Mar 11, 2014
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## On the strong and weak topologies:

So here’s the gist of it: normed vector spaces generate, via their norm, a natural metric topology which we call strong; $$(x_n) \rightarrow x$$ strongly if it converges in norm; when we look at functions rather than points, however, this topology’s too strong to capture (i.e. it can’t really see weaker but important modes of convergence, i.e. pointwise- or measure- or smooth- or compact-convergence; we’ll be extending the tools we’ve developed so far to a theory of topological vector spaces; two basic ones we’ll be looking at are the weak topology and the weak* topology; we’ll obtain via Tychonoff’s theorem (and its sequential counterpart) the Banach-Alaoglu theorem (and its sequential counterpart.)

Definition. A topological vector space is a vector space with a topology such that addition and scalar multiplication is continuous when viewed as a map from $$V \times V$$ (equipped with the product topology) $$\rightarrow V.$$

Well, that was easy, wasn’t it?

Okay, I’m goin’ to sleep. I’ll continue tomorrow.

Mar 11, 2014
1 note

## Hey,

I just found out I don’t have homework in 245B this week!

I guess I’ll TeX up some notes on chapter 1.9 from Tao’s book, brush up on some algebraic geometry, and call it a night.

Mar 11, 2014
1 note

## Defining the complex logarithm:

We begin by setting $\log z = \log r + i \theta,$ since what we really want is an inverse to $z = r e^{i \theta};$ of course, we sort of lose the periodicity of the exponential (but in turn we get a way to count the number of times we go around something, which is really nice.)

(Theorem 6.1, Chapter 3, S&S.) Suppose $$\Omega$$ is a simply connected region containing $$1$$ but not the origin; then there exists a branch (or sheet i.e. a restriction of the domain so that the logarithm actually makes any fuarkin’ sense) of the logarithm F(z) = \log_{\Omega}(z) such that $$F$$ is holomorphic over $$\Omega$$, that we recover $$z$$ via exponentation, and $$F(r) = \log r$$ whenever $$r \in \mathbb{R}$$ near $$1.$$

Proof. We construct $$F$$ as the primitive of $$\dfrac{1}{z}$$ (as one might expect, lols); by assumption $$f = \dfrac{1}{z}$$ is holomorphic over $$\Omega;$$ by simple-connectedness, we can unambiguously define $\log_{\Omega} = F = \int_{\gamma} f(\zeta) d \zeta$ where $$\gamma$$ connects $$1$$ with any $$z \in \Omega;$$ by simple-connectedness $$F$$ is holomorphic on $$\Omega$$ and F’ = \dfrac{1}{z} on $$\Omega;$$ for the second claim; it’ll be enough to show $\dfrac{z}{e^{F(z)}} = 1;$ differentiation gives us $e^{-F(z)} - z \cdot F’(z) \cdot e^{-F(z)} = e^{-F(z)} \cdot (1 - Z F’(z)) = 0,$ i.e. $$\dfrac{z}{e^{F(z)}}$$ is constant and so evaluation at $$z = 1$$ gives $$1.$$ For the final claim, we can find a path on the real line i.e. we can write $F = \int_{1}^r \dfrac{dx}{x} = \log r,$ and we’re done. $$_\square$$

Alrighty, then, onto functional analysis in $$L^p$$ spaces.

Mar 11, 2014
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## Rouche’s theorem, the open mapping theorem, and the maximum modulus principle:

You won’t win, Elias Stein. You won’t. -shakes fist angrily at heavens-

(Like most characters in most RPGs, I only have a limited range of emoticons. I may or may not be a figment of your imagination. It depends on how high you are.)

Okay, continuing.

Rouche’s theorem. Let $$f,g$$ holo on $$\bar{D} \subset \Omega;$$ if $|f| > |g|, \forall z \in \partial D$ then $$f$$ and $$f + g$$ have the same number of zeros in $$D.$$

Proof. Define, for $$t \in [0,1], f_t = f + t \cdot g.$$ Since $$f$$ doesn’t vanish on the circle and we haven’t got any poles anywhere, we have by the argument principle (denoting by $$n_t$$ the number of zeros of $$f_t$$) $n_t = \dfrac{1}{2 \pi i} \int_{\partial D} \dfrac{f_t’}{f_t} dz;$ it’ll suffice here to show that $$n_t$$’s a continuous function of $$t$$ since otherwise we could find by the intermediate value theorem a value of $$n_{t_0}$$ not in $$\mathbb{Z}$$, for some $$t_0$$.

(Okay, I don’t really grok the next part of the proof S&S give, but here:)

Since $$f’_t(z)$$ is jointly continuous on $$\mathbb{C} \times [0,1]$$ so is its primitive; since the denominator of $$\dfrac{1}{2 \pi i} \int_{\partial D} \dfrac{f_t’}{f_t} dz$$ doesn’t vanish on $$\partial D$$ then $$n_t$$ is continuous, integer-valued, and constant; then $$n_0 = n_1,$$ and we have our result, $$_\square$$

(Open mapping theorem.) If $$f$$ holo and not constant on $$\Omega$$ an open region then $$f$$ is an open mapping.

Proof. This’ll use Rouche’s theorem. Namely, let $$w_0 = f(\Omega)$$ be the image of some $$z_0;$$ set $$g = f - w$$ for $$w$$ near $$w_0$$ and write $g = (f - w_0) + (w_0 - w) = F + G;$ it’ll suffice to show, since $$w_0$$ was arbitrary, that all $$w$$ near $$w_0$$ are also in $$f(\Omega).$$ Choose a $$D = \delta$$-ball around $$z_0$$ with its closure contained in $$\Omega$$ such that $$f$$ is never $$w_0$$ on its boundary (where we can do this because if we assume otherwise, we obtain a convergent sequence on which $$f$$ is constant, which then forces everything in $$\Omega$$ to be constant); then choose $$\epsilon : |f - w_0| \geq \epsilon$$ over $$\partial D;$$ then $$\forall w \in B(w_0,\epsilon), |F| > G$$ over $$\partial D;$$ by Rouche’s theorem $$g$$ has a zero inside the circle i.e. $$f(z) = w$$ for some $$z \in D,$$ as desired. $$_\square$$

Finally, we have the maximum modulus principle: $$f$$ not constant holo over $$\Omega$$ implies $$f$$ doesn’t obtain its maximum in $$\Omega.$$

Proof. Suppose towards a contradiction $$\exists z_0 \in \Omega : \sup_{z \in \Omega} |f(z)| = f(z_0);$$ let $$D$$ a small disc centered at $$z_0;$$ then $$f(D)$$ contains a small disc centered at $$f(z_0;$$ pick a point further from the origin and you have a contradiction.

As an immediate corollary of this, we have $$\sup_{z \in \Omega} |f(z)| \leq \sup_{z \in \bar{\Omega} \backslash \Omega} |f(z)|._square$$

Mar 11, 2014
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## The argument principle:

Okay, I’m doin’ this and then Rouche/open-mapping/max-modulus, and then a bit about the logarithm, and then I’m starting my problem set for 245B.

First, begin with the observation $\dfrac{(\prod_{k \leq n} f_k)’}{\prod_{k \leq n} f_k} = \displaystyle \sum_{k \leq n} \dfrac{f’_k}{f_k}.$ Now, if $$f$$ holo with a zero of order $$n$$ at $$z_0,$$ we can write $f(z) = (z-z_0)^n \cdot g(z)$ for $$g$$ nonvanishing holomorphic near $$z_0;$$ compute $f’(z) = \sum_{i \leq n - 1}\dfrac{(z - z_0)’}{(z-z_0)} + \dfrac{g’}{g} = \dfrac{n}{z - z_0} + G$ where $$G = \dfrac{g’}{g};$$ observe that $$\dfrac{f’}{f}$$ has a pole with residue $$n$$ whenever $$f$$ has a zero of order $$n$$, at $$z_0;$$ an identical computation gives, in the case that $$f$$ has a pole of order $$n$$ at $$z_0$$, $\dfrac{f’}{f} = \dfrac{-n}{z - z_0} + H,$ where $$H = \dfrac{h’}{h}$$ for holomorphic nonvanishing $$h;$$ i.e. $$\dfrac{f’}{f}$$ has poles of order $$n$$ (resp. $$-n$$ corresponding to zeros (resp. poles) of order $$n$$ of $$f.$$

Neat, right?

This, combined with the residue formula, gives us the argument principle, namely that if $$f$$ is meromorphic on an open $$\Omega \supset \bar{D} : f \text{ neither blows up nor vanishes on } \partial D,$$ then $\dfrac{1}{2 \pi i} \int_{\partial D} \dfrac{f’}{f} dz$ is an integer that gives the number of zeros of $$f$$ in $$D$$ minus the number of poles of $$f$$ in $$D,$$ with multiplicity.

Mar 11, 2014
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## Meromorphic functions in the extended complex plane are rational:

(Theorem 3.4, Ch. 3, S&S)

Proof: Let $$f$$ meromorphic on the extended plane; then $$f$$ has finitely many poles in the plane, say $$z_1,…,z_n;$$ around each $$z_i$$ we can write $$f = f_i + g_i$$ where $$f_i$$ is the principal part of $$f$$ at $$z_i$$ and $$g$$ is holomorphic and nonvanishing (I think); in particular $$f_i$$ is a polynomial in $$\dfrac{1}{z - z_i};$$ similarly we can write $f \left ( \dfrac{1}{z} \right ) = \bar{f}_{\infty}(z) + \bar{g}_\infty(z)$ around the removable singularity/pole at infinity, where $$\bar{f}_\infty$$ is the principal part of $$f$$ applied to the multplicative inversion of $$\mathbb{C}$$ at zero and is hence a polynomial in $$\frac{1}{z}.$$ Then $H = f - f_\infty - \sum_{i \leq n} f_k$ (where $$f_\infty$$ is $$\bar{f}_\infty$$ composed with multiplicative inversion is entire, bounded, hence constant; rearranging we have $$f$$ is a rational function, as desired. $$_\square$$

Mar 11, 2014
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## Okay, intuitively,

we also see that (if we set $$F =$$) $f \left ( \dfrac{1}{z} \right )$ that $$\dfrac{1}{F}$$ can’t also be zero at the origin if there’s also a sequence of poles (read: zeroes) that approach the origin, because then by the post before the post before the post before the post before this post (modulo a post or two) $$\dfrac{1}{F}$$ would be zero on the unit disc, and that would break math.

Mar 10, 2014
1 note

## Wait,

poles can accumulate at infinity? Such confuse.

edit: Okay, poles can accumulate at infinity (say $$\dfrac{1}{\sin(z)}$$) but if they do then there can be no pole at infinity because if you apply $$f$$ to the multiplicative inversion of $$\mathbb{C}$$ i.e. $f \left( \dfrac{1}{z} \right )$ then there’s no neighborhood of the origin (minus the origin) on which $$f$$ is actually holomorphic.

Mar 10, 2014
0 notes

## Meromorphic functions and the Riemann sphere:

Say $$f$$ is meromorphic on $$\Omega$$ if there’s a sequence $$(z_n)$$ that doesn’t accumulate in $$\Omega$$ such that $$f$$ is holomorphic on $$\Omega$$ minus $$(z_n)$$ and $$f$$ has poles at each $$z_i.$$ We see by basic point-set topological definition-shuffling (compactness, etc) that the functions which are meromorphic in the extended (i.e. one-point compactified) complex plane are precisely those with only a finite number of poles, where we include the pole-at-infinity, i.e. those that blow up in every direction.

(In other words, no matter how slowly you make an unbounded sequence of poles grow, it’ll still ‘converge’ (in the Riemann-sphere sense) to infinity.)

We construct the Riemann sphere as follows: place a unit (it doesn’t matter, because infinite scaling) sphere $$\textbf{S}$$ on the origin of the complex plane and map $$\mathbb{C} \rightarrow \textbf{S}$$ by drawing a line from a point $$z \in \mathbb{C}$$ to the top of $$\textbf{S}$$ and mapping $$z$$ to wherever that line intersects the surface of $$\textbf{S}.$$ Note that the lattice of integers embedded in $$\mathbb{R}^2 \hookrightarrow \mathbb{C} \rightarrow \textbf{S}$$ becomes infinitely dense on the sphere as you move closer to the top; we call the antipode $$\mathcal{N}$$ of the origin the point at infinity.

Most importantly, we can now consider functions meromorphic on the extended complex planes to just be mappings of $$\textbf{S}$$ to itself, with poles being sent to $$\mathcal{N}.$$

Mar 10, 2014
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## Analytic continuation, 2:

By the above, if $$f, g$$ holo on $$\Omega$$ and agree on a sequence that accumulates in $$\Omega$$, then $$f = g$$ everywhere on $$\Omega.$$

Furthermore, if there’s a bigger $$\hat{\Omega}$$ over which some $$F$$ is holomorphic such that $$F = f$$ on some sequence of points that accumulates in $$\Omega$$ then $$F = f$$ on all of $$\Omega;$$ by the first paragraph $$F$$ is unique and we call it the analytic continuation of $$f$$.

(This is so cool. Everything’s so nice when you’re working with holomorphic functions :))

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