1, anda point or collection of points is a manifold of dimension 0. Also associated withany manifold is its tangent space at a point. In the case in which the manifoldis also a subset of some R n , this notion of tangent space corresponds closely to thenotion developed in calculus, except that because we want to think of the tangentspace to a manifold at a point as a vector space, we always think of our tangentspaces as going through the origin.

Let us consider the three examples above. In all cases, since the manifold underconsideration is a subset of R 2 , we will also think of the tangent space as a subspaceof R 2 . If p is a point of an open subset U of R 2 , the tangent space to U at p, denotedT p(U ), is just a copy of R 2 . For the point p = (1 / 2, 1/ 2) on the unit circle C ,since the equation of the tangent line is y1/ 2 = (x 1/ 2), the tangent spaceis given by T p(C ) = {(x, y) : y = x }. Finally, the tangent space to the manifoldconsisting of the single point p is T p({ p}) = {(0, 0)}.Exercise 6. Look up the denitions of a diffeomorphism, a smooth k-dimensional manifold M

R n , and the tangent space T p(M ) to M at p. Using these denitions,show that the unit circle C in R 2 is indeed a smooth 1-dimensional manifold and that (as claimed above) T (1 / 2,1/ 2) (C ) = {(x, y) : y = x }. A good source is[GP74], p.3 and p.9.

2.2. Tangent vectors and forms on R n . Consider R n as an n-dimensional man-ifold, or, more generally, consider an open subset U of R n . As discussed in theprevious subsection, at any point p, the tangent space T p(R n ) (resp. T p(U )) is justR n itself. We could denote the standard basis for this vector space by {e1 , . . . , en },but to stress that this copy of R n arises as the tangent space to a manifold at apoint, a more common notation for this basis is

{

x 1, . . . ,

x n }.

Denition 2. A 1-form at pR n (or U ) is a linear map : T p(R n ) R . Theset of all 1-forms at p is the dual space to T p(R n ) and is denoted T p (R n ).

The above is a special case of a standard construction in linear algebra:

Denition 3. Let V be a normed vector space over R . A linear functional is a linear map : V

R . The set of all continuous linear functionals on V is called

its dual space and is denoted by V .

It is a standard result from linear algebra that the space dual to a vector space(equipped with the obvious operations) is itself a vector space, and when the originalspace has nite dimension n, so does the dual space. We denote by {dx1 , . . . , d x n }the basis dual to the standard basis, so that the following relations hold:

dx i (

x j) = i,j .

More generally, we dene k-forms on R n .


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4 JENNIFER HALFPAP

Denition 4. A k-form on R n (thought of as T p(R n ) or T p(U )) is a function : (T p(R n ))k R that is multi-linear and anti-symmetric, i.e.,

(a) if L1 , . . . , L k are vectors and L i = M 1 + M 2 , then (L1 , . . . , M 1 +M 2 , . . . , L k ) = (L1 , . . . , M 1 , . . . , L k ) + (L1 , . . . , M 2 , . . . , L k ).

(b) (L1 , . . . , L i , . . . , L j , . . . , L k ) = (L1 , . . . , L j , . . . , L i , . . . , L k ).We denote the space of k-forms by k (T p (R n )) .

The rst essential remark is that you are already quite familiar with one exampleof the above.

Example 5. The determinant function det denes an n-form on R n . Indeed, toevery collection of n vectors in R n , it assigns a real number. This function is linear in each entry, and if two vectors are interchanged, the determinant changes sign.In fact, det is the unique n-form on R n for which det( x 1 , . . . ,

x n ) = 1 .

Not only does the determinant provide a familiar example of a form, but also insome sense all k-forms on R n can be thought of as generalizations of the determi-nant.

We dene an operation that allows us to combine forms to obtain a new form.

Denition 6. Let be a k-form on R n and an l-form. We dene :(T p(R n ))k + l R by (2)(L

1 , . . . , L k , Lk+1 , . . . , L k + l ) =S k + l

sgn ()(L (1) , . . . , L (k ) )(L (k+1) , . . . , L (k+ l) )

where the set S k+ l consists of all permutations of {1, . . . , k + l}.For example, if we consider the 1-forms dx1 and dx2 on R 2 , their wedge product

dx1

dx2 is a 2-form on R 2 . If L1 = a x 1 + c

x 2 and L2 = b x 1 + d

x 2 , then

dx1dx2(L1 , L2) = + dx1(L1)dx2(L2) dx1(L2)dx2(L1) = ad bc,

which we recognize as the determinant of the 2 2 matrixa bc d .

The following proposition contains properties of this wedge product.

Proposition 7. Let , be k-forms, let be an l-form, let be an r -form, and let , be scalars.

(1) is a k + l-form (i.e., it is multi-linear and antisymmetric).(2) ( + ) = () + ().(3) ( ) = ( ) . We may thus use the notation for either.

Exercise 7. Use the above denition to describe

(1) dx1dx1

(2) the relationship between dx1dx2 and dx2dx1 .(3) 123 where each i is a 1-form on R 2 .

State propositions generalizing the above for wedge products of forms on R n .

We make a few nal comments in light of general results suggested by the aboveexercise. The only interesting spaces k (T p (R n )) are for k n, so henceforth whenwe talk about k-forms on R n , we will be thinking of k n. Furthermore, there is anobvious vector space structure on k (T p (R n )), and {dx I := dx i 1dx i 2. . .dx i k :1 i1 < i 2 < . . . < i k n }is a basis.


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6 JENNIFER HALFPAP

2.4. The theorems. The result you know as Greens Theorem is probably some-thing close to the following:

Theorem 12 (Greens Theorem, First Version) . Let R 2 be an open, connected,

and bounded set whose boundary is piece-wise smooth and positively-oriented (i.e., as one traverses the boundary in the positive sense, lies to ones left). If P and Q are smooth throughout some open set containing , then

Pdx + Qdy = Qx P y dxdy.This can be expressed in a different way using forms and exterior derivatives.

For the 1-form = P dx + Qdy (using subscripts to denote partial derivatives)d = dP dx + dQdy

= ( P x dx + P y dy)dx + ( Qx dx + Qy dy)dy= P x dx

dx + P y dy

dx + Qx dx

dy + Qy dy

dy= P y dxdy + Qx dxdy.

We may thus restate Greens Theorem:

Theorem 13 (Greens Theorem, Second Version) . Let R 2 be an open, con-

nected, and bounded set whose boundary is piece-wise smooth and positively-oriented. If is a smooth 1-form on a neighborhood of ,

= d.Greens theorem is a special case of Stokes theorem, which says more generally

that for an oriented k-manifold M with boundary M having the induced orienta-tion, the integral of a k 1 form on the boundary equals the integral of its exteriorderivative over the manifold. Stating it more precisely would require us to extendour denitions of vector, form, smooth differential form, and exterior derivative tofunctions dened on manifolds. This is beyond the scope of this course.

We now return to the complex setting. Think of as a connected, open, boundedsubset of C with piece-wise smooth boundary. Consider = fdz for f C

1().Then

d(fdz ) = df dz

=f x

dx +f y

dy (dx + idy)

= if x

dxdy f y

dxdy

= 2 if z dxdy

=f z

dzdz.

Greens Theorem and this calculation thus give the following:

Corollary 14 (Cauchy Integral Theorem) . If and satisfy the same hypothesesas in Greens Theorem and if f is analytic in a neighborhood of , then

f (z) dz = 0 .


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Exercise 9. Prove the identity 2idxdy = dzdz used above.

Exercise 10. Let C denote the positively-oriented unit circle. For each n

Z ,evaluate (with justications) C z

n dz. Look up the denition of a simple closedcurve in C . What happens if C is replaced by a simple, closed, positively-oriented curve having the origin in its interior?

We can now state the Cauchy Integral Formula.

Theorem 15 (Cauchy Integral Formula) . If f C 1(),

(5) 2if ( ) = f (z)z dz + f z

z dzdz.

Exercise 11. Fill in the details of H ormanders proof of this theorem. In particular,include complete justication of the statements

lim0

2

0f ( + e i ) d = 2 f ( )

and

lim0 f zz dzdz =

f z

z dzdz.

2.5. H ormanders Theorem 1.2.2.

Theorem 16. Let be a measure with compact support K in C . The integral

(6) f ( ) = 1z d(z)denes an analytic C function on C \ K . Furthermore, in any open set whered = (2 i )1dzdz for C

k (), we have f C k () with f z = if k 1.

To summarize, this theorem tells us that (1) for not in the support of ,integrating against 1 / (z ) produces an analytic function and (2) for in thesupport of , if equals (2 i )1dzdz, then f has the same degree of smoothnessas and satises f = .

Understanding the theorem and its proof requires us to understand what a mea-sure is and how to differentiate a function dened as an integral.Measures. The theory of measure and integration is a course in itself. In this section,we give a denition of measure which should be enough to allow us to proceed.

Denition 17. Let X be a compact subset of R n and let C (X ) denote the vector space of complex-valued continuous functions on X , equipped with the sup-norm

||f || := sup xX |f (x)|. A complex measure on X is a continuous complex-valued linear functional on C (X ).If is a complex measure on X , we could denote its value on an f C (X ) by(f ), but it is more traditional to denote it by f (x) d(x). The following exercisehelps to explain why such notation is used.

Exercise 12. Suppose is itself a continuous function on the compact subset X of R n . Show that

f f (x)(x) dxdenes a complex measure on X . A common notation for this measure would be dx.


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Theorem 18. Let be a complex measure and let {f n }and f be integrable. Sup-pose for (almost) every x, limn f n (x) = f (x) and that there exists a function g with |g|d < such that |f

n (x)| |g(x)| for all n and for (almost) every x.Then limn f n (x) d = f d .In other words, this theorem says that the limit of a sequence of integrals is theintegral of the limit if the integrands converge pointwise (almost everywhere) tothe limit and if there is a single integrable function dominating all the functionsin the sequence.

We use the Dominated Convergence Theorem to prove the following theorem onthe continuity and differentiability of functions dened by integrals.

Theorem 19 ([Fol99], Theorem 2.27) . Suppose f : X (a, b) C ( < a


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10 JENNIFER HALFPAP

Exercise 15. Return to H ormanders Theorem 1.2.2. If = t + is , show that f tand f s can be obtained by differentiating under the integral sign and that

f = 0 on

C \K . (If the presence of the measure makes you uncomfortable, assume throughout that d = (2 i )1dzdz.)Completing the Proof of Theorem 1.2.2. The above discussion and exercise establishthe rst statement of the Theorem, that f ( ) = (z )1 d(z) is C and analyticon C \ K . We thus consider the second assertion.Furthermore, we treat only the case = R 2 , so that for all z, d(z) = (2 i )1(z)dzdz for in C k (R 2) with compact support K . Then

f ( ) =1

2i (z)z dzdz=

12i (z + )z dzdz.

For every = t + is , the integrand is the product of a compactly-supported functionand a locally-integrable function and is thus integrable. Since C

k with compactsupport K , the integrand is a continuous function of t and s with the integrableupper bound z1|||| K (z). By Theorem 19, f is continuous. If k 1, since anyderivative of up to order k is continuous and compactly supported, any derivativeof the integrand in t and s up to order k is bounded by an integrable function of zalone. It follows from Theorem 19 again that f C

k with derivatives of f obtainedby differentiating under the integral sign.

Thus if k 1,f

( ) =1

2i (z + )z dzdz= 1

2i z (z + )

zdzdz

=1

2i z (z)z dzdz.The last integral is over all of R 2 , but since has compact support K , we obtainthe same value if we integrate instead over an open disc B (0, R) large enough thatit contains K . Since is identically zero on B (0, R),

f

( ) =1

2i B (0 ,R ) z (z)z dzdz=

1

2i B (0 ,R )

(z)

z dz +

B (0 ,R )

z (z)

z dz

dz

= ( ),

where for the last equality we have used the Cauchy Integral Formula. Theorem1.2.2 is thus established when = R 2 .

2.6. Consequences of the Cauchy Integral Formula. It is not immediatelyobvious that the condition that an f in C 1() is also analytic is a particularlystrong one. After all, we have seen that this is merely equivalent to complex dif-ferentiability of f , i.e., the existence of lim zz 0

f (z )f (z 0 )zz 0 for every z0 . Just


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as real-valued functions can be once-differentiable but not twice-differentiable, itseems there must exist functions that are once complex-differentiable but not twice

complex-differentiable.This is, in fact, not the case. It turns out that the condition of complex dif-ferentiability is quite strong; analytic functions are remarkably well-behaved. Asevidence, we establish a number of consequences of H ormanders Theorems 1.2.1and 1.2.2.

Corollary 20. If f A(), f C () and f A().Proof. Fix . There exists r > 0 and 0 such that B ( 0 , r ) andB ( 0 , r ). By the Cauchy Integral Formula,

f ( ) =1

2i B ( 0 ,r ) f (z)z dz=

1

z (2i )1

f (z) B ( 0 ,r ) (z)dzdz.Since d(z) := (2 i )1f (z) B ( 0 ,r ) (z)dzdz is a compactly-supported measure(supported on B ( 0 , r )), by Theorem 1.2.2, f is C and analytic on C \ B ( 0 , r ),in particular at . Consider g := f . Then gC

1(B ( 0 , r )) andg

= 2f

=

f

= 0

for all B ( 0 , r ). Thus f is analytic at and the result follows.

Thus H ormander uses his quite general Theorem 1.2.2 to conclude that an ana-lytic f is in fact innitely-differentiable and that all of its complex derivatives areanalytic. One can obtain the result more directly:

Exercise 16 (Cauchy Integral Formula for Derivatives) . Let f be analytic on .For and B (, R ), show without appealing to Theorem 1.2.2 that for any jN ,

(8) f ( j ) ( ) =j !

2i B (,R ) f (z)(z )j +1 dz.The next exercise contains number of familiar and beautiful consequences of the

above exercise. Of course any undergraduate text in complex analysis will have theproofs; try to prove them on your own.

Exercise 17. Let f : C be analytic. Suppose 0 and that B ( 0 , R ).(1) (Cauchy Estimates)

(9)|f ( j ) (

0)|

j !

R jsup

| 0 |R |f ( )

|.

(2) (Liouvilles Theorem) If f is bounded and entire, f is constant.(3) (Fundamental Theorem of Algebra) If f is a non-constant polynomial

over C , there exists z0C such that f (z0) = 0 .

It is also natural to ask for what notion of limit is the limit of a sequence of analytic functions analytic? We will prove the following:

Proposition 21. If f n A() and f n f uniformly on compact subsets of ,then f A().


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12 JENNIFER HALFPAP

Exercise 18. You may wish to consult an undergraduate analysis text such as[Str00].

(1) Let {f n }be a sequence of continuous functions from [a, b] to R . For what mode of convergence is the limit function f guaranteed to be continuous? State such a theorem and give an example showing that if the hypothesis isviolated, the conclusion may fail.

(2) Let {f n }be a sequence of differentiable functions from [a, b] to R . Supposef n (x) f (x) for all x. Under what additional hypotheses is f differentiablewith f (x) = lim n f n (x)? State such a theorem and give an exampleshowing that if the hypothesis is violated, the conclusion may fail.

The proof of the proposition requires the next theorem on the size of derivativesof an analytic function.

Theorem 22 (Hormanders Theorem 1.2.4) . For every compact subset K of and every open neighborhood of K in , there exist constants C j such that for all f A(),

(10) supzK

|f ( j ) (z)| C j |f (z)|dxdy.Proof. Fix a compact subset K and a bounded open neighborhood of K with.Let C 0 () denote the set of all innitely-differentiable functions with compactsupport contained in . Let C 0 () with the additional property that 1on a neighborhood of K . Consider f A() and dene g := f . We want toapply the Cauchy Integral Formula to g on . Since g(z) 0 on and g z = f z ,

g( ) = f ( )( )

=1

2i f (z) z (z)z dzdz.Since is a smooth function with compact support, its rst partial derivatives arebounded functions. Thus there exists M satisfying |/ z| M on . Further-more, since / z 0 on , for K there exists d > 0 such that |z | d onthe support of the integrand. Thus for K

|f ( )| = |g( )|= f (z) z (z)z dxdy M d |f (z)|dxdy.The result follows for j = 0.

Exercise 19. Finish the proof by rst establishing that for j N and K ,

f ( j ) ( ) =a j

2i f (z) z (z)(z )j +1 dzdz.We now prove Proposition 21.


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Proof. Fix a compact set K in and an open neighborhood of K with compactclosure in . (This is expressed succinctly as K .) Applying the previous

theorem to f n f m givessupzK

|f n (z) f m (z)| C 1 |f n (z) f m (z)|dxdy.Since the sequence {f n }is uniformly convergent, it is uniformly Cauchy, and henceby the Dominated Convergence Theorem, the right-hand side of the above inequal-ity tends to zero. The sequence {f n /z }is thus uniformly convergent on K .Since the f n are analytic,

f n z

=12

f nx

+ if ny

= 0 ,

i.e., f nx = i f ny . Thus f nz = i f ny = f nx and the uniform convergence of

{f n /z }on K implies the uniform convergence of the sequences {f n /x }and{f n /y }. We now invoke the real-variable result to conclude that f x = lim n f nxand f y = lim n

f ny , so that

12 (

f x + i

f y ) = lim n

12 (

f nx + i

f ny ) = 0.

As a consequence, we obtain

Corollary 23. Suppose the power series f (z) = j =0 a j z j converges on B (0, R).Then f A(B (0, R)) .

Proof. We will show that for every 0 < r < R , the series j =0 a j z j convergesuniformly and absolutely on B (0, r ). Recall that to say that this series convergesuniformly on B (0, r ) means that the sequence {f n }of partial sums, dened byf n (z) :=

nj =0 a j z

j , converges uniformly to f on B (0, r ). Uniform convergence on

every closed ball B (0, r ) in B (0, R) would then imply that f n f uniformly oncompact subsets of B (0, R). Since the f n are polynomials in z, they are analytic,and so Proposition 21 would give the analyticity of the limit function f .

The proof that a power series whose convergence set includes B (0, R) convergesuniformly and absolutely on every B (0, r ) B (0, R) is left to the reader as anexercise.

Exercise 20. Go to it, reader! You may begin by recalling or proving that if wehave a series u j of functions all dened on a common ball B , and if M j isa convergent series of non-negative numbers, then if |u j | M j throughout B , theseries u j converges uniformly and absolutely on B. This is sometimes called theWeierstrass M-test .

We are now ready to prove the converse of the above, which is yet another resultillustrating our point in this section that the condition of analyticity is remarkablystrong.

Theorem 24 (Hormanders Theorem 1.2.8) . If f A(B (0, R)) , we have

(11) f (z) =

j =0

f ( j ) (0) j !

zj ,

with uniform convergence on every compact subset of B (0, R).


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14 JENNIFER HALFPAP

Proof. For 0 < r < < R , by the Cauchy Integral Formula (with a change innotation), for |z| r ,

f (z) = 12i | |= f ( ) z d.Observe,

1 z

=1

1(1 1z)

=1

j =0 j zj ,

where we have used the formula for the sum of a convergent geometric series.Dene gn (z, ) := f ( )

nj =0 j zj and g(z, ) :=

f ( ) z . Since

|gn (z, )| sup | |= |f ( )|n

j =0

j r j

=sup | |= |f ( )|

1

1 r,

gn g uniformly. The following is therefore legitimate:1

2i | |= f ( ) z d = 12i | |= limn gn (z, ) d = lim

n 1

2i | |= gn (z, ) d = limn

n

j =0

12i | |=

f ( ) j +1 d zj

=

j =0

f ( j ) (0) j !

zj .

For functions that are merely in C (), knowing the function in a neighborhoodof a single point tells us nothing about its behavior in other parts of the domain.The situation is radically different for those functions represented by power series.

Corollary 25 (Uniqueness of analytic continuation) . As usual, let C be a

connected open set. If f

A() and if there exists z0

such that f ( j ) (z0) = 0 for all j 0, then f 0 on .Proof. Let Z := {z : f ( j ) (z) = 0 , j = 0 , 1, 2, . . . }. Since Z = j =0 (f ( j ) )1({0})and each function f ( j ) is continuous, Z is a closed subset of .

On the other hand, by hypothesis, Z is not empty since z0Z . Fix Z andconsider g dened by g(z) := f (z + ). g is analytic in some neighborhood B (0, R)of the origin, and on this disc it is represented by the series f ( j ) (z0)zj /j !. Thusg 0 on B (0, R), i.e., f 0 on B (, R ). This proves that Z is open in . Since Z is both open and closed in and is connected, Z = .


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Exercise 21 (Laurent series) . Let 0 r < R and dene the annulus a(r, R ) :={z : r < |z| < R }. If f is analytic on this annulus, then

f (z) =

j = a j z j , aj =

12i | |= f ( ) j +1 d,

where r < < R . (Suggestion: Apply the Cauchy Integral Formula to a(r , R ) for r < r < R < R and use the idea of the proof of Theorem 24.)

2.7. Mean-Value and Maximum Modulus Properties. Whereas the conse-quences of the Cauchy Integral formula discussed in the previous subsection largelyconcern regularity properties of analytic functions, the results of this section beginto explore the extent to which, for f A(), we can relate its behavior on to itsvalues on .

To begin with, if f

A() and B (, R )

, then

f ( ) =1

2i B (,R ) f (z)z dz=

12i 20 f ( + Re i )Re i iRe i d

=1

2 20 f ( + Re i ) d.This is the mean value property for analytic functions: for a function analyticon B (, R ), its value at the center of the disc is equal to the average of its valueson the boundary. From this, the maximum modulus principle will follow.

Theorem 26 (Maximum Principle - local version) . Suppose f is analytic on B (, R ). If for all zB (, R ), |f (z)| |f ( )|, then f is constant on B (, R ).Proof. If f ( ) = 0, then |f (z)| |f ( )| on B (, R ) gives f 0 on B (, R ). Thus inthe remainder of the proof we may suppose f ( ) = 0. Fix 0 < r < R . By the meanvalue property,

f ( ) =1

2 20 f ( + re i ) d.Equivalently,

0 =1

2

2

0

f ( ) d 1

2f ( + re i ) d,

or, dividing by f ( ) and writing the right-hand side as a single integral,

0 =1

2 20 1 f ( + re i )f ( ) d.It follows that

(12) 0 =1

2 20 Re 1 f ( + re i )f ( ) d.


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16 JENNIFER HALFPAP

Write z = + re i and observe

2Re 1 f (z)f ( ) = 1 f (z)f ( )

+ 1 f (z)f ( )= 2

f (z)f ( ) f (z)f ( )|f ( )|2

=2|f ( )|2 f (z)f ( ) f (z)f ( )

|f ( )|2

|f ( )|2 f (z)f ( ) f (z)f ( ) + |f (z)|2

|f ( )|2= |f ( ) f (z)|2

|f ( )|2

0.Return to equation (12). The integrand is a continuous function of . Furthermore,the last calculation shows it to be non-negative. Thus it can only have integral zeroif it is zero for all , i.e., f ( + re i ) = f ( ) for all . Since r < R was arbitrary,the result follows.

Theorem 27 (Maximum Principle) . Suppose f is analytic on a domain C

and f extends continuously to . Then f attains its maximum modulus on .

Exercise 22. Prove Schwarzs Lemma : Let f be analytic on B (0, 1) with f (0) =

0 and |f (z)| 1 for all z B (0, 1). Then |f (0) | 1 and |f (z)| |z| for all z B (0, 1). If |f (0)| = 1 or if there exists z0 B (0, 1) such that |f (z0)| = |z0|,then there exists C with | | = 1 such that f (z) = z for all zB (0, 1).

3. Runges Theorem

This theorem is often skipped in a rst course in complex variables. If youwould like to read a more elementary and more extensive discussion of the theoremthan that given by H ormander, see Conways book [Con78]. In fact, we will followConway to some degree here.

Look again at Theorem 24. One way to restate it is to say that every analyticfunction on B (0, R) is the uniform limit on compact subsets of a sequence of poly-nomials. We consider generalizations. For instance, if we replace B (0, R) by anarbitrary open set G, is it true that for every f A(G) and every compact K G,f can be uniformly approximated on K by polynomials? As the next exampleshows, the answer is no.

Example 28. Take G := {z : 0 < |z| < 2}. Then f (z) = z1 is analytic on G.Suppose we had a sequence { pn }of polynomials such that pn f uniformly on compact subsets of G. In particular, since K := B (0, 1) is a compact subset of G,


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pn f uniformly on B (0, 1). Thus (if the circle is oriented counter-clockwise)

2i = B (0 ,1) z1

dz

= B (0 ,1) limn pn (z) dz= lim

n B (0 ,1) pn (z) dz= 0 .

Since this is a contradiction, no such sequence of polynomials exists.

We make two remarks concerning this example. First, neither Gc nor K c isconnected. Second, in the notation of Exercise 21, G = a(0, 2), and so everyanalytic function on G is a limit (uniform on compact subsets) of a sequence of rational functions with poles outside of G.

In light of the above, we might try to characterize those G with the property thatevery analytic function on G can be approximated uniformly on compact subsets bypolynomials. We might also look for a more general theorem on the approximationof analytic functions on a set G by rational functions with poles outside G.

Theorems of this sort vary in their precise formulation but tend to go by the nameof Runges Theorem. In many texts such as Conways [Con78] Runges Theorem isvery clearly about the uniform approximation of analytic functions on a compactset by polynomials, whereas in H ormanders text, no explicit mention is made of rational functions. We include both statements here so that the reader may comparethem. We will only prove the version stated by H ormander.

Denition 29. Consider C := C {}. We put a topology on this set by taking as a basis for the topology all open discs B (a, r ) in C together with sets of the form {z : |z| > r }{}. Because C with this topology is homeomorphic to the 2-sphereS 2 , we visualize C this way and call it the Riemann sphere .Theorem 30 (Runges Theorem [Con78]) . Let K be a compact subset of C and let E be an open subset of C \ K that meets each component of C \ K . Let bean open set containing K . If f A() and if > 0 is given, then there exists a rational function R, all of whose poles lie in E , such that sup zK |f (z) R(z)| < .Theorem 31 (Runges Theorem [Hor90]). Let be an open set in C and let K bea compact subset of . The following conditions on and K are equivalent:

(a) Every function analytic in a neighborhood of K can be approximated uniformly on K by functions in A().

(b) The open set \ K has no component which is relatively compact in .(c) For every \ K there exists f A() such that (13) |f ( )| > supK |f |.

If we take = C , we obtain a theorem on the uniform approximation of analyticfunctions by polynomials.

Corollary 32. Let K be a compact subset of C . Every function analytic in a neighborhood G of K can be uniformly approximated by polynomials on K if and only if K c is connected, or, equivalently, if and only if for every K

c there existsa polynomial p in z such that | p( )| > supK | p|.


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18 JENNIFER HALFPAP

Upon a rst reading, you may nd the statement of this theorem somewhatintimidating. This is a normal response. It is also likely that jumping right into

the proof will not help. Something useful to do at this point is to look at someexamples. Examples dont prove theorems, but they are the key way we buildintuition.

For simplicity, let us take to be all of C . Dene two compact subsets: K 1 :=a(1/ 2, 3/ 2) and K 2 := B (0, 1).

Consider rst \ K 1 . This has two components, namely B (0, 1/ 2) and {z :|z| > 3/ 2}. The rst component, call it O, has closure {z : |z| 1/ 2}, which is acompact subset of . Therefor the pair ( , K 1) violates (b). If the theorem is true,we should be able to see that (a) and (c) are violated as well. Indeed, the examplepreceding the statements of the theorems essentially addresses (a); the functionf (z) = z1 is analytic on C \{0}, which is a neighborhood of K 1 , but this functionis not the uniform limit on K 1 of a sequence of entire functions.

To see that (c) is also violated, we must nd

\K 1 such that for all f

A(),

|f ( )| supK 1 |f |. Take = 0 and any entire f (since = C ). Then since f isanalytic on any B (0, r ), by the maximum principle,

|f (0)| sup|z |= r |f (z)|,

and in particular

|f (0) | sup1/ 2|z |3/ 2 |f (z)| = supK 1 |f |.

On the other hand, \ K 2 = {z : |z| > 1}. This set is connected and its closureis {z : |z| 1}, which is certainly not compact in . Thus (b) is satised, andif the theorem is true, (a) and (c) must hold as well. Concerning (a), we havealready proved that if G is an open set containing B (0, 1), any function analytic on

G is the uniform limit on the compact subset B (0, 1) of the sequence of polynomialsconsisting of the partial sums of the power series representing the analytic function.To see that (c) holds, note that if \ K 2 , then | | > 1. Consider f (z) = z.This is entire and

| | = |f ( )| > 1 = supK 2 |f |.With these examples in mind, we turn to the proof of the theorem.

Exercise 23. H ormander begins by proving that (c) implies (b). Read this part of his proof, and write out a version of it for yourself in which you ll in any detailsyou need. Focus especially on the statement that, if O is a component of \K with compact closure in , then for every f A(),(14) sup

O|f | supK |f |.

Proof of (a)(b). This will be a proof by contradiction. Suppose (b) fails, and letO denote a component of \ K with compact closure in . Since (a) is valid, forany f analytic in a neighborhood of K , there exists a sequence {f n }of elements of A() such that f n f uniformly on K . Inequality (14) yieldssup

O |f n f m | supK |f n f m |.Thus {f n } is also uniformly convergent on O. Let F denote the limit function.Since F is the uniform limit on compact subsets of O of a sequence of analytic


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functions, F is analytic on O. As a uniform limit of continuous functions, it is alsocontinuous on O . Furthermore, since F and f are both continuous on O and are

the limit of the same sequence there, F = f on O .The above argument applies to any f analytic in a neighborhood of K . Thus if O, f (z) :=

1z is analytic in the neighborhood

C \ { }of K , and we obtainan associated F analytic in O and continuous on O with F (z) = 1z on O . Setg(z) := ( z )F (z). This function is analytic in O and is identically 1 on theboundary. Thus g 1 on O. Since O and g( ) = 0, this is a contradiction,proving that (a) implies (b).Exercise 24. Prove the assertion made in the penultimate sentence of the preceding paragraph: If O is a bounded open set and if g is analytic on O and continuous on O , then if g c on O , then g c on O.Proof of (b)(a). Hormander intends to apply the Hahn-Banach Theorem in this

part of the proof. See Rudins book [Rud66] for a statement of this theorem as wellas its application in the proof of Runges Theorem. We follow that developmenthere. Recall,

Theorem 33 (Hahn-Banach, [Rud66]) . If M is a subspace of a normed vector space X and if is a bounded linear functional on M , then can be extended to a bounded linear functional on all of X without increasing the norm, i.e.,

||||= ||||.The following is a consequence:

Proposition 34. Let M be a subspace of a normed vector space X and let x0X .Then x0 is in the closure M of M if and only if there exists no bounded linear functional on X with (x) = 0 for all x in M but (x0) = 0 .

Exercise 25. Prove Proposition 34.We are now prepared to understand H ormanders application of the Hahn-

Banach theorem. For K and as in part (b), consider C (K ), the vector spaceof continuous functions on K equipped with the sup-norm. Thus C (K ) plays therole of the normed vector space X above. Let M be the subspace consisting of therestrictions of elements of A(). Let f be analytic on a neighborhood of K . Thenf X , and the question of whether or not it is the uniform limit on K of elementsof A() is equivalent to the question of whether or not it is in M . By Proposition34, to prove that f M , it suffices to prove that for every continuous linear func-tional on X with (g) = 0 for every gA(), (f ) = 0 as well. Since the set of all continuous linear functionals on K is precisely the set of complex measures, wemust show that if is a measure such that

g d = 0 for all gA(),then f d = 0.Exercise 26. With this background, work through the rest of H ormanders proof of the Runge approximation theorem. Write a summary of the key steps of thearguments that (b) implies (a) and that (b) implies (c).

We restate Runges Theorem to motivate the next denition: Given a domain, some compact subsets have the right topological properties that guarantee that


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20 JENNIFER HALFPAP

analytic functions on them are close to analytic functions on whereas other com-pact subsets do not have this property. Given a compact subset K of without

this property, it would be nice to be able to enlarge it to satisfy the hypothesesof Runges Theorem. Moreover, condition (c) in Runges Theorem gives us an-other way to describe those points we must add to K . We thus make the followingdenition.

Denition 35. Let K be a compact subset of . We dene K , the A()-hull of K by

(15) K = K := {z : |f (z)| supK |f | for every f A() }.This set has the following properties:

Proposition 36. Let K be a compact subset of , K its A()-hull.

(a) K K .(b) K satises the hypotheses of Runges theorem.(c) d(K, c) = d( K,

c).(d) If ch (K ) denotes the convex hull of K , K ch (K ).

Exercise 27. Prove part (c) of Proposition 36. One direction is immediate. For the other, follow H ormanders suggestion: For

c , consider the function f (z) =(z )1 .Proof of (b). This simply requires us to unwind the denitions. To show that K satises the hypotheses of Runges Theorem, we must show that for all \ K ,there exists f A() with |f ( )| > sup K |f |. If this were not the case, we couldnd 0

\K such that for all f

A(),

|f ( 0)

| sup

K |f

|. But then such a 0

would be an element of K . This contradiction establishes the claim.Proof of (d). We discuss part (d) because it presents us with an opportunity to beginour exploration of notions related to convexity. We recall a number of denitions.

Denition 37. A subset E of R n is convex if for all x, y E , the line segment joining x and y is contained in E , i.e., (1 t)x + tyE for all t[0, 1].Denition 38. If E

R n , the convex hull of E , denoted ch (E ), is the intersec-tion of all convex sets containing E .

We can describe the convex hull of a set E in another way. We focus here on E closed since this is the situation we have in mind, though one can formulate similarresults for more general E . A hyperplane in R n is given by

{x : a j x j = c

}, a

R n . Such a hyperplane divides R n into two closed half-spaces H + := { a j x j c}and H := { a j x j c}and is said to be a support hyperplane for a set E if E iscontained in one of the half-spaces and if there exists yE such that a j yj = c.Exercise 28. Let E

R n be closed.(1) If y is exterior to ch (E ), then there exists a hyperplane a j x j = c such

that E H and a j yj > c . Such a hyperplane separates E and {y}.(2) If ch (E ) is not all of R n , then ch (E ) is the intersection of all support closed half-spaces containing it.


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With this background, we are ready to prove part (d) of Proposition 36. Takez0 = x0 + iy0

K and suppose that z0 is not in ch( K ). Then there exists ( a, b)

R 2

and cR such that ax + by c for all z = x + iyK but ax 0 + by0 > c . SinceRe[(a ib)(x + iy)] = ax + by, if we set = a + ib, we have

Re(z ) c but Re( z 0) > c.Now, for any complex , z ez is entire, hence in A() for any . Furthermore,for all zK

|ez | = eRe( z ) ec< e Re( z 0 )

= |ez 0 |.This contradicts the assumption that z0

K , hence K

ch(K ). This completesthe proof of part (d) of the proposition.The topological description of K . We close this section with a different descriptionof the set K which matches with the intuition that K lls in the holes of K .

Theorem 39 (Hormander, Theorem 1.3.3) . K is the union of K with the com-ponents of \ K that are relatively compact in .Proof. Suppose that O is a component of \ K that is relatively compact in .Then by inequality (14), O K . Thus if K 1 is the union of K with such rela-tively compact subsets of \ K , then K 1 K . We wish to establish the reversecontainment.

Since K 1 is the union of K with components of \ K , \ K 1 is open. It followsthat K 1 is closed and in fact compact. Furthermore, \

K 1 has no componentswhich are relatively compact in . Thus K 1 satises (b) in Runges theorem, andhence (c). Thus we conclude K 1 = K 1 . Since K K 1 , we conclude K K 1 , asdesired.

3.1. Meromorphic Functions and the Mittag-Leffler Theorem. This is an-other context in which H ormanders treatment differs substantially from what istypically done in a course on functions of one complex variable. Thus in this sectionwe begin by giving the more elementary denitions.

Here, we follow Section 1.4.2 of [Var11]. We begin by dening three kinds of singularities for functions analytic on a punctured disc.

Denition 40. Let f be analytic on a punctured disc B (, r ) := {z : 0 < |z | 1. (You probably know a result called the Residue Theorem that al lows you to dothis quickly. If you want to use that theorem, prove it rst. If you dont recall


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22 JENNIFER HALFPAP

that theorem, approach this problem by considering the integral over a contour that includes the circle of radius R together with circles of radius about the poles.)

Exercise 30. Suppose f is analytic on B (, r ).(1) If f is bounded in a neighborhood of , show that the Laurent series for f

about has no non-zero coefficients a j for j < 0.(2) Prove that f has a pole at if and only if limz |f (z)| = .

This last exercise shows that, for a function with a pole at , it makes sense toassign the value at , so that such a function could be viewed as a function intoC{}. The same can not be said of a function with an essential singularity at .With this background, we have the following standard denition of a mero-

morphic function.

Denition 41. Let C . f is meromorphic at if there is a punctured disc centered at on which f is analytic and if is not an essential singularity of

f . f is meromorphic on if it is meromorphic at every point of .

Since an analytic function h can only vanish to at most nite order at a point,it is clear that the ratio of analytic functions dened on a neighborhood of ismeromorphic at . We will eventually address the question of whether every mero-morphic functions is a ratio of analytic functions.

Now that we have reviewed the classical treatment of meromorphic functions,we discuss Hormanders formulation of the concept.

Fix z C . We consider the set of all functions which are analytic on some

neighborhood of z. We dene an equivalence relation on this set so that f g if there is a neighborhood of z on which f is identically equal to g. We let Az bethe set of equivalence classes under this relation, and denote the equivalence classdetermined by f by f z . With addition and multiplication dened in the obviousway, Az is a ring. Furthermore, it has no zero divisors, and we may form thequotient eld M z := {f z /g z : f z , gz Az , gz = 0 z }.Exercise 31. Let Az be as dened above.

(1) How should we dene f z + gz ? Show that (according to your denition)f z + gz is in fact well-dened.

(2) Give all the details of the argument that Az has no zero divisors.

We can now state H ormanders denition of a meromorphic function.

Denition 42. Let be an open subset of C . A meromorphic function on is a mapping : z M z such that for all z in , (z) M z and such that for all points z0 in there is a neighborhood and elements f, g A() such that (z) = f z /g z for all z. We denote the set of all meromorphic functions on by M () and write z instead of (z) for M ().

The aspect of this denition that matches our intuition for meromorphic func-tions is its explicit reference to quotients of (equivalence classes of) analytic func-tions. On the other hand, its formulation in terms of equivalence classes may makeyou wonder what connection these objects have with analytic functions.

The rst essential remark, then, is that A() can be identied with a subsetof M () in a natural way. Take F A(). For any z , since is open,F is indeed analytic in a neighborhood of z and hence determines an equivalence


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class F z . Since the ring Az is naturally embedded in M z , F z is identied with anelement of M z . Consider the map dened by z F z . In order for this to be ameromorphic function, there is another condition it must satisfy. Let z0 be a pointof and consider a neighborhood of z0 in . Since F A() as well, it is indeedthe case that there is a single pair of analytic functions F and 1 on for whichz := F z / 1z for all z. We conclude that the dened in this way is indeed ameromorphic function.

In order for A() to be identied with a subset of M (), we need to also knowthat distinct elements of A() are identied with different elements of M (). Thussuppose F, G A() and that these are not identically equal. Then there exists z0 such that F (z0) = G(z0). With respect to the equivalence relation associated withfunctions analytic in a neighborhood of z0 , these two functions are not equivalent,and so they determine different equivalence classes. That is, F z 0 = Gz 0 in Az 0 , andso F z 0 / 1z 0 = Gz 0 / 1z0 in M z 0 .

References[Con78] John B. Conway. Functions of One Complex Variable, second edition . Grauduate Texts

in Mathematics. Springer, 1978.[Fol99] Gerald Folland. Real Analysis: Modern Techniques and Their Applications . John Wiley

& Sons, Inc., 1999.[GP74] Victor Guillemin and Alan Pollack. Differential Topology . Prentice Hall, 1974.[Hor90] Lars H ormander. An Introduction to Complex Analysis in Several Variables, third edi-

tion . North Holland, 1990.[Rud66] Walter Rudin. Real and Complex Analysis . McGraw-Hill, 1966.[Str00] Robert Strichartz. The Way of Analysis . Jones and Bartlett, 2000.[Var11] Dror Varolin. Riemann Surfaces by Way of Complex Analytic Geometry . Graduate Stud-

ies in Mathematics. American Mathematical Society, 2011.

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