Giải tích 7 (MA4247) « Ngô Quốc Anh

Tháng Mười Hai 17, 2009

Schwarz’s Lemma, Schwarz-Pick theorem, and some applications involving inequalities

Chuyên mục: Các Bài Tập Nhỏ, Giải tích 7 (MA4247), Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 10:52

In mathematics, the Schwarz lemma, named after Hermann Amandus Schwarz, is a result in complex analysis about holomorphic functions defined on the open unit disk.

Schwarz’s Lemma: Let $D=\{z : |z|<1\}$ be the open unit disk in the complex plane $\mathbb C$ . Let $f : D \to \overline D$ be a holomorphic function with $f(0)=0$ . The Schwarz lemma states that under these circumstances $|f(z)| \leq |z|$ for all $z \in D$ , and $|f'(0)| \leq 1$ . Moreover, if the equality $|f(z)|=|z|$ holds for any $z \ne 0$ , or $|f'(0)|=1$ then $f$ is a rotation, that is, $f(z)=az$ with $|a=1$ .

This lemma is less celebrated than stronger theorems, such as the Riemann mapping theorem, which it helps to prove; however, it is one of the simplest results capturing the “rigidity” of holomorphic functions. No similar result exists for real functions, of course. To prove the lemma, one applies the maximum modulus principle to the function $\frac{f(z)}{z}$ .

Proof: Let $g(z)=\frac{f(z)}{z}$ . The function $g(z)$ is holomorphic in $D$ (excluding $0$ ) since $f(0)=0$ and $f$ is holomorphic. Let $D_r$ be a closed disc within $D$ with radius $r$ . By the maximum modulus principle,

$\displaystyle |g(z)| = \frac{|f(z)|}{|z|} \leq \frac{|f(z_r)|}{|z_r|} \le \frac{1}{r}$

for all $z$ in $D_r$ and all $z_r$ on the boundary of $D_r$ . As $r$ approaches $1$ we get $|g(z)| \leq 1$ . Moreover, if there exists a $z_0$ in $D$ such that $g(z_0)=1$ . Then, applying the maximum modulus principle to $g$ , we obtain that $g$ is constant, hence $f(z)=kz$ , where $k$ is constant and $|k|=1$ . This is also the case if $|f'(0)|=1$ .

A variant of the Schwarz lemma can be stated that is invariant under analytic automorphisms on the unit disk, i.e. bijective holomorphic mappings of the unit disc to itself. This variant is known as the Schwarz-Pick theorem (after Georg Pick):

Schwarz-Pick theorem: Let $f : D \to D$ be holomorphic. Then, for all $z_1, z_2 \in D$ ,

$\displaystyle\left|\frac{f(z_1)-f(z_2)}{1-\overline{f(z_1)}f(z_2)}\right| \le \frac{\left|z_1-z_2\right|}{\left|1-\overline{z_1}z_2\right|}$

and, for all $z \in D$

$\displaystyle\frac{\left|f'(z)\right|}{1-\left|f(z)\right|^2} \le \frac{1}{1-\left|z\right|^2}$ .

The expression

$\displaystyle d(z_1,z_2)=\tanh^{-1}\left(\frac{\left|z_1-z_2\right|}{\left|1-\overline{z_1}z_2\right|}\right)$

is the distance of the points $z_1, z_2$ in the Poincaré metric, i.e. the metric in the Poincaré disc model for hyperbolic geometry in dimension two. The Schwarz-Pick theorem then essentially states that a holomorphic map of the unit disk into itself decreases the distance of points in the Poincaré metric. If equality holds throughout in one of the two inequalities above (which is equivalent to saying that the holomorphic map preserves the distance in the Poincaré metric) , then $f$ must be an analytic automorphism of the unit disc, given by a Möbius transformation mapping the unit disc to itself.

An analogous statement on the upper half-plane $\mathbb H$ can be made as follows:

Let $f: \mathbb H \to \mathbb H$ be holomorphic. Then, for all $z_1, z_2 \in \mathbb H$ ,

$\displaystyle\left|\frac{f(z_1)-f(z_2)}{\overline{f(z_1)}-f(z_2)}\right| \le \frac{\left|z_1-z_2\right|}{\left|\overline{z_1}-z_2\right|}$ .

This is an easy consequence of the Schwarz-Pick theorem mentioned above: One just needs to remember that the Cayley transform

$\displaystyle W(z) = \frac{z-i}{z + i}$

maps the upper half-plane $\mathbb H$ conformally onto the unit disc $D$ . Then, the map $W \circ f \circ W^{-1}$ is a holomorphic map from $D$ onto $D$ . Using the Schwarz-Pick theorem on this map, and finally simplifying the results by using the formula for $W$ , we get the desired result. Also, for all $z \in \mathbb H$ ,

$\displaystyle\frac{\left|f'(z)\right|}{\mbox{Im }f(z)} \le \frac{1}{\mbox{Im }(z)}$ .

If equality holds for either the one or the other expressions, then $f$ must be a Möbius transformation with real coefficients. That is, if equality holds, then

$\displaystyle f(z)=\frac{az+b}{cz+d}$

with $a$ , $b$ , $c$ , $d$ being real numbers, and $ad-bc>0$ .

Proof: The proof of the Schwarz-Pick theorem follows from Schwarz’s lemma and the fact that a Möbius transformation of the form

$\displaystyle\frac{z-z_0}{\overline{z_0}z-1}$ where $|z_0|<1$

maps the unit circle to itself. Fix $z_1$ and define the Möbius transformations

$\displaystyle M(z)=\frac{z_1-z}{1-\overline{z_1}z}$ and $\displaystyle\phi(z)=\frac{f(z_1)-z}{1-\overline{f(z_1)}z}$ .

Since $M(z_1)=0$ and the Möbius transformation is invertible, the composition $\varphi(f(M^{-1}(z)))$ maps $0$ to $0$ and the unit disk is mapped into itself. Thus we can apply Schwarz’s lemma, which is to say

$\displaystyle |\phi(f(M^{-1}(z)))|=\left|\frac{f(z_1)-f(M^{-1}(z))}{1-\overline{f(z_1)}f(M^{-1}(z))}\right| \leq |z|$ .

Now calling $z_2=M^{-1}(z)$ (which will still be in the unit disk) yields the desired conclusion

$\displaystyle\left|\frac{f(z_1)-f(z_2)}{1-\overline{f(z_1)}f(z_2)}\right| \le \left|\frac{z_1-z_2}{1-\overline{z_1}z_2}\right|$ .

To prove the second part of the theorem, we just let $z_2$ tend to $z_1$ .

Application 1 (QE Berkeley Spring 1991). Let the function $f$ be analytic in the unit disc, with $|f(z)| \leq 1$ and $f(0)=0$ . Assume that there is a number $r \in (0,1)$ such that $f(r)=f(-r)=0$ . Prove that

$\displaystyle\left| {f\left( z \right)} \right| \leqslant \left| z \right|\left| {\frac{{{z^2} - {r^2}}} {{1 - {r^2}{z^2}}}} \right|$ .

Solution. Schwartz’s lemma implies that the function $f_1(z)=\frac{f(z)}{z}$ satisfies $|f_1(z)| \leq 1$ . The linear fractional map $z \mapsto \frac{z-r}{1-rz}$ sends the unit disc onto itself. Applying Schwartz’s lemma to the function

$\displaystyle f_2(z)=f_1\left( \frac{z-r}{1-rz} \right)$

we conclude that the function

$\displaystyle f_3(z)=\frac{f_1(z)}{\left(\frac{z-r}{1-rz}\right)}$

satisfies $|f_3(z)| \leq 1$ . Similarly, the map $z \mapsto \frac{z+r}{1+rz}$ sends the unit disc onto itself, and Schwartz’s lemma applied to the function

$\displaystyle f_4(z)=\frac{f_3(z)}{\left( \frac{z+r}{1+rz} \right)}$

implies that the function $f_5(z) =f_3\left( \frac{z+r}{1+rz} \right)$ satisfies $|f_5(z)| \leq 1$ . All together, then

$\displaystyle\left| {f\left( z \right)} \right| \leqslant \left| z \right|\left| {\frac{{z - r}} {{1 - rz}}} \right|\left| {\frac{{z + r}} {{1 + rz}}} \right|\left| {{f_5}\left( z \right)} \right| \leqslant \left| z \right|\left| {\frac{{z - r}} {{1 - rz}}} \right|\left| {\frac{{z + r}} {{1 + rz}}} \right|$ .

which is the desired inequality.

Application 2 (QE NUS Spring 2009). Suppose $f$ is analytic in $D := \{ z \in \mathbb C : |z|<1\}$ with $|f(z)|<1$ . Show that

$\displaystyle\frac{{\left| {f\left( 0 \right)} \right| - \left| z \right|}} {{1 - \left| {f\left( 0 \right)} \right|\left| z \right|}} \leqslant \left| {f\left( z \right)} \right| \leqslant \frac{{\left| {f\left( 0 \right)} \right| + \left| z \right|}} {{1 + \left| {f\left( 0 \right)} \right|\left| z \right|}}$

for all $z \in D$ .

Application 3 (QE NUS Fall 2009). Is there an analytic function $f$ on $\Delta$ (unit disk in the complex plane with center $0$ ) such that $|f(z)|<1$ for $|z|<1$ with $f(0)=\frac{1}{2}$ and $f'(0)=\frac{3}{4}$ ? If so, find such an $f$ . Is it unique?

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Tháng Mười 28, 2009

The weak and weak* topologies: A few words

Chuyên mục: Giải tích 7 (MA4247), Linh Tinh, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 2:48

The weak and weak* topologies are the weakest in which certain linear functionals are continuous.

We start with a normed linear space $X$ . The dual space of $X$ , denoted by $X'$ , is the collection of all continuous linear functionals, i.e., the set of all mapping $\ell : X \to \mathbb R$ satisfying

$\ell(ax)=a \ell (x)$ , $\ell(x+y)=\ell(x)+\ell(y)$

and

$\displaystyle\lim_{n \to \infty} \ell(x_n) = \ell(x)$ when $\displaystyle\lim_{n \to \infty} \|x_n - x\|=0$ .

Definition 1. In $X$ , the strong topology is the norm topology, i.e., we can talk about an open set of $X$ , for example $U$ in the following sense: $U \subset X$ is said to be open if and only if for each $x_0 \in U$ , there exists $\varepsilon>0$ such that $\{ x \in X: \|x-x_0\|<\varepsilon\} \subset U$ .

Claim 1. Bounded linear functionals are continuous in the strong topology.

Proof. We first recall that a linear functional $\ell$ is said to be bounded if there is a positive number $c$ such that $|\ell (x)| \leq c\|x\|$ for all $x \in X$ .

Now we assume $\ell$ is continuous but not bounded; then for any choice of $c=n$ , one has $\ell(x_n) > n \|x_n\|$ . Clearly, $x_n$ can be replaced by any multiple of $x_n$ ; if we normalize $x_n$ so that

$\displaystyle \|x_n\|=\frac{1}{\sqrt{n}}$

then $x_n \to 0$ but $\ell (x_n) \to \infty$ . This shows the lack of boundedness implies the lack of continuity.

Now we assume $\ell$ is bounded. For arbitray $x_n$ and $x$ , one gets

$|\ell(x_n)-\ell (x)| = |\ell (x_n-x)| \leq c\|x_n-x\|$ ;

this shows that boundedness implies continuity.

Definition 2. In $X$ , the weak topology is the weakest topology in which all bounded linear functionals are continuous.

The open sets in the weak topology are unions of finite intersections of sets of the form

$\{ x : a< \ell(x) < b\}$ .

Clearly, in an infinite-dimensional space the intersection of a finite number of sets of the above form is unbounded. This shows that every set that is open in the weak topology is unbounded. In particular, the balls

$\{ x : \|x\|<R\}$

opens in the strong topology, are not open is the weak topology.

Definition 3. In $X'$ the dual space of $X$ , the weak* topology is the crudest topology in which all linear functionals

$x: X' \to \mathbb R, x(\ell) := \ell(x)$

are continuous.

If $X'$ is nonreflexive, the weak* topology is genuinely coarser than the weak topology, as will be clear from the following theorem due to Alaoglu

Theorem (Alaoglu). The closed unit ball in $X'$ is compact in the weak* topology.

We end this topic by the following theorem

Theorem. The closed unit ball in $X$ is compact in the weak topology if and only if $X$ is reflexive.

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Tháng tám 29, 2009

On a polynomials of degree n, having all its zeros in the unit dis

Chuyên mục: Các Bài Tập Nhỏ, Giải tích 7 (MA4247), Linh Tinh, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 21:07

I found a very useful inequality involving a polynomials of degree $n$ , having all its zeros in the unit disk from the following paper, doi:10.1016/j.jmaa.2009.07.049, published in J. Math. Anal. Appl.

Statement. If $P(z)$ is a polynomial of degree $n$ , having all its zeros in the disk $|z| \leq 1$ , then

$\left| {zP'(z)} \right| \geq \displaystyle\frac{n} {2}\left| {P(z)} \right|$

for $|z|=1$ .

Proof. Since all the zeros of $P(z)$ lie in $latex|z| \leq 1$. Hence if $z_1, z_2,...,z_n$ are the zeros of $P(z)$ , then $|z_j| \leq 1$ for all $j =1,2,...,n$ . Clearly,

$\displaystyle\Re \frac{{{e^{i\theta }}P'\left( {{e^{i\theta }}} \right)}}{{P\left( {{e^{i\theta }}} \right)}} = \sum\limits_{j = 1}^n {\Re \frac{{{e^{i\theta }}}}{{{e^{i\theta }} - {z_j}}}} ,$

for every point $e^{i\theta}$ , $0 \leq \theta < 2 \pi$ which is not a zero of $P(z)$ . Note that

$\displaystyle\sum\limits_{j = 1}^n {\Re \frac{e^{i\theta }}{e^{i\theta } - z_j}}\geq\sum\limits_{j = 1}^n{\frac{1}{2}}=\frac{n}{2}.$

This implies

$\displaystyle\left| {\frac{{{e^{i\theta }}P'\left( {{e^{i\theta }}} \right)}}{{P\left( {{e^{i\theta }}} \right)}}} \right| \geq \Re \frac{{{e^{i\theta }}P'\left( {{e^{i\theta }}} \right)}}{{P\left( {{e^{i\theta }}} \right)}} \geq \frac{n}{2},$

for every point $e^{i \pi}$ , $0 \leq \theta < 2\pi$ . Hence

$\left| {zP'(z)} \right| \geq \displaystyle \frac{n}{2}\left| {P(z)} \right|$

for $|z|=1$ and this completes the proof.

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Tháng tám 14, 2009

How to find a conformal mapping between the quadrants and the semidisc

Chuyên mục: Các Bài Tập Nhỏ, Giải tích 7 (MA4247), Linh Tinh, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 21:34

In the previous topic I show you by the following map

$f : z \mapsto \frac{z+1}{z-1}$

maps $\{z : \Re z < 0\}$ onto $\{w : |w|<1\}$ , and is conformal. Therefore the map

$g : z \mapsto \frac{z+1}{1-z}$

maps $\{z : \Re z > 0\}$ onto $\{w : |w|<1\}$ , and is conformal. What I am going to do is to prove that

$\Re z > 0\quad \Leftrightarrow \quad \Re \left( {\frac{{z + 1}} {{1-z}}} \right) > 0.$

To this purpose, we assume z=x+iy , i.e., $\Re z = y$ . Now by a simple calculation

$\frac{{z + 1}} {{1 - z}} = \frac{{\left( {x + 1} \right) + iy}} {{\left( {1 - x} \right) - iy}} = \frac{{\left[ {\left( {x + ...$

which yields

$\Re \left( {\frac{{z + 1}} {{1 - z}}} \right) = \frac{{2y}} {{{{\left( {1 - x} \right)}^2} + {y^2}}}.$

Having this fact we can easily see that under the map the first and fourth quadrants maps to upper and lower semidisks, respectively.

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An easy way to construct a conformal mapping between upper half plane and the open unit disk

Chuyên mục: Các Bài Tập Nhỏ, Giải tích 7 (MA4247), Linh Tinh, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 20:54

The locus |z + 1|= | z -1| is the perpendicular bisector of the line segment joining to , that is, the imaginary axis. The set |z + 1|< | z -1| is then the set of points closer to than to , that is, the left half-plane $\Re z <0$ . Hence, $\Re z <0$ if and only if

$\frac{|z+1|}{|z-1|}<1.$

The map

$f : z \mapsto \frac{z+1}{z-1}$

maps $\{z : \Re z < 0\}$ onto $\{w : |w|<1\}$ , and is conformal as $f' \ne 0$ . The inverse map is easily seen to be

$w \mapsto z=\frac{w+1}{w-1}.$

The locus |z + i|= | z - i| is the perpendicular bisector of the line segment joining to , that is, the real axis. The set |z + i|< | z -i| is then the set of points closer to than to , that is, the lower half-plane $\Im z <0$ . Hence, $\Im z <0$ if and only if

$\frac{|z+i|}{|z-i|}<1.$

The map

$g : z \mapsto \frac{z+i}{z-i}$

maps $\{z : \Im z < 0\}$ onto $\{w : |w|<1\}$ , and is conformal as $g' \ne 0$ . The inverse map is easily seen to be

$w \mapsto z=\frac{w+1}{w-1}i.$

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Tháng tám 13, 2009

Two examples of analytic function on a punctured unit disk which has a removable singularity at the origin

Chuyên mục: Các Bài Tập Nhỏ, Giải tích 7 (MA4247), Linh Tinh — Ngô Quốc Anh @ 15:54

The following question was proposed in NUS under the QE in AY 2007-2008:

Consider the punctured disk $D=\{ z \in \mathbb C | 0 <|z| <1\}$ . Suppose $f : D \to \mathbb C$ is an analytic function such that

$|f''(z)| \leq \frac{2}{|z|^2}$

for all $z \in D$ . Is it true that has a removable singular point at z=0 ?

Proof. Denote the Laurent expansion of by

$f\left( z \right) = \sum\limits_{n = - \infty }^{ + \infty } {{a_n}{z^n}}$

where

${a_n} = \frac{1} {{2\pi i}}\int\limits_{\left| z \right| = r < 1} {\frac{{f\left( z \right)dz}} {{{z^{n + 1}}}}} .$

Then from

$f''\left( z \right) = \sum\limits_{n = - \infty }^{ + \infty } {\underbrace {\left( {n + 2} \right)\left( {n + 1} \right){a_...$

we get

${b_n} = \frac{1} {{2\pi i}}\int\limits_{\left| z \right| = r < 1} {\frac{{f''\left( z \right)dz}} {{{z^{n + 1}}}}} .$

Thus,

$\left| {{b_n}} \right| \leqslant \frac{1} {{2\pi }}\int\limits_{\left| z \right| = r < 1} {\left| {\frac{{f''\left( z \rig...$

When $n \leq -3$ , let $r \to 0$ we see that z=0 is a removable singularity of since b_n=0 ( $n \leq -3$ ) implies a_n=0 ( $n \leq -1$ ).

Remark. The second derivative can be replaced by an $\mathbb Z\ni m \geq 1$ and therefore $\frac{2}{|z|^2}$ should be $\frac{2}{|z|^m}$ . This is a question of UCLA QE in Winter 2007.

We also have a similar question proposed in a QE of Indiana University. It says that if $f : D \to \mathbb C$ is an analytic function such that

for all $z \in D$ . Then $f \equiv 0$ .

Proof. As above, one has

$\left| {{b_n}} \right| = \left| {\frac{1} {{2\pi i}}\int\limits_{\left| z \right| = r < 1} {\frac{{f\left( z \right)dz}} {...$

When n<0 , letting $r \to 0$ we have a_n= 0 for all $n \leq -1$ which implies z=0 is a removable singularity of . In other words, can be extended to an analytic function of the unit disk. Since $\log \frac{1}{|z|} =0$ when |z|=1 , by the Maximum Modules Principle we obtain $f \equiv 0$ .

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Tháng tám 4, 2009

Another point of view of the Argument Principle

Chuyên mục: Các Bài Tập Nhỏ, Giải tích 7 (MA4247), Linh Tinh, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 23:07

With Laurent series and the classication of singularities in hand, it is easy to prove the Residue Theorem. In addition to being a handy tool for evaluating integrals, the Residue Theorem has many theoretical consequences, for example, the Argument Principle, Rouché’s Theorem, the Local Mapping Theorem, the Open Mapping Theorem, the Hurwitz Theorem, the general Casorati-Weierstrass Theorem, and Riemann’s Theorem. This writeup presents the Argument Principle with an application.

The Argument Principle

Theorem (Argument Principle). Let $\gamma$ be a simple closed counterclockwise curve. Let be analytic and nonzero on $\gamma$ and meromorphic inside $\gamma$ . Let Z(f) denote the number of zeros of inside $\gamma$ , each counted as many times as its multiplicity, and let P(f) denote the number of poles of inside $\gamma$ , each counted as many times as its multiplicity. Then

$Z\left( f \right) - P\left( f \right) = \frac{1} {{2\pi i}}\int_\gamma {\frac{{df}} {f}} .$

The Winding Number

Definition. Let $\gamma : [0,1] \to \mathbb C$ be any closed rectifiable path. By the usual abuse of notation, let $\gamma$ also denote the corresponding subset of $\mathbb C$ . Consider a complex-valued function on the complement of the path,

$Ind\left( {\gamma , \cdot} \right): \mathbb{C} - \gamma \to \mathbb{C}, \quad Ind\left( {\gamma ,z} \right) = \frac{1} {{2\p...$

For any $z \in \mathbb C- \gamma$ , the function $Ind(\gamma, z)$ is the winding number of $\gamma$ about . For instance, if $\gamma$ is a circle traversed once counterclockwise about then its winding number about is .

The Argument Principle Again

With the winding number in hand, we can rephrase the Argument Principle in a way that explains its name.

Theorem (Argument Principle, second version). Let $\gamma$ be a simple closed counterclockwise curve. Let be analytic and nonzero on $\gamma$ and meromorphic inside $\gamma$ . Let Z(f) denote the number of zeros of inside $\gamma$ , each counted as many times as its multiplicity, and let P(f) denote the number of poles of inside $\gamma$ , each counted as many times as its multiplicity. Then

$Z(f)-P(f) = Ind(f \circ \gamma, 0).$

That is, the theorem is called the Argument Principle because the number of zeros minus poles of inside $\gamma$ is the number of times that the argument of $f\circ \gamma$ increases by $2\pi$ .

For example, consider the polynomial

f(z)=z^4-8z^3+3z^2+8z+3.

To count the roots of in the right half plane, let be a large disk centered at the orgin, large enough to contain all the roots of . Let $\gamma$ be the boundary of the right half of . Thus $\gamma$ is the union of a segment of the imaginary axis and a semicircle. The values of on the imaginary axis are

f(iy) = (y^4+3y^2+3) + i(8y^3+8y).

Therefore, Ref(iy) is always positive, and so takes the imaginary axis into the right half plane. On the semicircle, f(z) behaves qualitatively as z^4 . Therefore the path $\Gamma = f \circ \gamma$ winds twice around the origin, showing that has two roots in the right half plane.

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Tháng tám 1, 2009

Weierstrass infinite products

Chuyên mục: Các Bài Tập Nhỏ, Giải tích 7 (MA4247), Linh Tinh, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 23:56

We now turn to the construction of an entire function with prescribed zeros

Theorem (Weierstrass). Given any sequence $\{a_n\}$ of complex numbers with $|a_n| \to \infty$ as $n\to \infty$ , there exists an entire function that vanishes at all and nowhere else. Any other such entire function is of the form $f(z)e^{g(z)}$ , where is entire.

To begin the proof, note first that if f_1 and f_2 are two entire functions that vanish at all z = a_n and nowhere else, then $\frac{f_1}{f_2}$ has removable singularities at all the points an. Hence $\frac{f_1}{f_2}$ is entire and vanishes nowhere, so that there exists an entire function with $\frac{f_1(z)}{f_2(z)} = e^{g(z)}$ . Therefore $f_1(z) = f_2(z)e^{g(z)}$ and the last statement of the theorem is verified.

Hence we are left with the task of constructing a function that vanishes at all the points of the sequence $\{a_n\}$ and nowhere else. A naive guess, suggested by the product formula for $\sin \pi z$ , is the product

$\prod\limits_n {\left( {1 - \frac{z} {{{a_n}}}} \right)} .$

The problem is that this product converges only for suitable sequences $\{a_n\}$ , so we correct this by inserting exponential factors. These factors will make the product converge without adding new zeros.

For each integer $k \geq 0$ we define canonical factors by

E_0(z) = 1-z and $E_k(z) = (1-z)e^{z+\frac{z^2}{2}+ \cdots +\frac{z^k}{k}}$ , for $k \geq 1$ .

The integer is called the degree of the canonical factor.

Lemma. If $|z| \leq \frac{1}{2}$ , then $|1-E_k(z)| \leq c|z|^{k+1}$ for some .

Suppose that we are given a zero of order at the origin, and that a_1 , a_2 ,… are all non-zero. Then we define the Weierstrass product by

$f\left( z \right) = {z^m}\prod\limits_{n = 1}^\infty {{E_n}\left( {\frac{z} {{{a_n}}}} \right)} .$

We claim that this function has the required properties; that is, is entire with a zero of order at the origin, zeros at each point of the sequence $\{a_n\}$ , and vanishes nowhere else.

Fix R>0 , and suppose that belongs to the disc |z|<R . We shall prove that has all the desired properties in this disc, and since is arbitrary, this will prove the theorem.

We can consider two types of factors in the formula defining , with the choice depending on whether $|a_n| \leq 2R$ or |a_n| > 2R . There are only finitely many terms of the first kind (since $|a_n| \to \infty$ ), and we see that the finite product vanishes at all z = a_n with |a_n| < R . If $|a_n| \geq 2R$ , we have $\left|\frac{z}{a_n}\right| \leq \frac{1}{2}$ , hence the previous lemma implies

$\left| {1 - {E_n}\left( {\frac{z} {{{a_n}}}} \right)} \right| \leqslant c{\left| {\frac{z} {{{a_n}}}} \right|^{n + 1}} \leqsl...$

Note that by the above remark, does not depend on . Therefore, the product

$\prod\limits_{\left| {{a_n}} \right| \geqslant 2R} {{E_n}\left( {\frac{z} {{{a_n}}}} \right)}$

defines a holomorphic function when |z| < R , and does not vanish in that disc. This shows that the function has the desired properties, and the proof of Weierstrass’s theorem is complete.

Question (Harvard’s QE). Given a sequence of complex numbers such that $|z_n| \to \infty$ , does there exist an entire function with ?

Proof. By the Weierstrass theorem, we can construct an entire function g(z) with simple zeros z_1, z_2,... . Then we define

$f\left( z \right) = \sum\limits_{n = 1}^\infty {{u_n}\left( z \right)} = \sum\limits_{n = 1}^\infty {{e^{{\gamma _n}\left(...$

where $\gamma_n$ is chosen such that when $|z| < \frac{|z_n|}{2}$ ,

$\left| {{u_n}\left( z \right)} \right| = \left| {{e^{{\gamma _n}\left( {z - {z_n}} \right)}}\frac{{g\left( z \right)}} {{z - ...$

Because $|z_n| \to \infty$ , for any R>0 , there exists N>0 such that |z_n| > 2R when $n \geq N$ . Hence

$\left| {{u_n}\left( z \right)} \right| \leqslant \frac{1} {{{n^2}}}$

holds for all $|z| \leq R$ when $n \geq N$ . In other words,

$\sum\limits_{n = 1}^\infty {{u_n}\left( z \right)}$

converges uniformly for all $|z| \leq R$ , so that f(z) in analytic in $\{ |z| < R\}$ . Since can be arbitrarily large, f(z) is an entire function. It is easy to see that

$\mathop {\lim }\limits_{x \to {z_n}} {u_n}\left( z \right) = \mathop {\lim }\limits_{x \to {z_n}} \left( {{e^{{\gamma _n}\lef...$

while for $k \ne n$ ,

u_k(z_n)=0,

which implies that f(z) is an entire function satisfying the required condition.

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Tháng Bảy 31, 2009

Picard’s Theorem + Hadamard’s Theorem = ?

Chuyên mục: Các Bài Tập Nhỏ, Giải tích 7 (MA4247), Linh Tinh, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 2:51

Question. Let be an entire non-constant function that satisfies the functional equation

f(1 - z) = 1 - f(z)

for all $z \in \mathbb C$ . Show that $f(\mathbb C) = \mathbb C$ .

Solution. Assume by contradiction, then W.L.O.G. by the Picard’s Theorem we can assume that misses $a \in \mathbb C$ . By the Hadamard’s Theorem,

$f(z)-a = e^{p(z)}$

for some polynomial . Therefore,

$f(z) = a +e^{p(z)}$

for all $z \in \mathbb C$ . From the fact that

f(1-z)=1-f(z)

we get

$\underbrace {a + {e^{p\left( {1 - z} \right)}}}_{f\left( {1 - z} \right)} = \underbrace {1 - \left( {a + {e^{p\left( z \right...$

which yields

${e^{p\left( z \right)}} = 2a - 1 + {e^{p\left( {1 - z} \right)}}.$

Put z=0 and z=1 , we obtain

${e^{p\left( 0 \right)}} = 2a - 1 + {e^{p\left( 1 \right)}}, \quad {e^{p\left( 1 \right)}} = 2a - 1 + {e^{p\left( 0 \right)}}.$

Hence

${e^{p\left( 0 \right)}} = 2\left( {2a - 1} \right) + {e^{p\left( 0 \right)}}$

which implies $a=\frac{1}{2}$ . From the identity f(1-z)=1-f(z) put $z=\frac{1}{2}$ we then deduce that

$f\left( {\frac{1} {2}} \right) = \underbrace {\frac{1} {2}}_a$

a contradiction.

Note: I think I should post some applications of the Hadamard’s Theorem.

Corrigendum to the previous proof. Assume by contradiction, then takes all values except possibly some . Also note that f(1/2) = 1/2 . So we can assume $a \neq 1/2$ which means that $1-a \ne a$ . It then does take the value 1-a , and hence takes the value . The proof is now complete.

I thank Xu Wei Biao for pointing out a mistake in the previous solution.

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Tháng Bảy 30, 2009

A couple of complex integrals involving exp(itx) for a real parameter t

Chuyên mục: Các Bài Tập Nhỏ, Giải tích 7 (MA4247), Linh Tinh, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 17:14

In this turn, I will consider a couple of examples of complex contour integrals with respect to variable involving the following factor $e^{itx}$ where a real parameter.

Problem 1. Evaluate the integral

$I\left( t \right) = \int\limits_{ - \infty }^\infty {\frac{{{e^{itx}}}} {{{{\left( {x + i} \right)}^2}}}dx}$

where $-\infty < t<\infty$ .

Solution. Let

and consider first the case t>0 . Then |f_t(z)| is bounded in the upper half-plane by $\frac{1}{|z+i|^2}$ . For R>1 let

$C_R=\Gamma_R \cup [-R, R]$ ,

where $\Gamma_R$ is the semicircle centered at the origin joining and , oriented counterclockwise.

integrationofexpitx1

Then function f_t is holomorphic on C_R and its interior, so, by Cauchy’s Theorem, we have

The absolute value of the second summand on the right is at most $\frac{\pi R}{R^2}$ since $|f_t(z)| \leq \frac{1}{R^2}$ on $\Gamma_R$ . Taking the limit as $R \to \infty$ we obtain I(t)=0 .

Suppose now t<0 . Then |f_t| is bounded in the lower half-plane by $\frac{1}{|z+i|^2}$ . Let C'_R be the reflection of C_R with respect to the real axis, oriented clockwise.

integrationofexpitx2

By Residue Theorem, we have

$\int\limits_{{C_R'}} {{f_t}\left( z \right)dz} = - 2\pi i\mathop {Res}\limits_{z = - i} {f_t}\left( z \right).$

A calculation shows that the residue equals to ite^t , so

As $R \to \infty$ , the contribution to the last integral from the semicircle tends to since $|f_t| \leq (R-1)^2$ on the semicircle, giving

$I\left( t \right) = 2\pi t{e^t}.$

For t=0 , the integral is elementary as shown below

$\int\limits_{ - R}^R {{f_t}\left( x \right)dx} = \int\limits_{ - R}^R {\frac{1} {{{{\left( {x + i} \right)}^2}}}dx} = \frac...$

as $R \to \infty$ . Thus I(0)=0 .

Problem 2. Evaluate the integral

where $-\infty < t<\infty$ .

Solution. For t=0 , the integral is elementary and finally we get F(0)=0 . For t<0 , the function

is bounded in the upper half-plan and in fact is $O(|z|^{-3})$ there. In this case, one has F(t)=0 for all t<0 . For t>0 , by Residue Theorem,

Finally, we obtain

$F\left( t \right) = \pi i{t^2}{e^{ - t}}$

for t>0 .

Similar question. Estimate the limit

$\mathop {\lim }\limits_{A \to \infty } \int\limits_{ - A}^A {{{\left( {\frac{{\sin x}} {x}} \right)}^2}{e^{itx}}dx}$

for a real number .

Hint. For t=0 , the limit equals to $\pi$ which I had considered here. For $t \ne 0$ , we need to calculate the following residue

$\mathop {Res}\limits_{z = 0} \left[ {\frac{{1 - {e^{2iz}}}} {{{z^2}}}{e^{itz}}} \right]$

which equals to 2t+2 by using the following helpful formula

$\mathop {Res}\limits_{z = {z_0}} \frac{{\varphi \left( z \right)}} {{\phi \left( z \right)}} = \frac{{\varphi \left( {{z_0}} ...$

where $\varphi$ and $\phi$ are holomorphic and z_0 is a simple pole.

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Tháng Bảy 31, 2009

Picard’s Theorem + Hadamard’s Theorem = ?

Tháng Bảy 30, 2009

A couple of complex integrals involving exp(itx) for a real parameter t

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