Friday January 10

Recall that $${\mathbb{C}}$$ is a field, where \begin{align*}z = x + iy \implies \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu = x - iy\end{align*} and if $$z\neq 0$$ then \begin{align*}z^{-1}= \frac{\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu} {{\left\lvert {z} \right\rvert}^2}\end{align*}

Lemma (Triangle Inequality)
\begin{align*}{\left\lvert {z + w} \right\rvert} \leq {\left\lvert {z} \right\rvert} + {\left\lvert {w} \right\rvert}.\end{align*}
Proof
\begin{align*} ({\left\lvert {z} \right\rvert} + {\left\lvert {w} \right\rvert})^2 - {\left\lvert {z+w} \right\rvert}^2 = 2( {\left\lvert {z\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu} \right\rvert} - \Re z\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu ) \geq 0 .\end{align*}
Lemma (Reverse Triangle Inequality)
\begin{align*}{\left\lvert {{\left\lvert {z} \right\rvert} - {\left\lvert {w} \right\rvert}} \right\rvert} \leq {\left\lvert {z-w} \right\rvert}.\end{align*}
Proof
\begin{align*} {\left\lvert {z} \right\rvert} = {\left\lvert {z-w + w} \right\rvert} \leq {\left\lvert {z-w} \right\rvert} + {\left\lvert {w} \right\rvert} \implies {\left\lvert {w} \right\rvert} - {\left\lvert {z} \right\rvert} \leq {\left\lvert {z-w} \right\rvert} = {\left\lvert {w-z} \right\rvert} .\end{align*}
Fact
$$({\mathbb{C}}, {\left\lvert {{\,\cdot\,}} \right\rvert})$$ is a normed space.
Definition (Limits of Complex Sequences)
\begin{align*}\lim z_n = z \iff {\left\lvert {z_n - z} \right\rvert} \to 0 \in {\mathbb{R}}.\end{align*}
Definition (Complex Discs)
A disc is defined as $$D_r(z_0) \mathrel{\vcenter{:}}=\left\{{z\in{\mathbb{C}}{~\mathrel{\Big|}~}{\left\lvert {z-z_0} \right\rvert} < r}\right\}$$, and a subset is open iff it contains a disc. By convention, $$D_r$$ denotes a disc about $$z_0 = 0$$.
Definition (Convergence in $${\mathbb{C}}$$)
$$\sum_k z_k$$ converges iff $$S_N \mathrel{\vcenter{:}}=\sum_{{\left\lvert {k} \right\rvert} < N} z_k$$ converges.

Note that $$z_n \to z$$ and $$z_n = x_n + iy_n$$, and \begin{align*}{\left\lvert {z_n - z} \right\rvert} = \sqrt{(x_n - x)^2 - (y_n - y)^2} < \varepsilon \implies {\left\lvert {x - x_n} \right\rvert}, {\left\lvert {y - y_n} \right\rvert} < \varepsilon.\end{align*}

Since $${\mathbb{R}}$$ is complete iff every Cauchy sequence converges iff every bounded monotone sequence has a limit.

Note: This is useful precisely when you don’t know the limiting term.

Note that $$\sum_k z_k$$ thus converges if $${\left\lvert {\sum_{k=m}^n z_k} \right\rvert} < \varepsilon$$ for $$m, n$$ large enough, so sums converges iff they have small tails.

Definition (Absolute Convergence)
$$S_N = \sum^N z_k$$ converges absolutely iff $$\tilde S \mathrel{\vcenter{:}}=\sum^N {\left\lvert {z_k} \right\rvert}$$ converges.

Note that the partial sums $$\sum^N {\left\lvert {z_k} \right\rvert}$$ are monotone, so $$\tilde S_N$$ converges iff the partial sums are bounded above.

Definition (Power Series)
A sum of the form $$\sum_{k=0}^\infty a_k z_k$$ is a power series.

Examples:

\begin{align*} \sum x^k &= \frac 1 {1-x} \\ \sum (-x^2)^k &= \frac 1 {1+x^2} .\end{align*}

Note that both of these have a radius of convergence equal to 1, since the first has a pole at $$x=1$$ and the second as a pole at $$x = i$$.

Monday January 13th

Recall that $$\sum z_k$$ converges iff $$s_n = \sum_{k=1}^n z_k$$ converges.

Lemma
Absolute convergence implies convergence.

The most interesting series: $$f(z) = \sum a_k z^k$$, i.e. power series.

Lemma (Divergence)
If $$\sum z_k$$ converges, then $$\lim z_k = 0$$.
Corollary
If $$\sum z_k$$ converges, $$\left\{{z_k}\right\}$$ is uniformly bounded by a constant $$C > 0$$, i.e. $${\left\lvert {z_k} \right\rvert} < C$$ for all $$k$$.

Proposition: If $$\sum a_k z_k$$ converges at some point $$z_0$$, then it converges for all $${\left\lvert {z} \right\rvert} < {\left\lvert {z} \right\rvert}_0$$.

Note that this inequality is necessarily strict. For example, $$\sum \frac{z^{n-1}}{n}$$ converges at $$z=-1$$ (alternating harmonic series) but not at $$z=1$$ (harmonic series).

Proof

Suppose $$\sum a_k z_1^k$$ converges. The terms are uniformly bounded, so $${\left\lvert {a_k z_1^k} \right\rvert} \leq C$$ for all $$k$$. Then we have \begin{align*}{\left\lvert {a_k} \right\rvert} \leq C/{\left\lvert {z_1} \right\rvert}^k\end{align*}, so if $${\left\lvert {z} \right\rvert} < {\left\lvert {z_1} \right\rvert}$$ we have \begin{align*}{\left\lvert {a_k z^k} \right\rvert} \leq {\left\lvert {z} \right\rvert}^k \frac{C}{{\left\lvert {z_1} \right\rvert}^k} = C ({\left\lvert {z} \right\rvert} / {\left\lvert {z_1} \right\rvert} )^k.\end{align*} So if $${\left\lvert {z} \right\rvert} < {\left\lvert {z_1} \right\rvert}$$, the parenthesized quantity is less than 1, and the original series is bounded by a geometric series. Letting $$r = {\left\lvert {z} \right\rvert} / {\left\lvert {z_1} \right\rvert}$$, we have

\begin{align*} \sum {\left\lvert {a_k z^k} \right\rvert} \leq \sum c r^k = \frac{c}{1-r} ,\end{align*}

and so we have absolute convergence.

Exercise (future problem set)
Show that $$\sum \frac 1 k z^{k-1}$$ converges for all $${\left\lvert {z} \right\rvert} = 1$$ except for $$z = 1$$. (Use summation by parts.)

The radius of convergence of a series is the real number $$R$$ such that $$f(z) = \sum a_k z^k$$ converges precisely for $${\left\lvert {z} \right\rvert} < R$$ and diverges for $${\left\lvert {z} \right\rvert} > R$$.

We denote a disc of radius $$R$$ centered at zero by $$D_R$$. If $$R=\infty$$, then $$f$$ is said to be entire.

Proposition
Suppose that $$\sum a_k z^k$$ converges for all $${\left\lvert {z} \right\rvert} < R$$. Then $$f(z) = \sum a_k z^k$$ is continuous on $$D_R$$, i.e. using the sequential definition of continuity, $$\lim_{z\to z_0} f(z) = f(z_0)$$ for all $$z_0 \in D_R$$.

Recall that $$S_n(z) \to S(z)$$ uniformly on $$\Omega$$ iff $$\forall \varepsilon > 0$$, there exists a $$M\in {\mathbb{N}}$$ such that \begin{align*}n> M \implies {\left\lvert {S_n(z) - S(z)} \right\rvert} < \varepsilon\end{align*} for all $$z\in \Omega$$

Note that arbitrary limits of continuous functions may not be continuous. Counterexample: $$f_n(x) = x^n$$ on $$[0, 1]$$; then $$f_n \to \delta(1)$$. This uniformly converges on $$[0, 1-\varepsilon]$$ for any $$\varepsilon > 0$$.

Exercise

Show that the uniform limit of continuous functions is continuous.

Hint: Use the triangle inequality.

Proof (of proposition)

Write $$f(z) = \sum_{k=0}^N a_k z^k + \sum_{N+1}^\infty a_k z^k \mathrel{\vcenter{:}}= S_N(z) + R_N(z)$$. Note that if $${\left\lvert {z} \right\rvert} < R$$, then there exists a $$T$$ such that $${\left\lvert {z} \right\rvert} < T < R$$ where $$f(z)$$ converges uniformly on $$D_T$$.

Check!

We need to show that $${\left\lvert {R_N(z)} \right\rvert}$$ is uniformly small for $${\left\lvert {z} \right\rvert} < s < T$$. Note that $$\sum a_k z^k$$ converges on $$D_T$$, so we can find a $$C$$ such that $${\left\lvert {a_k z^k} \right\rvert} \leq C$$ for all $$k$$. Then $${\left\lvert {a_k} \right\rvert} \leq C/T^k$$ for all $$k$$, and so

\begin{align*} {\left\lvert {\sum_{k=N+1}^\infty a_k z^k} \right\rvert} &\leq \sum_{k=N+1}^\infty {\left\lvert {a_k} \right\rvert} {\left\lvert {z} \right\rvert}^k \\ &\leq \sum_{k=N+1}^\infty (c/T^k) s^k \\ &= c\sum {\left\lvert {s/T} \right\rvert}^k \\ &= c \frac{r^{N+!}}{1-r} &= C \varepsilon_n \to 0 ,\end{align*}

which follows because $$0 < r = s/T < 1$$.

So $$S_N(z) \to f(z)$$ uniformly on $${\left\lvert {z} \right\rvert} < s$$ and $$S_N(z)$$ are all continuous, so $$f(z)$$ is continuous.

There are two ways to compute the radius of convergence:

• Root test: $$\lim_k {\left\lvert {a_k} \right\rvert}^{1/k} = L \implies R = \frac 1 L$$.
• Ratio test: $$\lim_k {\left\lvert {a_{k+1} / a_k} \right\rvert} = L \implies R = \frac 1 L$$.

As long as these series converge, we can compute derivatives and integrals term-by-term, and they have the same radius of convergence.

Wednesday January 15th

See references: Taylor’s Complex Analysis, Stein, Barry Simon (5 volume set), Hormander (technically a PDEs book, but mostly analysis)

Good Paper: Hormander 1955

We’ll mostly be working from Simon Vol. 2A, most problems from from Stein’s Complex.

Topology and Algebra of $${\mathbb{C}}$$

To do analysis, we’ll need the following notions:

1. Continuity of a complex-valued function $$f: \Omega \to \Omega$$

2. Complex-differentiability: For $$\Omega \subset {\mathbb{C}}$$ open and $$z_0 \in \Omega$$, there exists $$\varepsilon > 0$$ such that $$D_\varepsilon = \left\{{z {~\mathrel{\Big|}~}{\left\lvert {z - z_0} \right\rvert} < \varepsilon}\right\} \subset \Omega$$, and $$f$$ is holomorphic (complex-differentiable) at $$z_0$$ iff \begin{align*}\lim_{h\to 0} \frac 1 h (f(z_0 + h) - f(z_0))\end{align*} exists; if so we denote it by $$f'(z_0)$$.

Example
$$f(z) = z$$ is holomorphic, since $$f(z+ h) - f(z) = z+h-z = h$$, so $$f'(z_0) = \frac h h = 1$$ for all $$z_0$$.
Example

Given $$f(z) = \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu$$, we have $$f(z+h)-f(z) = \mkern 1.5mu\overline{\mkern-1.5muh\mkern-1.5mu}\mkern 1.5mu$$, so the ratio is $$\frac{\mkern 1.5mu\overline{\mkern-1.5muh\mkern-1.5mu}\mkern 1.5mu}{h}$$ and the limit doesn’t exist.

Note that if $$h\in {\mathbb{R}}$$, then $$\mkern 1.5mu\overline{\mkern-1.5muh\mkern-1.5mu}\mkern 1.5mu = h$$ and the ratio is identically 1, while if $$h$$ is purely imaginary, then $$\mkern 1.5mu\overline{\mkern-1.5muh\mkern-1.5mu}\mkern 1.5mu = -h$$ and the limit is identically $$-1$$.

We say $$f$$ is holomorphic on an open set $$\Omega$$ iff it is holomorphic at every point, and is holomorphic on a closed set $$C$$ iff there exists an open $$\Omega \supset C$$ such that $$f$$ is holomorphic on $$\Omega$$.

Fact
If $$f$$ is holomorphic, writing $$h = h_1 + ih_2$$, then the following two limits exist and are equal:

\begin{align*} \lim_{h_1 \to 0} \frac{f(x_0 + iy_0 + h_1) - f(x_0 + iy_0)}{h_1} = {\frac{\partial f}{\partial x}\,}(x_0, y_0) \\ \lim_{h_2 \to 0} \frac{f(x_0 + iy_0 + ih_2) - f(x_0 + iy_0)}{ih_2} = \frac 1 i {\frac{\partial f}{\partial y}\,}(x_0, y_0) \\ \implies {\frac{\partial f}{\partial x}\,} = \frac 1 i {\frac{\partial f}{\partial y}\,} .\end{align*}

So if we write $$f(z) = u(x, y) + i v(x, y)$$, we have

\begin{align*} {\frac{\partial u}{\partial x}\,} + i {\frac{\partial v}{\partial x}\,} \mathrel{\Big|}_{(x_0, y_0)} = \frac 1 i \qty{ {\frac{\partial u}{\partial y}\,} + i {\frac{\partial v}{\partial y}\,} } \mathrel{\Big|}_{(x_0, y_0)} ,\end{align*}

and equating real and imaginary parts yields the Cauchy-Riemann equations:

\begin{align*} {\frac{\partial u}{\partial x}\,} + i {\frac{\partial v}{\partial x}\,} = -i {\frac{\partial u}{\partial y}\,} + {\frac{\partial v}{\partial y}\,} \\ \iff {\frac{\partial u}{\partial x}\,} = {\frac{\partial v}{\partial y}\,} \quad\text{and}\quad {\frac{\partial u}{\partial y}\,} = - {\frac{\partial v}{\partial x}\,} .\end{align*}

The usual rules of derivatives apply:

1. $$(\sum f)' = \sum f'$$
Proof
Direct.
1. $$(\prod f)' =$$ product rule
Proof
Consider $$(f(z+h)g(z+h) - f(z)g(z))/h$$ and use continuity of $$g$$ at $$z$$.
1. Quotient rule
Proof
Nice trick, write \begin{align*}q = \frac f g\end{align*} so $$qg = f$$, then $$f' = q'g + qg'$$ and $$q' = \frac {f'} g - \frac{fg'}{g^2}$$.
1. Chain rule
Proof

Use the fact that if $$f'(g(z)) = a$$, then \begin{align*}f(z+h) - f(z) = a h + r(z, h),\quad {\left\lvert {r(z, h)} \right\rvert} = o({\left\lvert {h} \right\rvert}) \to 0.\end{align*}

Write $$b = g'(z)$$, then \begin{align*}f(g(z + h)) = f(g(z) + b h + r_1 ) = f(g(z)) + f'(g(z))bh + r_2\end{align*} by considering error terms, and so \begin{align*}\frac 1 h (f(g(z+h)) - f(g(z))) \to f'(g(z)) g'(z)\end{align*}.

Friday January 17th

Antiholomorphic Derivative

Reference: See Lang’s Complex Analysis, there are plenty of solution manuals. Note: look for 13 statements equivalent to holomorphic: Springer GTM Lipman.

Let $$f; \Omega \to {\mathbb{C}}$$ be a complex-valued function. Recall that $$f$$ is complex differentiable iff the usual ratio/limit exists. Note that $$h = x+iy$$ and $$h\to 0 \iff x,y\to 0$$.

We can write \begin{align*}f'(z) = {\frac{\partial f}{\partial x}\,} = \frac 1 i {\frac{\partial f}{\partial y}\,}.\end{align*} This follows from Cauchy-Riemann since $$u_x = v_y$$ and $$u_y = -v_x$$.

We want to define $${\partial}, \mkern 1.5mu\overline{\mkern-1.5mu\partial\mkern-1.5mu}\mkern 1.5mu$$ operators. We have the identities

\begin{align*} x = \frac{z + \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu}{z} \quad y = \frac{z - \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu}{iz} .\end{align*}

We can then write

\begin{align*} dz &= dx + idy \\ d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu &= dx - i dy .\end{align*}

We define the dual operators by $${\left\langle {{\frac{\partial }{\partial z}\,}},~{dz} \right\rangle} = 1$$ and similarly $${\left\langle { {\frac{\partial }{\partial \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu}\,} },~{d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu} \right\rangle} = 1$$.

By the chain rule, we can write

\begin{align*} f_z &= {\frac{\partial f}{\partial x}\,} {\frac{\partial x}{\partial z}\,} + {\frac{\partial f}{\partial y}\,} {\frac{\partial y}{\partial z}\,} \\ &= \frac 1 2 {\frac{\partial f}{\partial x}\,} + {\frac{\partial f}{\partial y}\,} \frac{1}{2i} \\ &= \frac 1 2 \qty{{\frac{\partial }{\partial x}\,} + i {\frac{\partial }{\partial y}\,} }f ,\end{align*}

and similarly \begin{align*} f_{\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu} &= {\frac{\partial f}{\partial x}\,} {\frac{\partial x}{\partial \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu}\,} + {\frac{\partial f}{\partial y}\,} {\frac{\partial z}{\partial \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu}\,} \\ &= \frac 1 2 \qty{ {\frac{\partial }{\partial x}\,} - \frac{1}{2i} {\frac{\partial }{\partial y}\,} }f .\end{align*}

We thus find $${\partial}_x = {\partial}_z + {\partial}_{\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu}$$ and $${\partial}_y = i\qty{ {\partial}_z - {\partial}_{\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu} }$$, so define

\begin{align*} {\partial}f &\coloneqq{\frac{\partial f}{\partial z}\,} dz \\ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu f &\coloneqq{\frac{\partial f}{\partial \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu}\,} d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu \\ \implies df &= {\frac{\partial f}{\partial z}\,} dz + {\frac{\partial f}{\partial \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu}\,} d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu .\end{align*}

Definition (Holomorphic and Antiholomorphic Derivatives)
\begin{align*} {{\partial}}f &= \frac 1 2 \qty{{\frac{\partial }{\partial x}\,} + i {\frac{\partial }{\partial y}\,} }f\\ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5muf &= \qty{ {\frac{\partial }{\partial x}\,} - \frac{1}{2i} {\frac{\partial }{\partial y}\,} }f .\end{align*}
Proposition (Holomorphic Functions have vanishing antiholomorphic derivatives)
$$f$$ is holomorphic iff $$\mkern 1.5mu\overline{\mkern-1.5mu\partial\mkern-1.5mu}\mkern 1.5muf = 0$$.

This means that $$f$$ depends on $$z$$ alone and not $$\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu$$.

Proof
$$\mkern 1.5mu\overline{\mkern-1.5mu\partial\mkern-1.5mu}\mkern 1.5muf = 0$$ iff $$\frac 1 2 (f_x + if_y) = 0$$, so $$(u_x - v_y) + i (v_x + u_y) = 0$$.

Application to PDEs: we can write \begin{align*}u_{xx} = v_{xy} \quad u_{yy} = v_{yx}\end{align*} and so \begin{align*}u_{xx} + u_{yy} = 0 = v_{xx} + v_{yy}.\end{align*}

Thus $$\Delta f = 0$$, sp $$f$$ satisfies Laplace’s equation and is said to be harmonic.

Corollary (Holomorphic Functions Have Harmonic Components)
If $$f$$ is analytic, then $$u, v$$ are both harmonic functions.
Theorem (Chain Rule)

Let $$w = f(z)$$ and $$g(w) = g(f(z))$$. Then

\begin{align*} h_z &= g_w f_z + g_{\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu} \mkern 1.5mu\overline{\mkern-1.5muf\mkern-1.5mu}\mkern 1.5mu_z \\ h_{\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu} &= g_w f_{\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu} + g_{\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu} \mkern 1.5mu\overline{\mkern-1.5muf\mkern-1.5mu}\mkern 1.5mu_{\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu} .\end{align*}

If $$f, g$$ are holomorphic, $$f_{\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu} = g_{\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu} = 0$$, so $$h_{\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu} = 0$$ and $$h$$ is holomorphic and \begin{align*}h_z = g_w f_z.\end{align*}

Example

Given a power series $$f=\sum a_n (z- z_0)^n$$. Then

1. There exists a radius of convergence $$R$$ such that $$f$$ converges precisely on $$D_R(z_0)$$.
2. $$f$$ is continuous on $$D_R(z_0)^\circ$$.
3. By the root test, $$R = (\limsup {\left\lvert {a_n} \right\rvert}^{1/n})^{-1}= \liminf {\left\lvert {a_n/a_{n+1}} \right\rvert} = (\limsup {\left\lvert {a_{k+1}/a_k} \right\rvert})^{-1}$$.

Recall the ratio test: \begin{align*}\sum {\left\lvert {a_k} \right\rvert} <\infty \iff \limsup {\left\lvert {a_{k+1} / a_k} \right\rvert} < 1\end{align*}

Theorem (Holomorphic series can be differentiated term-by-term)
If $$f(z) = \sum_{n=0} a_n z^n$$ is holomorphic on $${\left\lvert {z} \right\rvert} < R$$ for $$R> 0$$ then \begin{align*}f'(z) = \sum_{n=1} a_n n z^{n-1}.\end{align*}
Proof

Given $${\left\lvert {z} \right\rvert} < R$$, fix $$r>0$$ such that $${\left\lvert {z} \right\rvert} < r < R$$. Suppose that $${\left\lvert {w-z} \right\rvert} < r - {\left\lvert {z} \right\rvert}$$, so $${\left\lvert {w} \right\rvert} < r$$.

We want to show \begin{align*} {\left\lvert {S} \right\rvert} = {\left\lvert {\frac{f(w) - f(z)}{w - z} - \sum_{n=1} a_n n z^{n-1}} \right\rvert} \to 0 \quad \text{as } w\to z .\end{align*}

Idea: write everything in terms of power series. Use the fact that $$a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + \cdots)$$, and so $${\left\lvert {(w^k-z^k)/(w-z)} \right\rvert} \leq k r^{k-1}$$.

\begin{align*} S &= \sum_{n=1} a_n \qty{ \frac{w^n - z^n}{w-z} - n z^{n-1} } \\ &= \sum a_n \qty{ w^{n-1} + w^{n-2}z + \cdots + z^{n-1} + nz^{n-1} } \\ &= \sum a_n \qty{ (w^{n-1} - z^{n-1}) + (w^{n-2} - z^{n-2})z + \cdots + (w-z) z^{n-2} } &= \sum a_n (w-z) \qty{ \cdots + z^{n-2} }\\ &\leq \sum_{n=2} {\left\lvert {a_n} \right\rvert} \frac 1 2 n(n-1) r^{n-2} {\left\lvert {z-w} \right\rvert} .\end{align*}

Exercise

Show $$\lim_n n^{\frac 1 n} = 1$$.

Also tricky: show $$\lim \sin(n)$$ doesn’t exist, and $$\sin(n)$$ is dense in $$[-1, 1]$$.

Proof
Consider $$\limsup {\left\lvert {a_n n} \right\rvert}^{\frac 1 n}$$.

Note that an analytic function is holomorphic in its domain of convergence, so analytic implies holomorphic. The converse requires Cauchy’s integral formula.

Next time: trying to prove holomorphic functions are analytic.

Wednesday January 22nd

Parameterized Curves

Note: multiple complex variables, see Hormander or Steven Krantz

Recall from last time that if \begin{align*}f(z) = \sum_{n=0}^\infty a_n z^n\end{align*} with $$z_0 \neq 0$$ has radius of convergence \begin{align*}R = (\limsup {\left\lvert {a_n} \right\rvert}^{1/n})^{-1}> 0\end{align*} then $$f'$$ exists and is obtained by differentiating term-by-term.

We know that $$f$$ analytic $$\implies$$ $$f$$ holomorphic (and smooth), and we want to show the converse. For this, we need integration.

Definition (Parameterized Curves)
A parameterized curve is a function $$z(t)$$ which maps a closed interval $$[a, b] \subset {\mathbb{R}}$$ to $${\mathbb{C}}$$.
Definition (Smooth Curves)
The curve is said to be smooth iff $$z'$$ exists and is continuous on $$[a,b]$$, and $$z'(t) \neq 0$$ for any $$t$$. At the boundary $$\left\{{a, b}\right\}$$, we define the derivative by taking one-sided limits.
Definition (Piecewise Smooth Curves)
A curve is said to be piecewise smooth iff $$z(t)$$ is continuous on $$[a, b]$$ and there are $$a < a_1 < \cdots < a_n = b$$ with $$z$$ smooth on each $$[a_k, a_{k+1}]$$.

Note that such a curve may fail to have tangent lines at $$a_i$$.

Definition (Equivalent Parameterizations)
Two parameterizations $$z: [a,b] \to {\mathbb{C}}, \tilde z: [c, d] \to {\mathbb{C}}$$ are equivalent iff there exists a $$C^1$$ bijection $$s: [c, d] \to [a, b]$$ where $$s \mapsto t(s)$$ such that $$s'>0$$ and $$\tilde z(s) = z(s(t))$$.

Note that $$s' > 0$$ preserves orientation and $$s'<0$$ reverses orientation.

Definition (Orientations of Curves)
A curve in reverse orientation is defined by \begin{align*} \gamma: [a, b] \to {\mathbb{C}}\implies \gamma^-: [a,b] &\to {\mathbb{C}}\\ t &\mapsto \gamma(a+b-t) .\end{align*}
Definition (Closed Curves)
A curve is closed iff $$z(a) = z(b)$$, and is simple iff $$z(t) \neq z_{t_1}$$ for $$t\neq t_1$$.
Definition (Positively Oriented Curves)
For $$C_r(z_0) \mathrel{\vcenter{:}}=\left\{{z{~\mathrel{\Big|}~}{\left\lvert {z-z_0} \right\rvert} = r}\right\}$$, the positive orientation is given by $$z(t) = z_0 + re^{2\pi i t}$$ for $$t\in [0, 1]$$.

Definition of the Integral

Definition (The Complex Integral)

The integral of $$f$$ over $$\gamma$$ is defined as

\begin{align*} \int_\gamma f ~dz = \int_a^b f(z(t)) z'(t)~dt .\end{align*}

Note: this doesn’t depend on parameterization, since if $$t = t(s)$$, then a change of variables yields \begin{align*} \int_\gamma f ~dz - \int_c^d f(z(t(s)))~z'(t(s))~t'(s) ~ds = \int_c^d f(\tilde z(s)) ~\tilde z'(s) ~ds .\end{align*}

Definition (Length of a Curve)
The length of $$\gamma$$ is defined as $${\left\lvert {\gamma } \right\rvert}= \int {\left\lvert {z'(t)} \right\rvert} ~dt$$.
Proposition
1. We can extend this definition to piecewise smooth curves by \begin{align*} \int_\gamma f~dz = \sum \int_{a_k}^{a_{k+1}} f ~dz \end{align*}

2. This integral is linear and $$\int_\gamma f = -\int_{\gamma^-} f$$.

3. We have an inequality

\begin{align*} {\left\lvert {\int_\gamma f} \right\rvert} \leq \max_{a\leq t \leq b} {\left\lvert {f(z(t))} \right\rvert} {\left\lvert {\gamma} \right\rvert} .\end{align*}

Definition (Primitive of a Function)
A function $$F$$ is a primitive for $$f$$ on $$\Omega$$ iff $$F$$ is holomorphic on $$\Omega$$ and $$F'(z) = f(z)$$ on $$\Omega$$.

Recall that in $${\mathbb{R}}$$, we have \begin{align*}F(x) =\int_a^x f(t)~dt\end{align*} as an antiderivative with $$F'(x) = f(x)$$, and $$\int f = F(b) - F(a)$$.

Theorem (Evaluating Integrals with Primitives)
If $$f$$ is continuous, has a primitive $$F$$ in $$\Omega$$, and $$\gamma$$ is a curve beginning at $$w_0$$ and ending at $$w_1$$, then $$\int_\gamma f = F(w_1) - F(w_0)$$.
Proof

Use definitions, write $$z(t)$$ where $$z(a) = w_1, z(b) = w_2$$. Then

\begin{align*} \int_\gamma f &= \int_a^b f(z(t)) z'(t) ~ dt \\ &= \int_a^b F'(z(t)) z'(t) ~dt \\ &= \int_a^b F_t ~dt \\ &= F(z(b)) - F(z(a)) \quad\text{by FTC}\\ &= F(w_1) - F(w_2) .\end{align*}

Note that if $$\gamma$$ is piecewise smooth, the sum of the integrals telescopes to yield the same conclusion.

Corollary (Functions with Primitives Integrate to Zero Along Loops)
If $$f$$ is continuous and $$\gamma$$ is a closed curve in $$\Omega$$, and $$f$$ has a primitive in $$\Omega$$, then \begin{align*}\oint f = 0.\end{align*}

Friday January 24th

Corollary
If $$\gamma$$ is a closed curve on $$\Omega$$ an open set and $$f$$ is continuous with a primitive in $$\Omega$$ (i.e. an $$F$$ holomorphic in $$\Omega$$ with $$F'=f$$) then $$\int_\gamma f ~dz = 0$$.
Proof (easy)
\begin{align*} \int_\gamma f ~dz = \int_\gamma F' = F'(z) z(t) ~dt = F(z(b)) - F(z(a)) = 0 .\end{align*}
Corollary
If $$f$$ is holomorphic with $$f'=0$$ on $$\Omega$$, then $$f$$ is constant.
Proof (easy)

Pick $$w_0 \in \Omega$$; we want to fix $$w_0 \in \Omega$$ and show $$f(w) = f(w_0)$$ for all $$w\in \Omega$$.

Take any path $$\gamma: w_0 \to w$$, then

\begin{align*} 0 = \int_\gamma f' = f(w) - f(w_0) .\end{align*}

Integral and Fourier Transform of $$e^{-x^2}$$

Example

Let $$f(z) = e^{-z^2}$$, this is holomorphic. Write \begin{align*} f(z) = \sum \frac{(-1)^nz^{2n}}{n!} ,\end{align*} so \begin{align*} \int f = \sum \frac{ (-1)^n z^{2n+1} }{ n! (2n+1) } .\end{align*}

Since $$f$$ is entire, $$\int f$$ is entire, and $$(\int f)' = f$$ so this function has a primitive. Thus $$\int_\gamma f(z) = 0$$ for any closed curve. So take $$\gamma$$ a rectangle with vertices $$\pm a , \pm a + ib$$.

So

\begin{align*} \int_\gamma f = \int_{-a}^a e^{-x^2} ~dx + \int e^{-(a+iy)^2} i ~dy - \int_{-a}^a e^{-(x+ib)^2} ~dx - \int_0^b e^{-(a+iy)^2} i dy = 0 .\end{align*}

We can do some estimates,

\begin{align*} e^{-(a+iy)^2} &= e^{-(a^2 + 2iay - y^2)} \\ &= e^{-a^2 + y^2} e^{2iay} \\ &\leq e^{-a^2 + y^2} \\ &\leq e^{-a^2 + b^2}, \\ \\ {\left\lvert {\int_0^b e^{-(a+ib)^2} i ~dy} \right\rvert} &\leq e^{-a^2 + b^2} \cdot b \\ \\ \int_{-a}^a e^{-(x^2 + 2ib x)-b^2} &= e^{b^2} \int_{-a}^a e^{-x^2} ( \cos(2bx) - i \sin(2bx) ) \\ \\ &\stackrel{\mathclap{\scriptscriptstyle{odd fn}}}{=} e^{b^2} \int_{-a}^a e^{-x^2} \cos(2bx) ~dx .\end{align*}

Now take $$a\to \infty$$ to obtain

\begin{align*} \int_{\mathbb{R}}e^{-x^2} ~dx = e^{b^2} \int_{\mathbb{R}}e^{-x^2} \cos(2bx) ~dx .\end{align*}

We can compute

\begin{align*} \int_{\mathbb{R}}e^{-x^2} = \left[ \qty{\int_{\mathbb{R}}e^{-x^2}}^2 \right]^{1/2} = \qty{ \int_0^{2\pi} \int_0^\infty e^{r^2} r~dr~d\theta} = \sqrt{\pi} .\end{align*}

and then conclude

\begin{align*} \int_{\mathbb{R}}e^{-x^2} \cos(2bx) = \sqrt{\pi} e^{-b^2} .\end{align*}

Make a change of variables $$2b = 2\pi \xi$$, so $$b = \pi \xi$$, then

\begin{align*} \int_{\mathbb{R}}e^{-x^2} \cos(2\pi \xi x) ~dx = \sqrt{\pi} e^{-\pi^2 \xi^2} .\end{align*}

Thus $${\mathcal{F}}(e^{-x^2}) = \sqrt{\pi} e^{-\pi^2 \xi^2}$$, allowing computation of the Fourier transform. Note that this can be used to prove the Fourier inversion formula.

Exercise
Show that this is an approximate identity and prove the Fourier inversion formula.
Exercise
Show $${\mathcal{F}}(e^{-ax^2}) = \sqrt{\pi/a} e^{-\pi^2/a \cdot \xi^2}$$, and thus taking $$a = \pi$$ makes $$e^{\pi x^2}$$ is an eigenfunction of $${\mathcal{F}}$$ with eigenvalue $$1$$.
Theorem (Holomorphic Integrals Vanish)
If $$f$$ has a primitive on $$\Omega$$ then $$F(z)$$ is holomorphic and $$\int_\gamma f = 0$$. If $$f$$ is holomorphic, then \begin{align*}\int_\gamma f = 0.\end{align*}
Theorem (Green’s)

Take $$\Omega \in {\mathbb{R}}^2$$ bounded with $${{\partial}}\Omega$$ piecewise smooth. If $$f, g\in C^1{\mkern 1.5mu\overline{\mkern-1.5mu\Omega\mkern-1.5mu}\mkern 1.5mu}$$, then

\begin{align*} \int_{{{\partial}}\Omega} f ~dx + g ~dy = \iint_{\Omega} \qty{g_x - f_y} ~dA .\end{align*}

Proof
Omitted.
Proof (that holomorphic integrals vanish)

Write $$\gamma = {{\partial}}\Gamma$$, and noting that $$f_z = f_x = \frac 1 i f_y$$ implies that $${\frac{\partial f}{\partial \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu}\,}$$, so

\begin{align*} \int_\gamma f ~dz &= \int_\gamma f(z) ~(dx + i dy) \\ &= \int f(z) ~dx + i f(z) ~dy \\ &= \iint_\Gamma \qty{if_x - f_y} ~dA \\ &= i \iint_\Gamma \qty{f_x - \frac 1 i f_y} ~dA \\ &= i \iint 0 ~dA \\ &= 0 .\end{align*}

Next up, we’ll prove that this integral over any triangle is zero by a limiting process.

Monday January 27th

Open question: does a PDE involving analytic functions always have solutions? Or does this hold with analytic replaced by smooth?

Green’s Theorem

Fix a connected domain $$\Omega$$ which is bounded with a piecewise $$C^1$$ boundary.

Theorem (Green’s)

Given $$f, g \in C^1 \mkern 1.5mu\overline{\mkern-1.5mu\Omega\mkern-1.5mu}\mkern 1.5mu$$, we can take a vector field $$F = \left\langle{f, g}\right\rangle$$ and have

\begin{align*} \int_{{{\partial}}\Omega} f~dx + g~dy &= \iint_{\Omega} \qty{ {\frac{\partial g}{\partial x}\,} - {\frac{\partial f}{\partial y}\,} } ~dA \\ \int_{{{\partial}}\Omega} - f~dx + g~dy &= \iint_{\Omega} \qty{ {\frac{\partial g}{\partial x}\,} + {\frac{\partial f}{\partial y}\,} } ~dA \\ \int_{{{\partial}}\Omega} f~dy - g~dy &= \iint_{\Omega} \qty{ {\frac{\partial f}{\partial x}\,} + {\frac{\partial g}{\partial y}\,} } ~dA \\ \int_{{{\partial}}\Omega} F \cdot \mathbf{n} ~ds &= \iint_{\Omega} \nabla \cdot F ~dA \\ \int_{{{\partial}}\Omega} \mathrm{curl}(F) ~ds &= \iint_{\Omega} \mathrm{div}(F) ~dA ,\end{align*}

where we take $$\mathbf{n}$$ to be orthogonal to $${{\partial}}\Omega$$. The quantities appearing on the RHS are referred to as the flux.

For $$f(z) \in C^1(\Omega)$$ holomorphic, we can then write

\begin{align*} \int_{{{\partial}}\Omega} f ~dz &= \int_{{{\partial}}\Omega} f ~(dx + idy) \\ &= \int_{{{\partial}}\Omega} f ~dx + if~dy \\ &= \iint_\Omega \qty{if_x - f_y} ~dA \\ &= 0 ,\end{align*}

which follows since $$f$$ holomorphic, we can write \begin{align*}f'(z) = f_x = \frac 1 i f_y,\end{align*} so $$i f_x = f_y$$ and thus $${\frac{\partial f}{\partial \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu}\,} = 0$$.

See Taylor’s Introduction to Complex Analysis

Theorem (Cauchy’s Integral Formula):

If $$f\in C^1(\mkern 1.5mu\overline{\mkern-1.5mu\Omega\mkern-1.5mu}\mkern 1.5mu)$$ and $$f$$ is holomorphic, then for any $$z\in \Omega$$

\begin{align*} f(z) = \frac{1}{2\pi i} \int_{{{\partial}}\Omega} \frac{d(\xi)}{\xi - z} ~d\xi .\end{align*}

Proof

Since $$z\in \Omega$$ an open set, we can find some $$r> 0$$ such that $$D_r(z) \subset \Omega$$. Then $$\frac{f(\xi)}{\xi - z}$$ is holomorphic on $$\Omega\setminus D_r(z)$$. Let $$C_r = {{\partial}}D_r(z)$$.

Claim: \begin{align*}\int_{{{\partial}}\Omega} \frac{f(\xi)}{\xi - z} ~d\xi = \int_{C_r} \frac{f(\xi)}{\xi - z} ~ d\xi.\end{align*}

If we can differentiate through the integral, we can obtain

\begin{align*} {\frac{\partial }{\partial z}\,} f(z) = \frac 1 {2\pi i} \int_{{{\partial}}\Omega} \frac{f(\xi)}{(\xi - z)^2} ~d\xi .\end{align*}

and thus inductively

\begin{align*} (D_z)^n f(z) = \frac{n!}{2\pi i} \int_{{{\partial}}\Omega} \frac{ f(\xi) ~d\xi }{ (\xi - z)^{n+1} } .\end{align*}

To prove rigorously, need to write

\begin{align*} \Delta_h f(z) = \frac 1 h \qty{ f(z+h) - f(z) } \\ = \frac 1 {2\pi i h} \int_{{{\partial}}\Omega} f(\xi) \qty{ \frac{1}{\xi - (z+h)} - \frac{1}{\xi - z} } ~d\xi = \frac 1 {2\pi i h} \int_{{{\partial}}\Omega} f(\xi) \qty{ \frac{1}{ (\xi - z- h)(\xi - z) } } ~d\xi ,\end{align*}

and show the integrand converges uniformly, where \begin{align*} \frac{1}{(\xi - z - h)(\xi - z)} \overset{u}\to \frac{1}{(\xi - z)^2} .\end{align*}

Continuing inductively yields the integral formula.

Proof (of claim used in main proof)

Use the parameterization of $$C_r$$ given by $$\xi = z + re^{i\theta}$$. Then

\begin{align*} \frac{1}{2\pi i} \int_{C_r} \frac{f(\xi)}{\xi - z} ~d\xi &= \frac{1}{2\pi i} \int_0^{2\pi} \frac{f(z + re^{i\theta})}{re^{i\theta}} ~ird\theta \\ &= \frac{1}{2\pi} \int_0^{2\pi} f(z + re^{i\theta}) ~d\theta \\ &\overset{r \to 0}\to \frac{1}{2\pi} \int_{{{\partial}}\Omega} \frac{f(\xi)} {\xi - z} .\end{align*}

where we use the fact that \begin{align*}f(z + re^{i\theta}) = f(z) + f'(z)re^{i\theta} + o(r) \overset{r\to 0}\to f(z)\end{align*}

Letting \begin{align*}F(\xi) = \frac{ f(\xi)}{\xi - z},\end{align*} this is holomorphic on $$\Omega\setminus D_r(z)$$. Let $$\Omega_r = {{\partial}}\Omega \cup(-C_r)$$. Take the following path integral:

Then \begin{align*} 0 = \int_{{{\partial}}\Omega_r} F(\xi) ~d\xi = \int_{{{\partial}}\Omega} F(\xi) ~d\xi - \int_{C_r} F(\xi) ~d\xi ,\end{align*}

which forces these integrals to be equal.

Corollary ($$C^1$$ implies smooth)
If $$f$$ is holomorphic, then $$f\in C^1(\Omega)$$ implies that $$f \in C^\infty(\Omega)$$.
Theorem (Holomorphic implies analytic)
If $$f$$ is holomorphic in $$\Omega$$, then $$f$$ is equal to its Taylor series (i.e. $$f(z_0$$ is analytic.)
Proof

Fix $$z_0 \in \Omega$$ and let $$r = {\left\lvert {z- z_0} \right\rvert}$$.

\begin{align*} \frac{1}{\xi - z} &= \frac{1}{ \xi - z_0 - (z-z_0) } \\ &= \frac{1}{\xi - z_0} \frac{1}{1 - \qty{ \frac{z-z_0}{\xi - z_0} } } \\ &= \frac{1}{\xi - z_0} \sum_n \qty{ \frac{z - z_0}{\xi - z_0} }^n \quad\text{for } {\left\lvert {z - z_0} \right\rvert} < {\left\lvert {\xi - z_0} \right\rvert} .\end{align*}

Note that $$\sum z^n$$ converges uniformly for any $${\left\lvert {z} \right\rvert} < \delta < 1$$.

Thus

\begin{align*} f(z) &= \frac{1}{2\pi i} \int_{\xi \in {{\partial}}\Omega} f(\xi) \sum \frac{(z-z_0)^n}{(\xi - z_0)^{n+1}} ~d\xi \\ &= \sum \qty{ \frac {1}{2\pi i} \int \frac{f(\xi) }{(\xi - z_0)^{n+1}} ~d\xi} (z-z_0)^n \\ &= \sum \frac{f^{(n)} (z_0)}{n!} (z-z_0)^n .\end{align*}

Corollary
$$f$$ is holomorphic iff $$f$$ is analytic.

Counterexample to keep in mind:

\begin{align*} f(x) = \begin{cases} x^2 & x > 0 \\ 0 & x\leq 0 \end{cases} .\end{align*}

In the case of $${\mathbb{R}}$$, smooth and analytic are very different categories of functions.

Wednesday January 29th

Cauchy’s Integral Formula

Theorem (Cauchy’s Integral Formula)

Let $$f: \Omega \to {\mathbb{C}}$$ be holomorphic, so $$f\in C^1(\mkern 1.5mu\overline{\mkern-1.5mu\Omega\mkern-1.5mu}\mkern 1.5mu)$$. Then for any $$z\in \Omega$$,

\begin{align*} f(z) = \frac{1}{2\pi i} \int_{{{\partial}}\Omega} \frac{f(\xi)}{\xi - z} ~d\xi .\end{align*}

In general, \begin{align*} f^{(n)}(z) = \frac{n!}{2\pi i} \int_{{{\partial}}\Omega} \frac{f(\xi)}{\qty{\xi - z}^{n+1}} ~d\xi .\end{align*}

This implies that $$f$$ is analytic, i.e.  \begin{align*} f(z) = \sum a_n (z-z_0)^n \quad\text{where}\quad a_n = \frac{f^{(n)}(z_0) }{n!} .\end{align*} Thus $$f$$ is holomorphic iff $$f$$ is analytic,

and

\begin{align*} \int_{{{\partial}}\Omega} f = 0 \implies \int_{{{\partial}}\Omega_\gamma} \frac{f(\xi)}{\xi - z}~d\xi = 0 .\end{align*}

where $$\Omega_r = \Omega\setminus D_r(z)$$, and $${{\partial}}\Omega_r = {{\partial}}\Omega \cup(-{{\partial}}D_r)$$.

We can thus shrink integrals:

\begin{align*} \int_{{{\partial}}\Omega} \frac{f(\xi)} {\xi - z} ~d\xi = \int_{C_r} \frac{f(\xi)} {\xi - z} ~d\xi .\end{align*}

Proposition (Homotopy Invariance)
Let $$f\in C^1(\Omega)$$ be holomorphic on $$\Omega$$. Let $$\gamma_s(t)$$ be a family of smooth curves in $$\Omega$$; then $$\int_{\gamma_s} f$$ is independent of $$s$$.
Proof

Write \begin{align*}\gamma_s(t) = \gamma(s, t): [a, b] \times[0, 1] \to \Omega.\end{align*}

We have $$\gamma_s(0) = \gamma_s(1)$$ so $${\frac{\partial \gamma}{\partial s}\,}(s, 0) = {\frac{\partial \gamma}{\partial s}\,}(s, 1)$$. Then

\begin{align*} {\frac{\partial \gamma}{\partial s}\,} &= \int_0^1 \qty{ f'(r(s, t)) {\frac{\partial r}{\partial s}\,} {\frac{\partial r}{\partial t}\,} + f(r(s, t)) \frac{\partial^2 \gamma}{\partial s \partial t} } ~dt \\ &= \int_0^1 \qty{ f'(r(s, t)) {\frac{\partial r}{\partial s}\,} {\frac{\partial r}{\partial t}\,} + f(r(s, t)) \frac{\partial^2 \gamma}{\partial \mathbf{t} \partial \mathbf{s}} } ~dt \\ &= \int_0^1 {\frac{\partial }{\partial t}\,} \qty{ f(\gamma(s, t)) \gamma_s } \\ &= f(\gamma(s, 1))\gamma_s(s, 1) - f(\gamma(s, 0)) \gamma_s(s, 0) \\ &= 0 .\end{align*}

where we can just take the paths $$\gamma(s, t) = z_0 \in \Omega$$ for all $$s, t$$.

Proposition (Pointwise Limit of Locally Uniform is Locally Uniform)
Let $$\Omega \subset {\mathbb{C}}$$ be open and $$f_v: \Omega \to {\mathbb{C}}$$. Suppose that each $$f_v$$ is holomorphic, $$f_v \to f$$ pointwise, and locally uniform, i.e. $$f_v \to f$$ uniformly on every compact $$K \subset \Omega$$. Then $$f$$ is holomorphic in $$\Omega$$ and $$f$$ is locally uniform.
Proof

Given a compact set $$K \subset \Omega$$, pick an $$O$$ with smooth boundary such that $$K \subset O \subset \mkern 1.5mu\overline{\mkern-1.5muO\mkern-1.5mu}\mkern 1.5mu \subset \Omega$$. We have

\begin{align*} f_v(z) &= \frac{1}{2\pi i} \int_{{{\partial}}O} \frac{f_v(\xi)}{\xi - z}~d\xi \\ f_v^{(n)}(z) &= \frac{n!}{2\pi i} \int_{{{\partial}}O} \frac{f_v(\xi)}{\qty{ \xi - z}^{n+1} }~d\xi \\ .\end{align*}

Then on $${{\partial}}O$$, we have uniform convergence

\begin{align*} \frac{f_v(\xi)}{\qty{\xi - z}^{n+1}} \overset{u}\to \frac{f(\xi)}{\qty{\xi - z}^{n+1}} .\end{align*}

By moving the limits inside, we obtain

\begin{align*} f(z) &= \frac{1}{2\pi i} \int_{{{\partial}}O} \frac{f(\xi)}{\xi - z}~d\xi \\ f^{(n)}(z) &= \frac{n!}{2\pi i} \int_{{{\partial}}O} \frac{f(\xi)}{\qty{ \xi - z}^{n+1} }~d\xi \\ .\end{align*}

Theorem (Cauchy’s Inequality)

Given $$z_0\in \Omega$$, pick the largest disc $$D_R(z_0) \subset \Omega$$ and let $$C_R = {{\partial}}D_R$$. Using the integral formula, defining $${\left\lVert {f} \right\rVert}_{C_R} = \max_{{\left\lvert {z-z_0} \right\rvert} = R} {\left\lvert {f(z)} \right\rvert}$$

\begin{align*} {\left\lvert { f^{(n)}(z_0) } \right\rvert} \leq \frac{n!}{2\pi} \int_0^{2\pi} \frac{{\left\lVert {f} \right\rVert}_{C_R}}{R^{n+1}} R~d\theta = \frac{n! {\left\lVert {f} \right\rVert}_{C_R}}{R^{n}} .\end{align*}

Corollary (Liouville’s Theorem)
If $$f$$ is entire and bounded, then $$f$$ is constant.
Proof
For all $$z_0 \in {\mathbb{C}}$$, there exists an $$M$$ such that $${\left\lvert {f(z)} \right\rvert} \leq M$$. Then $${\left\lvert {f'(z_0)} \right\rvert} \leq \frac{M}{R}$$ for any $$R> 0$$. Taking $$R\to \infty$$ yields $$f'(z_0) = 0$$, so $$f$$ is constant.
Corollary (Weak Fundamental Theorem of Algebra)
Every non-constant polynomial $$p(z) = a_0 + a_1z + \cdots a_n z^n$$ has a root in $${\mathbb{C}}$$.

Remark: A general proof technique is when proving something for $$f(z)$$, consider $$\frac{1}{f(z)}$$ and $$f(\frac 1 z)$$.

Proof

Suppose $$p$$ is nonconstant and does not have a root, $$\frac 1 p$$ is entire. Assume that $$a_n \neq 0$$, then

\begin{align*} \frac{p(z)}{z^n} = a_n \qty{ \frac{a_{n-1}}{z} + \cdots + \frac{a_0}{z^n} } \coloneqq a_n + y \end{align*}

We can note that $$\lim_{z\to \infty} \frac{a_{n-k}}{z^k} \to 0$$, so there exists an $$R>0$$ such that

\begin{align*} {\left\lvert { \frac{p(z)}{z^n} } \right\rvert} \geq \frac 1 2 {\left\lvert {a_n} \right\rvert} \quad \text{ for } {\left\lvert {z} \right\rvert} > R \\ \implies {\left\lvert {p(z)} \right\rvert} \geq \frac 1 2 {\left\lvert {a_n} \right\rvert} {\left\lvert {z} \right\rvert}^n \geq \frac 1 2 {\left\lvert {a_n} \right\rvert} R^n .\end{align*}

Since $$p(z)$$ is continuous and has no root in the disc $${\left\lvert {z} \right\rvert} \leq R$$, $${\left\lvert {p(z)} \right\rvert}$$ is bounded from below in this disc. Since $$p(z)$$ is continuous on a compact set, it attains a minimum, and so $${\left\lvert {p(z)} \right\rvert} \geq \min_{{\left\lvert {z} \right\rvert} \leq R} {\left\lvert {p(z)} \right\rvert} = c_2 \neq 0$$. Then $${\left\lvert {p(z)} \right\rvert} \geq A = \min(C_2, \frac 1 2 {\left\lvert {a_n} \right\rvert}R^n)$$, so $$\frac{1}{p}$$ is bounded. Then $$f$$ is constant, a contradiction.

Friday January 31st

Fundamental Theorem of Algebra

Recall that if $$f$$ is holomorphic, we have Cauchy’s integral formula.

Corollary (Weak Fundamental Theorem of Algebra)
If $$P(z)$$ is a polynomial in $${\mathbb{C}}$$ then $$P$$ has a root in $${\mathbb{C}}$$.
Proof
See previous notes.
Corollary (Fundamental Theorem of Algebra)
Every polynomial of degree $$n$$ has precisely $$n$$ roots in $${\mathbb{C}}$$.
Proof

By induction on the degree of $$P$$. From the first corollary, $$P$$ has a root $$w_1$$, so write $$z = z-w_1 + w_1$$. Then

\begin{align*} p(z) &= p(z - w_1 + w_1) \\ &= \sum_k^n a_k(z -w_1 + w_1)^k \\ &= \sum_k^n a_k \sum_{j}^k {k\choose j} w_1{k-j} (z-w_1)^j \\ &= \sum_k^n \sum_j^k a_k {k\choose j} w_1^{k-j} (z-w_1)^j \\ &= \sum_j^n \qty{\sum_{k\geq j} a_k {k\choose j}}(z-w_1)^j \\ &= b_0 + b_1(z-w_1) + \cdots + b_n(z-w_1)^n .\end{align*}

Since $$P(w_1) = 0$$, we must have $$b_0 = 0$$, and thus this equals

\begin{align*} b_1(z-w_1) + \cdots + b_n(z-w_1)^n &= (z-w_1) \qty{ b_1 + \cdots + b_n (z-w_1)^{n-1} } \\ &\coloneqq(z-w_1) \phi(z) ,\end{align*}

where $$\phi(z)$$ is degree $$n-1$$, which has $$n-1$$ roots by induction.

Definition (Characterizations of Limit Points)

For a sequence $$\left\{{z_n}\right\}$$, TFAE

1. $$z$$ is a limit point.
2. There exists a subsequence $$\left\{{z_{n_k}}\right\}$$ converging to $$z$$.
3. For every $$\varepsilon> 0$$, there are infinitely many $$z_i$$ in $$D_\varepsilon(z)$$.
Theorem (Only the zero function vanishes on a sequence in a domain (Stein 4.8))
Suppose $$f$$ is holomorphic on a bounded connected region $$\Omega$$ and $$f$$ vanishes on a sequence of distinct points with a limit point in $$\Omega$$. Then $$f$$ is identically zero.
Proof

WLOG by restricting to a subsequence, suppose that $$\left\{{w_k}\right\} \in \Omega$$ with $$f(w_i) = 0$$ for all $$i$$ and $$z_0$$ is a limit point of $$\left\{{w_i}\right\}$$. Let $$U = \left\{{ z\in \Omega {~\mathrel{\Big|}~}f(z) = 0 }\right\}$$. Then

1. $$U$$ is nonempty since $$f(w_k) = f(z_0) = 0$$.
2. Since holomorphic functions are continuous, if $$w_k \to z$$ then $$z\in U$$, so $$U$$ is closed.
3. (To prove) $$U$$ is open.

Since $$U$$ is closed and open, $$U = \Omega$$.

We will first show that $$f(z) \equiv 0$$ in a disk containing $$z_0$$. Choose a disc $$D$$ containing $$z_0$$ and contained in $$\Omega$$. Since $$f$$ is holomorphic on $$D$$, we can write \begin{align*}f(z) = \sum a_nn (z-z_0)^n.\end{align*} Since $$f(z_0) = 0$$, we have $$a_0 = 0$$.

Suppose $$f\not\equiv 0$$. Then there exists a smallest $$n\in {\mathbb{Z}}^+$$ such that $$a_n \neq 0$$, so $$f(z) = a_n(z-z_0)^n + \cdots$$. Since $$a_n \neq 0$$, we can factor this as $$a_n(z-z_0)^n \qty{ 1 + g(z-z_0) }$$ where \begin{align*}g(z-z_0) = \sum_{k=n+1}^\infty \frac{a_k}{a_n} (z-z_0)^{k-n}.\end{align*} Note that $$g$$ is holomorphic, and $$g(z_0 - z_0) = 0$$.

Choose some $$w_k$$ such that $$f(w_k) = 0$$ and $${\left\lvert {g(w_k - z_0)} \right\rvert} \leq \frac 1 2$$ by continuity of $$g$$. Then \begin{align*}{\left\lvert {1 + g(w_k - z_0)} \right\rvert} > 1 - \frac 1 2 = \frac 1 2.\end{align*}

So \begin{align*}{\left\lvert {f(w_k)} \right\rvert} = {\left\lvert { a_n(w_k - z_0)^n \qty{1 + g(w_k - z_0) } } \right\rvert} > {\left\lvert {a_n} \right\rvert} {\left\lvert {w_k - z_0} \right\rvert}^n \frac 1 2 > 0,\end{align*} a contradiction. So $$U$$ is open, closed, and nonempty, so $$U = \Omega$$.

Corollary
Suppose $$f, g$$ are holomorphic in a region $$\Omega$$ with $$f(z_k) = g(z_k)$$ where $$\left\{{z_k}\right\}$$ has a limit point. Then $$f(z) \equiv g(z)$$.
Theorem (Mean Value)

Let $$z_0$$ be a point in $$\Omega$$ and $$C_\gamma$$ the boundary of $$D_r(z_0)$$. Then

\begin{align*} f(z_0) &= \frac{1}{2\pi i} \int_{C_\gamma} f(z)/(z-z_0) dz \\ &= \frac {1}{2\pi i} \int_0^{2\pi} f(z_0 + re^{i\theta})/re^{i\theta} r i e^{i\theta} ~d\theta \quad\text{by } z = z_0 + re^{i\theta} \\ &= \frac{1}{2\pi} \int_0^{2\pi} f(z_0 + re^{i\theta})~d\theta \\ &= \frac{1}{2\pi r} \int_0^{2\pi} f(z_0 + re^{i\theta})~rd\theta \\ &= \frac{1}{{\left\lvert {C_\gamma} \right\rvert}} \int_0^{2\pi} f(z) ~ds ,\end{align*}

which is the average value of $$f$$ on the circle.

Note that there is another formula that averages over the disc (see book for derivation?)

\begin{align*} f(z_0) &= \frac{1}{D_s(z_0)} \int_{P_s} \int_{D_s} f(z) ~dA .\end{align*}

These imply the maximum modulus principle, since the average can not be the max or min unless $$f$$ is constant. Note that $${\left\lvert {f(z)} \right\rvert}$$ is continuous!

Next time: maximum modulus principle.

Monday February 3rd

Mean Value Theorem

Theorem (Mean Value for Holomorphic functions)
\begin{align*}f(z_0) = \frac{1}{\pi r^2} \iint_{D_r(z_0)} f(z) ~dA\end{align*}
Proof (of MVT?)

Let $$f: \Omega \to {\mathbb{C}}$$ be holomorphic where $$\Omega$$ is open and connected. Then by Cauchy’s integral formula, we have $$f(z_0) = \frac{1}{2\pi} \int_0^{2\pi} f(z_0 + re^{i\theta}) ~d\theta$$ for any $$z_0 \in \Omega$$.

We can consider $$D_r(z_0)$$, in which case we have for all $$0 < s < r$$,

\begin{align*} f(z_0) &= \frac{1}{2\pi} \int_0^{2\pi} f(z_0 + se^{i\theta}) ~d\theta \\ \implies s\cdot f(z_0) &= \frac{1}{2\pi} \int_0^{2\pi} s\cdot f(z_0 + se^{i\theta}) ~d\theta \\ \implies \cdot f(z_0) \int_0^r s ~ds &= \frac{1}{2\pi} \int_0^{2\pi} \int_0^r f(z_0 + se^{i\theta})\cdot s ~ds~d\theta \\ \implies \frac 1 2 r^2 f(z_0) &= \frac{1}{2\pi} \iint_{D_r(z_0)} f(z) ~dA\\ \implies f(z_0) &= \frac{1}{\pi r^2} \iint_{D_r(z_0)} f(z) ~dA \\ \implies f(z_0) &= \frac{1}{2\pi} \int_0^{2\pi} f(z_0 + re^{i\theta}) ~d\theta .\end{align*}

Proposition (Maximum in Interior Implies Constant)

Let $$f$$ be holomorphic on $$\Omega$$ be open and connected, and suppose that there is a $$z_0 \in \Omega$$ such that \begin{align*}{\left\lvert {f(z_0)} \right\rvert} = \sup_{z\in \Omega} {\left\lvert {f(z)} \right\rvert},\end{align*} i.e. $$z_0$$ is a maximal point of $$f$$. Then $$f$$ is constant on $$\Omega$$.

If $$\Omega$$ is additionally bounded, then $$f$$ is continuous on $$\mkern 1.5mu\overline{\mkern-1.5mu\Omega\mkern-1.5mu}\mkern 1.5mu$$, then \begin{align*}\sup_{z \in \mkern 1.5mu\overline{\mkern-1.5mu\Omega\mkern-1.5mu}\mkern 1.5mu} {\left\lvert {f(z)} \right\rvert} = \max_{z\in\mkern 1.5mu\overline{\mkern-1.5mu\Omega\mkern-1.5mu}\mkern 1.5mu} {\left\lvert {f(z)} \right\rvert}.\end{align*}

Proof

Since $${\left\lvert {f} \right\rvert}$$ is continuous and $$\mkern 1.5mu\overline{\mkern-1.5mu\Omega\mkern-1.5mu}\mkern 1.5mu$$ is compact, $${\left\lvert {f} \right\rvert}$$ attains a maximum at some point in $$\mkern 1.5mu\overline{\mkern-1.5mu\Omega\mkern-1.5mu}\mkern 1.5mu$$. We want to show that if $${\left\lvert {f(z_0)} \right\rvert} = \sup_{z\in \Omega} {\left\lvert {f(z)} \right\rvert}$$, then $$f$$ is constant.

Assume that there exists a $$z_0 \in \Omega$$ such that $$f(z) = f(z_0)$$. Let $$O = \left\{{ \xi\in \Omega {~\mathrel{\Big|}~}f(\xi) = f(z_0) }\right\}$$.

Claim
1. $$O$$ is not empty, since $$z_0 \in O$$.
2. $$O$$ is closed, since if $$\xi_n \to \xi$$ then $$f(\xi_n) = f(z_0)$$ implies $$f(\xi) = f(z_0)$$ since $$f$$ is continuous.
3. (Claim) $$O$$ is open.

Suppose $$\xi_0 \in O$$, then there exists a disc $$D_\rho(\xi_0) \subset \Omega$$ such that \begin{align*}f(\xi_0) = \frac{1}{\pi\rho^2} \int_{D_\rho(\xi_0)} f(z) dA.\end{align*} Then (claim) $${\left\lvert {f(\xi_0)} \right\rvert} \geq {\left\lvert {f(z)} \right\rvert}$$ for all $$z\in D_\rho(\xi_0)$$, which forces $$f(z) = f(\xi_0)$$ for all $$z\in D_\rho(\xi_0)$$.

Proof (of the claim):

Suppose that $$\sup_{a\in \Omega} {\left\lvert {f(z)} \right\rvert} = {\left\lvert {f(\xi_0)} \right\rvert}$$ and write $$f(\xi_0) = Be^{i\alpha}$$ for $$B>0$$ and $$\alpha \in {\mathbb{R}}$$. Then define $$g(z) = f(z) e^{-i\alpha}$$; then $$g(\xi_0) = B$$ is real, and thus

\begin{align*} 0 = g(\xi_0) - B = \frac{1}{\pi\rho^2} \iint_{D_\rho(\xi_0)} \Re( g(z) - B ) ~dA .\end{align*}

Note that $$\Re(g(z) - B) \leq 0$$ implies that $$\Re(g(z) - B) \equiv 0$$ on $$D_\rho(z_0)$$, so we can write $$g(z) = B + iI(z)$$ for some real-valued function $$I$$.

But then $${\left\lvert {g(z)} \right\rvert}^2 = B^2 + I(z)^2 = B^2$$ by the previous statement, and so $$I(z) = 0$$, forcing $$g(z) = B$$ and thus $$f(z) = Be^{i\alpha}$$. This shows that $$O$$ is open, and thus $$O = \Omega$$.

Biholomorphisms of the Open Disc

Proposition (Biholomorphisms of the Open Disc are Contractions (Stein 2.1))

Suppose $$f$$ is holomorphic on $$D_1(0)$$ and $${\left\lvert {f(z)} \right\rvert} \leq 1$$ for all $${\left\lvert {z} \right\rvert} < 1$$ with $$f(0) = 0$$. Then $${\left\lvert {f(z)} \right\rvert} \leq {\left\lvert {z} \right\rvert}$$ for all $${\left\lvert {z} \right\rvert} < 1$$.

Moreover, there is a point $$z_0\in D_1(0)$$ such that $${\left\lvert {f(z_0)} \right\rvert} = {\left\lvert {z_0} \right\rvert}$$ iff $$f(z) = c(z)$$ for some $$c \in S^1$$.

Proof

Define

\begin{align*} g(z) = \begin{cases} \frac{f(z)}{z} & z\neq 0 \\ f'(0) & z = 0 \end{cases} .\end{align*}

Then $$g$$ is holomorphic on $$D_1(0)$$ and $${\left\lvert {g(z)} \right\rvert} \leq \frac{1}{\rho}$$ for all $${\left\lvert {z} \right\rvert} < \rho < 1$$. Now apply the maximum principle: since this is true for all $$\rho < 1$$, consider the limit $$\rho\to 1^-$$.

Then $${\left\lvert {g(z)} \right\rvert} \leq 1$$, so $${\left\lvert {\frac{f(z)}{z}} \right\rvert} \leq 1$$ and $${\left\lvert {f(z)} \right\rvert} \leq {\left\lvert {z} \right\rvert}$$. If $${\left\lvert {f(z_0)} \right\rvert} = {\left\lvert {z_0} \right\rvert}$$ for any point, then $${\left\lvert {g(z_0)} \right\rvert} = 1$$ implies $$g(z_0) = c$$ and $$c\in S^1$$.

Thus $$f(z) = cz$$ for some $$c\in S^1$$.

Corollary (Characterization of Biholomorphisms of the Disc)

Recall that \begin{align*}\Phi_a(z) \mathrel{\vcenter{:}}=\frac{z-a}{1-az}.\end{align*}

If $$f: D_1(0) \to D_1(0)$$ is a biholomorphism, then \begin{align*}f(z) = c \Phi_a(z) = e^{i\theta} \Phi_a(z)\end{align*} So every such function is a rotated form of $$\Phi_a$$.

Let $$\Omega$$ be a connected open domain and $$f: \Omega \to {\mathbb{C}}$$ holomorphic with $$f\in C^1$$. Then \begin{align*}\int_\gamma f(z) ~dz = 0\end{align*} for every closed curve $$\gamma \subset \Omega$$, which implies that $$f^{(k)} (z)$$ exists for all $$k\in {\mathbb{N}}$$ and $$f$$ is smooth/holomorphic.

Morera

Theorem (Morera, Partial Converse to Cauchy’s Integral Theorem)

Suppose $$g: \Omega \to {\mathbb{C}}$$ is continuous and $$\int_\gamma g(z)~dz = 0$$ whenever $$\gamma = {{\partial}}R$$ for some rectangle $$R\subset \Omega$$ with sides parallel to the axes:

Then $$g(z)$$ is holomorphic in $$\Omega$$.

Proof

Fix a point $$\alpha = a + ib$$ and given $$z = x+iy$$, construct a rectangle $$R$$ containing $$z$$. Then by assumption, $$\int_{{{\partial}}R} g(z) ~dz = 0$$. Let $$\gamma_{az}$$ be the path given by traversing the bottom edge of $$R$$, and $$\sigma_{az}$$ by the top path.