Challenge Math Challenge - August 2020

[benorin]

Summary:: functional analysis, operator theory, topology, measure theory, calculus
Authors: Math_QED (MQ), Infrared (IR), Wrobel (WR), fresh_42 (FR).

8. Let F be a meromorphic function (holomorphic up to isolated poles) in C with the following properties:
(1) F is holomorphic (complex differentiable) in the half plane H(0)={z∈C:ℜ(z)>0}.
(2) zF(z)=F(z+1).
(3) F is bounded in the strip {z∈C:1≤ℜ(z)≤2}.
Show that F(z)=F(1)Γ(z). (FR)

Spoiler: problem #8

By (3) $\exists M :| F(x+iy) |< M,\forall x\in \left[ 1,2\right]$ and in particular $| F(1) | <M$, also by (2) $F(z)=\tfrac{1}{z} F(z+1)$ and hence F(z) is bounded in the half-plane $H(0)$. Also by (2) $\lim_{z\to 0} F(z)=\lim_{z\to 0}\tfrac{1}{z}F(z+1)=F(1)\lim_{z\to 0}\tfrac{1}{z}=\infty$ but $\lim_{z\to 0}zF(z)=\lim_{z\to 0}F(z+1)=F(1)$ so $z=0$ is a simple pole of $F(z)$ and continuing by (2) we see that $a_k=-k,\forall k\in\mathbb{N}$ are simple poles of $F(z)$.
Define $f(z):=\tfrac{1}{F(z+1)}$ so that $f(z)$ is entire, then by the Weierstrass' Factor Theorem
$$f(z)=f(0)e^{\tfrac{f(0)}{f^\prime (0)}z}\prod_{k=1}^\infty \left\{\left(1-\tfrac{z}{a_k}\right)e^{\tfrac{z}{a_k}}\right\}\quad \text{(eqn 1)}$$

and we have from the definition of $f(z)$ that $f(0)=\tfrac{1}{F(1)}$ and that $f^\prime (0) =-\tfrac{1}{\left[ F^\prime (1)\right] ^2}$ substituting all known values into (eqn 1) we get

$$\tfrac{1}{zF(z)}=\tfrac{1}{F(z+1)}=\tfrac{1}{F(1)}e^{-\tfrac{F(1)}{\left[ F^\prime (1)\right] ^2}z}\prod_{k=1}^\infty \left\{\left(1+\tfrac{z}{k}\right)e^{-\tfrac{z}{k}}\right\}$$

multiply by $z$ on both sides

$$\tfrac{1}{F(z)}=\tfrac{z}{F(1)}e^{-\tfrac{F(1)}{\left[ F^\prime (1)\right] ^2}z}\prod_{k=1}^\infty \left\{\left(1+\tfrac{z}{k}\right)e^{-\tfrac{z}{k}}\right\}\quad (eqn 2)$$

set $z=1$ in the above

$$\tfrac{1}{F(1)}=\tfrac{1}{F(1)}e^{-\tfrac{F(1)}{\left[ F^\prime (1)\right] ^2}}\prod_{k=1}^\infty \left\{\left(1+\tfrac{1}{k}\right)e^{-\tfrac{1}{k}}\right\}$$

do some algebra to get

$$\tfrac{F(1)}{\left[ F^\prime (1)\right] ^2}=\lim_{N\to\infty}\sum_{k=1}^N\left[\log \left(1+\tfrac{1}{k}\right)-\tfrac{1}{k}\right]=-\gamma$$

by the definition of Euler's constant $\gamma$, substitute this into (eqn 2) and compare to the Weierstrass product definition of the Gamma function and we have arrived at the required result.

Edit: See below for my next post for a more well explained proof.

 

[nuuskur]

Oh brother, this simply connected business for \mathbb C^* is simply not necessary. Let's go again.

Spoiler: Problem 9

Suppose g:\mathbb C^* \to \mathbb C is continuous with e^{g(z)} \equiv z. Then g is holomorphic, because the exponential map is holomorphic and
<br /> \lim _{z\to z_0} \frac{g(z) - g(z_0)}{z-z_0} = \lim _{g(z)\to g(z_0)} \frac{g(z)-g(z_0)}{e^{g(z)} - e^{g(z_0)}}.<br />
By differentiating we get 1 = g&#039;(z) e^{g(z)} = zg&#039;(z), which implies g&#039;(z) = \frac{1}{z} on \mathbb C^*. But now by Cauchy's integral theorem, we get
<br /> 0 = \oint _{|z|=1} g&#039;(z) dz = \oint_{|z|=1} \frac{1}{z} dz = 2i\pi.<br />
Thus, such a g cannot exist.

Maybe I'm hunting a fly with a 50 cal rifle with armor-piercing rounds..

 

[fresh_42]

Spoiler: problem #8

By (3) $\exists M :| F(x+iy) |< M,\forall x\in \left[ 1,2\right]$ and in particular $| F(1) | <M$, also by (2) $F(z)=\tfrac{1}{z} F(z+1)$ and hence F(z) is bounded in the half-plane $H(0)$. Also by (2) $\lim_{z\to 0} F(z)=\lim_{z\to 0}\tfrac{1}{z}F(z+1)=F(1)\lim_{z\to 0}\tfrac{1}{z}=\infty$ but $\lim_{z\to 0}zF(z)=\lim_{z\to 0}F(z+1)=F(1)$ so $z=0$ is a simple pole of $F(z)$ and continuing by (2) we see that $a_k=-k,\forall k\in\mathbb{N}$ are simple poles of $F(z)$.
Define $f(z):=\tfrac{1}{F(z+1)}$ so that $f(z)$ is entire, then by the Weierstrass' Factor Theorem
$$f(z)=f(0)e^{\tfrac{f(0)}{f^\prime (0)}z}\prod_{k=1}^\infty \left\{\left(1-\tfrac{z}{a_k}\right)e^{\tfrac{z}{a_k}}\right\}\quad \text{(eqn 1)}$$

and we have from the definition of $f(z)$ that $f(0)=\tfrac{1}{F(1)}$ and that $f^\prime (0) =-\tfrac{1}{\left[ F^\prime (1)\right] ^2}$ substituting all known values into (eqn 1) we get

$$\tfrac{1}{zF(z)}=\tfrac{1}{F(z+1)}=\tfrac{1}{F(1)}e^{-\tfrac{F(1)}{\left[ F^\prime (1)\right] ^2}z}\prod_{k=1}^\infty \left\{\left(1+\tfrac{z}{k}\right)e^{-\tfrac{z}{k}}\right\}$$

multiply by $z$ on both sides

$$\tfrac{1}{F(z)}=\tfrac{z}{F(1)}e^{-\tfrac{F(1)}{\left[ F^\prime (1)\right] ^2}z}\prod_{k=1}^\infty \left\{\left(1+\tfrac{z}{k}\right)e^{-\tfrac{z}{k}}\right\}\quad (eqn 2)$$

set $z=1$ in the above

$$\tfrac{1}{F(1)}=\tfrac{1}{F(1)}e^{-\tfrac{F(1)}{\left[ F^\prime (1)\right] ^2}}\prod_{k=1}^\infty \left\{\left(1+\tfrac{1}{k}\right)e^{-\tfrac{1}{k}}\right\}$$

do some algebra to get

$$\tfrac{F(1)}{\left[ F^\prime (1)\right] ^2}=\lim_{N\to\infty}\sum_{k=1}^N\left[\log \left(1+\tfrac{1}{k}\right)-\tfrac{1}{k}\right]=-\gamma$$

by the definition of Euler's constant $\gamma$, substitute this into (eqn 2) and compare to the Weierstrass product definition of the Gamma function and we have arrived at the required result.

Ok, I see "some algebra" but could you at least elaborate the last lines? Especially as you nowhere mentioned which definition of the Gamma function you used. I have the impression that it takes me more effort to make your proof readable than it takes to give another one. Also Weierstrass needed a bit more explanation. I doubt that an average member can understand how you applied it to what.

 

[benorin]

Spoiler: problem #8

By (3) $\exists M :| F(x+iy) |< M,\forall x\in \left[ 1,2\right]$ and in particular $| F(1) | <M$, also by (2) $F(z)=\tfrac{1}{z} F(z+1)$ and hence F(z) is bounded in the half-plane $H(0)$. Also by (2) $\lim_{z\to 0} F(z)=\lim_{z\to 0}\tfrac{1}{z}F(z+1)=F(1)\lim_{z\to 0}\tfrac{1}{z}=\infty$ but $\lim_{z\to 0}zF(z)=\lim_{z\to 0}F(z+1)=F(1)$ so $z=0$ is a simple pole of $F(z)$ and continuing by (2) we see that $a_k=-k,\forall k\in\mathbb{N}$ are simple poles of $F(z)$.
Define $f(z):=\tfrac{1}{F(z+1)}$ so that $f(z)$ is entire, then by the Weierstrass' Factor Theorem
$$f(z)=f(0)e^{\tfrac{f(0)}{f^\prime (0)}z}\prod_{k=1}^\infty \left\{\left(1-\tfrac{z}{a_k}\right)e^{\tfrac{z}{a_k}}\right\}\quad \text{(eqn 1)}$$

and we have from the definition of $f(z)$ that $f(0)=\tfrac{1}{F(1)}$ and that $f^\prime (0) =-\tfrac{1}{\left[ F^\prime (1)\right] ^2}$ substituting all known values into (eqn 1) we get

$$\tfrac{1}{zF(z)}=\tfrac{1}{F(z+1)}=\tfrac{1}{F(1)}e^{-\tfrac{F(1)}{\left[ F^\prime (1)\right] ^2}z}\prod_{k=1}^\infty \left\{\left(1+\tfrac{z}{k}\right)e^{-\tfrac{z}{k}}\right\}$$

multiply by $z$ on both sides

$$\tfrac{1}{F(z)}=\tfrac{z}{F(1)}e^{-\tfrac{F(1)}{\left[ F^\prime (1)\right] ^2}z}\prod_{k=1}^\infty \left\{\left(1+\tfrac{z}{k}\right)e^{-\tfrac{z}{k}}\right\}\quad (eqn 2)$$

set $z=1$ in the above

$$\tfrac{1}{F(1)}=\tfrac{1}{F(1)}e^{-\tfrac{F(1)}{\left[ F^\prime (1)\right] ^2}}\prod_{k=1}^\infty \left\{\left(1+\tfrac{1}{k}\right)e^{-\tfrac{1}{k}}\right\}$$

do some algebra to get

$$\tfrac{F(1)}{\left[ F^\prime (1)\right] ^2}=\lim_{N\to\infty}\sum_{k=1}^N\left[\log \left(1+\tfrac{1}{k}\right)-\tfrac{1}{k}\right]=-\gamma$$
edit begins here:
by the definition of Euler's constant $\gamma$, substitute this into (eqn 2) and invert to get

$$\begin{gathered} \boxed{ F(z)=F(1)\cdot\tfrac{1}{z} e^{-\gamma z}\prod_{k=1}^\infty \left\{\left(1+\tfrac{z}{k}\right) ^{-1} e^{\tfrac{z}{k}}\right\} \\ =F(1)\cdot \Gamma (z) } \\ \end{gathered}$$

since the Weierstrass' product definition of the Gamma function is

$$\Gamma (z) =\tfrac{1}{z} e^{-\gamma z}\prod_{k=1}^\infty \left\{\left(1+\tfrac{z}{k}\right) ^{-1} e^{\tfrac{z}{k}}\right\} $$

we have arrived at the required result.

 

[TeethWhitener]

Spoiler: Problem 2

Maybe I'm misunderstanding the problem, but if we're only worried about continuity (and not differentiability), can't we just do something simple? I just built a parallelogram and let $f$ and $g$ take paths along opposite sides:

Let $f(\frac{3}{2})=\frac{3}{2}, g(\frac{3}{2})=\frac{3}{2}, f(\frac{1}{2})=\frac{1}{2}, g(\frac{1}{2})=\frac{1}{2}$. Notice that $\frac{1}{2}x+\frac{3}{4}$ and $2x-\frac{3}{2}$ both go through $(\frac{3}{2},\frac{3}{2})$, and $\frac{1}{2}x+\frac{1}{4}$ and $2x-\frac{1}{2}$ both go through $(\frac{1}{2},\frac{1}{2})$. We see that $2x-\frac{1}{2}$ intersects $\frac{1}{2}x+\frac{3}{4}$ at $(\frac{5}{6},\frac{7}{6})$, and $2x-\frac{3}{2}$ intersects $\frac{1}{2}x+\frac{1}{4}$ at $(\frac{7}{6},\frac{5}{6})$.

So now we just define:
$$f(x)=\begin{cases}
x &\text{if } 0\leq x < \frac{1}{2} \text{ or }\frac{1}{2} \leq x \leq 2\\
2x-\frac{1}{2} & \text{if } \frac{1}{2} \leq x < \frac{5}{6} \\
\frac{1}{2}x+\frac{3}{4}& \text{if } \frac{5}{6} \leq x < \frac{3}{2} \\
\end{cases}
$$
and
$$g(x)=\begin{cases}
x &\text{if } 0\leq x < \frac{1}{2} \text{ or }\frac{1}{2} \leq x \leq 2\\
\frac{1}{2}x+\frac{1}{4}& \text{if } \frac{1}{2} \leq x < \frac{7}{6} \\
2x-\frac{3}{2} & \text{if } \frac{7}{6} \leq x < \frac{3}{2} \\
\end{cases}
$$

 

[nuuskur]

Spoiler: Problem 1

Let K\subseteq \mathbb C be non-empty compact. Suppose \{\gamma_n \in K \mid n\in\mathbb N\} \subseteq K is dense in K\ \ (\mathbb C is separable). Let D : e_n\mapsto \gamma_ne_n be a diagonal operator (in the infinite case H\cong \mathbb C^\omega, otherwise H\cong \mathbb C^N).

For any n\in\mathbb N we have \gamma_n\in\sigma (D), because D-\gamma_n\mathrm{id} is not injective. If we look at the kernel, we have
<br /> \sum \lambda _k(D-z_n\mathrm{id})(e_k) = \sum _{k\neq n} (\lambda _k\gamma_k - \gamma _n)e_k + (\lambda_n -1)\gamma _n e_n = 0.<br />
We can clearly see \lambda _k \equiv 0 is not forced.

Since \sigma (D) is closed, we also have K\subseteq \sigma (D). On the other hand, suppose \lambda\notin K. Then
F \left ( \sum \lambda _ke_k\right ) := \sum\frac{\lambda _k}{\gamma _k-\lambda}e_k
is the inverse of D-\lambda \mathrm{id}, thus \lambda\notin \sigma (D). Indeed, we can check for basis elements. We have
<br /> (F \circ (D-\lambda \mathrm{id}))(e_k) = F((\gamma _k-\lambda)e_k) = \frac{\gamma _k-\lambda}{\gamma _k-\lambda}e_k = e_k<br />
and
<br /> ((D-\lambda \mathrm{id}) \circ F)(e_k) = (D-\lambda \mathrm{id}) \left ( \frac{1}{\gamma _k-\lambda}e_k \right ) = e_k.<br />
Altogether \sigma (D) = K.

 

[member 587159]

Spoiler: Problem 1

Let K\subseteq \mathbb C be non-empty compact. Suppose \{\gamma_n \in K \mid n\in\mathbb N\} \subseteq K is dense in K\ \ (\mathbb C is separable). Let D : e_n\mapsto \gamma_ne_n be a diagonal operator (in the infinite case H\cong \mathbb C^\omega, otherwise H\cong \mathbb C^N).

For any n\in\mathbb N we have \gamma_n\in\sigma (D), because D-\gamma_n\mathrm{id} is not injective. If we look at the kernel, we have
<br /> \sum \lambda _k(D-z_n\mathrm{id})(e_k) = \sum _{k\neq n} (\lambda _k\gamma_k - \gamma _n)e_k + (\lambda_n -1)\gamma _n e_n = 0.<br />
We can clearly see \lambda _k \equiv 0 is not forced.

Since \sigma (D) is closed, we also have K\subseteq \sigma (D). On the other hand, suppose \lambda\notin K. Then
F \left ( \sum \lambda _ke_k\right ) := \frac{\lambda _k}{\gamma _k-\lambda}e_k
is the inverse of D-\lambda \mathrm{id}, thus \lambda\notin \sigma (D). Indeed, we can check for basis elements. We have
<br /> (F \circ (D-\lambda \mathrm{id}))(e_k) = F((\gamma _k-\lambda)e_k) = \frac{\gamma _k-\lambda}{\gamma _k-\lambda}e_k = e_k<br />
and
<br /> ((D-\lambda \mathrm{id}) \circ F)(e_k) = (D-\lambda \mathrm{id}) \left ( \frac{1}{\gamma _k-\lambda}e_k \right ) = e_k.<br />
Altogether \sigma (D) = K.

Seems to work! Working with diagonal operators was also what I had in mind. Here is a more sophisticated approach using the theory of $C^*$-algebras:

Consider the $C^*$-algebra $C(K)$. Then the inclusion $i: K \to \Bbb{C}$ has spectrum $K$. Next, choose a Hilbert space $H$ and an injective $*$-homomorphism $\varphi: C(K) \to B(H)$ (GNS-construction). Then $\varphi(i)$ is a bounded operator on $H$ with spectrum $K$.

I will look at the other solution attempt soon.

 

[Infrared]

@TeethWhitener If I take $a=3/2$ and $b=1/2$ in your example, then $f(3/2)-f(1/2)=1$ and $g(3/2)-g(1/2)=1$ are both integers.

 

[nuuskur]

Spoiler: Problem 3

Suppose V/W is complete. Let x_n\in V,\ n\in\mathbb N, be a Cauchy sequence. Let \pi :V\to V/W be the projection, then \pi (x_n),\ n\in\mathbb N, is Cauchy in V/W. For some x\in V we have \pi (x_n) \xrightarrow[]{}\pi (x). By definition of infimum pick z_n\in V such that \|x_n-x-z_n\| \leq \|\pi (x_n-x)\| + \frac{1}{n},\ n\in\mathbb N. Then \|x_n-(x+z_n)\| \to 0. So we would have x_n \to x+\lim z_n. This is justified, because by triangle equality
<br /> \|z_n-z_m\| \leq \|x_n-x-z_n\| + \|x_n-x_m\| + \|x_m-x-z_m\| \xrightarrow[m,n\to\infty]{}0.<br />

 

[TeethWhitener]

@TeethWhitener If I take $a=3/2$ and $b=1/2$ in your example, then $f(3/2)-f(1/2)=1$ and $g(3/2)-g(1/2)=1$ are both integers.

I must have misread the question then. Are you looking for a proof that it's true for all $f,g$ given the problem constraints? I thought the question was just asking for an example.

 

[nuuskur]

I must have misread the question then. Are you looking for a proof that it's true for all $f,g$ given the problem constraints? I thought the question was just asking for an example.

Given arbitrary continuous maps f,g on [0,2] we are to show such a\neq b exist. I'm thinking Intermediate value theorem might help. Not sure, right now.

 

[nuuskur]

Spoiler: Problem 4

For every x\in\mathbb R we have \mathbb P\{X\leq x\}\in \{0,1\}. Let F be the distribution function for X. F is right continuous and we have \lim _{x\to\infty} F(x) = 1 and \lim _{x\to -\infty}F(x) = 0. This implies there exists c\in\mathbb R such that F = I_{[c,\infty]}. Now \mathbb P\{X&lt;c\} = F(c-) = 0, therefore X=c a.s.

Spoiler: Problem 4 #2 Kolmogorov 0-1

We have \mathbb P\{X\leq x\} \in\{0,1\} for every x\in\mathbb R. One readily verifies X is independent of itself. Put c:= \inf \{x\in\mathbb R \mid X\leq x\}. It is finite, because \lim _{x\to \infty} \mathbb P\{X\leq x\} = 1, therefore for sufficiently large x_0 we have \mathbb P\{X\leq x_0\}=1. By Kolmogorov's 0-1 law \mathbb P\{X=c\}=1.

Since X is almost surely constant, the probability it takes any other value is zero, hence its variance is zero. Therefore \mathbb E(X-\mathbb E(X))^2 = 0 implies X=\mathbb E(X) a.s. The first moment exists, because variance is finite.

.. by triangle equality

Triangle equality I meant triangle inequality.

 

[benorin]

Summary:: functional analysis, operator theory, topology, measure theory, calculus
Authors: Math_QED (MQ), Infrared (IR), Wrobel (WR), fresh_42 (FR).

9. Show that there exists no continuous function g:C∖{0}→C such that eg(z)=z for all z∈C∖{0} (i.e. there is no continuous logarithm on C∖{0}). (MQ)
(Corrected version, credits go to @PeroK for pointing out and solving the original problem.)

I beg to differ: A Course of Modern Analysis, 3rd ed. Whittaker & Watson pg 589 (heading A6) says $\log z = \Log |z| + i \arg z$ is a continuous multi-valued function of $z$ since $|z|$ is a continuous function of $z$.

 

[PeroK]

I beg to differ: A Course of Modern Analysis, 3rd ed. Whittaker & Watson pg 589 (heading A6) says $\log z = \Log |z| + i \arg z$ is a continuous multi-valued function of $z$ since $|z|$ is a continuous function of $z$.

I suspect you can make log continuous by taking the domain to be a set of complex planes disconnected at the branch cuts.

These are, of course, Riemann surfaces.

In any case it's $\arg z$ that is discontinuous.

 

[nuuskur]

Oh brother, this simply connected business for \mathbb C^* is simply not necessary. Let's go again.

Spoiler: Problem 9

Suppose g:\mathbb C^* \to \mathbb C is continuous with e^{g(z)} \equiv z. Then g is holomorphic, because the exponential map is holomorphic and
<br /> \lim _{z\to z_0} \frac{g(z) - g(z_0)}{z-z_0} = \lim _{g(z)\to g(z_0)} \frac{g(z)-g(z_0)}{e^{g(z)} - e^{g(z_0)}}.<br />
By differentiating we get 1 = g&#039;(z) e^{g(z)} = zg&#039;(z), which implies g&#039;(z) = \frac{1}{z} on \mathbb C^*. But now by Cauchy's integral theorem, we get
<br /> 0 = \oint _{|z|=1} g&#039;(z) dz = \oint_{|z|=1} \frac{1}{z} dz = 2i\pi.<br />
Thus, such a g cannot exist.

Maybe I'm hunting a fly with a 50 cal rifle with armor-piercing rounds..

We can kill a fly with a fly swatter. Don't need the rifle.

Spoiler: Problem 9

Suppose g:\mathbb C^* \to\mathbb C is continuous and satisfies the identity e^{g(z)} = z. Since g is right inverse to \exp, it is injective, thus g(\mathbb C^*) \cong \mathbb C^*. But now \exp is forced to be injective on \mathbb C^*. Indeed, for any u,v\neq 0 we have
e^u = e^v \Leftrightarrow e^{g(z)} = e^{g(w)} \Leftrightarrow z=w \Rightarrow u=v.
But that's impossible.

 

[Infrared]

@nuuskur Where did you use continuity (this is definitely a necessary condition)? Anyway, you seem to be claiming that $g$ is surjective in saying $g(\mathbb{C}^*)=\mathbb{C}^*$. How do you know this? If you only mean that $g$ gives a bijection from $\mathbb{C}^*$ to its image, then how are you writing $u=g(z), v=g(w)$?

 

[member 587159]

If we try to make $\beta$ continuous, then without loss of generality we can take $\beta(0) = 0$, hence $\beta = \theta$ and we get a discontinuity at $\theta = 0$. More generally, for any $\phi$ we could could take $\beta(\phi) = \phi + 2n\pi$ and get the discontinuous branch cut at $\theta = \phi$.

Hmm, I need a little more details here. Explain why the discontinuities arise. I find your argument a bit too handwavy.

 

[nuuskur]

@nuuskur Where did you use continuity (this is definitely a necessary condition)? Anyway, you seem to be claiming that $g$ is surjective in saying $g(\mathbb{C}^*)=\mathbb{C}^*$. How do you know this?

Continuity is implicit. I'm working in the category with continuous maps. No, g is injective, therefore it's an isomorphism onto its image.

 

[Infrared]

Continuity is implicit.

Which part of your argument fails if $g$ is not required to be continuous?

No, g is injective, therefore it's an isomorphism onto its image.

Please explain this line.

e^u = e^v \Leftrightarrow e^{g(z)} = e^{g(w)} \Leftrightarrow z=w \Rightarrow u=v.

What are $z$ and $w$ in relation to $u$ and $v$? It looks to me like you're assuming that $g$ is surjective here (by saying that you can write $u=g(z), v=g(w)$)

 

[member 587159]

Continuity is implicit. I'm working in the category with continuous maps. No, g is injective, therefore it's an isomorphism onto its image.

You really need that $g(\Bbb{C}^*) = \Bbb{C}^*$ for your argument to work. Also, when using the symbol $\cong$, explain what you mean with the symbol. I assume that you mean isomorphism in the category of continuous maps, but how are you even sure the inverse on the image is continuous as well?

 

[member 587159]

This implies there exists c\in\mathbb R such that F = I_{[c,\infty]}.

Explain this line.

 

[PeroK]

Hmm, I need a little more details here. Explain why the discontinuities arise. I find your argument a bit too handwavy.

Effectively it boils down to the discontinuity of $\arg z$. I thought that was trivial and well established!

 

[nuuskur]

We can kill a fly with a fly swatter. Don't need the rifle.

Spoiler: Problem 9

Suppose g:\mathbb C^* \to\mathbb C is continuous and satisfies the identity e^{g(z)} = z. Since g is right inverse to \exp, it is injective, thus g(\mathbb C^*) \cong \mathbb C^*. But now \exp is forced to be injective on \mathbb C^*. Indeed, for any u,v\neq 0 we have
e^u = e^v \Leftrightarrow e^{g(z)} = e^{g(w)} \Leftrightarrow z=w \Rightarrow u=v.
But that's impossible.

Ok, here are the details. Firstly, since g is injective, it means g(\mathbb C^*) \cong \mathbb C^* as sets. That means any u\neq 0 can be written uniquely as u=g(z). But we also have a homemorphism with the obvious choice ( this is forced, really) f(g(z)) := z,\ z\neq 0. For continuity of f:g(\mathbb C^*) \to \mathbb C^* suppose g(z_n) \to g(z), then continuity of exponential map implies e^{g(z_n)} \to e^{g(z)} which by assumption is the same as z_n\to z so f is continuous.

What fails if g is not continuous: it wouldn't be a morphism in the category with continuous maps, so the above won't apply.

 

[nuuskur]

Explain this line.

By definition of limit. As x\to -\infty we must have A&gt;0 such that x\leq -A implies F(x) = \mathbb P\{X\leq x\}=0. Since such A are bounded from below, take the infimum i.e c := \inf \{x\in\mathbb R \mid X\leq x\}. Now, whatever happens for x&gt;c must occur with probability 1 (otherwise c wouldn't be the infimum). There can be no in-betweens so F = I_{[c,\infty]} is forced due to right continuity of F i.e F(c+) = 1.

 

[nuuskur]

Spoiler: Problem 7

Suffices to show C(f) = \{x\in X \mid f\text{ is continuous at }x\} is a Borel set, it can also be empty. Then C(f)^c = D(f) is a Borel set. Let's chase epsilons from the definition of continuity. f is continuous at x\in X if and only if
<br /> \forall n\in\mathbb N,\ \exists \delta &gt;0,\ \forall z\in X,\quad z\in B(x,\delta) \Rightarrow f(z) \in B\left (f(x), n^{-1}\right ).<br />
In other words we have \{x\}= \bigcap _{n\in\mathbb N} \{U\subseteq X \mid x\in U,\ U\text{ is open and }f(U) \subseteq B(f(x), n^{-1}) \}. Do this for all points of continuity, then we have the G_\delta set
<br /> \bigcap _{n\in\mathbb N} \bigcup \left\{ U\subseteq X \mid U\text{ is open, }\ \sup_{u,v\in U} d_Y(f(u),f(v)) \leq n^{-1} \right \} = C(f).<br />
The equality should be clear.

 

[Infrared]

That means any u\neq 0 can be written uniquely as u=g(z).

Why?

 

[nuuskur]

Why?

Oops, that's bad expression by me. I really had in mind that u is identified uniquely with g(z). But now e^u = e^v \Rightarrow e^{g(z)} = e^{g(w)} might break. Thanks for noticing. The homemorphism part still works, but usable if g is an inclusion as @Math_QED said.

I'll stick to my rifle at #32 for now. Yet I'm not convinced continuity is a required assumption. I get a feeling e^{g(z)} =z forces g to be continuous and in this case, holomorphic.

 

[Infrared]

Yet I'm not convinced continuity is a required assumption. I get a feeling e^{g(z)} =z forces g to be continuous and in this case, holomorphic

It doesn't. $g(z)=\ln|z|+\arg(z)$ is a counterexample

 

[nuuskur]

It doesn't. $g(z)=\ln|z|+\arg(z)$ is a counterexample

Of course *smacks forehead*. I had a thought maybe |z|\leq |e^z|, but that's not true in \mathbb C.

 

[PeroK]

Hmm, I need a little more details here. Explain why the discontinuities arise. I find your argument a bit too handwavy.

We have a function $\beta (\theta)$ for $0 \le \theta < 2\pi$ with $\cos \beta (\theta) = \cos \theta$ and $\sin \beta (\theta) = \sin \theta$. We know that $\beta(0) = 2\pi n$ for some $n$. As adding a constant does not effect the continuity of a function we can, wlog, take $\beta(0) = 0$.

The technical point outstanding is that if $\beta$ is continuous, then $\beta(\theta) = \theta$.

In general, we have $\beta(\theta) = \theta + 2\pi n(\theta)$ with $n(0) = 0$, as above.

The only continuous functions of the form $2\pi n(\theta)$ are constant functions. I'll spare everyone an epsilon-delta proof of that. Therefore, we must have $g(z) = i\theta = i\arg z$ for $0 \le \theta < 2\pi$.

The second technical point is that $\arg(z)$ is discontinuous on the unit circle. I'll quote that as a known result.

In any case, that proves that $g$ cannot be continuous.