I'm having a difficult time finding any theory on an inverse problem I've come up against. Let's say I have an unknown function $f:[0,1] \rightarrow \mathbb{R}$, and I know $\int_{a}^{b} f$ for some collection $A$ of pairs $(a,b)\in[0,1]^2$. I'm looking for pointers to any material that discusses condtions on $f$ and $A$ that are sufficient recover (all of? some of?) the values of $f$. Google searches just keep turning up elementary-calculus-help-type pages. I'm a beginning graduate student, if it matters. Thanks.

**Edit:** I'm actually looking for something broader than I asked for. I know that recovering the values of $f$ is a lot to ask and is very unlikely unless $A=[0,1]^2$; I'm also looking for approximations to $f$, anything that can be said about its properties/behaviour, etc. when $A\subsetneqq [0,1]^2$.

Find the slope and the y-intercept of f(x)=8-(5/6)x

I got this wrong on my HW. I understand how the Y-intercept is (0,y) and slope is (y2-y1)/(x2-x1). I also know that whatever I plug into "x" is the domain and "y" will be range.

What I did not understand was which variables do I pick for "x"? If I start with x=0 then would (0,8) be the y-intercept? If that is the case then what's the next variable I pick to begin the slope formula?

Let $H < G$ be a subgroup of a finite group $G$. Let $X:=G/H$ and $\mathcal{F} \in Sh_G(X)$ be an equivariant sheaf on $X$ (w.r.t. left multiplication) associated to a finite dimensional representation $\pi$ of $H$ with character $\chi$. Let $Ind_H^G(\chi)$ be the character of $Ind_H^G(\pi)$ and for any $g \in G$ define $X_g=\{x\in X :gx=x \}$.

Recently I found out that the basic **formula for the induced character** $Ind_H^G(\chi)$ can be interpreted very naturally from the point of view equivariant sheaves giving the following elegant equation:

$$Ind_H^G(\chi)(g)=\Sigma_{x\in X_g} Tr(g^* ,\mathcal{F_x})$$

Where $Tr(g^*,\mathcal{F}_x)$ is the trace of the induced action of $g$ on stalk of $\mathcal{F}$ at $x \in X_g$.

The formula above is surprisingly elegant compared to the one I derived it from (which involved either choosing representatives for cosets or dividing by the order of $H$ while here we avoided both).

**Question 1:** Is there a reasonably geometric argument for why this formula is true?

**Question 2:** In what kind of generality does this formula hold? (continuous representations, locally compact groups, distributional characters etc...).

Let $\Lambda \stackrel{F}{\to} \Omega \stackrel{G}{\leftarrow} \Gamma$ be a diagram of groupoids and functors and $\Gamma \times_\Omega \Lambda$ the homotopy pullback. We will regard all these groupoids as spaces and compute the cohomology with coefficients in some field.

There should be a map $$ C^*(\Gamma) \stackrel{\mathbb{L}}{\otimes}_{C^*(\Omega)} C^*(\Lambda) \to C^* (\Gamma \times_\Omega \Lambda)$$ from the derived tensor product to the cohomology of the homotopy pullback.

Is this map an equivalence?

Let $G=(V,A)$ be a simple directed acyclic graph. A set consisting of two vertices is called *dependent* if there is a directed path from one of the vertices to the other. The question is to find directed paths $\{P_1,\ldots,P_m\}$ of minimum length $m$ such that any pair of dependent vertices belongs to at least one path $P_i$ for some $i$.

I have no idea how to solve this or how to transform this problem to a known problem.

Suppose $w$ is a solution of $$\frac{d^2}{dx^2}w+\{u(x)+k^2\}w=0$$ with asymptotic condition $$\lim_{x\rightarrow \infty}w(x)e^{ikx}=1$$ and $u\in L^1_1(\mathbb{R})=\{f:\int_\mathbb{R}(1+|x|)|f|dx<\infty\}$, and $k>0$

Is the solution unique?

It is well-known that the conformal vector field $\nabla_{S^n}x^{n+1}$ on the unit sphere $S^n$ has its smooth conformal extension to the unit ball $B^{n+1}$, where $n \geq 3$ and $x$ is the position vector in $\mathbb{R}^{n+1}$. I wonder whether for a compact manifold with umbilical boundary (even you can assume the boundary is totally geodesic) there exists a similar conformal vector field on the boundary such that it has smooth conformal extension to the whole manifold.

In a field, if we require that all polynomials have at least one root, then it's algebraically closed and all polynomials factor completely. In a *ring*, the same requirement implies that it's an ACF, because linear polynomials need a root and this ensures multiplicative inverses. This raises the question, is there some other way a ring can be "almost" algebraically closed, without turning completely into a field?

So my question: Is there a ring, in which all polynomials of degree $\ge 2$ have a root, that is not also a field?

Naturally we can't require that all polynomials of degree $\ge 2$ have as many roots as their degree, otherwise one of the two roots of $$ax^2 - (a-1)x + 1 = (x-1)(ax-1)$$ would give us an inverse of $a$.

Let $(V,\pi)$ be an irreducible, admissible, supercuspidal representation of $G = \operatorname{GL}_n(F)$ for $F$ a $p$-adic field. Let $B = TU$ be the usual Borel subgroup, maximal torus, and unipotent radical of $G$. Let $f_{v^{\ast},v}(g) = \langle v^{\ast}, \pi(g)v \rangle$ be a matrix coefficient for $v \in V$ and $v^{\ast} \in V^{\ast \infty}$. Then $f = f_{v^{\ast},v}$ is locally constant and compactly supported modulo the center $Z$ of $G$. If $\chi$ is a generic character of $U$, the integral

$$\int\limits_U f(ug)\chi(u^{-1}) du $$

can be shown to converge absolutely for every $g \in G$. Also, a change of variables shows that for fixed $v^{\ast}$,

$$v \mapsto \int\limits_U f_{v^{\ast},v}(u)\chi(u^{-1})du \tag{1} $$

defines a linear functional $\lambda: V \rightarrow \mathbb{C}$ which satisfies $\lambda(\pi(u_1)v) = \chi(u_1)\lambda(v)$ for all $u_1 \in U$. However, this linear functional might be the zero functional.

In general, $\pi$ is called *generic* if there exists a nonzero linear functional satisfying the property of the previous paragraph for $\chi$. If $\pi$ is generic for $\chi$, it is also generic for every other generic character.

1 . Is every irreducible, admissible supercuspidal representation of $G$ generic?

2 . If $\pi$ is generic, does there exist a smooth linear functional $v^{\ast}$ such that the map (1) is not the zero map? In other words, if there is a nonzero Whittaker functional, can it always be defined by an integral?

Consider a classical space $M_k(N)$ of elliptic modular forms of weight $k$ for $\Gamma_0(N)$. The definition of an unramified Hecke operator $T_{p^m}$ in terms of double cosets is the disjoint union of double cosets $$\Gamma_0(N) \begin{pmatrix} a&b \\ c&d \end{pmatrix} \Gamma_0(N),$$ where $a, b, c, d \in \mathbb Z$ with $ad-bc = p^m$ and $c \equiv 0$ mod $N$. The usual definition for a ramified Hecke operator $T_{p^m}$ ($p | N$), the union of double cosets as above, now with the additional restriction that $p \nmid a$.

On the other hand, Eichler in his work on the basis problem worked with ramified Hecke operators without this restriction $p \nmid a$.

To me, the usual definition seems rather ad hoc, and the one Eichler uses seems more natural from the point of view of orders and ideals. (Note these definitions give different operators.) Of course, the usual definition is nice because it acts nicely on Fourier coefficients.

**Question:** Where were these ramified Hecke operators first defined?

For bonus credit: how did these definitions actually come about?

I'm studying a paper (see citation below) on numerical analysis, and came across this estimate. I am unable to figure out what was done in the final step.

**Preliminary information**: $\Delta{t}$ is the time step, $n$ is the iteration count from $0$ to final iteration $(N+1)$ such that $T=N\Delta{t}$ is the final time at $n=N+1$, $\vec{u}$ is a solution vector, $\vec{f}$ is a vector of forcing functions, and $E_k$ is the energy at iteration $n = k$. We also have various positive constants $C_1, C_2, C_3, C_a$.

**Here is the estimate in the text**:

$E_n + \frac{C_a\Delta{t}^2(C_1+C_2)}{1+(C_1+C_2)\Delta{t}}\|\nabla\vec{u}^{n+1}\|^2+\frac{C_a\Delta{t}^2(C_1+C_2)}{3(1+(C_1+C_2)\Delta{t})}\|\nabla\vec{u}^n\|^2$

$\le C_3\|\vec{f}^{n+1}\|^2\Delta{t} + (1 + (C_1 + C_2)\Delta{t})E_{n-1}$

$\le e^{(C_1+C_2)T}E_0 + \frac{C_3}{C_1+C_2}e^{(C_1+C_2)T} \max\limits_{n}\|\vec{f}^{n+1}\|^2$

**Here is my attempt** at what happened at the last step, before the final inequality:

For clarity, take $X_n = \frac{C_a\Delta{t}^2(C_1+C_2)}{1+(C_1+C_2)\Delta{t}}\|\nabla\vec{u}^{n+1}\|^2+\frac{C_a\Delta{t}^2(C_1+C_2)}{3(1+(C_1+C_2)\Delta{t})}\|\nabla\vec{u}^n\|^2$

Since $X_n \ge 0$, then $E_n \le E_n + X_n$ so

$E_n \le C_3\|\vec{f}^{n+1}\|^2\Delta{t} + (1 + (C_1 + C_2)\Delta{t})E_{n-1}$

**Problem:** Ignoring the first term on the right, I figured the second term can be bounded as:

$E_n \le \left(1 + (C_1 + C_2)\Delta{t}\right)E_{n-1}$

$\quad\: \le \left(1 + (C_1 + C_2)\Delta{t}\right)^{n-1}E_0$

$\quad\: \le e^{(1 + (C_1 + C_2)T}E_0 \qquad$ since $e^{nx} \ge (1+x)^n\ \forall\ n,x\in\mathbb{R}_+$

The parts I can't figure out:

- what happened to the first term i.e. $C_3\|\vec{f}^{n+1}\|^2\Delta{t}$.
- How to re-combine the results with the omitted $X_n$.

Thanks.

Paper: *Chen, Wenbin; Gunzburger, Max; Sun, Dong; Wang, Xiaoming*, **Efficient and long-time accurate second-order methods for the Stokes-Darcy system**, SIAM J. Numer. Anal. 51, No. 5, 2563-2584 (2013). ZBL1282.76094.

I was wondering about the distribution of $\sqrt{p}$ mod $1$ this morning, as one does while brushing one's teeth. I remembered the paper of Elkies and McMullen (Duke Math. J. 123 (2004), no. 1, 95–139.) about $\sqrt{n}$ mod $1$, but hadn't really thought about it before.

Question 1: is $\sqrt{p}$ equidistributed mod $1$, as $p$ varies over all prime numbers? Is this known? Within range of current techniques?

Question 2: What about subtler statistics of $\sqrt{p}$ (and $\sqrt{n}$) mod $1$?

I made three plots, giving histograms of $\sqrt{n}$ mod $1$ (for natural numbers up to 100,000) and $\sqrt{p}$ mod $1$ (for primes up to 1 million) and (for comparison) a histogram of 100,000 samples drawn uniformly at random from {0,1,...,999}. Here they are for your enjoyment.

There's some wild stuff going on, I think!

Question 2(a): What's up with these sharp peak/valleys at rational numbers, in the distribution of $\sqrt{n}$ mod $1$? They are especially prominent at fractions of the form $a / 2^{e}$. How tall are these peaks near rational numbers? They persist when sampling from the primes, i.e., in the distribution of $\sqrt{p}$ mod $1$ too.

Question 2(b): Outside of those funky spots in 2(a), the distribution of $\sqrt{n}$ mod $1$ is far *flatter* than one would expect, e.g., from samples drawn uniformly at random as displayed in the bottom histogram. This must have been noticed and quantified before... what's the relevant quantitative result here?

Question 2(c): The distribution of $\sqrt{p}$ mod 1 displays the same funky spots near rational numbers, but otherwise seems much closer (in noise-volume) to the random samples at the bottom. Maybe for a larger sample, the funkiness goes away... I don't know. Explanations or conjectures are welcome.

Question 3: These seem like natural images to look at. If you know a reference where others have drawn such pictures or studied similar phenomena, I'd love to take a look!

-------------Update after answers below-----------------

It looks like the answer to Question 1 is YES. Lucia's answer below explains this, and also some of the flatness evident in the $\sqrt{n}$ distribution mod 1.

Igor and Aaron discuss the "spikes" around rational numbers. This seems related to binning: if our bins have width 1/1000, we see spikes at multiples of 1/2, 1/4, 1/5, 1/10, etc., related to divisors of 1000. Here's a new picture, which might help us understand the behavior of the distribution of $\sqrt{n}$ mod 1 near rational numbers. I've intentionally drawn the bins so that their endpoints lie on rational numbers with denominator up to 60. (I call this Farey-binning). This seems to bring the "spikes" around rational numbers down to the same size (independent of denominator).

I think I'll accept Lucia's answer soon, because it answers the most direct Question 1. But more insights are welcome.

Given an $l_p$ norm $\| x \|_p = (\sum_{i=1}^n |x_i|^p)^{1/p},$ we call a convex differentiable function $f: \mathbb{X} \rightarrow \mathbb{R}$ both strongly convex and smooth if there exists $m, M > 0$, such that

$$ \frac{m}{2}\| x - y\|_p^2 \le f(x) - f(y) - \nabla f(y)^T(x-y) \le \frac{M}{2}\|x - y\|_p^2,$$
for all $x, y \in \mathbb{X}$. Specifically, I would like examples where the condition number $\kappa = \frac{M}{m}$ is not dependent on the dimension $n$.

If $p=2$, there are quite a few examples of both strongly convex and smooth functions, in fact, these are the class of functions where gradient descent achieves linear convergence rate.

For $p \ne 2$ I don't know whether there are analogous examples. The only half-examples that I know of are:

For $p=1$ and $\mathbb{X}$ being the probability simplex, the negative entropy function $f(x) = \sum_{i=1}^n x_i \log x_i$ is 1-strongly convex (due to Pinsker's inequality). This fact is widely used in the discussion of mirror descent algorithms.

For $p \in (2, +\infty)$, $\|x\|_p^2$ is $\max\{2, (p-1)\}$-smoothed in $l_p$ norm by Nemirovski (https://www2.isye.gatech.edu/~nemirovs/LargeDev2004.pdf) and for $p \in (1, 2)$, $\|x \|_p^2$ is $\min\{ \frac{1}{2}, (p-1) \} $strongly convex in $l_p$ norm by Kakade et al (http://ttic.uchicago.edu/~shai/papers/KakadeShalevTewari09.pdf).

I am interested in any example where a function is both strongly convex and smooth for $l_p$ norm where $p \ne 2$ with dimension independent $\kappa$.

The following statements suggests $B(H)$-Egoroff's theorem when $H$ is a separable Hilbert space. Probably it will be hold even if a von Neumann algebra $M$ whose predual is separable is replaced. I have no clear proof for this claim. Any comment on this claim?

**Operator-valued Egoroff's Theorem.** Assume that $H$ is a separable Hilbert space and $(\Omega,M,\mu)$ is a finite measure space. Let $\varphi_n:\Omega\to B(H)$ be a sequence of operator-valued measurable functions on $\Omega$ converging pointwise-strongly to $\varphi$. Then for any arbitrary positive trace class operator $\omega$ and positive real numbers $\epsilon$ and $\delta$, there exist a measurable set $ E\subseteq\Omega$ and a projection $p\in B(H)$ satisfying $ \mu(E)<\epsilon$ and $ \omega(1-p)<\delta$,

such that

$$ \sup_{_{t\in E^c}}\|(\varphi_n(t)-\varphi(t))p\|\longrightarrow 0.$$

**Proof.** Note that $\omega=\sum_{i=1}^\infty\alpha_i e_i\otimes e_i$ where $\{\alpha_i\}_{_{i\in \mathbb{N}}}\in \ell^1$ ($\alpha_i\geq0$) and $\{e_i\}_{_{i\in \mathbb{N}}}$ is an orthonormal set of $H$. Let $N\in \mathbb{N}$ with $\sum_{i=N+1}^\infty\alpha_i<\delta.$ Without loss of generality, we may assume that $\{e_i\}_{_{i\in \mathbb{N}}}$ is an orthonormal basis for $H$. It is concluded $\omega(1-p)<\delta$ where $p=\sum_{i=1}^N e_i\otimes e_i$. For any $1\leq i\leq N$, we consider the following sequence of positive functions:

$$g_i^n(t)=\|(\varphi_n(t)-\varphi(t))e_i\|
=(\sum_{j=1}^\infty|\langle(\varphi_n(t)-\varphi(t))e_i,e_j\rangle|^2)^{\frac{1}{2}}$$
All functions $g_i^n$'s are measurable. Moreover, for every $1\leq i\leq N$ the sequence

$\{g_i^n\}_{n\in\mathbb{N}}$ is pointwise convergent to zero, since $\varphi_n\stackrel{p.s}{\longrightarrow}\varphi$. By the classical Egoroff's

theorem, there exists a measurable set $ E_i\subseteq\Omega$ with $\mu(E_i)<\frac{\epsilon}{2^i}$ such that $g_i^n$ converging uniformly to 0 on $E_i^c$, for $1\leq i\leq N$. It means that

$$ \sup_{_{t\in E_i^c}}\|(\varphi_n(t)-\varphi(t))e_i\|\longrightarrow 0.$$

By taking $E=\bigcup_{i=1}^N E_i$, we have then $\mu(E)<\epsilon$. One may see that$$\sup_{_{t\in E^c}}\|(\varphi_n(t)-\varphi(t))p\|\longrightarrow 0.$$

Consider a smooth curve $\gamma$ of finite length in the unit square $[0,1]\times[0,1]$. Is the following statement correct?

There exists a Lebesgue null set $N$ such that for all $x\in [0,1]\setminus N$ the set $\{y\in[0,1]:(x,y)\in\gamma\}$ has only finitely many points.

Generalization:

Consider a smooth manifold $M$ in the $n$ - cube $[0,1]^n$. Assume that the $n-1$ dimensional Hausdorff measure ${\cal{H}}^{n-1}(M)$ is finite. Is the following statement correct?

There exists a set $N$ with ${\cal{L}}^{n-1}(N)=0$ such that for all $x\in [0,1]^{n-1}\setminus N$ the set $\{y\in[0,1]:(x,y)\in M\}$ has only finitely many points. Here ${\cal{L}}^{n-1}$ denotes the $n-1$ dimensional Lebesgue measure.

I guess that the number of integers $x$ which satisfy the condition $x*p(x) \leq n$ is $O(n^{2/3})$ or $O(n^{3/4} / \ln n$), but I cannot prove it. I just write a program to count the number. The results are listed.

Note: $p(x)$ means the largest prime factor of $x$.

n the numbers

$10^5$ 1894

$10^6$ 9108

$10^7$ 44948

$10^8$ 228102

Crossposted from https://math.stackexchange.com/questions/2605895/asymptotics-for-partial-sum-of-product-of-binomial-coefficients

For some $c<n$ and $2c\leq x\leq 2n$, are there references or previous results for determining the asymptotics (as $n\to\infty$) of the partial sum $$ \sum_{k=x-c}^c\binom{n}{k}\binom{n}{x-k} $$ or equivalently if $c=n\lambda_1$ and $x=2n\lambda_2$, for constants $0<\lambda_2\leq\lambda_1<1$ $$ \sum_{k=2n\lambda_2-n\lambda_1}^{n\lambda_1}\binom{n}{k}\binom{n}{2n\lambda_2-k} $$

I don't think I can just apply Stirling's approximations to the binomial coefficients individual and take the sum and product.

**EDIT**

Could someone comment if this is a valid attempt?

Using @robjohn's solution in this post, let $$ a_k=\binom{n}{k}\binom{n}{2n\lambda_2-k} $$ Then letting $k=n\lambda_2+j$, $$ \log\left(\frac{a_{k+1}}{a_k}\right)=-\frac{2j}{n\lambda_2(1-\lambda_2)}+O(n^{-1}) $$ Thus, $$ a_k=a_{n\lambda_2}\exp\left(-\frac{2j^2}{n\lambda_2(1-\lambda_2)}+O(j/n)\right) $$ Estimating $$ a_{n\lambda_2}\sim C(\lambda_2)=\frac{1}{2\pi n\lambda_2(1-\lambda_2)}(1-\lambda_2)^{-2n}\left(\frac{1-\lambda_2}{\lambda_2}\right)^{2n\lambda_2} $$ by Stirling's formula and using Riemann integral for the exponential, $$ \sum_{j=-n(\lambda_1-\lambda_2)}^{n(\lambda_1-\lambda_2)}\exp\left(-\frac{2j^2}{n\lambda_2(1-\lambda_2)}+O(j/n)\right)=\sqrt{n\lambda_2(1-\lambda_2)}\int_{-\infty}^{\infty}\exp\left(-2t^2\right)dt(1+O(1/n)) $$ we have \begin{eqnarray} \sum_{k=2n\lambda_2-n\lambda_1}^{n\lambda_1}\binom{n}{k}\binom{n}{2n\lambda_2-k}&\sim& C(\lambda_2)\sqrt{n\lambda_2(1-\lambda_2)}\sqrt{\pi/2}\\ &=&\frac{1}{2\sqrt{2\pi n\lambda_2(1-\lambda_2)}}(1-\lambda_2)^{-2n}\left(\frac{1-\lambda_2}{\lambda_2}\right)^{2n\lambda_2} \end{eqnarray} Substituting back $c=n\lambda_1$ and $x=2n\lambda_2$, and noticing Stirling's formula for $\binom{2n}{x}$, we get $$ \sum_{k=x-c}^c\binom{n}{k}\binom{n}{x-k}\sim\frac{1}{\sqrt{2}}\sqrt{\frac{2n}{2\pi x(2n-x)}}\left(\frac{2n}{2n-x}\right)^{2n}\left(\frac{2n-x}{x}\right)^x\sim \frac{1}{\sqrt{2}}\binom{2n}{x} $$ To me this is very interesting that it doesn't involve $c$, which disappeared when estimating with the Riemann integral above. However, after plugging in a couple of values in Mathematica, the approximation on the right hand side doesn't always give an accurate approximation to the partial sum.

**QUESTION 2**
Is there a way to figure out how far this partial sum is from the upper bound of $\binom{2n}{x}$?

**EDIT 2**

It turns out that $$ \sum_{k=x-c}^c\binom{n}{k}\binom{n}{x-k}=\binom{2n}{x}-2\sum_{k=0}^{x-c-1}\binom{n}{k}\binom{n}{x-k} $$

I guess, then I'm interested in showing if $$ 2\sum_{k=0}^{x-c-1}\binom{n}{k}\binom{n}{x-k}=o\left(\binom{2n}{x}\right) $$ How would I go about showing this?

I have a broad question. I know nothing about perfectoid spaces and I just have read What is ... a Perfectoid Space? by Bhatt. However, I am very curious if these spaces could have a conceivable application in physics, especially quantum field theory and string theory.

The setting is a convex cone $C$ in $\mathbb{R}^d$ with the property that if you cut it with $S^d$ the volume of the cut is greater than or equal to $\operatorname{Vol}(S^d)/d+1$. That is, the volume of the cone (i.e. the volume of its cut with $S^d$) is greater than or equal to the volume of the cone from the standard partition from $\mathbb{R}^d$ into $d+1$ cones.

The cut of the boundary of $C$ and $S^d$ determines a $d-1$ hyperplane. It is stated now, that the distance from this hyperplane to the origin is less than or equal to $\frac{1}{2}$.

In dimension $d=2$, this fact is easy to confirm by simple triangle properties. I would be very grateful though for an aproach in arbitrary dimensions.

Are the Atiyah-Bott-Shapiro Orientation and the Anderson-Brown-Peterson Splitting compatible in any sense?

The first guess is that the ABS-Orientation is related to the projections on the $BO\langle 4(n(J))\rangle$, respectively $BO\langle 4(n(J))-2\rangle$ factors.