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246B, Notes 1: Zeroes, poles, and factorisation of meromorphic functions

23 December, 2020 in 246B - complex analysis, math.CV | Tags: , , , | by | 65 comments

Previous set of notes: 246A Notes 5. Next set of notes: Notes 2.

— 1. Jensen’s formula —

Suppose {f} is a non-zero rational function {f =P/Q}, then by the fundamental theorem of algebra one can write

\displaystyle f(z) = c \frac{\prod_\rho (z-\rho)}{\prod_\zeta (z-\zeta)}

for some non-zero constant {c}, where {\rho} ranges over the zeroes of {P} (counting multiplicity) and {\zeta} ranges over the zeroes of {Q} (counting multiplicity), and assuming {z} avoids the zeroes of {Q}. Taking absolute values and then logarithms, we arrive at the formula

\displaystyle \log |f(z)| = \log |c| + \sum_\rho \log|z-\rho| - \sum_\zeta \log |z-\zeta|, \ \ \ \ \ (1)

 as long as {z} avoids the zeroes of both {P} and {Q}. (In this set of notes we use {\log} for the natural logarithm when applied to a positive real number, and {\mathrm{Log}} for the standard branch of the complex logarithm (which extends {\log}); the multi-valued complex logarithm {\log} will only be used in passing.) Alternatively, taking logarithmic derivatives, we arrive at the closely related formula

\displaystyle \frac{f'(z)}{f(z)} = \sum_\rho \frac{1}{z-\rho} - \sum_\zeta \frac{1}{z-\zeta}, \ \ \ \ \ (2)

 again for {z} avoiding the zeroes of both {P} and {Q}. Thus we see that the zeroes and poles of a rational function {f} describe the behaviour of that rational function, as well as close relatives of that function such as the log-magnitude {\log|f|} and log-derivative {\frac{f'}{f}}. We have already seen these sorts of formulae arise in our treatment of the argument principle in 246A Notes 4.

Exercise 1 Let {P(z)} be a complex polynomial of degree {n \geq 1}.
  • (i) (Gauss-Lucas theorem) Show that the complex roots of {P'(z)} are contained in the closed convex hull of the complex roots of {P(z)}.
  • (ii) (Laguerre separation theorem) If all the complex roots of {P(z)} are contained in a disk {D(z_0,r)}, and {\zeta \not \in D(z_0,r)}, then all the complex roots of {nP(z) + (\zeta - z) P'(z)} are also contained in {D(z_0,r)}. (Hint: apply a suitable Möbius transformation to move {\zeta} to infinity, and then apply part (i) to a polynomial that emerges after applying this transformation.)

There are a number of useful ways to extend these formulae to more general meromorphic functions than rational functions. Firstly there is a very handy “local” variant of (1) known as Jensen’s formula:

Theorem 2 (Jensen’s formula) Let {f} be a meromorphic function on an open neighbourhood of a disk {\overline{D(z_0,r)} = \{ z: |z-z_0| \leq r \}}, with all removable singularities removed. Then, if {z_0} is neither a zero nor a pole of {f}, we have

\displaystyle \log |f(z_0)| = \int_0^1 \log |f(z_0+re^{2\pi i t})|\ dt + \sum_{\rho: |\rho-z_0| \leq r} \log \frac{|\rho-z_0|}{r} \ \ \ \ \ (3)

\displaystyle - \sum_{\zeta: |\zeta-z_0| \leq r} \log \frac{|\zeta-z_0|}{r}

where {\rho} and {\zeta} range over the zeroes and poles of {f} respectively (counting multiplicity) in the disk {\overline{D(z_0,r)}}.

One can view (3) as a truncated (or localised) variant of (1). Note also that the summands {\log \frac{|\rho-z_0|}{r}, \log \frac{|\zeta-z_0|}{r}} are always non-positive.

Proof: By perturbing {r} slightly if necessary, we may assume that none of the zeroes or poles of {f} (which form a discrete set) lie on the boundary circle {\{ z: |z-z_0| = r \}}. By translating and rescaling, we may then normalise {z_0=0} and {r=1}, thus our task is now to show that

\displaystyle \log |f(0)| = \int_0^1 \log |f(e^{2\pi i t})|\ dt + \sum_{\rho: |\rho| < 1} \log |\rho| - \sum_{\zeta: |\zeta| < 1} \log |\zeta|. \ \ \ \ \ (4)

We may remove the poles and zeroes inside the disk {D(0,1)} by the useful device of Blaschke products. Suppose for instance that {f} has a zero {\rho} inside the disk {D(0,1)}. Observe that the function

\displaystyle B_\rho(z) := \frac{\rho - z}{1 - \overline{\rho} z} \ \ \ \ \ (5)

 has magnitude {1} on the unit circle {\{ z: |z| = 1\}}, equals {\rho} at the origin, has a simple zero at {\rho}, but has no other zeroes or poles inside the disk. Thus Jensen’s formula (4) already holds if {f} is replaced by {B_\rho}. To prove (4) for {f}, it thus suffices to prove it for {f/B_\rho}, which effectively deletes a zero {\rho} inside the disk {D(0,1)} from {f} (and replaces it instead with its inversion {1/\overline{\rho}}). Similarly we may remove all the poles inside the disk. As a meromorphic function only has finitely many poles and zeroes inside a compact set, we may thus reduce to the case when {f} has no poles or zeroes on or inside the disk {D(0,1)}, at which point our goal is simply to show that

\displaystyle \log |f(0)| = \int_0^1 \log |f(e^{2\pi i t})|\ dt.

Since {f} has no zeroes or poles inside the disk, it has a holomorphic logarithm {F} (Exercise 46 of 246A Notes 4). In particular, {\log |f|} is the real part of {F}. The claim now follows by applying the mean value property (Exercise 17 of 246A Notes 3) to {\log |f|}. \Box

An important special case of Jensen’s formula arises when {f} is holomorphic in a neighborhood of {\overline{D(z_0,r)}}, in which case there are no contributions from poles and one simply has

\displaystyle \int_0^1 \log |f(z_0+re^{2\pi i t})|\ dt = \log |f(z_0)| + \sum_{\rho: |\rho-z_0| \leq r} \log \frac{r}{|\rho-z_0|}. \ \ \ \ \ (6)

This is quite a useful formula, mainly because the summands {\log \frac{r}{|\rho-z_0|}} are non-negative; it can be viewed as a more precise assertion of the subharmonicity of {\log |f|} (see Exercises 60(ix) and 61 of 246A Notes 5). Here are some quick applications of this formula:

Exercise 3 Use (6) to give another proof of Liouville’s theorem: a bounded holomorphic function {f} on the entire complex plane is necessarily constant.

Exercise 4 Use Jensen’s formula to prove the fundamental theorem of algebra: a complex polynomial {P(z)} of degree {n} has exactly {n} complex zeroes (counting multiplicity), and can thus be factored as {P(z) = c (z-z_1) \dots (z-z_n)} for some complex numbers {c,z_1,\dots,z_n} with {c \neq 0}. (Note that the fundamental theorem was invoked previously in this section, but only for motivational purposes, so the proof here is non-circular.)

Exercise 5 (Shifted Jensen’s formula) Let {f} be a meromorphic function on an open neighbourhood of a disk {\{ z: |z-z_0| \leq r \}}, with all removable singularities removed. Show that

\displaystyle \log |f(z)| = \int_0^1 \log |f(z_0+re^{2\pi i t})| \mathrm{Re} \frac{r e^{2\pi i t} + (z-z_0)}{r e^{2\pi i t} - (z-z_0)}\ dt \ \ \ \ \ (7)

\displaystyle + \sum_{\rho: |\rho-z_0| \leq r} \log \frac{|\rho-z|}{|r - \rho^* (z-z_0)|}

\displaystyle - \sum_{\zeta: |\zeta-z_0| \leq r} \log \frac{|\zeta-z|}{|r - \zeta^* (z-z_0)|}

for all {z} in the open disk {\{ z: |z-z_0| < r\}} that are not zeroes or poles of {f}, where {\rho^* = \frac{\overline{\rho-z_0}}{r}} and {\zeta^* = \frac{\overline{\zeta-z_0}}{r}}. (The function {\Re \frac{r e^{2\pi i t} + (z-z_0)}{r e^{2\pi i t} - (z-z_0)}} appearing in the integrand is sometimes known as the Poisson kernel, particularly if one normalises so that {z_0=0} and {r=1}.)

Exercise 6 (Bounded type)
  • (i) If {f} is a holomorphic function on {D(0,1)} that is not identically zero, show that {\liminf_{r \rightarrow 1^-} \int_0^{2\pi} \log |f(re^{i\theta})|\ d\theta > -\infty}.
  • (ii) If {f} is a meromorphic function on {D(0,1)} that is the ratio of two bounded holomorphic functions that are not identically zero, show that {\limsup_{r \rightarrow 1^-} \int_0^{2\pi} |\log |f(re^{i\theta})||\ d\theta < \infty}. (Functions {f} of this form are said to be of bounded type and lie in the Nevanlinna class for the unit disk {D(0,1)}.)

Exercise 7 (Smoothed out Jensen formula) Let {f} be a meromorphic function on an open set {U}, and let {\phi: U \rightarrow {\bf C}} be a smooth compactly supported function. Show that

\displaystyle \sum_\rho \phi(\rho) - \sum_\zeta \phi(\zeta)

\displaystyle = \frac{-1}{2\pi} \int\int_U ((\frac{\partial}{\partial x} + i \frac{\partial}{\partial y}) \phi(x+iy)) \frac{f'}{f}(x+iy)\ dx dy

\displaystyle = \frac{1}{2\pi} \int\int_U ((\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y}^2) \phi(x+iy)) \log |f(x+iy)|\ dx dy

where {\rho, \zeta} range over the zeroes and poles of {f} (respectively) in the support of {\phi}. Informally argue why this identity is consistent with Jensen’s formula.

When applied to entire functions {f}, Jensen’s formula relates the order of growth of {f} near infinity with the density of zeroes of {f}. Here is a typical result:

Proposition 8 Let {f: {\bf C} \rightarrow {\bf C}} be an entire function, not identically zero, that obeys a growth bound {|f(z)| \leq C \exp( C|z|^\alpha)} for some {C, \alpha > 0} and all {z}. Then there exists a constant {C'>0} such that {D(0,R)} has at most {C' R^\alpha} zeroes (counting multiplicity) for any {R \geq 1}.

Entire functions that obey a growth bound of the form {|f(z)| \leq C_\varepsilon \exp( C_\varepsilon |z|^{\rho+\varepsilon})} for every {\varepsilon>0} and {z} (where {C_\varepsilon} depends on {\varepsilon}) are said to be of order at most {\rho}. The above theorem shows that for such functions that are not identically zero, the number of zeroes in a disk of radius {R} does not grow much faster than {R^\rho}. This is often a useful preliminary upper bound on the zeroes of entire functions, as the order of an entire function tends to be relatively easy to compute in practice.

Proof: First suppose that {f(0)} is non-zero. From (6) applied with {r=2R} and {z_0=0} one has

\displaystyle \int_0^1 \log(C \exp( C (2R)^\alpha ) )\ dt \geq \log |f(0)| + \sum_{\rho: |\rho| \leq 2R} \log \frac{2R}{|\rho|}.

Every zero in {D(0,R)} contribute at least {\log 2} to a summand on the right-hand side, while all other zeroes contribute a non-negative quantity, thus

\displaystyle \log C + C (2R)^\alpha \geq \log |f(0)| + N_R \log 2

where {N_R} denotes the number of zeroes in {D(0,R)}. This gives the claim for {f(0) \neq 0}. When {f(0)=0}, one can shift {f} by a small amount to make {f} non-zero at the origin (using the fact that zeroes of holomorphic functions not identically zero are isolated), modifying {C} in the process, and then repeating the previous arguments. \Box

Just as (3) and (7) give truncated variants of (1), we can create truncated versions of (2). The following crude truncation is adequate for many applications:

Theorem 9 (Truncated formula for log-derivative) Let {f} be a holomorphic function on an open neighbourhood of a disk {\{ z: |z-z_0| \leq r \}} that is not identically zero on this disk. Suppose that one has a bound of the form {|f(z)| \leq M^{O_{c_1,c_2}(1)} |f(z_0)|} for some {M \geq 1} and all {z} on the circle {\{ z: |z-z_0| = r\}}. Let {0 < c_2 < c_1 < 1} be constants. Then one has the approximate formula

\displaystyle \frac{f'(z)}{f(z)} = \sum_{\rho: |\rho - z_0| \leq c_1 r} \frac{1}{z-\rho} + O_{c_1,c_2}( \frac{\log M}{r} )

for all {z} in the disk {\{ z: |z-z_0| < c_2 r \}} other than zeroes of {f}. Furthermore, the number of zeroes {\rho} in the above sum is {O_{c_1,c_2}(\log M)}.

Proof: To abbreviate notation, we allow all implied constants in this proof to depend on {c_1,c_2}.

We mimic the proof of Jensen’s formula. Firstly, we may translate and rescale so that {z_0=0} and {r=1}, so we have {|f(z)| \leq M^{O(1)} |f(0)|} when {|z|=1}, and our main task is to show that

\displaystyle \frac{f'(z)}{f(z)} - \sum_{\rho: |\rho| \leq c_1} \frac{1}{z-\rho} = O( \log M ) \ \ \ \ \ (8)

for {|z| \leq c_2}. Note that if {f(0)=0} then {f} vanishes on the unit circle and hence (by the maximum principle) vanishes identically on the disk, a contradiction, so we may assume {f(0) \neq 0}. From hypothesis we then have

\displaystyle \log |f(z)| \leq \log |f(0)| + O(\log M)

on the unit circle, and so from Jensen’s formula (3) we see that

\displaystyle \sum_{\rho: |\rho| \leq 1} \log \frac{1}{|\rho|} = O(\log M). \ \ \ \ \ (9)

 In particular we see that the number of zeroes with {|\rho| \leq c_1} is {O(\log M)}, as claimed.

Suppose {f} has a zero {\rho} with {c_1 < |\rho| \leq 1}. If we factor {f = B_\rho g}, where {B_\rho} is the Blaschke product (5), then

\displaystyle \frac{f'}{f} = \frac{B'_\rho}{B_\rho} + \frac{g'}{g}

\displaystyle = \frac{g'}{g} + \frac{1}{z-\rho} - \frac{1}{z-1/\overline{\rho}}.

Observe from Taylor expansion that the distance between {\rho} and {1/\overline{\rho}} is {O( \log \frac{1}{|\rho|} )}, and hence {\frac{1}{z-\rho} - \frac{1}{z-1/\overline{\rho}} = O( \log \frac{1}{|\rho|} )} for {|z| \leq c_2}. Thus we see from (9) that we may use Blaschke products to remove all the zeroes in the annulus {c_1 < |\rho| \leq 1} while only affecting the left-hand side of (8) by {O( \log M)}; also, removing the Blaschke products does not affect {|f(z)|} on the unit circle, and only affects {\log |f(0)|} by {O(\log M)} thanks to (9). Thus we may assume without loss of generality that there are no zeroes in this annulus.

Similarly, given a zero {\rho} with {|\rho| \leq c_1}, we have {\frac{1}{z-1/\overline{\rho}} = O(1)}, so using Blaschke products to remove all of these zeroes also only affects the left-hand side of (8) by {O(\log M)} (since the number of zeroes here is {O(\log M)}), with {\log |f(0)|} also modified by at most {O(\log M)}. Thus we may assume in fact that {f} has no zeroes whatsoever within the unit disk. We may then also normalise {f(0) = 1}, then {\log |f(e^{2\pi i t})| \leq O(\log M)} for all {t \in [0,1]}. By Jensen’s formula again, we have

\displaystyle \int_0^1 \log |f(e^{2\pi i t})|\ dt = 0

and thus (by using the identity {|x| = 2 \max(x,0) - x} for any real {x})

\displaystyle \int_0^1 \log |f(e^{2\pi i t})|\ dt \ll \log M. \ \ \ \ \ (10)

 On the other hand, from (7) we have

\displaystyle \log |f(z)| = \int_0^1 \log |f(e^{2\pi i t})| \mathrm{Re} \frac{e^{2\pi i t} + z}{e^{2\pi i t} - z}\ dt

which implies from (10) that {\log |f(z)|} and its first derivatives are {O( \log M )} on the disk {\{ z: |z| \leq c_2 \}}. But recall from the proof of Jensen’s formula that {\frac{f'}{f}} is the derivative of a logarithm {\log f} of {f}, whose real part is {\log |f|}. By the Cauchy-Riemann equations for {\log f}, we conclude that {\frac{f'}{f} = O(\log M)} on the disk {\{ z: |z| \leq c_2 \}}, as required. \Box

Exercise 10
  • (i) (Borel-Carathéodory theorem) If {f: U \rightarrow {\bf C}} is analytic on an open neighborhood of a disk {\overline{D(z_0,R)}}, show that

    \displaystyle \sup_{z \in D(z_0,r)} |f(z)| \leq \frac{2r}{R-r} \sup_{z \in \overline{D(z_0,R)}} \mathrm{Re} f(z) + \frac{R+r}{R-r} |f(z_0)|.

    (Hint: one can normalise {z_0=0}, {R=1}, {f(0)=0}, and {\sup_{|z-z_0| \leq R} \mathrm{Re} f(z)=1}. Now {f} maps the unit disk to the half-plane {\{ \mathrm{Re} z \leq 1 \}}. Use a Möbius transformation to map the half-plane to the unit disk and then use the Schwarz lemma.)
  • (ii) Use (i) to give an alternate way to conclude the proof of Theorem 9.

A variant of the above argument allows one to make precise the heuristic that holomorphic functions locally look like polynomials:

Exercise 11 (Local Weierstrass factorisation) Let the notation and hypotheses be as in Theorem 9. Then show that

\displaystyle f(z) = P(z) \exp( g(z) )

for all {z} in the disk {\{ z: |z-z_0| < c_2 r \}}, where {P} is a polynomial whose zeroes are precisely the zeroes of {f} in {\{ z: |z-z_0| \leq c_1r \}} (counting multiplicity) and {g} is a holomorphic function on {\{ z: |z-z_0| < c_2 r \}} of magnitude {O_{c_1,c_2}( \log M )} and first derivative {O_{c_1,c_2}( \log M / r )} on this disk. Furthermore, show that the degree of {P} is {O_{c_1,c_2}(\log M)}.

Exercise 12 (Preliminary Beurling factorisation) Let {H^\infty(D(0,1))} denote the space of bounded analytic functions {f: D(0,1) \rightarrow {\bf C}} on the unit disk; this is a normed vector space with norm

\displaystyle \|f\|_{H^\infty(D(0,1))} := \sup_{z \in D(0,1)} |f(z)|.

  • (i) If {f \in H^\infty(D(0,1))} is not identically zero, and {z_n} denote the zeroes of {f} in {D(0,1)} counting multiplicity, show that

    \displaystyle \sum_n (1-|z_n|) < \infty

    and

    \displaystyle \sup_{1/2 < r < 1} \int_0^{2\pi} | \log |f(re^{i\theta})| |\ d\theta < \infty.

  • (ii) Let the notation be as in (i). If we define the Blaschke product

    \displaystyle B(z) := z^m \prod_{|z_n| \neq 0} \frac{|z_n|}{z_n} \frac{z_n-z}{1-\overline{z_n} z}

    where {m} is the order of vanishing of {f} at zero, show that this product converges absolutely to a holomorphic function on {D(0,1)}, and that {|f(z)| \leq \|f\|_{H^\infty(D(0,1)} |B(z)|} for all {z \in D(0,1)}. (It may be easier to work with finite Blaschke products first to obtain this bound.)
  • (iii) Continuing the notation from (i), establish a factorisation {f(z) = B(z) \exp(g(z))} for some holomorphic function {g: D(0,1) \rightarrow {\bf C}} with {\mathrm{Re}(g(z)) \leq \log \|f\|_{H^\infty(D(0,1)}} for all {z\in D(0,1)}.
  • (iv) (Theorem of F. and M. Riesz, special case) If {f \in H^\infty(D(0,1))} extends continuously to the boundary {\{e^{i\theta}: 0 \leq \theta < 2\pi\}}, show that the set {\{ 0 \leq \theta < 2\pi: f(e^{i\theta})=0 \}} has zero measure.

Remark 13 The factorisation (iii) can be refined further, with {g} being the Poisson integral of some finite measure on the unit circle. Using the Lebesgue decomposition of this finite measure into absolutely continuous parts one ends up factorising {H^\infty(D(0,1))} functions into “outer functions” and “inner functions”, giving the Beurling factorisation of {H^\infty}. There are also extensions to larger spaces {H^p(D(0,1))} than {H^\infty(D(0,1))} (which are to {H^\infty} as {L^p} is to {L^\infty}), known as Hardy spaces. We will not discuss this topic further here, but see for instance this text of Garnett for a treatment.

Exercise 14 (Littlewood’s lemma) Let {f} be holomorphic on an open neighbourhood of a rectangle {R = \{ \sigma+it: \sigma_0 \leq \sigma \leq \sigma_1; 0 \leq t \leq T \}} for some {\sigma_0 < \sigma_1} and {T>0}, with {f} non-vanishing on the boundary of the rectangle. Show that

\displaystyle 2\pi \sum_\rho (\mathrm{Re}(\rho)-\sigma_0) = \int_0^T \log |f(\sigma_0+it)|\ dt - \int_0^T \log |f(\sigma_1+it)|\ dt

\displaystyle + \int_{\sigma_0}^{\sigma_1} \mathrm{arg} f(\sigma+iT)\ d\sigma - \int_{\sigma_0}^{\sigma_1} \mathrm{arg} f(\sigma)\ d\sigma

where {\rho} ranges over the zeroes of {f} inside {R} (counting multiplicity) and one uses a branch of {\mathrm{arg} f} which is continuous on the upper, lower, and right edges of {C}. (This lemma is a popular tool to explore the zeroes of Dirichlet series such as the Riemann zeta function.)

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