8_InfiniteProducts

Chapter 8 Infinite Products As is amply demonstrated by power series expansions, a highl y useful technique in complex analysis is to express an analytic function as an i nfinite sum of much simpler functions. Likewise, it can be useful to express an a nalytic function as an infinte product of simpler functions. This is especially t rue in the study of the zeroes of analytic functions. For example, a polynomial pof degree ncan be written as a product p(z) =an/productdisplay j=1(z−zj), where {z1, z2,·, zn}are the zeroes of p. It turns out that product expansions of a similar type (but with infinitely many factors) are possibl e for other analytic functions. Since the exponential function converts sums to products, w e can expect that the theory of infinite products will be closely related to the theory of infinite sums. 8.1 Convergence of Infinite Products In the following discussion, log will be the principal branc h of the log function. Definition 8.1.1. If{uk}is a sequence of complex numbers and pn=n/productdisplay k=1uk, (8.1.1) then we will say that the infinite product ∞/productdisplay k=1uk (8.1.2) converges to the complex number pif lim n→∞pn=p. 255 256 CHAPTER 8. INFINITE PRODUCTS Theorem 8.1.2. If{uk}is a sequence of complex numbers, then the infinite product (8.1.2) converges to a non-zero number pif and only if the infinite sum ∞/summationdisplay k=1loguk (8.1.3) converges to a number λ. In this case, p=eλ. Furthermore, if the infinite series converges absolutely, then the infinite product is unchange d by a rearrangement of the factors. Proof. We have to be careful here, because the log function converts products to sums only up to ±2πi. However, it is true that log uv= log u+ log vifu andvhave positive real part since, in this case, log uand log vhave imaginary parts in ( −π/2, π/2) and log uvhas imaginary part in ( −π, π). Thus, log uvand logu+ logvcannot differ by a non-zero multiple of 2 πi. We define the partial products pnas in (8.1.1). If p= lim n→∞pnexists and is non-zero, then lim n→∞log(pn/p) = 0, since log is continuous at 1. In particular, there is an Nsuch that −π/4<Im(log( pn/p))< π/4 whenever n≥N. It follows that pn/pm= (pn/p)(pm/p)−1is in the right half plane for n, m≥N. In particular, un+1=pn+1/pnis in the right half plane for n≥N. Thus, log(pn+1/pN) = log(( pn/pN)un+1) = log( pn/pN) + log un+1 whenever n≥N. This equation and an induction argument beginning with n=Nshow that log(pn/pN) =n/summationdisplay k=N+1loguk for all n > N . Since the left side of this equality converges as n→ ∞ so does the right side. This implies the convergence of the series (8 .1.3). Conversely, if this series converges and we let λn=n/summationdisplay k=1logun, be its nth partial sum, then the sequence {λn}converges to a number λ. Since pn= eλn, and the exponential function is continuous, the sequence {pn}converges to eλ. If the series (8.1.3) converges absolutely, then each of its rearrangements converges to the same number. It follows that each rearrange ment of the infinite product (8.1.2) also converges to the same number. 8.1. CONVERGENCE OF INFINITE PRODUCTS 257 Uniform Convergence of Products We will be primarily interested in infinite products of analy tic functions. In this situation, whether or not the product converges unifor mly is of critical importance. We say that an the infinite product of a sequence o f functions {uk} converges uniformly on a set Sif the sequence pnof partial products converges uniformly on S. Theorem 8.1.3. Letukbe a sequence of complex valued functions defined and bounded on a set S. If the series ∞/summationdisplay k=1loguk(z) converges uniformly to λ(z)onS, then the infinite product ∞/productdisplay k=1uk(z) converges uniformly to eλ(z)onS. Proof. Letλn(z) be the nth partial sum of the infinite sum and pn(z) thenth partial product of the infinite product. Then the uniform con vergence of the series on Simplies that λn(z)−λ(z) converges uniformly to 0 on S. Since the exponential function is continuous at 0, this implies that pn(z) p(z)= eλn(z)−λ(z) converges uniformly to 1. The fact that each λnis bounded on Sand the convergence is uniform implies thatλis bounded on Sand, hence, that p(z) = e−λ(z)is also bounded on S. Hence, pn= (pn/p)pconverges uniformly to ponS. Theorem 8.1.4. Let{ak(z)}is a sequence of complex valued functions defined on a set S. If the series ∞/summationdisplay k=1|ak(z)| (8.1.4) converges uniformly on S, then the infinite product ∞/productdisplay k=1(1 +ak(z)) (8.1.5) converges uniformly on S. Each rearrangement of the infinite product converges to the same function. If the infinite product converges to p(z), then each zero ofp(z)is a zero, with the same order, of some finite product of the fac tors 1 +ak(z). 258 CHAPTER 8. INFINITE PRODUCTS Proof. If|w|<1/2, then (Exercise 8.1.1) 2 3|w| ≤ |log(1 + w)| ≤2|w|. (8.1.6) If the series (8.1.4) converges uniformly on S, then there is a Ksuch that |ak(z)| ≤1/2 fork≥Kand for all z∈S. If we use (8.1.6) with w=ak, it follows that one of the two series ∞/summationdisplay k=K|log(1 + ak(z))|and∞/summationdisplay k=K|ak(z)| converges uniformly on Sif and only if the other one does also. Hence, if (8.1.4) converges uniformly then ∞/summationdisplay k=Klog(1 + ak(z)) converges uniformly and absolutely. By the previous two the orems, this is im- plies the uniform convergence of ∞/productdisplay k=K(1 +ak(z)) to a function on Swith no zeroes. It follows that (8.1.5) converges uniformly onS, the limit is unaffected by by rearrangements of the factors, and each of its zeroes is a zero, with the same order, of the product of the factors 1 + ak(z) fork < K . Example 8.1.5. Prove that the infinite product ∞/productdisplay k=1(1−z2/k2) (8.1.7) converges uniformly on each bounded subset of C. Solution: We have | −z2/k2| ≤R2/k2for all zin the disc DR(0). Since the positive termed series ∞/summationdisplay k=1R2 k2 converges, it follows that the series ∞/summationdisplay k=1|z2| k2 converges uniformly on DR(0). Hence, by the previous theorem, the infinite product (8.1.7) also converges uniformly on DR(0) for each Rand, hence, on each bounded subset of C. 8.1. CONVERGENCE OF INFINITE PRODUCTS 259 Logarithmic Derivative of a Product If an analytic function fon an open set Uhas an analytic logarithm gonU– that is, if f= egonUwithganalytic – then g′=f′/f. The expression f′/f is independent of which logarithm is chosen for f. Furthermore, as long as f is not identically zero on any component of U,f′/fexists (as a meromorphic function on U) even if fdoes not have an analytic logarithm on U. Note that f cannot have an analytic or even a meromorphic logarithm in an y neighborhood of a point where it has the value zero (Exercise 8.1.4). Definition 8.1.6. Letfbe an analytic function on an open set Uand suppose thatfis not identically 0 on any component of U. Then the meromorphic function f′/fis called the logarithmic derivative offonU. Logarithmic derivative is quite a well behaved notion. The l ogarithmic derivative of the product of two functions is the sum of there logrithmic deriva- tives (Exercise 8.1.5). Furthermore, the following theore m states that logarith- mic derivative is preserved by uniform limits. The proof is l eft to the exercises (Exercise 8.1.6). Theorem 8.1.7. Let{fn}be a sequence of analytic functions on a connected open set U. If this sequence converges uniformly to fonUthen the sequence {f′ n/fn}converges uniformly to f′/fon compact subsets of U\S, where Sis the set of zeroes of f. When applied to infinite products, this immediately implies the following corollary. Corollary 8.1.8. Let{uk}be a sequence of analytic functions on a connected open set U. If the product f(z) =∞/productdisplay k=1uk(z) converges uniformly on compact subsets of Uto a function fwhich is not iden- tically 0, then the infinite sum ∞/summationdisplay k=1u′ k(z) uk(z) converges uniformly to f′/fon compact subsets of U\S, where Sis the set of zeroes of f. Example 8.1.9. Show that the function f(z) =πz∞/productdisplay k=1/parenleftbigg 1−z2 k2/parenrightbigg has a logarithmic derivative which can be written as f′(z) f(z)=1 z+∞/summationdisplay k=12z z2−k2(8.1.8) 260 CHAPTER 8. INFINITE PRODUCTS or as f′(z) f(z)= lim n→∞n/summationdisplay k=−n1 z−k(8.1.9) Solution Note that the infinite product in the expression for fconverges uniformly on each compact disc in the plane by Example 8.1.5. By the previous theorem, f′(z) f(z)=1 z+∞/summationdisplay k=1−2z/k2 1−z2/k2=1 z+∞/summationdisplay k=12z z2−k2. This proves (8.1.8). Since 2z z2−k2=1 z−k+1 z+k, thenth partial sum of the series (8.1.8) can be re-written as the s um that appears in (8.1.9). The logarithmic derivative of f, as computed in the above example, will be used in the problem set to prove that f(z) = sin πz. That is, sin(πz) =πz∞/productdisplay k=1/parenleftbigg 1−z2 k2/parenrightbigg . (8.1.10) Exercise Set 8.1 1. Prove that if wis a complex number with |w| ≤1/2, then 2 3|w| ≤ |log(1 + w)| ≤2|w|. 2. Does the infinite product ∞/productdisplay k=1/parenleftbigg 1 +1 k/parenrightbigg converge? How about the product ∞/productdisplay k=1/parenleftbigg 1 +1 k3/2/parenrightbigg ? 3. Show that the infinite product ∞/productdisplay k=1/parenleftBig 1−z k/parenrightBig ez/k converges uniformly on compact subsets of the plane. 8.2. WEIERSTRASS PRODUCTS 261 4. Prove that if fis analytic on Uand has a zero at z0∈U, then there is no meromorphic function gdefined in a neighborhood Vofz0such that f= egonV. 5. Prove that the logarithmic derivative of the product fgof two analytic functions is the sum of the logarithmic derivative of fand the logarithmic derivative of g. Also prove the analogous statement for the quotient f/g. 6. Prove Theorem 8.1.7. Hint: first prove that it is true on any disc in Uon which fhas no zeroes. 7. Prove that the logarithmic derivative of a meromorphic fu nction fonC is also a meromorphic function on Cand is odd (even) if fis odd (even). 8. Iffis the function defined in Example 8.1.9, prove that the logar ithmic derivative of fis an odd meromorphic function which is periodic of pe- riod 1. Observe that the logarithmic derivative of sin πzhas the same properties. 9. Prove that if fis the function of the previous exercise, and we set g(z) =sin(πz) f(z), thengis an entire function with no zeroes and, hence, has a logarit hmh which is entire. Then, sin( πz) =f(z)eh(z). 10. Prove that if f,gandhare the functions of the previous exercise, then the logarithmic derivative of gish′(z) =πcotπz−f′(z)/f(z). 11. With has above, prove that h′is bounded on the strip 0 ≤Re (z)≤1 (use (8.1.8)). Show that this implies it is bounded on the ent ire plane and, hence, is constant. 12. With has above, prove that h′(0) = 0 and, hence, that h′is identically 0 andhis a constant. Then use the fact that lim z→0z−1sinz= 1 to show that this constant is 0. Conclude that sin(πz) =πz∞/productdisplay k=1/parenleftbigg 1−z2 k2/parenrightbigg . 8.2 Weierstrass Products In this section we will show that, given any sequence of point s of an open set U⊂C, with no limit point in U, there is an analytic function on Uwith exactly the points of this sequence as its zeroes, with each zero havi ng order equal to the number of times it appears in the sequence. The analytic f unction will be constructed as an infinite product of certain simple functio ns, each of which has exactly one zero. These simple functions are constructed as follows. 262 CHAPTER 8. INFINITE PRODUCTS Forp= 0,1,2,···we define entire functions Ep(z) byE0(z) = 1−zand Ep(z) = (1 −z)ez+z2/2+···+zp/pforp >0. Note that z+z2/2 +···+zp/pis the pth partial sum for the power seriies expansion of −log(1−z) about z= 0 and so, although Ep(1) = 0, the sequence Ep(z) will converge uniformly to (1 −z)(1−z)−1= 1 on each disc of radius less than 1 centered at 0. More precisely: Theorem 8.2.1. EachEp(z)is an entire function with the following properties: (a) the only zero of Ep(z)occurs at z= 1; (b) if |z| ≤1, then |Ep(z)−1|<|z|p+1. Proof. Part (a) is obvious. To prove Part (b), we note that the deriva tive of 1−Ep(z) is (Exercise 8.2.1) (1−Ep(z))′=−E′ p(z) =zpez+z2/2+···+zp/p. (8.2.1) Since this has a zero of order patz= 0, the function 1 −Ep(z) has a zero of order p+ 1 at z= 0. The function (8.2.1) has a power series expansion about 0 wit h all of its coef- ficients non-negative real numbers, since this is true of the exponential function and the function z+z2/2 +···+zp/p. It follows that the function h(z) =1−Ep(z) zp+1, also has non-negative real numbers as coefficients for its pow er series expansion about 0. This implies that the maximum value achieved by |h(z)|for|z| ≤1 is h(1) = 1. That is,/vextendsingle/vextendsingle/vextendsingle/vextendsingle1−Ep(z) zp+1/vextendsingle/vextendsingle/vextendsingle/vextendsingle≤1 for |z| ≤1. Part (b) follows from this. Iffis an analytic function on U, then we will say that a sequence {zk} ⊂U is a list of the zeroes of fcounting multiplicity if each zkis a zero of f, and if each zero woffoccurs m(w) times in this sequence, where m(w) is the order of the zero w. Let{zk}be a sequence of non-zero complex numbers converging to ∞. The next theorem show how to use scaled versions of the functions Epto construct an entire function with this sequence as a list of its zeroes c ounting multiplicity. The resulting product is called a Weierstrass product . Theorem 8.2.2. LetAbe a subset of C. If{zk}is a sequence of non-zero complex numbers and {pk}is a sequence of integers such that ∞/summationdisplay k=1/vextendsingle/vextendsingle/vextendsingle/vextendsingler zk/vextendsingle/vextendsingle/vextendsingle/vextendsinglepk+1 <∞for all r >0, (8.2.2) 8.2. WEIERSTRASS PRODUCTS 263 then the Weierstrass product f(z) =∞/productdisplay k=1Epk(z/zk), (8.2.3) converges uniformly on compact subsets of Cto an entire function which has {zk}as a list of its zeroes counting multiplicity. Proof. By part (b) of the previous theorem, we have |Epk(z/zk)−1| ≤/vextendsingle/vextendsingle/vextendsingle/vextendsinglez zk/vextendsingle/vextendsingle/vextendsingle/vextendsinglepk+1 if|z| ≤ |zk|. The condition |z| ≤ |zk|must be satisfied for all sufficiently large kif the series (8.2.2) converges. The theorem follows by applying Theorem 8.1.4 with ak(z) = Epk(z/zk)−1. The Weierstrass Theorem Theorem 8.2.3. If{zk}is any sequence of complex numbers converging to infinity, then there is an entire function with {zk}as a list of its zeroes counting multiplicity. Proof. Suppose mof the zkare equal to 0 ( mmight be 0). We may as well assume these are the first mterms of the sequence. Then {zk}∞ k=m+1is a sequence of non-zero complex numbers. IfR >0, then, since zk→ ∞, there is a K > m such that |zk|>2Rfor all k≥K. Then the series /summationdisplay k=m+1/vextendsingle/vextendsingle/vextendsingle/vextendsinglez zk/vextendsingle/vextendsingle/vextendsingle/vextendsinglek converges uniformly on |z| ≤Rby comparison with the geometric series with ratio 1 /2. Thus, the hypotheses of the previous theorem are satisfied if we choose pk=k−1 for each k. The resulting Weierstrass product ∞/productdisplay k=m+1Ep(z/zk) converges uniformly on compact subsets of Cto an entire function which has {zk}∞ k=m+1as a list of its zeroes counting multiplicity. Then f(z) =zm∞/productdisplay k=m+1Ep(z/zk) is an entire function with {zk}∞ k=1as a list of its zeroes counting multiplicity. 264 CHAPTER 8. INFINITE PRODUCTS Example 8.2.4. Find an entire function which has a zero of order kat each positive integer k. Solution: We construct a sequence 1,2,2, ,3,3,3,4,4,4,4,··· in which each nappears ktimes and the terms are arranged in increasing order. Then, for this sequence {zk}and a given positive integer p, we have ∞/summationdisplay k=11 |zk|p+1=∞/summationdisplay k=1k1 kp+1=∞/summationdisplay k=11 kp If we choose p= 2, then the right side is the convergent series ∞/summationdisplay k=11 k2. The Weierstrass product for the sequences {zk}and{pk= 2}is ∞/productdisplay k=1/parenleftBig (1−z/k)ez/k+z2/(2k2)/parenrightBigk . By Theorem 8.2.2 this infinite product converges to an entire function with the required zeroes. Weierstrass Factorization The Weierstrass Theorem for the plane leads immediately to t he Weierstrass Factorization Theorem for entire functions: Theorem 8.2.5. Letfbe an entire function which is not identically zero. Let mbe the order of the zero of fat0, and let {zk}be a list of the non-zero zeroes offcounting multiplicity. Then there exists non-negative int egers p1, p2,··· and an entire function hsuch that f(z) =eh(z)zm∞/productdisplay k=1Epk(z/zk). The sequence {pk}may be chosen in any way which satisfies (8.2.2). Proof. The product g(z) =zm∞/productdisplay k=1Epk(z/zk) converges uniformly on compact sets if {pk}is chosen such that (8.2.2) holds (pk=k−1 is one choice which always works, but there may be better cho ices for a given f). Furthermore, the resulting function ghas the same zeroes as f 8.2. WEIERSTRASS PRODUCTS 265 with the same multiplicities. Thus, fg−1is an entire function with no zeroes (after removable singularities are removed). It follows th at fg−1= eh for some entire function h. The theorem follows from this. In many cases, the sequence {pk}can be chosen to be constant. Example 8.2.6. Find a Weierstrass factorization for sin( πz). Solution: This function has a zero of order 1 at each integer and has no other zeroes. Since ∞/summationdisplay k=11 k2<∞, the condition (8.2.2) holds if we choose pk= 1 for every k. Then the above theorem tells us that sin(πz) = eh(z)z/productdisplay k/negationslash=0E1(z/k) = eh(z)z/productdisplay k/negationslash=0(1−z/k)ez/k, where the product is over all non-zero integers k. Note that if the factors for k and−kin this product are paired, the result is (1−z/k)ez/k(1 +z/k)e−z/k= 1−z2/k2. We conclude from Exercise 8.1.12 that eh(z)=π, that sin(πz) =πz/productdisplay k/negationslash=0(1−z/k)ez/k is a Weierstrass factorization of sin( πz), and that this factorization is equivalent to the factorization sin(πz) =πz∞/productdisplay k=1(1−z2/k2). The General Weierstrass Theorem IfCis replaced by an arbitrary non-empty, proper open subset of S2, the ana- logue of Theorem 8.2.3 holds with only a slightly more compli cated proof. Theorem 8.2.7. LetUbe a non-empty, proper open subset of S2. If{zk}is any sequence of points of Uwith no limit points in U, then there is an analytic function fonUwith{zk}as a list of its zeroes counting mulltiplicity. Proof. Either Uor its image under some linear fractional transformation wi ll contain ∞. Thus, we may as well assume ∞ ∈U. Then the complement of U inS2is a compact subset Kof the plane. 266 CHAPTER 8. INFINITE PRODUCTS Since {zk}has no limit point in U, the distance between zkandKmust approach 0 as k→ ∞. It follows that we may choose a sequence {wk}of points ofKsuch that lim |zk−wk|= 0. We set f(z) =∞/productdisplay n=1Ek/parenleftbiggzk−wk z−wk/parenrightbigg . The product converges uniformly on compact subsets of U, since lim |zk−wk|= 0 implies the uniform convergence on compact subsets of Uof ∞/summationdisplay k=1/vextendsingle/vextendsingle/vextendsingle/vextendsinglezk−wk z−wk/vextendsingle/vextendsingle/vextendsingle/vextendsinglek+1 . The function fis analytic in Uas has {zk}as a list of its zeroes counting multiplicity. Meromorphic Functions On a connected open set U, the set of analytic functions forms an integral domain – that is, it is a commutative ring with the property th at the product of two elements is zero if and only if one of them is zero. The set o f meromorphic functions of Uforms a field – that is, a commutative ring in which every non-z ero element has an inverse. The next theorem shows that the field o f meromorphic function is actually the quotient field of the ring of analyti c functions. That is, every meromorphic function is the quotient f/gof two analytic functions. Theorem 8.2.8. IfUis a connected open subset of C, then each meromorphic function on Uhas the form f/g, where fandgare analytic on Uandgis not identically zero. Proof. Lethbe a meromorphic function on Uand let {zk}be a sequence con- sisting of the poles of h, with each zklisted as many times as the order of the pole at zk. By Theorem 8.2.7 there is an analytic function gonUwith{zk} as a list of its zeroes counting multiplicity. Then, after re moving removable singularities, f=ghis an analytic function on U. Thus, h=f/gwithfandg analytic on U. The Mittag-Leffler Theorem The Weierstrass Theorem (Theorem 8.2.7) gives the existenc e of an analytic function with a specified list of zeroes counting multiplici ty. The Mittag-Leffler Theorem is a companion theorem. It gives the existence of a me romorphic function with a specified list of poles and principal parts. W e prove it only for discs, although it is true for general open sets. Theorem 8.2.9. LetRbe a positive number or ∞. Let Sbe a discrete set of points of DR(0)and{hw:w∈S}a set of polynomials with no constant terms. Then there exists a meromorphic function fwith a pole at wwith principal part hk((z−w)−1)for each w∈Sand with no other poles. 8.2. WEIERSTRASS PRODUCTS 267 Proof. We choose an increasing sequence of radii {rn}withrn→Rand we let S1be the subset of Swhich lies in Dr1(0) and, for n >1, and let Sn={w∈S:rn−1<|w| ≤rn. Then, for each n, gn(z) =/summationdisplay w∈Snhk((z−w)−1) is a meromorphic function on the plane with a pole at wwith the required principal part for each w∈Snand with no other poles. We might hope to construct the function we are after by simply taking the infinite sum of the functions gn. Unfortunately, there is no reason to think this sequence should converge on DR(0). However, we can modify each gn, without changing its poles and pricipal parts, in such a way as to end u p with an infinite series which does converge. For each n >1,the function gnis analytic on an open set containing the closed disc Drn−1(0). Hence, it is the uniform limit on this closed disc of its power series at 0. It follows that there is a polynomial pnsuch that |gn(z)−pn(z)|<2−nfor|z| ≤rn−1. If we set f1=g1andfn=gn−pnforn >1, then, for each m >1, the series ∞/summationdisplay n=m+1fn(z) converges uniformly to an analytic function on Drm(0). This means that f(z) =∞/summationdisplay n=1fn(z) is defined as a meromorphic function on Drm(0) and has the required poles and principal parts at those points of Swhich lie in this disc. Since this is true for eachm, and lim rm=R,fis meromorphic on all of DR(0) and has the required poles and principal parts. Exercise Set 8.2 1. Show that the derivative of Ep(z) iszpez+z2/2+···+zp/p. 2. Compute the logarithmic derivative of Ep(z). 3. Find an entire function (given by a Weierstrass product) t hat has a zero of order 1 at√nforn= 1,2,3,···and no other zeroes. 4. Find an entire function (given by a Weierstrass product) t hat has a zero of order 2 at√nforn= 1,2,3,···and has no other zeroes. 268 CHAPTER 8. INFINITE PRODUCTS 5. Find an entire function (given by a Weierstrass product) t hat has a zero of order natn2forn= 1,2,3,···and has no other zeroes. 6. Iffis an entire function, show that f=gnfor some entire function gif and only if the order of each zero of fis divisible by n. 7. Suppose fis an entire function such that {zk}is a list of its non-zero zeroes counting multiplicity and suppose that ∞/summationdisplay k=11 |zk|<∞. Describe the simplest Weierstrass factorization of f. 8. Suppose fis an odd entire function, the order of the zero at 0 is m, and {zk}is a list of the other zeroes of fcounting multiplicity. Show that m is positive and odd. If ∞/summationdisplay k=11 |zk|2<∞, prove that fhas a factorization of the form f(z) =zmeh(z)∞/productdisplay k=1/parenleftbigg 1−z2 z2 k/parenrightbigg where his an entire function. 9. Show that, if Uis any non-empty open subset of the plane, then there is an analytic function on Uwhich cannot be extended to be analytic on any larger open set. Hint: Use the general Weierstrass Theorem t o construct an analytic function on Uwith a lot of zeroes. 10. Prove that, given a sequence {zk}of complex numbers converging to in- finity and a sequence {nk}of integers, there is an entire function fwith given values for fand its derivatives up to order nkatzkfor each k. Hint: Use the Mittag-Leffler and Weierstrass Theorems together. 11. Prove that if f1andf2are two entire functions with no common zeroes, then there exist entire functions g1andg2such that g1f1+g2f2= 1. Hint: Use the Mittag-Leffler Theorem to show that an entire fun ction g2 can be chosen so that, at each zero of f1, the function 1 −g2f2has a zero of order at least as large. 12. Let f1, f2,···fnbe entire functions. Show that there are entire functions h1, h2,·, hn, and usuch that fj=uhjforj= 1,···, nand the func- tions h1, h2,···, hnhave no common zeroes. Hint: Use the Weierstrass Theorem. 8.3. ENTIRE FUNCTIONS OF FINITE ORDER 269 13. Let f1, f2,···fnbe entire functions with no common zero. Use induction and the preceding two exercises to show that there are entire functions g1, g2,···, gnsuch that g1f1+g2f2+···+gnfn= 1. 14. Those who are familiar with commutative ring theory may w ant to do this exercise. Let Ebe the ring of entire functions. Show that the following ring theoretic properties of Eare consequences of the preceding two exercises: (a) every finitely generated ideal of Eis a principle ideal; (b) every finitely generated maximal ideal of Eis of the form Mw={f∈E:f(w) = 0}for some w∈C. 15. The conclusions of the last four exercises actually hold for the ring of analytic functions on any open subset of the plane. However, to prove them all in this generality would require a stronger form of t he Mittag- Leffler Theorem than the one proved here. Prove these results f or the largest class of open sets that you can using the machinery de veloped in this text. 8.3 Entire Functions of Finite Order Definition 8.3.1. An entire function fis said to be of finite order if there is a number tsuch that |f(z)| ≤e|z|t for all zwith|z|sufficiently large. The infimum of all such numbers tis called theorder off. For each non-negative integer p, the function ezpis an entire function of finite order p. More generally: Example 8.3.2. Show that eh(z)is an entire function of finite order pifhis a polynomial of degree p. Solution: Ift > p, then lim z→∞|z|−t|h(z)|= 0. This implies that there is anR >0 such that |h(z)|<|z|tfor|z|> R. Then |eh(z)| ≤e|z|tfor|z|> R. (8.3.1) Since such a statement is true for all t > p, by definition eh(z)has finite order at most p. On the other hand, if t < p, then lim z→∞|z|−t|h(z)|= +∞. Hence, there is noRfor which (8.3.1) holds. We conclude that the order of fis at least pand, hence, is equal to p. 270 CHAPTER 8. INFINITE PRODUCTS Non-vanishing Entire Functions of Finite Order It turns out that the functions eh(z)of the preceding example are the only entire functions of finite order which are non-vanishing. To prove this, we will need the following theorem of Borel-Ca rath´ eodory relating the growth of the real part of an analytic function t o the growth of the the absolute value of the function. Theorem 8.3.3. Suppose 0< r < R and let gbe a function analytic on an open set containing DR(0). Then |g(z)| ≤2r R−rsup{Re(g(w)) :|w|=R}+R+r R−r|g(0)|if|z| ≤r. Proof. We suppose first that g(0) = 0. We set m= sup {Re(g(w)) :|w|=R}. Note that the Mean Value Theorem for harmonic functions impl ies that m≥0. If|w|=Randu= Re ( g(w)), then u≤mand u−2m≤u≤2m−u. Thus, |u| ≤ |2m−u|, from which it follows that |g(w)| ≤ |2m−g(w)|, since the numbers g(w) and 2 m−g(w) have the same imaginary parts and have real parts uand 2m−u, respectively. We conclude from the above, that the function h(z) =g(z) z(2m−g(z)) satisfies the inequality |h(w)| ≤1 Rfor|w|=R. Since the analytic function hhas a removable singularity at 0, this inequality holds throughout the disc DR(0) by the Maximum Modulus Theorem. Thus, |g(z)| r|2m−g(z)|≤1 Rwhenever |z|=r, which implies |g(z)| ≤r R(2m+|g(z)|). If we collect terms involving |g(z)|on the left and divide by 1 −r/R, the result is |g(z)| ≤2r R−rmfor|z| ≤r. This concludes the proof in the case where g(0) = 0. This general case follows from applying this result to the function g0(z) =g(z)−g(0). The details are left to the exercises. 8.3. ENTIRE FUNCTIONS OF FINITE ORDER 271 Theorem 8.3.4. An entire function fwith no zeroes has finite order pif and only if pis a non-negative integer and fhas the form f(z) =eh(z), where his a polynomial of degree p. Proof. In view of Example 8.3.2, we need only show that every non-van ishing entire function fof finite order phas the above form. Since fhas no zeroes and the plane is simply connected, there is an en tire function hsuch that f(z) = eh(z)for all z∈C. Since fhas finite order p, for each t > p there is an M >0 such that eRe(h(z))=|f(z)| ≤e|z|tfor|z| ≥M. This implies Re (h(z))≤ |z|tfor|z| ≥M. We apply the previous theorem with r > M andR= 2rto conclude |h(z)| ≤2|z|t+ 3|h(0)|if|z|=r. Since this is true for all r > M , Exercise 3.3.9 implies that hmust be a poly- nomial of degree at most t. Since twas an arbitrary number greater than p, we conclude that his a polynomial of degree at most p. If it were a polynomial of degree less than p, then fwould have order less than p. Hence, the degree of the polynomial his exactly p. This, of course, implies that pis a non-negative integer. Canonical Products Given a sequence {zk}, we let µbe the inf of the numbers tsuch that ∞/summationdisplay k=11 |zk|t<∞. (8.3.2) If there is no such t, then we set µ=∞. The number µis called the exponent of convergence for the sequence {zk}. If{zk}has finite exponent of convergence µ, then we can write down a convergent Weierstrass product (8.2.3), using {zk}, in which the sequence {pk} is a constant p. We choose pto be the smallest integer such that µ < p + 1. Then the condition∞/summationdisplay k=11 |zk|p+1<∞, (8.3.3) is satisfied. Hence, by Theorem 8.2.2, the Weierstrass produ ct f(z) =∞/productdisplay k=1Ep(z/zk) (8.3.4) 272 CHAPTER 8. INFINITE PRODUCTS converges. This is called the canonical product for the sequence {zk}. The significance of the choice of pmade for the canonical product is that, with this choice, the resulting product is an entire functio n with order λequal to the exponent of convergence µof the sequence {zk}. The next theorem yields part of what is needed to prove this. The remainder of the proo f will come in the next section. Theorem 8.3.5. The canonical product for a sequence {zk}, with finite expo- nent of convergence µ, is an entire function of finite order λ≤µ. Proof. We choose pto be the smallest integer such that µ < p + 1, and let tbe any number in the range µ < t < p + 1. We claim that there is a positive constant Asuch that |Ep(z)| ≤eA|z|t(8.3.5) for all z. If|z| ≤1/2, this follows from (8.1.6) with w=Ep(z)−1 and Theorem 8.2.1. These combine to show that |logEp(z)| ≤2|z|p+1≤2|z|t, and this implies (8.3.5) holds with A= 2. If|z|>1/2, then |z|k≤2t−k|z|t, and so log|Ep(z)|= log|1−z|+p/summationdisplay k=1Re/parenleftbig zk/parenrightbig k ≤ |z|+p/summationdisplay k=1|z|k≤(p+ 1)2t|z|t. Thus, (8.3.5) holds with A= (p+ 1)2tin this case. To prove the theorem, we note that, if fis given by the canonical product (8.3.4), then by (8.3.5), |f(z)| ≤∞/productdisplay k=1eA|z/zk|t= eB|z|t, where B=A∞/summationdisplay k=11/|zk|t. The series in this expression converges because tis larger than the exponent of convergence µ. Since for any s > t, we have B|z|t≤ |z|sfor|z|sufficiently large, it follows thatfhas finite order at most t. Since twas an arbitrary number strictly between µandp+ 1, we conclude that fhas order at most µ. 8.3. ENTIRE FUNCTIONS OF FINITE ORDER 273 One might guess, based on Theorem 8.3.4 that the order of an en tire function of finite order must be a non-negative integer. This is not the case, as is shown by the following example. Example 8.3.6. Find an entire function with finite order 1 /2. Solution: The function sinπz πz=∞/productdisplay k=1/parenleftbigg 1−z2 k2/parenrightbigg has order 1 (Exercise 8.3.3). It seems reasonable that if we r eplace z2byzin this product that the result would be an entire function of or der 1/2. In fact, the resulting function has a zero of order 1 at k2for each positive integer kand ∞/summationdisplay k=11 (k2)t<∞ for every t >1/2 and for no smaller values of t. Hence, the sequence {1/k2} has exponent of convergence 1 /2. Since 0 is the smallest integer psuch that 1/2< p+ 1, the preceding theorem implies that the canonical produc t f(z) =∞/productdisplay k=1/parenleftBig 1−z k2/parenrightBig is an entire function of finite order at most 1 /2. In fact, it is easy to directly compute the order of fif we note that f(z) =sinπ√z π√z. This expression on the right is entire and is independent of t he choice of the square root function because the function ( πz)−1sinπzis an even function. It is easy to see from this that, since ( πz)−1sinπzhas order 1, fhas order 1 /2 (Exercise 8.3.5). Exercise Set 8.3 1. Finish the proof of Theorem 8.3.3 by showing that, if it is t rue in the case where g(0) = 0, then it is true in general. 2. Show that a polynomial has finite order 0. 3. Show that sin z,z−1sinz, and cos zall have finite order 1. 4. Iffis an entire function of order λ(f),kis a non-negative integer, and g(z) =f(zk), then prove that λ(g) =kλ(f), where λ(g) is the order of g. 5. Prove that if g(z) is an even entire function of finite order λandf(z) = g(√z), then fis an entire function of finite order λ/2. In particular, show that cos√zhas order 1 /2. 274 CHAPTER 8. INFINITE PRODUCTS 6. Prove that the order of the sum or product of two entire func tions is less than or equal to the maximum of the orders of the two functions . 7. What is the order of the entire function esinz? 8. Suppose fis an entire function which satisfies the inequality |f(z)| ≤ |z||z| for|z|sufficiently large. Prove that fhas finite order at most 1. 9. Find the exponent of convergence of the following sequenc es:{2k},{kr} (r >0),{logk}. 10. Given an arbitrary non-negative real number µ, show that there is a se- quence of complex numbers {zk}with exponent of convergence µ. 11. Does the order of an entire function necessarily have to b e the same as the exponent of convergence of its sequence of zeroes? Justify y our answer. 8.4 Hadamard’s Factorization Theorem Our goal in this section is to complete the characterization of entire functions of finite order λ. We will prove a theorem of Hadamard which asserts that every such function factors as a power of ztimes a canonical product of order at most λtimes the exponential of a polynomial of degree at most λ. The key ingredient in the proof is Jensen’s Formula relating the density of the z eroes of an entire function to the rate of growth at infinity of the function. Jensen’s Formula Theorem 8.4.1. Iffis analytic in an open set containing the disc Dr(0),f has no zeroes on the boundary of this disc, f(0)/\e}atio\slash= 0, and z1, z2,···znare the zeroes, counting multiplicity, of finDr(0), then log/parenleftbigg|f(0)|rn |z1| · |z2| · ···| zn|/parenrightbigg =1 2π/integraldisplay2π 0log(|f(reiθ)|)dθ. Proof. We first prove this in the case where r= 1. We divide fby a product of linear fractional transformations which preserve the unit circle and have zeroes at the points zi. This yields a function g(z) =f(z)1−z1z z−z11−z2z z−z2···1−znz z−zn. This function is analytic and non-vanishing in an open set co ntaining the closed unit disc D, and has the same modulus on the unit circle as does f. Thus, g has an analytic logarithm in an open set containing D. Then log |g(x)|is the real part of an analytic function in this set and, hence, is ha rmonic. The Mean Value Theorem for harmonic functions implies that log/parenleftbigg|f(0)| |z1| · |z2| · ···| zn|/parenrightbigg = log|g(0)|=1 2π/integraldisplay2π 0log|f(eiθ)|dθ. (8.4.1) 8.4. HADAMARD’S FACTORIZATION THEOREM 275 To prove the theorem for general r, it suffices to apply (8.4.1) with freplaced by the function f(rz). Iffhas zeroes at z1, z2,···, znin the disc Dr(0), then f(rz) has zeroes z1/r, z2/r,···, zn/rin the unit disc D. Thus, the equation of the theorem follows directly from (8.4.1) applied to f(rz). This leads to the following estimate on the number of zeroes o f an entire function inside a disc Dr(0). Theorem 8.4.2. Iffis an entire function with |f(0)|= 1,n(r)is the number of zeroes of finside a disc Dr(0), and M(2r)is the supremum of |f(z)|on the boundary of D2r(0), then n(r)≤logM(2r) log 2. Proof. Letn=n(r) and m=n(2r), and let z1, z2,···zmbe the zeroes of f inside the disc D2r(0) ordered so that |zj| ≤ |zk|forj≤k. Then Jensen’s Theorem with rreplaced by 2 rimplies that log/vextendsingle/vextendsingle/vextendsingle/vextendsinglef(0)2r z12r z2···2r zn···2r zm/vextendsingle/vextendsingle/vextendsingle/vextendsingle≤logM(2r). Since 2<2r |zj|ifj≤nand 1 <2r |zj|ifj > n, this implies that log|f(0)2n| ≤logM(2r), or log|f(0)|+nlog2≤logM(2r). The theorem follows from this, since log |f(0)|= 0. Zeroes of Functions of Finite Order The preceding theorem has the following consequence for ent ire functions of finite order. Theorem 8.4.3. Letfbe an entire function of finite order λand with f(0)/\e}atio\slash= 0. Let{zk}be a list of the zeroes of f, counted according to multiplicity and indexed in order of increasing modulus, and let µbe the exponent of convergence of {zk}, thenµ≤λ. Proof. We claim that, for each t > λ, there are constants N,C >0 and q >1 such that |zk|t≥Ckqfor all k≥N. (8.4.2) Assuming this, we conclude that the series ∞/summationdisplay k=11 |zk|t 276 CHAPTER 8. INFINITE PRODUCTS converges for all t > λ, by comparison with the series ∞/summationdisplay k=11 kq, which converges for q >1. This, in turn, implies the exponent of convergence µis at most λ. To complete the proof, we must verify the claim concerning (8 .4.2). In doing this, we may as well assume that |f(0)|= 1, since, if this is not so, we may make it so by replacing fbyfdivided by a constant times a power of z. Such a replacement will have no effect on whether the above cla im is true. Letrk=|zk|. Since the zeroes are indexed in such a way that the modulus is a non-decreasing function of k, there are at least kzeroes of fwith modulus less than or equal to rk. By Theorem 8.4.2, k≤logM(2rk) log 2, where M(2rk) is the sup of |f(z)|on the circle |z|= 2rk. We choose swithλ < s < t . Since fhas order λ, there is an Rsuch that rk≥Rimplies M(2rk)≤e(2rk)s. Hence, for rk≥R, k≤(2rk)s log 2. This implies rt k≥(log 2)t/s 2tkt/s=Ckq, where C=(log 2t/s) 2tand q=t s>1. This is true provided rk=|zk|> R. However, since lim zk=∞, there is an N such that k > N implies |zk|> R. This completes the proof The above theorem, when combined with Theorem 8.3.5, yields the following corollary. Corollary 8.4.4. The canonical product for a sequence {zk}with exponent of convergence µhas finite order λ=µ. Hadamard’s Theorem In the proof of the next theorem, we will need the following es timates on the size of the inverse E−1 p(z) of the function Ep(z). 8.4. HADAMARD’S FACTORIZATION THEOREM 277 Lemma 8.4.5. Ifpis a non-negative integer, p≤t≤p+ 1, and z∈C, then there is a constant Asuch that 1 |Ep(z)|≤eA|z|t(8.4.3) if|z| ≥2or|z| ≤1/2. Proof. If|z| ≥2, then |1−z| ≥1 and |zk/k| ≤ |z|tfork≤p. Hence, |E−1 p(z)|=|1−z|−1|e−z−z2/2−···− zp/p| ≤ep|z|t and so (8.4.3) holds with A=pin this case. On the other hand, if |z| ≤1/2, then logEp(z) = log(1 −z) +p/summationdisplay k=1zk/k=−∞/summationdisplay k=p+1zk/k, and so |logEp(z)| ≤ |z|p+1∞/summationdisplay j=0|z|k≤2|z|p+1≤2|z|t Thus, (8.4.3) holds with A= 2 in this case. If we choose A= max {2, p}, then (8.4.3) holds in both cases. We are now in a position to prove Hadamard’s Theorem characte rizing entire functions of finite order. This will be used in the proof of the Prime Number Theorem in the next chapter. Theorem 8.4.6. Iffis an entire function of order λ, and pis the smallest integer such that p+ 1> λ, then ffactors as f(z) =zmeh(z)∞/productdisplay k=1Ep(z/zk), (8.4.4) where mis the order of the zero of fat0,{zk}is a list of the other zeroes of f counting multiplicity, and h(z)is a polynomial of degree at most p. Proof. According to the Weierstrass Factorization Theorem (Theor em 8.2.5) f has a factorization of the form (8.4.4), where his an entire function. Thus, the only thing to be proved is that his a polynomial of degree at most p. This will follow from Theorem 8.3.4 if we can show that the function g(z) = eh(z)=f(z) zm/producttext∞ j=1Ep(z/zk) has finite order at most λ. 278 CHAPTER 8. INFINITE PRODUCTS Lettbe any number with λ < t ≤p+ 1 and let r≥1 be any radius which is not one of the numbers |zk|. We factor g(z) asg(z) =g1(z)g2(z), where g1(z) =f(z)z−m/productdisplay |zk|≤2rE−1 p(z/zk), (8.4.5) and g2(z) =/productdisplay |zk|>2rE−1 p(z/zk). (8.4.6) Suppose |z|= 4r=R. Then |z/zk| ≥2 for all kwith|zk| ≤2r. By the previous lemma, there is a positive constant A1such that |g1(z)| ≤ |f(z)|/productdisplay |zk|≤2reA1|z/zk|t. Since fhas finite order λ, for sufficiently large rwe have |f(z)| ≤e|z|t and, hence, |g1(z)| ≤eB1rt(8.4.7) where B1= 4t/parenleftBigg 1 +A1∞/summationdisplay k=11 |zk|t/parenrightBigg . The infinite series in this expression converges by Theorem 8 .4.3. Since g1(z) is an entire function (once the removable singularities at the zkwith|zk|<2rare removed), if the inequality (8.4.7) holds for |z|= 4r=Rit must hold for all zin the disc |z| ≤R, by the maximum modulus principle. In particular, this inequality holds for all zwith|z|=r. Also if |z|=r, then |z/zk|<1/2 if|zk|>2r, and the previous lemma implies that there is a constant A2such that |g2(z)| ≤/productdisplay |zk|>2reA2|z/zk|t≤eB2rt, (8.4.8) where B2=A2∞/summationdisplay k=11 |zk|. If we set B=B1+B2and combine (8.4.7) and (8.4.8), we obtain |g(z)| ≤eB|z|t. Since tis an arbitrary number larger than λand less than or equal to p+ 1,g has order at most λ. This completes the proof. 8.4. HADAMARD’S FACTORIZATION THEOREM 279 Exercise Set 8.4 1. What does Hadamard’s Factorization Theorem say about an e ntire func- tion of order λ <1? 2. If a non-constant entire function of finite order λhas zeroes at the points i√nwhat are the possible values for λ? 3. Show that an even entire function of order 1 has the form Czm/productdisplay k/parenleftbigg 1−z2 z2 k/parenrightbigg , where mis even, Cis a non-zero constant, and the sequence {z1,−z1, z2,−z2,···, zk,−zk,···} is a list of the zeroes of fcounting multiplicity. 4. What is the exponent of convergence for the sequence of zer oes in the preceding exercise. 5. State and prove the analogues of the previous two exercise s for odd entire functions of order 1. 6. Prove that if fis an entire function of order λandλis not an integer, thenfhas infinitely many zeroes. 7. Under the hypotheses of the preceding exercise, prove tha tftakes on every complex value infinitely many times. 8. Prove that if fandgare entire functions of finite order λand if f(zk) = g(zk) on a sequence which satisfies /summationdisplay k=11 |zk|t=∞ for some t > λ, then f(z) =g(z) identically. 9. Use the previous exercise to prove that if two functions of finite order agree at the points of the sequence {logn}∞ n=1}, then they agree identically. Thus, ezis the only entire function of finite order which has the value n at the point log nforn= 1,2,···. 10. Find an entire function which has zeroes at the points of t he sequence {logn}∞ n=1. Does it have finite order? 11. Suppose fis an entire function of finite order λandµis the exponent of convergence of the list of zeroes of f. Prove that if µ < λ , then λis an integer. 12. Is there an entire function of order 3 /2 which has the integers as its list of zeroes, counting multiplicity?