the problem of extending the Lebesgue integral to more general functions

My question originated from here, and since my reputation is not high enough, I have opened a new question.

The author mentioned that this example obstructs the desire to have both the Fubini theorem and the Newton-Leibniz theorem hold simultaneously when extending the Lebesgue integral.

Such a counterexample would be an obstruction to the extension of the Lebesgue integral to more general functions, such that the Fubini theorem and the Newton-Leibniz theorem both remain true.

I would like to ask for a detailed explanation and related references on this topic.

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Why is the exponent $p^+$ used to bound $|u_m|^{p(x)-2}u_m$ in variable exponent PDEs?

I am reading a paper dealing with a parabolic PDE with a variable exponent nonlinearity and the Faedo--Galerkin method. The nonlinear term is $|u_m|^{p(x)-2}u_m$.

In the proof, the authors estimate

$$ \int_{\Omega}\Big||u_m|^{p(x)-2}u_m\Big|^{\frac{p^+}{p^+-1}}dx $$

and split the domain into two parts

$$ \Omega_1=\{x\in\Omega:|u_m(x,t)|\le 1\}, \qquad \Omega_2=\{x\in\Omega:|u_m(x,t)|\ge 1\}. $$

Then they obtain a bound

$$ \int_{\Omega}\Big||u_m|^{p(x)-2}u_m\Big|^{\frac{p^+}{p^+-1}}dx \le C. $$

From this estimate they deduce that $|u_m|^{p(x)-2}u_m$ is bounded in

$$ L^\infty(0,T;L^{(p^+)'}(\Omega)). $$

Therefore there exists a function $\chi$ such that

$$ |u_m|^{p(x)-2}u_m \rightharpoonup \chi \quad \text{in } L^\infty(0,T;L^{(p^+)'}(\Omega)) \text{ weakly*}, $$

and

$$ |u_m|^{p(x)-2}u_m \rightharpoonup \chi \quad \text{in } L^2(0,T;L^{(p^+)'}(\Omega)) \text{ weakly}. $$

My question is:

Why do the authors choose the exponent $p^+$ (the maximal value of $p(x)$) in this estimate, instead of $p(x)$ or $p^-$?

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Chain rule for distributional derivative

Let $V \subset H \subset V^*$ be a Gelfand triple (eg. $H^1 \subset L^2 \subset H^{-1}$).

Let $u \in L^2(0,T;V)$ have a distributional derivative $u' \in L^2(0,T;V^*)$. So $\int_0^T u(t)\varphi'(t) = -\int_0^T u'(t)\varphi(t)$ holds for all $\varphi \in C_c^\infty(0,T)$ as an equality in $V^*$.

Let $f:\mathbb{R} \to \mathbb{R}$ be a differentiable function with $f'$ bounded.

Suppose $u' \in L^2(0,T;L^2)$; then the chain rule formula for the weak derivative $(f(u))' = f'(u)u' \in L^2(0,T;L^2)$ makes sense.

But if $u' \in L^2(0,T;H^{-1})$ only, how to make sense of $(f(u))'$? Is it right to define the derivative as $$\langle (f(u))', v \rangle := \langle u', f'(u)v \rangle$$ if for all $v \in L^2(0,T;H^1)$, $f'(u)v \in L^2(0,T;H^1)$?

Where may I find more details about this sort of thing? I'd like to avoid BV spaces and measures because this is simpler. Thank you.

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Nonharmonic solutions of Laplace's equation

Let $f \colon U \to \mathbb{R}$ be a twice differentiable function, where $U$ is an open subset of $\mathbb{R}^n$. Here twice differentiable means that all the second partial derivatives $\frac{\partial}{\partial x_i} (\frac{\partial}{\partial x_j} f)$ exist (however they are not necessarily continuous). Suppose $\Delta f = 0$, i.e. $\sum_{i=1}^n \frac{\partial^2}{\partial x_i^2} f = 0$. Does this imply that $f$ is harmonic, i.e. that $f$ is twice continuously differentiable? (For $n = 1$ it does, so let's assume that $n \ge 2$.)

Remark: I have read somewhere that if $f$ is weakly harmonic, then it is harmonic. However I think the second derivatives of $f$ here are not necessarily the same as the weak second derivatives. So my guess is that there is a counterexample, but I was not able to find one. Such a counterexample would be an obstruction to the extension of the Lebesgue integral to more general functions, such that the Fubini theorem and the Newton-Leibniz theorem both remain true.

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Authorship in Mathematics

In a mathematical research scenario, you are a junior or a student who is given a project to work on and is given some advice and ideas on it to work with. For 2-3 months you work on it and give some improvements of the known results from ideas of your supervisor and sometimes yourself. Then, eventually, after 0.5-1.5 more months, your advisor told you some more ideas to work on, and it completely surpassed what you had done, and so the final result is now stronger and doesn't need your input at all.

How do you determine the authorship in this case? Should you retract yourself from the project and as an author of that? Or is it still appropriate for you to be included as a co-author at all because the final result doesn't contain any of your significant contribution?

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A weighted sum over squarefree numbers involving Bernoulli numbers: bounded or divergent?

This may feel like a somewhat arbitrary series, but I am trying to understand the growth of the Bernoulli numbers as they appear in a summation I am working with. Let $S_X$ be the set of square free numbers less than $X$ and $\omega(n)$ be the number of distinct prime factors in $n$. Let,

\begin{equation} F(X) = \sum_{n \in S_X} \frac{B_{\omega(n)}}{n} \end{equation}

where $B_m$ is the $m$th Bernoulli Number (let $B_1 = +\frac{1}{2})$.

Question: Is $F(X)$ bounded by constants $F(X) \in [c_1, c_2]$ as $X \to \infty$? If so, are there ways of estimating these constants?

Attempt/Intuition: Below is a plot of $F(X)$ for $X<10^8$ and some of my notes. The reason I suspect there is some boundedness is that $\frac{B_{\omega(n)}}{n}$ decays very rapidly, and $\frac{B_{\omega(n)}}{n}$ oscillates with the Bernoulli numbers.

For squarefree $n$ with $\omega(n)=m$, we have

\begin{equation} B_m \sim (-1)^{m/2+1}\frac{2m!}{(2\pi)^m}. \end{equation}

Also, among squarefree integers with $\omega(n)=m$, the smallest possible $n$ is the primorial

\begin{equation} p_m^\#:=\prod_{j\le m} p_j, \end{equation}

so in general

\begin{equation} n\ge p_m^\#. \end{equation}

Hence

\begin{equation} \left|\frac{B_{\omega(n)}}{n}\right| \le \frac{|B_m|}{p_m^\#} \sim \frac{2m!}{(2\pi)^m\,p_m^\#}. \end{equation}

Now use Stirling's formula

\begin{equation} m!\sim \sqrt{2\pi m}\left(\frac{m}{e}\right)^m \end{equation}

and the prime number theorem in the form

\begin{equation} \log p_m^\#=\vartheta(p_m)\sim p_m\sim m\log m, \end{equation}

so

\begin{equation} p_m^\#=\exp\bigl((1+o(1))m\log m\bigr)=m^{\,m+o(m)}. \end{equation}

Therefore

\begin{equation} \frac{2m!}{(2\pi)^m\,p_m^\#} \sim 2\sqrt{2\pi m}\, \frac{(m/e)^m}{(2\pi)^m\,m^{m+o(m)}} = 2\sqrt{2\pi m}\,(2\pi e)^{-m}m^{-o(m)}. \end{equation}

So, along the extremal case $n=p_m^\#$,

\begin{equation} \left|\frac{B_{\omega(n)}}{n}\right| = \left|\frac{B_m}{p_m^\#}\right| = \exp\bigl(-(1+o(1))\,m\log\log m\bigr), \end{equation}

more precisely

\begin{equation} \left|\frac{B_m}{p_m^\#}\right| \asymp \sqrt{m}\,(2\pi\log m)^{-m}. \end{equation}

Thus $B_{\omega(n)}/n$ decays extremely rapidly as $\omega(n)\to\infty$, and its largest possible order for fixed $\omega(n)=m$ occurs at the primorial.

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Divide angles by coefficients relate to Fibonacci sequence

• In the left Figure, consider a right triangle $OPA$ with $\angle {AOP} = 90^\circ$. Let $\ell$ be the reflection of $PO$ in $PA$ and $\ell$ meets $OA$ at $A_1$. Let $O_1$ be the center of the circle $(PAA_1)$, the line $PO_1$ meets $OA$ at $A_2$. Let $O_2$ be the center of the circle $(PA_1A_2)$, the line $PO_2$ meets $OA$ at $A_3$,.....Let $O_n$ be the center of the circle $(PA_{n-1}A_{n})$, the line $PO_n$ meets $OA$ at $A_{n+1}$ for $n=3, 4, 5,\ldots$. Denote $\angle {OPA} = \angle {APA_1} = \alpha_1$, $ \angle A_i PA_{i+1}=\alpha_{i+1}$ for $i=\overline{1,n}$

• Easily to show that $\alpha_2=\alpha_1$ and $\alpha_{n+1}=\alpha_{n}+\alpha_{n-1}$ for $n=2,3,4,...$ this is Fibonacci sequence.

• Now in Cartesian coordinates, let $P=(0,1)$, $O=(0,0)$, $A=(x,0)$. What is the locus equation of some circumcenters $O_1$, $O_2$, $O_3$...when we move $A$ on $Ox$? If we can find these locus equations, we can divide any angles by coefficients relate to Fibonacci sequence.

• Example: In right Figure, let $AOB$ be a right triangle with $A(0,1)$, $O(0,0)$, $B(x,0)$. The locus of $O_1$ (blue curve) meets $AB$ at $O_1$, the circle $(O_1, O_1B)$ meets $OB$ at $A_1, A_2$ then $\angle OAA_1 = \angle A_1AA_2 = \angle A_2AB = \frac{\angle OAB}{3}$.

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Two-parameter “$\varepsilon$-$\delta$ filtration” given a function between metric spaces

Let $X,Y$ be metric space and $f : X \to Y$ a (not necessarily continuous) function. I'm interested in the two-parameter filtration $(X_{\varepsilon, \delta})_{{\varepsilon, \delta} > 0}$ where $X_{{\varepsilon, \delta}}$ is the clique complex of the graph $(X,V)$ such that that $\left\{x,y\right\} \in V$ iff $d(f(x), f(y)) < \varepsilon$ and $d(x,y) < \delta$.

The motivation is the following: if $X = [m] \times [n]$ is a discrete grid of pixels and $Y = [0,1]$, so that $f$ represents a greyscale image, then the persistent homology of $X_{\bullet\ \bullet}$ computes a topological segmentation of the image that is robust against gradients and shading.

I would be grateful for any references to previous work on this idea or other ideas in the same vein.

Many thanks.

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Square of the integral of a positive definite function

I imagine the answer is out there, but search engines are just miserable. If someone can point to a link or article that answers this I'd be more than happy.

I have an unknown function $g(t)$ that is being integrated. The function is positive definite over its entire domain. I will eventually be applying numerical methods, but for now I'm trying to simplify a larger expression and I get the term

$$\left[ \int_{t_0}^t g(\tau)d\tau \right] \cdot \left[ \int_{t_0}^t g(\tau)d\tau \right]=\left[ \int_{t_0}^t g(\tau)d\tau \right]^2$$

I want to reduce the integral such that the square disappears and can be represented as a repeated integral, and allow me to use the order reduction formula

$$\int_{t_0}^t f(\tau) d\tau^n=\int_{t_0}^t \frac{(t-\tau)^{n-1}}{(n-1)!}f(\tau)d\tau$$

The classic strategy, for example $e^{-x^2}$ over the reals, is done by multiplying by the same integral with a different variable and converting to polar coordinates. Since I don't know g, this isn't an option as far as I'm aware. However, it seems like I would be allowed to say $$\left[ \int_{t_0}^t g(\tau)d\tau \right]^2=\int_{t_0}^t \int_{t_0}^q g(\tau) g(q') d\tau dq'$$

It would seem since g is the same, the interval is the same, I should just be able to say $$\int_{t_0}^{t} g^2(\tau)d\tau^2$$

However this clearly isn't correct. I even verified it myself using the gaussian integral over a finite domain. Wolfram alpha shows

$$\left[ \int_{0}^{10} e^{-x^2}dx\right]^2 \approx 0.785398...$$

If I nievely think it just becomes the square of the function integrated twice, using the order reduction formula gives.

$$ \int_{0}^{10} (10-x)e^{-2x^2}dx \approx 6.01657...$$

So is there much of anything I can do about my integral since its function is unknown? This is specifically time integration. So I don't even know if attempting the polar coordinate integration would work or even make physical sense.

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Cohomological bounds for scalar curvature of an extremal Kähler metric

There is an interesting trick used in Chen-LeBrun-Weber's paper on the extremal Kähler metrics of $\mathbb{CP}^2\#2\overline{\mathbb{CP}^2}$, and I would like to know whether it can be (has been?) exploited further.

The trick applies to certain Kähler classes $\mathfrak{k}$ of a complex manifold $M$, and allows one to give upper and lower bounds for the scalar curvature of an extremal Kähler metric in $\mathfrak{k}$, if such a metric exists. This is important later in the paper -- in dimension 2 it allows one to control the Sobolev constant of such a metric, and thus to control the degenerations of sequences of such metrics.

The trick appears at the end of Section 3. I will describe it in a more general setting than Chen-LeBrun-Weber's (hopefully not making any mistakes in this generalization). Let $M$ be a complex manifold, fix a maximal connected compact subgroup $K$ of the group of automorphisms of $M$, and let $\mathfrak{k}$ be a Kähler class of $M$. According to Futaki-Mabuchi there is a canonically determined holomorphic vector field $\Xi_\mathfrak{k}$ which is the extremal Kähler vector field of a $K$-invariant extremal Kähler metric $\omega\in\mathfrak{k}$, if any exists. Suppose that $\mathfrak{k}$ satisifes

Important Property. The holomorphic vector field $\Xi_\mathfrak{k}$ generates a closed 1-parameter subgroup ($\cong S^1$) of automorphisms of $M$.

(Remark: in Chen-LeBrun-Weber's setting, all bilaterally symmetric Kähler classes of $\mathbb{CP}^2\#2\overline{\mathbb{CP}^2}$ have this property.)

Then the range (max minus min) of the scalar curvature of $\omega$ is equal to $\lambda \ \mathfrak{k}\cdot [F]/4\pi$, where $F$ is a rational curve in $M$ which is the closure of a ``generic'' orbit of the group ($\cong\mathbb{C}^*$) of automorphisms generated by $\Xi_\mathfrak{k}$ and its complexification, and where the real number $\lambda$ is such that $\lambda^{-1}\Xi_\mathfrak{k}$ is the standard generator of the $S^1$.

Since the average of the scalar curvature of $\omega$ is cohomologically determined by $\mathfrak{k}$, this then gives upper and lower bounds on the scalar curvature of $\omega$.

Here are my questions (assuming that my slight generalization of Chen-LeBrun-Weber's statement is correct).

• First, is there any general way to determine which Kähler classes $\mathfrak{k}$ satisfy the Important Property?
• Second, if not, what are the known examples of Kähler classes with the Important Property? (The examples I can think of are (1) Chen-LeBrun-Weber's; (2) all Kähler classes on a manifold for which $K=S^1$.)
• And thirdly, in any known examples (other than Chen-LeBrun-Weber's) with the Important Property, has it been computed/can someone compute the explicit bounds on scalar curvature that are obtained?

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Solving SDE with sign function in drift term?

Consider the following SDE with $X_0 = 1$, $$ dX_t = X_t\operatorname{sign}(X_t) \, dt + X_t \, dW_t, $$ where $\operatorname{sign}(x) = \mathbb{1}\{x \ge 0\}$. How am I supposed to solve this SDE?

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Cluster algebras of type A and X

I will base my question on Fock and Goncharov's paper Dual Teichmüller and lamination spaces.

Let $S$ be a surface with boundaries, marked points on such boundaries, punctures and boundaries without marked points. We collectively call punctures and unmarked boundaries holes.

To $S$ we can associate two Teichmüller spaces, let me call them $T_A(S)$ and $T_X(S)$. The first is the decorated Teichmüller space, where above each puncture and marked point is made a choice of a horocycle. In this case all holes must be punctures. The second is the usual Teichmüller space, and we now allow un-marked boundaries.

We define a triangulation $T$ of $S$ to be a triangulation of the surface obtained by replacing each hole with a puncture.

Associated to a triangulation $T$ of $S$ there are coordinate systems for both $T_A(S)$ and $T_X(S)$. They are positive numbers associated to the edges $e$ of $T$ and to the non-boundary edges of $T$ respectively. They are given by $\lambda$-lengths for $T_A(S)$ (indicated by $\lambda_e$) and by shear coordinates for $T_X(S)$ (indicated by $x_e$).

There is a map $T_A(S) \to T_X(S)$, which I will call the shear map, which is given in coordinates by $x_e = \frac{\lambda_a \lambda_c}{\lambda_b \lambda_d}$, for each inner edge $e$ belonging to triangles $[e,a,b]$ and $[e,c,d]$ of $T$.

This map lands in the subspace $\prod_{e \in p} x_e = 1$, where the product is taken over all edges $e$ incident to a hole $p$. This is a subspace of $T_X(S)$ because in general this product gives the exponential of the hyperbolic length of the hole.

QUESTION

Is it possible to define a "natural" map $T_A(S) \to T_X(S)$ that doesn't land in this subspace? By "natural" I mean that it does not depend on any choice of triangulation.

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Abelian p groups counting

Suppose $G, K$ are subgroups of $\mathbb{Z}_p^n$. Consider the Haar measure on $M_n(\mathbb{Z}_p)$, then I want to consider the following question:

Given a Haar-random matrix $M\in M_n(\mathbb{Z}_p)$ and fix a partition $\mu\in \mathbf{Par}_{n\times \infty}$, what is the probability that $\mathrm{im}(M)+G=\mathbb{Z}_p^n$, $\mathrm{cok}(M)$ has type $\mu$ and $\mathrm{im}(M)\supseteq K$?

Ideally, I hope that this answer depends only on the type and cotype of $\mathrm{im}(M)$ along the quotient $\mathbb{Z}_p^n\twoheadrightarrow \mathbb{Z}_p^n/K$.

1. If $K=0$, then this question resolves into the classical Cohen Lenstra Measure result about the cokernel of a random matrix in $M_n(\mathbb{Z}_p)$.

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How can I find the defender movement angle that minimizes the quarterback’s safe region in a rectangle?

I am studying a geometric optimization problem in the plane.

Let $R=[0,W]\times[0,H]$ be the rectangle with corners $(0,0)$, $(W,0)$, $(W,H)$, and $(0,H)$.

Let the quarterback and defender start at $q=(x_q,y_q)$ and $d=(x_d,y_d)$.

Assume they have the same speed.

The defender chooses a direction $\psi\in[0,2\pi)$ and moves a fixed distance $s>0$, so the defender’s new position is

$$ d'(\psi)=(x_d+s\cos\psi,\; y_d+s\sin\psi). $$

For each angle $\psi$, define the quarterback’s safe region to be the set of points in the rectangle that are at least as close to the quarterback as to the defender’s new position:

$$ S(\psi)=\{(x,y)\in R:\|(x,y)-q\|\le \|(x,y)-d'(\psi)\|\}. $$

In other words, $S(\psi)$ is the set of destinations the quarterback can reach no later than the defender after the defender moves in direction $\psi$.

Because the speeds are equal, $S(\psi)$ is the part of the rectangle on the quarterback’s side of the perpendicular bisector of $q$ and $d'(\psi)$.

The defender’s goal is to choose $\psi$ so that this safe region is as small as possible.

So if

$$ A(\psi)=\operatorname{Area}(S(\psi)), $$

then the optimization problem is

$$ \psi^*=\arg\min_\psi A(\psi). $$

My question is whether there is a standard exact or piecewise-analytic method to determine $\psi^*$ without brute-force angular sampling.

In particular, is there a standard way to:

1. describe $A(\psi)$ piecewise according to how the perpendicular bisector intersects the rectangle, and
2. solve for the minimizing angle exactly, or at least reduce the problem to finitely many analytic cases?

I am mainly interested in an exact or piecewise-analytic approach, not a purely numerical approximation, unless no exact method is known.

A schematic example is shown below: image.

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Is non-symmetric monoidal $\infty$-category $\mathcal{V}$ a monoid in the $2$-category of $\mathcal{V}$-bienriched categories?

Let $\mathcal{V}$ be a presentably monoidal $\infty$-category. If $\mathcal V$ is moreover symmetric, the $(\infty,2)$-category ${}_{\mathcal V} \mathsf{Cat}$ of $\mathcal V$-enriched categories inherits a monoidal structure and $\mathcal{V}$ is naturally a monoid in ${}_{\mathcal V} \mathsf{Cat}$. This does no longer hold once we drop the symmetry.

My question: can we make up for this deficiency by considering instead the $2$-category ${}_{\mathcal V} \mathsf{Cat}_{\mathcal V}$ of $\mathcal{V}$-bienriched categories (in the sense of Gepner and Heine)? It seems to me that this $2$-category should inherit a monoidal structure once again in such a way that $\mathcal V$ becomes a monoid therein, but I cannot find a reference in the literature. I am mostly interested in the case where $\mathcal{V}$ is $\mathsf{Cat}_{(\infty,2)}$ with the (lax) Gray tensor product.

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Verifying the action of the Collatz function on a partition of odd integers

I am studying a partition of the odd positive integers into three disjoint sets (B, C, D) in the context of the Collatz conjecture. The sets are defined as follows:

\begin{aligned} B &:= \left\{ \frac{4^a - 1}{3} + 2 \cdot 4^a \cdot n \;\Big|\; a \in \mathbb{N},\; n \in \mathbb{N}_0 \right\}, \\[4pt] C &:= \left\{ 3 + 4n \mid n \in \mathbb{N}_0 \right\}, \\[4pt] D &:= \left\{ \frac{10 \cdot 4^a - 1}{3} + 4^{a+1} \cdot n \;\Big|\; a \in \mathbb{N},\; n \in \mathbb{N}_0 \right\}. \end{aligned}

I have computed the action of the Collatz function

f(x) = \begin{cases} x/2, & \text{if } x \text{ is even},\\[4pt] 3x+1, & \text{if } x \text{ is odd} \end{cases}

on these sets and obtained the following results:

• For any b ϵ B with parameters (a) and (n),

  f(b) = 1 + 6n.

• For any c ϵ C with (c = 3+4n),

  f(c) = 5 + 6n.

• For any d ϵ D with parameters (a) and (n),

  f(d) = 5 + 6n.

My derivations are elementary, but I would like to be absolutely certain that there is no mistake, especially concerning the exact power of two removed in the (3x+1) step. I have checked many examples numerically, but a formal verification would be reassuring.

Questions:

1. Are the formulas above correct for all elements of the respective sets?
2. Could there be any hidden counterexample where the function behaves differently?
3. Does this partition help in understanding why numbers in C increase under ⨍ while those in B and D decrease? (For B and D, the value clearly drops because (1+6n) is much smaller than the original numbers; for C, (5+6n > 3+4n) for all (n).

Any comments, references, or insights would be greatly appreciated.

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Does Bayes theorem imply that if $P_{X|\Theta}$ is a regular cond. distr., that then also $P_{\Theta|X}$ is a regular cond.distr.?

In the book Theory of Statistics, Schervish states in Theorem 1.31 the Bayes theorem very rigorously. To this end let $(\Omega, \mathcal{A}, P)$ be an underlying probability space, let $\Theta:(\Omega, \mathcal{A})\rightarrow (\vartheta, \mathcal{F})$ and $X:(\Omega, \mathcal{A})\rightarrow (\mathcal{X}, \mathcal{G})$ be random variables taking values in two (standard) Borel spaces. Then for $P_{X|\Theta}\ll\nu$ for all $\theta\in\vartheta$, Bayes theorem is given by

$$\frac{dP_{\Theta|X}}{d P_{\Theta}}(\theta|x)=\frac{\frac{P_{X|\Theta}}{d\nu}(x|\theta)}{\int_{\vartheta} \frac{P_{X|\Theta}}{d\nu}(x|t)P_{\Theta}(dt)} $$

Now in a footnote, Schervish writes: "In fact the proof applies even if $(\vartheta, \mathcal{F})$ is not a [standard] Borel space. In this last case, a regular conditional distribution is explicitely constructed without knowing in advance that one will exist."

So of course in the setting that both $(\mathcal{X}, \mathcal{G})$ and $(\vartheta, \mathcal{F})$ are Borel spaces we can always pick a regular version of the corresponding conditional distributions, i.e., we can assume without loss of generality that $P_{X|\Theta}$ and $P_{\Theta|X}$ are regular conditional distributions. Now if $(\vartheta, \mathcal{F})$ is not assumed to be standard Borel then we can not assume that $P_{\Theta|X}$ will be a regular conditional distribution. His footnote, however, seems to imply that via Bayes theorem, we construct a regular version of $P_{\Theta|X}$. Therefore, my question is: Is it correct that it suffices that we have that $P_{X|\Theta}$ is a regular conditional distribution and we can then apply Bayes theorem and obtain a regular conditional distribution $P_{\Theta|X}$? And if so, is it also possible the other way round as well: I.e, that only $P_{\Theta|X}$ is a rcd by assumption and via the above form of Bayes theorem we would obtain that $P_{X|\Theta}$ is a rcd as well?

I had previously posted this question on mathstackexchange where I didnt receive any answers nor comments. Thus I deleted it there to ask it here.

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How to draw knots with LaTeX?

I am writing an exam for my students, and the topic is intro knots theory. I have no idea how to put knots into the file, but I know many MO users who can draw amazing diagrams in their papers.

Can someone please provide some hints on what can be used, preferably with some example codes? I do not need complicated diagrams, just some simple knots and links with few crossings. Thanks in advance.

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The first comprehensive and conclusive proof of the Collatz Conjecture [closed]

The attached historic paper contains a conclusive proof of the Collatz Conjecture through the discovery of a pattern in the behavior of odd numbers under the Collatz function. These numbers are then partitioned into three distinct sets, B, C, and D.

A detailed study is carried out on the crucial set V, which consists of numbers congruent to 5 modulo 12, and it is shown that this set satisfies the relation . This relation serves as the launching point for a simple yet decisive mathematical induction, demonstrating that all odd numbers (except multiples of 3) converge to 1.

The result is then easily generalized to multiples of 3 and to even numbers.https://vixra.org/pdf/2510.0073v5.pdf

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Multidimensional intersection property

Consider the multidimensional annulus $\{(p,\theta)\} = \mathbb R^n\times\mathbb T^n$ endowed by the $1$-form $\omega=p\,d\theta$. A diffeomorphism $A$ of this annulus onlo itself is said to be exact symplectic if the $1$-form $A^{\ast}\omega-\omega$ is exact, and the $n$-torus of the form $p=f(\theta)$ is said to be Lagrangian if the $1$-form $f(\theta)\,d\theta$ is closed.

The following very easy fact is well known: if a diffeomorphism A is exact symplectic and close to the multidimensional rotation by a varying angle: $(p,\theta)\mapsto(p,\theta+\partial S(p)/\partial p)$, then any Lagrangian torus $p=f(\theta)$ close to $p=\mathrm{const}$ intersects its $A$-image (in fact, more general statements hold).

For each $n\geqslant 2$ it is also very easy to construct an exact symplectic diffeomorphism $A:(p,\theta)\mapsto(p,\theta+\mathrm{const})$ (the multidimensional rotation by a constant angle) and a non-Lagrangian torus $p=f(\theta)$ that is arbitrarily close to $p=\mathrm{const}$ and does not intersect its $A$-image.

The question: are such examples known in the literature? If yes, I'd like to have a relevant reference.

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