Some results in real analysis

Proposition. Let $f: [a, b] \rightarrow \mathbb{R}$ be a function such that left-hand limit $f(x-)$ and right-hand limit $f(x+)$ of $f$ exist at every $x \in [a, b]$. Then the number of discontinuities of $f$ is at most countable.

Proof: Define a real-valued function $g$ on $[a, b]$ by $g(x) = \max\{|f(x) - f(x+)|, |f(x) - f(x-)|\}$. Then $f$ is continuous at $x$ iff $g(x) = 0$. Let $D_n = \{x \in [a, b]: g(x) \geq 1/n\}$. The points of discontinuity of $f$ is exactly $\bigcup_{n = 1}^\infty D_n$. We prove that each $D_n$ is finite.

Suppose on the contrary that $D_n$ is infinite. Then there is a sequence $\{x_n\}$ of distinct points in $D_n$ which converges to $x \in [a, b]$. Since $f(x+), f(x-)$ exist, we can choose some $\delta > 0$ such that $|f(y) - f(x+)| < 1/4n$ whenever $0 < y - x < \delta$ and $|f(y) - f(x-)| < 1/4n$ whenever $0 < x - y < \delta$. Therefore $|f(y) - f(z)| < 1/2n$ whenever $y, z \in (x - \delta, x)$ or $y, z \in (x, x + \delta)$. We can find some $k$ such that $0 < |x_k - x| < \delta$. If $x_k > x$, then $|f(y) - f(x_k)| < 1/2n$ whenever $y \in (x, x + \delta)$, let $y \rightarrow x_k$ from both left and right we get $|f(x_k+) - f(x_k)| \leq 1/2n$ and $|f(x_k-) - f(x_k)| \leq 1/2n$, so $g(x) \leq 1/2n < 1/n \leq g(x)$, contradiction. If $x_k < x$ then similarly we also obtain a contradiction. Hence $D_n$ must be finite. QED.

Proposition. Let $f_n: U \rightarrow \mathbb{R}$ be a sequence of continuous functions on $U \subseteq \mathbb{R}^d$ such that $f_1 \leq f_2 \leq \cdots$ and $\{f_n\}$ converges pointwise to a continuous function $f$ on $U$. Then $f_n \rightarrow f$ uniformly on compact subsets.

Proof: Note that $f_1 \leq \cdots \leq f_n \leq \cdots \leq f$. Let $K \subseteq U$ be compact and $\epsilon > 0$. For any $x \in K$, since $f_n(x) \rightarrow f(x)$, there is $N_x \in \mathbb{N}$ such that $f(x) - f_n(x) < \epsilon/3$ whenever $n \geq N_x$. For any $n \in \mathbb{N}$, by uniform continuity of $f$ and $f_n$ on $K$, there is $\delta_n > 0$ such that for any $x, y \in K$, $|f(x) - f(y)| < \epsilon/3$ and $|f_n(x) - f_n(y)| < \epsilon/3$ whenever $|x - y| < \delta_n$. For every $x \in K$, if $y \in B(x, \delta_{N_x})$, then
$\begin{align*} &f(y) - f_n(y) \leq f(y) - f_{N_x}(y) \\ &= [f(y) - f(x)] + [f(x) - f_{N_x}(x)] + [f_{N_x}(x) - f_{N_x}(y)] \\ &< \epsilon/3 + \epsilon/3 + \epsilon/3 = \epsilon \end{align*}$
whenever $n \geq N_x$. Since $\{B(x, \delta_{N_x}): x \in K\}$ is an open cover of $K$, then there are $x_1, \ldots, x_k \in K$ such that $K = \bigcup_{j = 1}^k B(x_j, \delta_{N_{x_j}})$. Take $N = \max\{N_{x_1}, \ldots, N_{x_k}\}$. For $n \geq N$, if $x \in K$, then $x \in B(x_j, \delta_{N_{x_j}})$ for some $1 \leq j \leq k$, so $f(x) - f_n(x) < \epsilon$. Hence $f_n \rightarrow f$ uniformly on $K$. QED.


(To be continued)

Conformal equivalence of annuli

For $0 < r < R$, let $A(r, R) = \{z \in \mathbb{C}: r < |z| < R\}$, the annulus centered at the origin with inner radius $r$ and outer radius $R$. We are going to show that the conformal equivalence classes of such anuuli in the complex plane are parametrized by the ratio $R/r$.

Theorem. $A(r_1, R_1)$ is conformally equivalent to $A(r_2, R_2)$ if and only if $R_1/r_1 = R_2/r_2$.
Proof: Suppose $R_1/r_1 = R_2/r_2$. Then there exists $k > 0$ such that $R_2 = k R_1$ and $r_2 = k r_1$. The map $f(z) = kz$ gives a conformal equivalence from $A(r_1, R_1)$ onto $A(r_2, R_2)$.
Conversely, let $f$ be a biholomorphic map from $A(r_1, R_1)$ onto $A(r_2, R_2)$. By scaling if necessary,  we may assume $r_1 = r_2 = 1$. Let $A_1 = A(1, R_1)$ and $A_2 = A(1, R_2)$. Fix some $1 < r < R_2$.
Let $C = \{z \in \mathbb{C}: |z| = r\}$.
Since $f^{-1}$ is continuous, $f^{-1}(C)$ is compact, so we can find some $\delta > 0$ such that $A(1, 1 + \delta) \cap f^{-1}(C) = \emptyset$. Let $V = f(A(1, 1 + \delta))$. Since $f$ is continuous, $V$ is connected, so either $V \subseteq A(1, r)$ or $V \subseteq A(r, R_2)$. By replacing $f$ with $R_2/f$, we may assume the first case holds.
Claim: $|f(z_n)| \rightarrow 1$ whenever $|z_n| \rightarrow 1$.
Proof of claim: Let $\{z_n\} \subseteq A(1, 1 + \delta)$. Note that $\{f(z_n)\}$ does not have a limit point in $A_2$ since otherwise $\{z_n\}$ would have a limit point in $A_1$ by continuity of $f^{-1}$, contradicting $|z_n| \rightarrow 1$. Hence $|f(z_n)| \rightarrow 1$ or $|f(z_n)| \rightarrow R_2$ (it must converge by continuity). The latter case is ruled out as $f(z_n) \in V \subseteq A(1, r)$.
Similarly, we also have:
Claim:  $|f(z_n)| \rightarrow R_2$ whenever $|z_n| \rightarrow  R_1$.
Now set $\alpha = \log R_2/ \log R_1$. Define $g: A_1 \rightarrow \mathbb{R}$ by $$g(z)  = \log |f(z)|^2 - \alpha \log |z|^2 = 2(\log |f(z)| - \alpha \log |z|).$$ We know that $\log |h|$ is harmonic whenever $h$ is holomorphic and nonzero, so $g$ is a harmonic function. By the two claims, $g$ extends to a continuous function on $\overline{A_1}$ vanishing on $\partial A_1$. This forces $g$ to vanish identically on $A_1$. In particular, $$0 = \frac{\partial g}{\partial z} = \frac{f'(z)}{f(z)} - \frac{\alpha}{z}.$$ Take some $1 < c < R_1$ and let $\gamma(t) = ce^{it}, t \in [0, 2\pi]$. We have $$\alpha = \frac{1}{2 \pi i}\int_\gamma \frac{\alpha}{z} dz = \frac{1}{2 \pi i}\int_\gamma \frac{f'(z)}{f(z)} dz = \mathrm{Ind}_{f \circ \gamma}(0),$$ so $\alpha > 0$ is an integer. Observe that $$\frac{d}{dz}(z^{-\alpha}f(z)) = z^{-\alpha - 1}(-\alpha f(z) + zf'(z)) = 0,$$ thus $f(z) = Kz^{\alpha}$ for some nonzero constant $K$. Since $f$ is injective, $\alpha = 1$. Therefore $R_2 = R_1$.

Reference:
Rudin - Real and Complex Analysis (3rd Ed.)

Koebe's 1/4-Theorem

Notation: $B(a, r) = \{z \in \mathbb{C}: |z - a| < r \}, B = B(0, 1), \mathbb{S}$ is the Riemann sphere.

Let $\mathcal{S}$ be the collection of all injective analytic function $f$ on the unit disk with  $f(0) = 0$ and $f'(0) = 1$. We know from the Open Mapping Theorem that $f(B)$ must contain a disk $B(0, r_f)$. For the class $\mathcal{S}$, there is a universal $r > 0$ such that $f(B)$ contains $B(0, r)$ for all $f \in S$.

Theorem (Koebe's 1/4-Theorem). If $f \in \mathcal{S}$ then $B(0, 1/4) \subseteq f(B)$.

We need to the following result, which is of independent interest.

Theorem (The Area Theorem). Let $g:  B \backslash \{0\} \rightarrow  \mathbb{C}$ be an injective analytic function with Laurent series expansion at 0 $$g(z) = \frac{1}{z} + c_0 + c_1 z + c_2 z^2 + \cdots.$$ Then $$\sum_{n = 1}^\infty n|c_n|^2 \leq 1.$$
Proof: Note that $g$ is a conformal equivalence between $B$ and $g(B) \subseteq \mathbb{S}$ and $\infty \in g(B)$. Fix any $0 < r < 1$. Consider $U_r = \mathbb{S} \backslash \overline{B(0, r)}$. It is an open set in $\mathbb{C}$ with boundary $\gamma$ parametrized by $\theta \mapsto h(re^{i\theta})$ where $\theta$ runs from $2\pi$ to 0. The area of $U_r$ is given by $\begin{align*} \mathrm{Area}(U_r) &= \frac{1}{2i}\int_\gamma \bar{w} dw = -\frac{1}{2i} \int_0^{2\pi} \overline{h(re^{i\theta})}h'(re^{i\theta})ire^{i\theta}d\theta \\ &= - \frac{1}{2i} \int_0^{2\pi}  [ \left(-\frac{1}{r^2} + |c_1|^2 + 2|c_2|^2 + 3|c_3|^2 + \cdots \right)r^2 \\ &{\,\,} + \textrm{ terms with nonzero integral powers of } e^{i\theta} ]  \\ &= \pi \left(\frac{1}{r^2} - \sum_{n = 1}^\infty n|c_n|^2 r^2 \right). \end{align*}$
Since area is always non-negative, let $r \rightarrow 1$ and we are done.

Proof of Koebe's Theorem: Let $f \in \mathcal{S}$. Then $f$ omits some $w_0 \in \mathbb{C}$ (or otherwise $f^{-1}$ is a bounded entire function, which is then constant by Liouville's Theorem, contradiction), so $h = 1/f$ omits 0 and $1/w_0$. Observe that the Taylor series of $f$ at 0 takes the form $f(z) = z + a_2 z^2 + a_3 z^3 + \cdots$, so $$h(z) = \frac{1}{z + a_2 z^2 + \cdots} = \frac{1}{z} + c_0 + c_1 z + c_2 z^2 + \cdots.$$  Let $z_0 = 0 \textrm{ or } 1/w_0$. We have $$g(z) = \frac{1}{h(z) - z_0} =  \frac{z}{1 + (c_0 - z_0)z + c_1 z^2 + \cdots} = z + (z_0 - c_0)z^2 + \cdots.$$ In particular, $g(0) = 0$ and $g'(0) = 1$. Also note that $g$ is injective. Thus $g \in \mathcal{S}$.
We claim that there exists some $u \in \mathcal{S}$ such that $u(z)^2 = g(z^2)$. Since $g$ is injective and $g(0) = 0$, the analytic function $g(z)/z$ omits 0 on $B$, we can find some analytic function $\varphi$ on $B$ such that $\varphi(z)^2 = g(z)/z$ and $\varphi(0) = 1$. Let $u(z) = z\varphi(z^2)$. Then $g(z^2) = z^2 \cdot g(z^2)/z^2 = z^2\varphi(z^2)^2 = u(z)^2$. Since $\varphi(0) = 1$ and $\varphi'(0) = \frac{1}{2}(g(z)/z)^{-1/2}\frac{d}{dz}(g(z)/z)|_{z = 0} = \frac{1}{2} (z_0 - c_0)$, $$u(z) = z\varphi(z^2) = z + \frac{1}{2}(z_0 - c_0)z^3 + \cdots.$$ Hence $u(0) = 0$ and $u'(0) = 1$. To show that $u \in \mathcal{S}$ it remains to prove that $u$ is injective. Suppose $u(z) = u(w)$. Then $g(z^2) = g(w^2)$. By injectivity of $g$, $z^2 = w^2$ or $z = \pm w$. If $z = -w$, then $u(w) = u(z) = u(-w) = -u(w)$, so $g(w^2) = u(w)^2 = 0$ and $w^2 = 0$ by injectivity of $g$ again, i.e. $w = z = 0$. In any case we have $z = w$.
Now consider the Laurent expansion of $1/u$ at 0, we have $$\frac{1}{u(z)} = \frac{1}{z + \frac{1}{2}(z_0 - c_0)z^3 + \cdots} = \frac{1}{z}(1 - \frac{1}{2}(z_0 - c_0)z^2 + \cdots) = \frac{1}{z} - \frac{1}{2}(z_0 - c_0)z + \cdots.$$ By the Area Theorem, $|-\frac{1}{2}(z_0 - c_0)| \leq 1$, or $|z_0 - c_0| \leq 2$. Therefore $|1/w_0| \leq |1/w_0 - c_0| + |0 - c_0| \leq 2 + 2 = 4$, or $|w_0| \geq 1/4$. It follows that $\mathbb{C} \backslash f(B) \subseteq \{w \in \mathbb{C}: |w| \geq 1/4\}$, so $B(0, 1/4) \subseteq f(B)$. QED.


Reference:
Andersson - Topics in Complex Analysis

Weak*-closed subspaces as dual spaces

Let $X$ be a Banach space and $M$ be a subspace of $X^*$ closed under the weak* topology. We show that M can be identified as the dual space of some Banach space.

Let $M_\perp = \{x \in X: f(x) = 0 \,\, \forall f \in M\} = \bigcap_{f \in M} \ker(f)$. This is a (weakly) closed subspace of $X$. Then we can form the quotient Banach space $X/M_\perp$ with the norm $\|x + M_\perp\| = \inf_{m \in M_\perp} \|x - m\|$.

Theorem. $M$ is isometrically isomorphic to $(X/M_\perp)^*$.
Proof: Define a map $\Phi: M \rightarrow (X/M_\perp)^*$ by $[\Phi(f)](x + M_\perp) = f(x)$ for all $f \in M$. Firstly, $\Phi(f)$ is a well-defined linear functional on $X/M_\perp$ because $f$ vanishes on $M_\perp$ whenever $f \in M$. Secondly, $\Phi(f)$ is bounded as $$|\Phi(f)(x + M_\perp)| = |f(x)| = |f(x - m)| \leq \|f\|\|x - m\| \,\, \forall m \in M_\perp,$$  so $|\Phi(f)(x + M_\perp)| \leq \|f\|\|x + M_\perp\|$. Hence $\Phi$ is well-defined.
Clearly $\Phi$ is linear. From the above we have $\|\Phi(f)\| \leq \|f\|$. Let $\epsilon > 0$. Choose some $x \in X$ with $\|x\| \leq 1$ and $|f(x)| > \|f\| - \epsilon$. Then $\|x + M_\perp\| \leq 1$ and $|[\Phi(f)](x + M_\perp)| = |f(x)|  > \|f\| - \epsilon$. Thus $\|\Phi(f)\| = \|f\|$. It follows that $\Phi$ is an isometry.
It remains to show that $\Phi$ is surjective. Let $\ell \in (X/M_\perp)^*$ be given. Define $f \in X^*$ by $f(x) = \ell(x + M_\perp)$, i.e. $f = \ell \circ \pi$ where $\pi$ is the quotient map. Then $f$ vanishes on $M_\perp$ by definition, hence $f \in M$ by the following lemma and it is clear that $\Phi(f) = \ell$.

Lemma. If $f \in X^*$ vanishes on $M_\perp$, then $f \in M$.
Proof: Suppose $f \in X^* \backslash M$. By Hahn-Banach Theorem and weak*-closedness we can find some $x \in X$ such that $f(x) = 1$ and $g(x) = 0$ whenever $g \in M$. But then $x \in M_\perp$ and so $f(x) = 0$, contradiction.

This result implies the following important observation in the theory of operator algebras.

Theorem. Every von Neumann algebra is the dual space of some Banach space.
Proof: Let $M$ be a von Neumann algebra on a Hilbert space $\mathcal{H}$. Then $M$ is a $\sigma$-weakly closed subalgebra of $B(\mathcal{H})$, i.e. it is a weak*-closed subalgebra of $B(\mathcal{H})$ where the weak* topology of $B(\mathcal{H})$ comes from the fact that the dual space of the trace-class operators $L^1(\mathcal{H})$ on $\mathcal{H}$ can be identified with $B(\mathcal{H})$. By our main theorem, $M \cong (L^1(\mathcal{H})/M_\perp)^*$.

A predual of a von Neumann algebra $M$ is a Banach space $X$ such that $M = X^*$. The above result says that every von Neumann algebra has a predual. In fact, the predual is unique by a result of Sakai. Furthermore, the property that having a predual characterizes von Neumann algebras among the class of C*-algebras.

Theorem (Sakai).  Suppose $A$ is a W*-algebra, i.e. $A$ is a C*-algebra and the dual space of some Banach space. Then $A$ is *-isomorphic to a von Neumann algebra over some Hilbert space.

Open mapping theorem in terms of norm estimates

Here we present a simple but useful corollary of the Open Mapping Theorem in functional analysis.

Notation: $B_X(a, r) = \{x \in X: \|x - a\| < r\}$.

Proposition. Suppose $X$ and $Y$ are Banach spaces. Let $f: X \rightarrow Y$ be a surjective bounded linear map. Then there exists some $\delta > 0$ such that $\|f(x)\| \geq \delta\|x\|$ for all $x \in X$.

Proof: By the Open Mapping Theorem, $f$ is an open map. Hence we can find some $\delta > 0$ such that $B_Y(0, \delta) \subseteq f(B_X(0, 1))$, so $f(X \backslash B(0, 1)) \subseteq Y \backslash B(0, \delta)$, i.e. $\|f(x)\| \geq \delta$ whenever $\|x\| \geq 1$. Therefore $\|f(x)\| = \|x\|\|f(x/\|x\|)\| \geq \delta\|x\|$.

Here is an application of the above result in the theory of Fourier series.

Theorem. The Fourier transform $\mathcal{F}: L^1(\mathbb{T}) \rightarrow c_0(\mathbb{Z})$ is not surjective.

Proof: Suppose not. By the Proposition, there exists $\delta > 0$ such that $\|\hat{f}\|_\infty \geq \delta\|f\|_1$. But if $\{D_N\}_{N = 0}^\infty$ is the Dirichlet kernel, i.e.

$$D_N(x) = \sum_{n = -N}^N e^{inx}$$

then we have $\|D_N\|_1 \rightarrow \infty$ and $\|\widehat{D_N}\|_\infty \geq \delta\|D_N\|_1 \rightarrow \infty$, which is absurd.

Boundedness of weakly convergent sequences

Let X be a Banach space. A net $\{x_\alpha\}$ in X is said to weakly converge to $x$ if $\ell(x_\alpha) \rightarrow \ell(x)$ for all $\ell \in X^*$. Clearly, every norm convergent net is weakly convergent (with the same limit). We know that a norm convergent sequence is necessarily bounded. It is natural to ask boundedness still hold for weakly convergent sequences.

Theorem. Every weakly convergent sequence in X is bounded.
Proof: Let $\{x_n\}$ be a weakly convergent sequence in X. Let $T_n \in X^{**}$ be defined by $T_n(\ell) = \ell(x_n)$ for all $\ell \in X^*$. Fix an $\ell \in X^*$. For any $n \in \mathbb{N}$, since the sequence $\{\ell(x_n)\}$ is convergent, $\{T_n(\ell)\}$ is a bounded set. By Uniform Boundedness Principle,
$$\sup_{n \in \mathbb{N}} \|x_n\| = \sup_{n \in \mathbb{N}} \|T_n\| < \infty,$$  i.e. $\{x_n\}$ is bounded.

However, the above result is no longer true when sequences are replaced by nets. To give a counterexample, let $X = L^1(\mathbb{R})$. Denote the characteristic function of a subset A by $\chi_A$. Let $\mathcal{O}$ be the collection of all open neighborhoods of 0 in $\mathbb{R}$ of finite Lebesgue measure with the ordering given by $U \prec V$ iff $U \supseteq V$. Consider the net $\{\chi_U: U \in \mathcal{O}\}$ in X. It is not bounded since $\|\chi_{(-n, n)}\|_1 = 2n \rightarrow \infty$ as $n \rightarrow \infty$. On the other hand, it converges weakly to 0. Fix any $f \in X^* = L^\infty(\mathbb{R})$. Choose M > 0 such that $|f| \leq M$ almost everywhere. For every $\epsilon > 0$, we can choose some $V \in \mathcal{O}$ with $m(V) < \epsilon/M$, then $|\int f\chi_U dm| \leq M m(U) \leq M m(V) < \epsilon$ whenever $U \succ V$.