Math Challenge - August 2020

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  • #26
etotheipi
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Assuming that this is correct, is this ##g## continuous?
Well ##\text{arctan}## will throw out a value between ##-\frac{\pi}{2}## and ##\frac{\pi}{2}##, so you will have a problem crossing over the negative ##x## axis. If you change ##g## so that the argument is taken between ##0## and ##2\pi##, now you have a problem crossing the positive ##x## axis. In any case, I guess not?
 
  • #27
Math_QED
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Well ##\text{arctan}## will throw out a value between ##-\pi## and ##\pi##, so you will have a problem crossing over the negative ##x## axis. If you change ##g## so that the argument is taken between ##0## and ##2\pi##, now you have a problem crossing the positive ##x## axis. In any case, I guess not?
Yes, the idea is indeed that on the x-axis trouble arise, but I don't think what you wrote is rigorous enough.
 
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  • #28
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Let ##u = x + \frac{1}{x}##, then ##x = \frac{u \pm \sqrt{u^2-4}}{2} := \alpha(u)## and$$I = \int_{n + n^{-1}}^{n + n^{-1}} f(u) \frac{\log(\alpha(u))}{\alpha(u)(1-\frac{1}{\alpha^2(u)})} du = 0$$because the limits are equal...
Doesn't ##a = b## work?
I have a bit the impression that you tend to ignore the little details, e.g. here when you define two different substitutions and "forget" to manage them. This attitude is o.k. as you concentrate on the core of a problem, and it is wide spread among especially talented students. However, there will be occasions when this becomes a boomerang: exams and publications. This here isn't either, but a playground to practice. Maybe you try to practice being especially correct here. Please don't change your nature, just practice to be pedantic occasionally when it is necessary.
 
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  • #29
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A cat's been trapped in a box for 85 years and you think it might still be alive?
Well, we cannot know for sure. The cat wasn't specified any further ...
Is there an explanation of this for non-poker-playing high-schoolers?
https://howtoplaypokerinfo.com/poker-101/texas-holdem-rules/

A tiny bit of research can be expected, me thinks. Today more than ever we have to be prepared to consult the internet to look up details.
 
  • #30
PeroK
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There are multiple approaches. First, you need to say what convention you use: I guess you write ##z=r e^ {i \theta}##. What domain do you allow for ##\theta##?

Then, why does ##g## take that form and why does it have a discontinuity?
Okay, it makes sense to look for ##e^{g(z)} =z##.

Consider the points on the unit circle, which can be written ##z = e^{i\theta} = \cos \theta + i\sin \theta##.

Let ##g(z) = \alpha(\theta) + i\beta(\theta)## with ##e^{g(z)} = e^{\alpha}(\cos \beta + i\sin \beta) = z = e^{i\theta}##

We have ##\alpha(\theta) = 0##, ##\cos \theta = \cos \beta## and ##\sin \theta = \sin \beta##.

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##.
 
  • #31
benorin
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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)
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.
 
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  • #32
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Oh brother, this simply connected business for [itex]\mathbb C^*[/itex] is simply not necessary. Let's go again.
Suppose [itex]g:\mathbb C^* \to \mathbb C[/itex] is continuous with [itex]e^{g(z)} \equiv z[/itex]. Then [itex]g[/itex] is holomorphic, because the exponential map is holomorphic and
[tex]
\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)}}.
[/tex]
By differentiating we get [itex]1 = g'(z) e^{g(z)} = zg'(z)[/itex], which implies [itex]g'(z) = \frac{1}{z}[/itex] on [itex]\mathbb C^*[/itex]. But now by Cauchy's integral theorem, we get
[tex]
0 = \oint _{|z|=1} g'(z) dz = \oint_{|z|=1} \frac{1}{z} dz = 2i\pi.
[/tex]
Thus, such a [itex]g[/itex] cannot exist.
Maybe I'm hunting a fly with a 50 cal rifle with armor-piercing rounds..
 
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  • #33
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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.
 
  • #34
benorin
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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.
 
  • #35
TeethWhitener
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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
$$g(x)=\begin{cases}
x &\text{if } 0\leq x < \frac{1}{2} \text{ or }\frac{1}{2} \leq x \leq 2\\
\frac{1}{2}x+\frac{1}{4}& \text{if } \frac{1}{2} \leq x < \frac{7}{6} \\
2x-\frac{3}{2} & \text{if } \frac{7}{6} \leq x < \frac{3}{2} \\
\end{cases}
$$
 
  • #36
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Let [itex]K\subseteq \mathbb C[/itex] be non-empty compact. Suppose [itex]\{\gamma_n \in K \mid n\in\mathbb N\} \subseteq K[/itex] is dense in [itex]K\ \ [/itex] ([itex]\mathbb C[/itex] is separable). Let [itex]D : e_n\mapsto \gamma_ne_n[/itex] be a diagonal operator (in the infinite case [itex]H\cong \mathbb C^\omega[/itex], otherwise [itex]H\cong \mathbb C^N[/itex]).

For any [itex]n\in\mathbb N[/itex] we have [itex]\gamma_n\in\sigma (D)[/itex], because [itex]D-\gamma_n\mathrm{id}[/itex] is not injective. If we look at the kernel, we have
[tex]
\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.
[/tex]
We can clearly see [itex]\lambda _k \equiv 0[/itex] is not forced.

Since [itex]\sigma (D)[/itex] is closed, we also have [itex]K\subseteq \sigma (D)[/itex]. On the other hand, suppose [itex]\lambda\notin K[/itex]. Then
[tex]F \left ( \sum \lambda _ke_k\right ) := \sum\frac{\lambda _k}{\gamma _k-\lambda}e_k[/tex]
is the inverse of [itex]D-\lambda \mathrm{id}[/itex], thus [itex]\lambda\notin \sigma (D)[/itex]. Indeed, we can check for basis elements. We have
[tex]
(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
[/tex]
and
[tex]
((D-\lambda \mathrm{id}) \circ F)(e_k) = (D-\lambda \mathrm{id}) \left ( \frac{1}{\gamma _k-\lambda}e_k \right ) = e_k.
[/tex]
Altogether [itex]\sigma (D) = K[/itex].
 
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  • #37
Math_QED
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Let [itex]K\subseteq \mathbb C[/itex] be non-empty compact. Suppose [itex]\{\gamma_n \in K \mid n\in\mathbb N\} \subseteq K[/itex] is dense in [itex]K\ \ [/itex] ([itex]\mathbb C[/itex] is separable). Let [itex]D : e_n\mapsto \gamma_ne_n[/itex] be a diagonal operator (in the infinite case [itex]H\cong \mathbb C^\omega[/itex], otherwise [itex]H\cong \mathbb C^N[/itex]).

For any [itex]n\in\mathbb N[/itex] we have [itex]\gamma_n\in\sigma (D)[/itex], because [itex]D-\gamma_n\mathrm{id}[/itex] is not injective. If we look at the kernel, we have
[tex]
\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.
[/tex]
We can clearly see [itex]\lambda _k \equiv 0[/itex] is not forced.

Since [itex]\sigma (D)[/itex] is closed, we also have [itex]K\subseteq \sigma (D)[/itex]. On the other hand, suppose [itex]\lambda\notin K[/itex]. Then
[tex]F \left ( \sum \lambda _ke_k\right ) := \frac{\lambda _k}{\gamma _k-\lambda}e_k[/tex]
is the inverse of [itex]D-\lambda \mathrm{id}[/itex], thus [itex]\lambda\notin \sigma (D)[/itex]. Indeed, we can check for basis elements. We have
[tex]
(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
[/tex]
and
[tex]
((D-\lambda \mathrm{id}) \circ F)(e_k) = (D-\lambda \mathrm{id}) \left ( \frac{1}{\gamma _k-\lambda}e_k \right ) = e_k.
[/tex]
Altogether [itex]\sigma (D) = K[/itex].
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.
 
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  • #38
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@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.
 
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  • #39
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Suppose [itex]V/W[/itex] is complete. Let [itex]x_n\in V,\ n\in\mathbb N,[/itex] be a Cauchy sequence. Let [itex]\pi :V\to V/W[/itex] be the projection, then [itex]\pi (x_n),\ n\in\mathbb N,[/itex] is Cauchy in [itex]V/W[/itex]. For some [itex]x\in V[/itex] we have [itex]\pi (x_n) \xrightarrow[]{}\pi (x)[/itex]. By definition of infimum pick [itex]z_n\in V[/itex] such that [itex]\|x_n-x-z_n\| \leq \|\pi (x_n-x)\| + \frac{1}{n},\ n\in\mathbb N.[/itex] Then [itex]\|x_n-(x+z_n)\| \to 0[/itex]. So we would have [itex]x_n \to x+\lim z_n[/itex]. This is justified, because by triangle equality
[tex]
\|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.
[/tex]
 
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  • #40
TeethWhitener
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@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.
 
  • #41
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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 [itex]f,g[/itex] on [itex][0,2][/itex] we are to show such [itex]a\neq b[/itex] exist. I'm thinking Intermediate value theorem might help. Not sure, right now.
 
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  • #42
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For every [itex]x\in\mathbb R[/itex] we have [itex]\mathbb P\{X\leq x\}\in \{0,1\}[/itex]. Let [itex]F[/itex] be the distribution function for [itex]X[/itex]. [itex]F[/itex] is right continuous and we have [itex]\lim _{x\to\infty} F(x) = 1[/itex] and [itex]\lim _{x\to -\infty}F(x) = 0[/itex]. This implies there exists [itex]c\in\mathbb R[/itex] such that [itex]F = I_{[c,\infty]}[/itex]. Now [itex]\mathbb P\{X<c\} = F(c-) = 0[/itex], therefore [itex]X=c[/itex] a.s.
We have [itex]\mathbb P\{X\leq x\} \in\{0,1\}[/itex] for every [itex]x\in\mathbb R[/itex]. One readily verifies [itex]X[/itex] is independent of itself. Put [itex]c:= \inf \{x\in\mathbb R \mid X\leq x\}[/itex]. It is finite, because [itex]\lim _{x\to \infty} \mathbb P\{X\leq x\} = 1[/itex], therefore for sufficiently large [itex]x_0[/itex] we have [itex]\mathbb P\{X\leq x_0\}=1[/itex]. By Kolmogorov's 0-1 law [itex]\mathbb P\{X=c\}=1[/itex].
Since [itex]X[/itex] is almost surely constant, the probability it takes any other value is zero, hence its variance is zero. Therefore [itex]\mathbb E(X-\mathbb E(X))^2 = 0[/itex] implies [itex]X=\mathbb E(X)[/itex] a.s. The first moment exists, because variance is finite.


.. by triangle equality
Triangle equality 😂 I meant triangle inequality.
 
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  • #43
benorin
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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##.
 
  • #44
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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 continous 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.
 
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  • #45
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Oh brother, this simply connected business for [itex]\mathbb C^*[/itex] is simply not necessary. Let's go again.
Suppose [itex]g:\mathbb C^* \to \mathbb C[/itex] is continuous with [itex]e^{g(z)} \equiv z[/itex]. Then [itex]g[/itex] is holomorphic, because the exponential map is holomorphic and
[tex]
\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)}}.
[/tex]
By differentiating we get [itex]1 = g'(z) e^{g(z)} = zg'(z)[/itex], which implies [itex]g'(z) = \frac{1}{z}[/itex] on [itex]\mathbb C^*[/itex]. But now by Cauchy's integral theorem, we get
[tex]
0 = \oint _{|z|=1} g'(z) dz = \oint_{|z|=1} \frac{1}{z} dz = 2i\pi.
[/tex]
Thus, such a [itex]g[/itex] 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.
Suppose [itex]g:\mathbb C^* \to\mathbb C[/itex] is continuous and satisfies the identity [itex]e^{g(z)} = z[/itex]. Since [itex]g[/itex] is right inverse to [itex]\exp[/itex], it is injective, thus [itex]g(\mathbb C^*) \cong \mathbb C^*[/itex]. But now [itex]\exp[/itex] is forced to be injective on [itex]\mathbb C^*[/itex]. Indeed, for any [itex]u,v\neq 0[/itex] we have
[tex]e^u = e^v \Leftrightarrow e^{g(z)} = e^{g(w)} \Leftrightarrow z=w \Rightarrow u=v.[/tex]
But that's impossible.
 
  • #46
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@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)##?
 
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  • #47
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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.
 
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@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, [itex]g[/itex] is injective, therefore it's an isomorphism onto its image.
 
  • #49
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Continuity is implicit.
Which part of your argument fails if ##g## is not required to be continuous?

No, [itex]g[/itex] is injective, therefore it's an isomorphism onto its image.
Please explain this line.

[tex]e^u = e^v \Leftrightarrow e^{g(z)} = e^{g(w)} \Leftrightarrow z=w \Rightarrow u=v.[/tex]
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)##)
 
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  • #50
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Continuity is implicit. I'm working in the category with continuous maps. No, [itex]g[/itex] 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?
 

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