UMD Analysis Qualifying Exam/Aug08 Complex

Solution 2
We will compute the general case:

$$ \int_0^\infty \frac{1}{x^n+1}dx \!\,$$

Find Poles of f(z)
The poles of $$ f(z)= \frac{1}{z^n+1} \!\,$$ are just the zeros of $$ z^n+1   \!\,$$, so we can compute them in the following manner:

If $$z=re^{i \theta}   \!\,$$ is a solution of $$ z^n+1=0   \!\,$$,

then $$ z^n=r^ne^{i\theta n} = -1  \!\,$$

$$  \Rightarrow r=1   \!\,$$ and $$ i n \theta  = i (\pi + 2\pi k )    \!\,$$

$$  \Rightarrow \theta = \frac{\pi + 2\pi k}{n}\!\,$$ , k=0,1,2,...,n-1.

Thus, the poles of $$ f(z)  \!\,$$ are of the form $$ z=e^{\frac{\pi+2\pi k}{n}}   \!\,$$ with $$k=0,1,...,n-1    \!\,$$

Choose Path of Contour Integral
In order to get obtain the integral of $$ f(x)   \!\,$$ from 0 to $$    \infty \!\,$$, let us consider the path $$  \gamma  \!\,$$ consisting in a line $$  A  \!\,$$ going from 0 to $$  R  \!\,$$, then the arc $$ B   \!\,$$ of radius $$ R   \!\,$$ from the angle 0 to $$  \frac{2 \pi}{n}  \!\,$$ and then the line $$ C   \!\,$$ joining the end point of $$B    \!\,$$ and the initial point of $$    A\!\,$$,



where $$ R  \!\,$$ is a fixed positive number such that

the pole $$ z_0 = e^{i \frac{\pi}{n}}  \!\,$$ is inside the curve $$    \gamma \!\,$$. Then, we need to estimate the integral

$$ \int_{\gamma} f(z) = \underbrace{\int_{0}^R f(z)}_{A} + \underbrace{\int_{S(R)} f(z)}_{B} + \underbrace{\int_{Re^{i\frac{2\pi}{n}}}^0 f(z) }_{C} \!\,$$

Compute Residues of f at z0= exp{i\pi /n}
$$ \begin{align}

Res(f,z_0)&= \left. \frac{1}{(z^n+1)' } \right|_{z=z_0}  \\ &= \frac{1}{nz_0^{n-1}} \\ &= \frac{1}{ne^{\frac{i\pi}{n}(n-1)}} \\ &= \frac{e^{\frac{i\pi}{n}}}{ne^{i \pi}} \\ &= \frac{-1}{n}e^{\frac{i\pi}{n}}

\end{align}

\!\,$$

Bound Arc Portion (B) of Integral
$$ \begin{align} &\leq \int_{S(R)}\frac{dz}{|z^n+1|} \\ &= \int_{S(R)} \frac{dz}{|R^n+1|} \\ &\leq \frac{1}{R^n} \int_{S(R)} dz \\ &=    \frac{1}{R^n} \frac{2 \pi}{n} R \\ &= \frac{2 \pi}{nR^{n-1}} \end{align} $$
 * B|                             &=\left| \int_{S(R)} \frac{dz}{z^n+1} \right|  \\

Hence as $$R \rightarrow \infty \!\,$$, $$|B| \rightarrow 0 \!\,$$

Parametrize (C) in terms of (A)
Let $$z=re^{i\frac{2\pi}{n}} \!\,$$ where $$r \!\,$$ is real number. Then $$ dz=e^{i\frac{2\pi}{n}}dr \!\,$$

$$ \begin{align} C   &= \int_{Re^{i\frac{2\pi}{n}}}^0 \frac{dz}{1+z^n}\\ &= -\int_0^{Re^{i\frac{2\pi}{n}}}\frac{dz}{1+z^n} \\ &= -\int_0^R \frac{e^{i\frac{2\pi}{n}}}{1+(re^{i\frac{2\pi}{n}})^n}\\ &= -\int_0^R \frac{e^{i\frac{2\pi}{n}}}{1+r^n}\\ &= -e^{i\frac{2\pi}{n}} \underbrace{\int_0^R \frac{dr}{1+r^n}}_{A} \\ \end{align} $$

Apply Cauchy Integral Formula
From Cauchy Integral Formula, we have,


 * $$ A+B+C = 2\pi i \frac{-e^{i\frac{\pi}{n}}}{n } \!\,$$

As $$R \rightarrow \infty \!\,$$, $$B \rightarrow 0 \!\,$$. Also $$C \!\,$$ can be written in terms of $$A \!\,$$. Hence

$$ \begin{align} A+B+C &= A+C \\ &= (1-e^{i\frac{2\pi}{n}}) A \end{align} $$

We then have,

$$ \begin{align} A    &= \frac{2\pi i }{n }\frac{e^{i\frac{\pi}{n}}}{e^{i\frac{2\pi}{n}}-1} \\ &= \frac{\pi}{n} \frac{2i e^{i\frac{\pi}{n}}}{e^{i\frac{\pi}{n}}(e^{i\frac{\pi}{n}}-e^{-i\frac{\pi}{n}})}\\ &= \frac{\pi}{n} \frac{1}{\sin(\frac{\pi}{n})} \end{align} $$

Lemma: Two fixed points imply identity
Lemma. Let $$ f \!\,$$ be analytic on the unit $$ D \!\,$$, and assume that $$|f(z)| < 1 \!\,$$ on the disc. Prove that if there exist two distinct points $$a\!\,$$ and $$b\!\,$$ in the disc which are fixed points, that is, $$                 f(a)=a\!\,$$ and $$ f(b)=b  \!\,$$, then $$   f(z)=z               \!\,$$.

Proof Let $$h:D \rightarrow D                  \!\,$$ be the automorphism defined as

$$ h(z)= \frac{a-z}{1-\overline{a}z} \!\,$$

Consider now $$ F(z)= h \circ f \circ h^{-1} (z)   \!\,$$. Then, F has two fixed points, namely

$$ F(0)=   h \circ f \circ h^{-1} (0)= h \circ f (a) = h  (a) =0              \!\,$$

$$F\left(\frac{a-b}{1-\overline{a}b}\right) = h \circ f \circ h^{-1} \left(\frac{a-b}{1-\overline{a}b}\right) = h \circ f (b) = h (b) = \frac{a-b}{1-\overline{a}b}        \!\,$$.

Since $$ F(0)=0 \!\,$$,

$$\frac{a-b}{1-\overline{a}b} \neq 0 \!\,$$ (since $$ a \!\,$$ is different to $$ b \!\, $$), and

$$ \left|F\left(\frac{a-b}{1-\overline{a}b}\right)\right|=\left|\frac{a-b}{1-\overline{a}b}\right| \!\,$$,

by Schwarz Lemma,

$$F(z)= \alpha z    \!\,$$.

But, replacing $$\frac{a-b}{1-\overline{a}b} \!\, $$ into the last formula, we get $$ \alpha =1 \!\,$$.

Therefore,

$$   h \circ f \circ h^{-1} (z)  = z           \!\,$$,

which implies

$$        f(z)=z         \!\,$$

Shift Points to Create Fixed Points
Let $$f(z)=g(z)-1 \!\,$$. Then $$f(0)=0 \!\,$$ and $$f(-1)=-1 \!\,$$.

Notice that $$S \!\,$$ is an infinite horizontal strip centered around the real axis with height $$\pi \!\,$$. Since $$f(z) \!\,$$ is a unit horizontal shift left, $$f(S) \subset S \!\,$$.

Use Riemann Mapping Theorem
From the Riemann mapping theorem, there exists a biholomorphic (bijective and holomorphic) mapping $$h \!\,$$, from the open unit disk $$D \!\,$$ to $$S \!\,$$.

Define Composition Function
Let $$F=h^{-1}\circ f \circ h \!\,$$. Then $$ F\!\,$$ maps $$D \!\,$$ to $$D \!\,$$.

From the lemma, since $$F(z) \!\,$$ has two fixed points, $$F(z)=z \!\,$$ which implies $$ f(z)=z \!\,$$ which implies $$g=z+1 \!\,$$.

Choose any compact set K in D
Choose any compact set $$K \!\, $$ in the open unit disk $$ D\!\, $$. Since $$K \!\, $$ is compact, it is also closed and bounded.

We want to show that for all $$f \in \mathcal{F} \!\, $$ and all $$z \in K \!\, $$, $$|f(z)| \!\, $$ is bounded i.e.


 * $$|f(z)| < B_K \!\, $$

where $$B_K \!\, $$ is some constant dependent on the choice of $$K \!\, $$.

Apply Maximum Modulus Principle to find |f(z0)|
Choose $$z_0 \!\, $$ that is the shortest distance from the boundary of the unit disk $$D \!\, $$. From the maximum modulus principle, $$|f(z_0)|=\max_{z \in K} |f(z|  \!\, $$.

Note that $$z_0 \!\, $$ is independent of the choice of $$f \in \mathcal{F}\!\, $$.

Apply Cauchy's Integral Formula to f^2(z0)
We will apply Cauchy's Integral formula to $$f^2(z_0) \!\, $$ (instead of $$ f(z_0)\!\, $$) to take advantage of the hypothesis.

Choose sufficiently small $$r_0 > 0\!\, $$ so that $$ D(z_0,r_0) \!\, \in D$$

$$ \begin{align} f^2(z_0) &= \frac{1}{2\pi i} \int_{|z-z_0|=r_0} \frac{f^2(z)}{z-z_0}dz \\ &= \frac{1}{2\pi i} \int_0^{2\pi} \frac{f^2(z_0+r_0e^{i\theta})ir_0e^{i\theta}}{(z_0+r_0e^{i\theta})-z_0} d\theta\\ &= \frac{1}{2\pi} \int_0^{2\pi} f^2(z_0+r_0e^{i\theta})d\theta \end{align} $$

Integrate with respect to r
$$ \begin{align} \int_0^{r_0} r f^2(z_0)dr &= \int_0^{r_0} \int_0^{2\pi} f^2(z_0+re^{i\theta})rdr d\theta \\ &= \frac{1}{2\pi}\int\int_{D(z_0,r)}f^2(x+iy)dxdy \end{align} $$

Integrating the left hand side, we have

$$ \int_0^{r_0} r f^2(z_0)dr =   \frac{r_0^2}{2}f^2(z_0)     \!\,$$

Hence,

$$\frac{r_0^2}{2}f^2(z_0) = \frac{1}{2\pi}\int\int_{D(z_0,r)}f^2(x+iy)dxdy    \!\,$$

Bound |f(z0)| by using hypothesis
$$ \begin{align} \left| \frac{r_0^2}{2} f^2(z_0) \right| &= \left| \frac{1}{2 \pi} \int\int_{D(z_0,r_0)} f^2(x+iy)dxdy \right| \\ &\leq   \frac{1}{2 \pi} \int\int_{D(z_0,r_0)} |f^2(x+iy)|dxdy  \\ &\leq \frac{1}{2 \pi} \int\int_{D} |f^2(x+iy)|dxdy \\ &\leq \frac{1}{2\pi} \\ \\ \mbox{Then    } \\ \left| \frac{r_0^2}{2} f^2(z_0) \right| &\leq \frac{1}{2 \pi}   \\ \\ \mbox{This implies}\\ \\ \end{align} $$
 * f(z_0)| &\leq \frac{1}{r_0 \sqrt{\pi}}

Apply Montel's Theorem
Then, since any $$ f \in \mathcal{F} \!\,$$ is uniformly bounded in every compact set, by Montel's Theorem, it follows that $$            \mathcal{F}\!\,$$ is normal