We have been told that the Riesz potential in $\mathbb{R}^d$, $I_{\alpha}(f)$, defined by $$I_{\alpha}(f)(x):= C\int_{\mathbb{R}^d} \frac{f(y)}{\left| x-y \right|^{d-\alpha}}\,\mathrm{d}y $$ has the following regularizing property.

For a function $f\in C^{\beta}$ (the Hölder class), we have $I_{\alpha}(f) \in C^{\alpha+\beta}$, but we can neither prove it nor find a reference for this. Does it sound familiar to any of you? Do you know any reference?

Many thanks for considering my request.

Suppose that P is a Borel subset of Baire $\times$ Baire, such that for every pair $x,x'$ of reals in the horizontal copy of Baire,

**if:** $x,x'$ are $E_0$-equivalent (that is, $x(n)=y(n)$ for all but finite $n$)

**then:** the vertical cross-sections $P_x$ and $P_{x'}$ coincide.

Does this imply that $P_x=P_{x'}$ for all $x,x'$ in Baire (or at least for all $x,x'$ in a set $X$ large in the sense of measure or category) ?

I need the following result (which I believe to be true though I was too lazy to write down a complete proof).

Let $f$ be a function of two complex variables analytic at the origin and $a\not\in\mathbb{N}$. Then the differential equation $$z'=\frac{az}{x}+f(x,z)$$ has an unique solution $z(x)$ which is analytic at $x=0$.

Can anyone tell me a reference for this?

Let $(X,\omega)$ be a complete noncompact Kahler manifold of finite volume. Suppose $X$ is can be compactified to a compact projective manifold $M$ so that $D=M-X$ is a divisor of simple normal crossings. My question is:

Is the Kahler metric $\omega$ on $X$ extendable (in certain natural way) to the compactification $M$?

When such extensions exist, what can one say about such metric? For example, volume, curvature, etc.

Here are two somewhat strange sums using the shifted decimal forms of the powers of $3.$

$\begin{equation*}\begin{array}{ccccccc} &1&&&&&& \\ &&3&&&&& \\ &&&9&&&&\\ &&&2&7&&&\\ &&&&8&1&&\\ &&&&2&4&3&\\ &&&&&7&2&9\\ &-&-&-&-&-&-&-\\ &1&4&2&8&5&7&\cdots \end{array}\end{equation*}\ \ \ \ \ \ $ $\begin{equation*}\begin{array}{ccccccccc} &&&&&&&&1& \\ &&&&&&&3&& \\ &&&&&&9&&&\\ &&&&2&7&&&&\\ &&&8&1&&&&\\ &2&4&3&&&&&\\ 7&2&9&&&&&&\\ -&-&-&-&-&-&-&-&-\\ \cdots&2&4&1&3&7&9&3&1 \end{array}\end{equation*}$

The one on the left turns out to repeat the pattern $142857\ 142857\cdots$ if every power of $3$ is included. So putting a decimal point in front We get $\frac{1}{7}.$ This is easy to establish:

If $$x=\frac1{10}+\frac{3}{100}+\frac{9}{1000}+\cdots$$ then $$3x=\frac3{10}+\frac{9}{100}+\frac{27}{1000}+\cdots=10x-1$$ So $7x=1.$

The second sum shifts to the right. If all the powers of $3$ are used is there a periodic pattern and, if so, what does the repeating decimal with that pattern equal? I will give two somewhat unsatisfactory explanations why it is

$$\frac{10}{109}=0.\mathbf{0344827586206896551724137931}0344827586206896551724137931\cdots$$

This is a phenomenon that occurs for every integer sequence given by a linear recurrence relation. Most famously $$\frac1{89}=0.\mathbf{01123595505617977528089887640449438202247191}011235\cdots$$

and

$$\frac{10}{109}=0.\mathbf{091743119266055\cdots238532110}0917431\cdots$$

For the shifting off to the left we obtain the repeating period from left to right in a familiar order. On the right we work backwards from the "end."

Below are two calculations. What is a better way to explain what is going on in the first one? The second is a candidate, but not a pleasing one.

Approach 1: **10-adic integers**. The $2$-adic and $5$-adic integers are integral domains. Their direct sum is the $10$-adic integers (not an integral domain) which can be thought of as possibly infinite decimal integers $\sum_0^{\infty}a_i10^i$ with the $a_i\in \{0,1,2,\cdots,9\}.$ They are well understood. I'll skimp on justifications for the following. First I claim that $\cdots99999.=-1$ (proof: look what happens when you add $1$.) Of course $0.999\cdots=1$ so $\cdots9999.9999\cdots=0.$ Many infinite decimal integers have no rational value, however I claim that if $q$ is a rational with periodic decimal $q=0.abcdabcdabcd\cdots$ (length $4$ chosen for illustration) then the $10$-adic integer $\cdots abcdabcdabcd.$ is equal to $-q.$ To see this, multiply $\cdots abcdabcdabcd.abcdabcd\cdots$ by $\frac1q$ turning it into $\cdots9999.9999\cdots=0.$ Hence the integer part is the additive inverse of the fractional part. Now I have a clear way to start my period at the end. The $10$-adic integer $$y=\cdots 0344827586206896551724137931\mathbf{0344827586206896551724137931}.$$ satisfies $30y=y-1$ hence $y=\frac{-1}{29}$ So the thing I want is the additive inverse $\frac1{29}.$ That is a tidy calculation and ends up with the desired result. It could be more fully justified, but seems like the wrong way to go at it.

Approach 2:

It seems possible that the thing we want should be $z=\frac{1/3}{10}+\frac{1/9}{100}+\frac{1/27}{1000}+\cdots.$ And there $30z=1+z.$ So yes $z=\frac1{29}$ and the method worked. But is it justified? Why is it clear that adding $0.0\mathbf{3}33\cdots+0.00\mathbf{1}11\cdots+.000\mathbf{037}037\cdots +\cdots$ corresponds to the sifted sum on the right?

To use this approach and get $\frac{10}{109}$ for the right shifted Fibonacci decimals recall that the full sequence is $\cdots 13,-8,5,-3,2,-1,1\ | \ ,0,1,1,2,3,5,\cdots$ So

$$z=\frac{1}{10}+\frac{-1}{100}+\frac{2}{1000}+\frac{-3}{10000}+\cdots$$ satisfies $$10z=1+\frac{-1}{10}+\frac{2}{100}+\frac{-3}{1000}+\frac{5}{10000}+\cdots$$ and $$10z+z=1+\frac{0}{10}+\frac{1}{100}+\frac{-1}{1000}+\frac{2}{10000}+\cdots$$

Thus $11z=1+\frac{z}{10}$ and $110z=10+z.$

Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on a complex Hilbert space $F$.

Let $T,S\in\mathcal{B}(F)$. Assume that there exists $\lambda\in \mathbb{C}^*$ such that $TS=\lambda ST$. I want to establish a necessary and sufficient condition under which $\lambda=1$ i.e. $[T,S]:=TS-ST=0$.

If $TS\geq 0$ and $ST\geq 0$, then $\lambda\geq 0$. Moreover since $\|TS\|=\|ST\|$, then $\lambda=1$.

I try to do question I found, but I don't know how to do

Consider the ring homomorphism $\varphi : Z[x] \to R$ defined by $\varphi(x) = \sqrt{5}$. Let $I = \{f \in Z[x]\mid \varphi(f) = 0\}$.

First prove that $I$ is an ideal in $Z[x]$. Then find $g \in Z[x]$ such that $I = (g)$.

thanks u, i stuck

Is there a minimum $\beta^\star\in(0,1)$ such that

if $\beta\in[\beta^\star,1]$ then for every $N>0$ and for every choice of $a,b,c,d,m,n$ chosen with $a,b$ coprime perfect squares and $c,d$ coprime with $m,n\in(N/2,N)$ and $a,b,c,d\in(N,2N)$ and $\gcd(am+bn,cm+dn)=t$ a prime close to $N^{2\beta}$ we will always have $ad-bc$ is not invertible $\bmod t$

if $\beta\in(0,\beta^\star)$ then there is an $N_\beta>0$ such that for every $N>N_\beta$ there are always choices of $a,b,c,d,m,n$ chosen with $a,b$ coprime perfect squares and $c,d$ coprime with $m,n\in(N/2,N)$ and $a,b,c,d\in(N,2N)$ and $\gcd(am+bn,cm+dn)=t$ a prime close to $N^{2\beta}$ with $ad-bc$ invertible $\bmod t$?

Note if $am+bn=cm+dn$ a prime (case $\beta\rightarrow1$) then $ad-bc=0$ holds (degenerate case of dependent linear equations) and I feel $\beta^\star=\frac12$ is possible.

Fix a set of integer weights $h_1,\dots,h_t$ with $|h_i|\in(n,\alpha\cdot n)$ for $\alpha\in(1,2)$ with $t$ even and half of all weights positive and half negative. Choose a random vector of variables $a_i$ uniformly from integers in $(n^{\beta},2n^\beta)$ for a fixed $0<\beta<1$.

What is the distribution of magnitude of the sum of $h_ia_i$?

What is the probability that the sum has magnitude $<n^\gamma$ for some fixed $\gamma\in(0,1+\beta)$?

Is there a $\beta$ below which there exists a set of weights such that for almost all choices of $a_i$ magnitude of sum is $O(n^\beta)$?

Call a number *abnormal* if its decimal expansion doesn't feature every digit an infinite number of times. Call a triangle in ${\Bbb R}^2$ *abnormal* if at least one of its angles spans an abnormal fraction of $2\pi$ radians.

Assuming the Continuum Hypothesis, by the obvious transfinite induction, one gets a set $M$ so small that no three of its points form an abnormal triangle, but so large that every point of ${\Bbb R}^2$ either falls in $M$ or else occurs as the midpoint of two points in $M$.

Does such an $M$ exist in ZFC?

let $\Omega:=\left( a,b\right) \subset\mathbb{R}$, and suppose
$f:\Omega\rightarrow\left[ 0,\infty\right) $ is a bounded (Lebesgue)
measurable function, with $f\not \equiv 0$ almost everywhere. It is *not* true
in general that there exists some interval $I\subset\Omega$ such that
*essinf*$_{I}f>0$.

My question is: is the assertion true adding the hypothesis $|($supp $f)\setminus\Omega_{+}|=0$ ? Here, supp $f$ is the support of $f$, $\Omega_{+}$ is the largest open subset of $\Omega$ where $f>0$ almost everywhere, and $\left\vert Z\right\vert $ denotes the Lebesgue measure of $Z\subset\mathbb{R}$.

Thanks in advance.

Uriel

Let $G$ be a finite simple group of type ${\rm L}_{n}^{\epsilon}(q)$ with the following conditions:

$n$ and $\dfrac{q^{n}-\epsilon}{(q-\epsilon)(n,q-\epsilon)}$ both prime, $n\geqslant3$ and $(n,q,\epsilon)\neq(3,4,+),(3,3,-),(3,5,-),(5,2,-)$

Show that:

1- Every minimal subgroup $L$ of order $s=\dfrac{q^{n}-\epsilon}{(q-\epsilon)(n,q-\epsilon)}$ is contained properly in a maximal subgroup $M$ isomorphic to $\mathbb{Z}_{s}\rtimes\mathbb{Z}_{n}$

2- The maximal subgroup $M$ is the only proper subgroup of $G$ which contains properly $L$, i.e. every minimal subgroup of order $s$ has a unique overgroup.

As far as I checked the "ATLAS" the above propositions both are true.

An *overring* of an integral domain is a domain lying between it and its quotient field. Is it possible to have an overring $S$ of an integral domain $R$ such that $S$ has two maximal ideals $\mathfrak{m}$ and $\mathfrak{n}$ with $R \cap \mathfrak{m} \subsetneq R \cap \mathfrak{n}$? I think this should be possible but I can't come up with an example. Such an overring $S$ of $R$ cannot be flat or integral over $R$.

I feel like it should be straightforward to come up with an example or prove that there can't be such an example, but thus far all of my attempts have failed. I posted the question first on Math StackExchange but didn't get any replies.

Let A,B are two positive semidefinite matrixes, can we prove that $A(I+BA)^{-1}$ is positive semidefinite matrix?

Motivated by the quantisation of the symmetric laws in physics, the category of quadratic algebras has been endowed with two tensor products by Manin in his Montreal lectures notes. These products have been extended to the category of quadratic operads by Ginzburg and Kapranov. A long time ago in my master thesis, (I have started to work on this topic in June 1995 and I have defended my master's thesis in September 1995), I have defined the notion of quadratic category which is a category endowed with two tensor products, quadratic algebras and quadratic operads are examples of quadratic categories. More precisely:

A quadratic category $(C,\bullet, \circ)$ is a category $C$ endowed with two tensor products $\bullet$ and $\circ$ such that:

$I_{\circ}$ and $I_{\bullet}$ are the respective neutral elements of $(C,\circ)$ and $(C,\bullet)$.

We denote by $c^{\bullet}:(A\bullet B)\bullet C\rightarrow A\bullet (B\bullet C)$ the associative constraint of $\bullet$.

For every $A\in C$, there exists $A^!$ in $C$,

morphisms:

$b_A:I_{\bullet}\rightarrow A\circ A^!$

$d_A:A^!\bullet A\rightarrow I_{\circ}$

two natural morphisms:

$f^1_{A,B,C}:(A\circ B)\bullet C\rightarrow A\circ (B\bullet C)$

$f^2_{A,B,C}: A\bullet (B\circ C)\rightarrow (A\bullet B)\circ C$

which verifies:

$f^1_{A\bullet B,C,D}(f^2_{A,B,C}\bullet id_D)=f^2_{A,B,C\bullet D}(Id_A\bullet f^1_{B,C,D})c^{\bullet}_{A,B\circ C,D}$

$(Id_A\circ f^2_{B,C,D})f^1_{A,B,C\circ D}=c^{\circ}_{A,B\bullet C,D}(f^1_{A,B,C}\circ Id_D)f^2_{A\circ B,C,D}$

and

$(Id_A\circ d_A)f^1_{A,A^!,A}(b_A\bullet Id_A)=Id_A$

$(Id_{A^!}\circ d_A)f^2_{A^!,A,A^!}(Id_{A^!}\bullet b_A)=Id_{A^!}$.

We have the following result:

Theorem.

Let $C$ be a quadratic category and $B,D$ two objects of $C$, the functor $A\rightarrow Hom_C(A\bullet B,D)$ is representable by $D\circ B^!$.

Questions.

I would like to know if there exist other examples of such quadratic categories related or not related to the theory of quantum groups ?

In a recent note, Manin studies the interaction between quadratic algebras, quadratic operad, a notion of enriched category due to Kelly and quantum cohomology? Can these relations be interpreted with this framework of quadratic category ?

Reference.

V. Ginzburg, M. Kapranov. Koszul duality for operads. Duke Math 1994.

Yu. Manin Higher structures, quantum group and genus zero operad

https://arxiv.org/pdf/1802.04072.pdf

Yu. Manin. Quantum groups and non–commutative geometry. Publ. de CRM, Universit´e de Montr´eal (1988),

Tsemo Aristide M\'emoire de D.E.A 1995.

Every continuous simple closed curve in the plane contains four points that are the vertices of a square.

I thinking about possibility of creating counter example to this. What if we create curve that can't have this property?

We can start from isosceles right-angled triangle $A$ because as far I know it can only have two squares in in it. Right now for this triangle conjecture hold but what happens when we alter locally this triangle a bit?

First observation is that 3 points of square determine position of 4th point, then we can create for set all possible $a$, $b$, $c$ points from triangle $A$ that create on they own new isosceles right-angled triangle $B_{a,b,c}$. Now for each this triangle we can construct square that match 3 vertex of this triangle and that determine 4th $d_{a,b,c} = f(a,b,c)$ vertex ($a = f(a, b, d_{a,b,c})$ e.t.c.).

Now when we create set $C_{A}$ of this $d_{a,b,c}$ points and if it intersect with $A$ then this conjecture holds.

Another observation is if we alter some curve $E$ locally then set $C_{E}$ do not change in same local area because it depend on other parts of $E$ (one thing to check is how much other parts of $C_{E}$ change).

With this if we can made that all points of $C_{E}$ of some curve $E$ that intersect with $E$ are isolated from rest of $C_{E}$ then we can alter $E$ to new curve $E^\prime$ that avoid points of $C_{E}$ and $C_{E^\prime}$.

Now when we back to out triangle $A$, set $C_{A}$ consists of line segments that interact with $A$, this intersection AFAIK define two squares. However some of this lines end on intersection points this allow us to change locally $A$ in that way that that $A^\prime$ it will not intersect with $C_{A}$ or $C_{A^\prime}$.

Do this logic have any apparent flaws? Right now I can easy "destroy" using this method two squares in that are inscribed on our triangle $A$ but I don't know yet if doing this I create new one in other place.

Legend:

black - triangle $A$

green - two squares inscribed in $A$

red - set $C_{A}$

A topological space $(X,\tau)$ is said to be *homogeneous* if for $x,y\in X$ there is a homeomorphism $\varphi: X\to X$ such that $\varphi(x) = y$. Let us call a topological space $(X,\tau)$ *homogenizable* if there is a homogeneous space $X_h$ and a continous map $e_X: X \to X_h$ such that for any homogeneous space $Z$ and continous map $f: X\to Z$, there is a continous map $g: X_h\to Z$ such that $f = g\circ e_X$.

What is an example of a non-homogenizable topological space?

My question relates, at least superficially, to this old one:

The value $\pm 1$ for the square root of Wilson's theorem, ((p-1)/2)! mod p

When $p\equiv 1 \mod 4$, if $x=((p-1)/2)!$, then $x^2 = -1 \mod p$.

For what primes does $x \in \{1,\ldots,(p-1)/2\}$ (the elements of the set regarded as residues $\mod p$)?

One gets "yes" for $5,13,29,41,53,61,73,89,97,\ldots$ and "no" for $17,37,101,\ldots$. Despite the slow start for "no", the counts substantially even out, say, when looking at primes up to 100000. Can one prove that the ratio approaches $1/2$?

Consider the wave equation with a potential: $$ E_{tt} = \Delta_x E - q(x) E, \quad (x,t) \in \mathbb R^3 \times (0,+\infty), \\ E(x,0) = 0, \quad E_t(x,0) = \delta_{x_0}(x), $$ under the assumption that $q \in L^\infty_\text{comp}(\mathbb R^3,\mathbb R)$. Then the solution $E$ can be represented in the form $$ E(x,x_0,t) = \frac{\delta(t-|x-x_0|)}{4\pi|x-x_0|} + \widetilde E(x,x_0,t), $$ where $\widetilde E(x,x_0,t) = 0 $ for $t<|x-x_0|$ and is sufficiently regular for $t>|x-x_0|$.

I was told that the function $\widetilde E(x,x_0,t)$ at fixed $x \neq x_0$ must decay exponentially as $t\to+\infty$, as well as its derivatives with respect to $x$ and $t$ up to order two. Is this result well-known?

In the formalism of Lawvere metric spaces, we have that the distance in the hom-space $[X,Y]$ is given by: $$ d(f,g) = \sup_{x\in X} d(f(x),g(x)) . $$ Therefore, a sequence of functions $f_n:X\to Y$ converges to $f:X\to Y$ in $[X,Y]$ if and only if it converges uniformly.

How can we talk about *pointwise* convergence in this setting, instead?