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1

This answer should perhaps be posted as a comment, since the whole topic was already debated in this MathOverflow Q&A. However i briefly restate the main result here and leave the above link for further detals: a definitive answer is Cafiero's convergence theorem (see ) which, roughly states that$\DeclareMathOperator{\Dm}{\operatorname{d\!}}$ $$\... 4 \newcommand\ep\varepsilonThe conjunction of the following conditions is enough: The f_n's are uniformly bounded: |f_n|\le M for some real M>0 and all n; X is Polish; f_n\to f uniformly on every compact K\subseteq X; \mu_n\to\mu weakly for some \mu. Indeed, take any real \ep>0. By Prokhorov's theorem and in view of conditions 2 ... 1 Using the Radon-Nikodym Theorem this problem can be reduced to the case for a fixed measure. Suppose none of the \mu_n is the zero-measure, i.e. \mu_n(X)>0 for all n \in \mathbb{N}. For a series (\alpha_n)_{n \in \mathbb{N}} \subset (0,1) with \sum\limits_{n=1}^\infty \alpha_n = 1 define the probability measure$$ \mu = \sum\limits_{n=1}^\infty \...

0

Since this got bumped to the front page... Let $\sigma_i$ be the singular values of $A$ and $\tilde{\sigma}_i$ be those of $A+E$. Then, $|\sigma_i - \tilde{\sigma_i}| \leq \|E\|$, by Weyl's inequalities. That gives you 1-Lipschitz continuity.

0

The box dimension is not countably stable, so even one exceptional point could drive it up. For example, given $\epsilon>0$, the function $f(x)=x^{1/2} \sin(1/x)$ on $[0,1]$, with $f(0)=0$, is locally Lipschitz on $(0,1]$, yet the graph of $f$ has box dimension at least $1.25$. Indeed, for $j<k$, to cover the graph above $[1/(2\pi(k+j)), 1/(2\pi(k+j+1))... 1 The following is obviously not optimal, but at least it shows that the optimal$c$satisfies$c \leqslant 4$. On the other hand,$c \geqslant 2 / \log 2 \approx 2.885$, so if$c$is to be integer, it remains to show that in fact$c < 4$and hence necessarily$c = 3$. Edit: As remarked in the comments, the following argument works for continuous decreasing ... 3 The inequality$(2)$(even with factor$\frac12$in the r.h.s.) follows from the inequality quoted in this answer: $$M_a - M_g \leq \frac1{2n\min_k x_k} \sum_{i=1}^n (x_i - M_a)^2.$$ First, we notice that \begin{split} (x_i - M_a)^2 &\leq \max_k x_k\cdot |x_i-M_a|\\ &= \max_k x_k\cdot \left|\frac1n \sum_{j=1}^n (x_i - x_j)\right| \\ & \leq \frac{\... 2 Edit: In the first part,$C^{k,\alpha}$is understood locally. For a global result, see the final part. The equation reads $$f(t) = \int_{-\infty}^t e^{-(t - s)\theta} g(s) ds,$$ is equivalent to $$e^{t \theta} f(t) = \int_{-\infty}^t e^{s \theta} g(s) ds,$$ so if$g$is$C^{k-1,\alpha}$, then$s \mapsto e^{s \theta} g(s)$is$C^{k-1,\alpha}$, its integral$...

2

$\newcommand\de\delta\newcommand\ol\overline$Your goal cannot be attained in general. Indeed, suppose that $\de\in(0,1/2)$. Take any interval $E$ of length $\de_1:=|E|\in(\de,2\de)$. Then for the closure $\ol E$ of $E$ and some closed intervals $I_1$ and $I_2$ of lengths $|I_1|=\de$ and $|I_2|=\de_1-\de\le\de$ we have $E=I_1\cup I_2$, so that $$\mu_\de^s(E)\... 4 \newcommand\R{\mathbb R}\newcommand\N{\mathbb N}Here are answers to your three questions (the latter two of them partial). Answer 1: Yes, for any real a and any k\in\{0,1,\dots\}, \begin{equation*} \text{if h does not have all the derivatives at a, then f^{(k)}(a)=0.}\tag{1} \end{equation*} Indeed, say that h is bad at a (or, ... 1 Your condition can be rewritten as the system of three equations:$$au(x)+bu(x+1/2)=0\ \forall x\in(0,1/2), \tag{1}au(x)+cu(x-1/2)=0\ \forall x\in(1/2,1), \tag{2} au(1/2)=0. \tag{3} $$In turn, (2) can be rewritten as$$cu(x)+au(x+1/2)=0\ \forall x\in(0,1/2). \tag{2a} $$So, if the determinant a^2-bc of the system (1)--(2a) of linear equations ... 1 Claim: Assuming that L is the Laplace transform, there is no complex p_0 such that for all y\in L^2(0,1) we have the implication L(\tilde y)(p_0)=0\implies y=0 on (0,1/2). So, your conditions can never be fulfilled, and therefore they imply any statement, be it true or false. Proof of the Claim: Take any complex p_0=a+ib, where a and b are ... 1 I think the following is a counter-example for the equivalence of (2) and (3): Consider the Baire space \mathbf N^\mathbf N and let f_n be the following function: If your sequence x starts with k zeroes followed by the entry n then f_n(x) = 1/(k+1), otherwise f_n(x) = 0. The f_n are continuous: away from the sequence that is constantly 0, ... 3 \newcommand\tk{\tilde k}\newcommand\ip{\langle #1,#2\rangle}Let k be a reproducing kernel of a reproducing kernel Hilbert space (RKHS) H:=\mathcal H of real-valued functions on a set X. Then$$\ip f{k_x}=f(x)\tag{1}$$and$$k(x,y)=\ip{k_x}{k_y}=k_x(y)\tag{2}$$for all f\in H and all x and y in X, where \ip\cdot\cdot is the inner product ... 11 Let g(x) = e^{-x} f(x), so that f(x) = e^x g(x). For a given x, f'(x) exists if and only if g'(x) exists, and g'(x) = e^{-x} (f'(x) - f(x)). In particular, f'(x) = f(x) if and only if g'(x) = 0. It follows that f is necessarily of the form f(x) = e^x g(x), where g satisfies g'(x) = 0 almost everywhere. 2 \newcommand{\R}{\mathbb{R}}\newcommand{\Z}{\mathbb{Z}}\newcommand{\de}{\delta}\newcommand{\ol}{\overline}The property of \R that you want to prove is that the topological space \R is normal. All metric spaces are perfectly normal and hence normal. This proves your desired result. That all metric spaces are normal immediately follows e.g. from Theorems ... 2 Yes, there is an expansion, but it is infinite. To simplify, set x=a/b, f(x)=x^n-1. This is an analytic function on the positive ray, so it has a Taylor expansion at x=1:$$f(x)=\sum_{k=1}^\infty\frac{n(n-1)\ldots(n-k+1)}{k!}(x-1)^k.$$And to answer your second question, yes, f(x) can be estimated in terms of x-1, since the series begins with the ... 2 The following papers provide conditions for a vNM utility function on a topological space to be continuous. Grandmont, 1972. "Continuity properties of a von Neumann-Morgenstern utility". Journal of Economic Theory 4, pp. 45–57. Miyake, M., 1990. "Continuous representation of von Neumann–Morgenstern preferences". Journal of ... 2 There exists a continuous, bounded utility function if and only if the relation is continuous in the stronger sense of being closed in P(X) \times P(X), using the weak{}^* topology on each factor. See Section 3.3 of this paper, for example. (The point of that paper is to find Lipschitz utility functions, but we give an overview of the general situation.) 7 Claim. The thief T can escape if C is a circle, with a simple strategy of dribbling left and right each policeman at a time in such a way that he is left out of reach of the thief no matter what the future dribbles will be. Proof. The key insight is due to Pietro Majer: the thief can approach C in such a way that its shadow S (closest point) on C ... 0 The cops should always win. Here is a sketch of a proof. One cop (for short) trying to catch the thief crossing a segment: the cop can run on the segment at top speed towards the thief (provided the thief starts far enough and the cop's reaction time is 0). Clearly they will meet. Likewise a dense set of cops on a convex curve C: they can run towards the ... 1 It seems to me that some useful information you can find in paragraph 1.5 of "Differentiable Functions on Bad Domains" by V. G. Mazia, S. Pobozchi. 0 The answer is yes. Suppose g is orthogonal to the image of S, and let v be the solution of the Dirichlet problem \Delta v=g\delta(\partial D_1) on D_2, where \delta(\partial D_1) is a delta function localized on \partial D_1. We find$$\int_{\partial D_1} gu\,dS=\int_{D_2}v\Delta u\,dx.$$Now suppose this is true for every u for which \Delta ... 1 The answer is yes: take any smooth function g_0 on \partial D_1 and solve the Dirichlet problem$$ \begin{cases} \Delta g = 0 & \text{ on } D_1\\ g = g_0 & \text{ on } \partial D_1. \end{cases} $$Now extend g to a smooth function on \mathbb{R}^n. Multiply by a smooth cutoff function \eta which is 1 on D_1 and compactly supported on ... 4 Let$$f_n(x) = \sup_{r \in {\mathbb Q} \cap [\frac{1}{n+1},\frac{1}{n})} \frac{|B(x,r)\setminus E|}{|B(x,r)|}\,,$$so that f_n(x) \to 0 for a.e. x \in E. By Egorov's theorem , for every \epsilon>0 there is a subset \tilde{E} \subset E with |E \setminus \tilde{E}| <\epsilon, such that f_n(x) \to 0 uniformly on \tilde{E}. It follows ... 1 \newcommand\R{\mathbb R}Your claim is incorrect. E.g., let \mu be the Lebesgue measure over X:=\R. Let$$\Phi:=\{f_n\colon\, n\in\mathbb N\},$$where$$f_n(x):=\frac1{1+(x-n)^2}$$for real x. Then \Phi is uniformly bounded in L^1, does not escape to vertical infinity, and does not escape to width infinity; however, \Phi is neither tight nor ... 33 A rescaling is needed for a nontrivial limit. As discussed in Iteration of Sine and Related Power Series by C. Towse (2014), denoting the n-th iterate by \sin^{\circ n}x, one has the limit$$\lim_{n\rightarrow\infty}\sqrt n\sin^{\circ n}(x/\sqrt n)=\frac{x}{\sqrt{1+x^2/3}}.$$The graph (from the cited paper) shows that the limit is attained quite rapidly.... 1 First, according to paper you cited, \mathfrak{M}_E is the [not "a"] family of all nonempty bounded subsets of E. The examples of MNC's given in the paper you cited -- namely, the Kuratowski MNC \alpha, the ball measure \chi, and the MNC \mu defined in Theorem 1 of that paper -- are all with the maximum property and nonsingular. Perhaps,... 10 If f is bounded on the imaginary line, (and has exponential type) then f has completely regular growth in the sense of Levin-Pfluger, with indicator c|\cos\theta|. This implies that density of zeros on the positive ray must be zero. Moreover, density of zeros in any angle |\arg z|<\pi/2-\epsilon and in the vertical angle is zero. Boundedness on ... 1 \newcommand\C{\mathscr C}\newcommand\ep{\varepsilon}\newcommand\A{\mathscr A}\newcommand\N{\mathbb N}Yes, \mu is a measure. Indeed, let \A:=A. Say that a partition of a set B\in\A is measurable if all its pieces are in \A. For any B\in\A, let P(B) denote the set of all countable measurable partitions of B. Then$$\mu(B)=\mu_Y(B):=\sup\Big\{\...

2

$\newcommand\si\sigma\newcommand\om\omega\newcommand\z{\mathbf z}\newcommand\R{\mathbb R}$Let us assume that your $d\Omega$ means $d\z$, so that $$I_\si(s):=\int_{\R^k} p_{Z}(z-s) f(z)\, dz,$$ and we shall assume that this definition holds for all $s\in\R^k$. So, $$I_\si(s)=Ef(s+\si Z),$$ where $Z$ is a standard normal random vector in $\R^k$. Let $\si\... 3$\DeclareMathOperator\E{E}\DeclareMathOperator\Var{Var}\DeclareMathOperator\P{P}$Note that $$\int_{[0,1)^d}\|x\|_2^p\,dx =\E S_d^{p/2},\tag{1}\label{1}$$ where$S_d:=\sum_1^d U_j^2$and the$U_j$'s are iid random variables uniformly distributed on the interval$[0,1]$. Note next that$\E S_d=d/3$and$\Var S_d=4d/45<d/10$. So, by Cantelli's inequality, $$\... 4 Here you prescribe in addition the sequence of signs of the rearranged series in the Riemann-Dini theorem to be alternating, but note that any non-stationary binary sequence of signs does as well. More precisely: Let (a_k)_{k\in\mathbb N} be an infinitesimal sequence of non-zero real numbers such that \sum_{k\ge0}a^+_k=\sum_{k\ge0}a^-_k=+\infty. Let \... 2 You have the integral representation$${}_2F_1\left(\begin{matrix}11/2,1\\5\end{matrix};-x\right)=\frac 4{x^4}\int_0^x\frac{(x-t)^3}{(1+t)^{11/2}}\,dt,$$which follows by expanding 1/(1+t)^{11/2} using the binomial theorem and integrating termwise. This should prove the positivity. Note that this integral is a remainder term in the Taylor expansion of 1/(... 1 I believe the answer is yes. Let a_{i_k} and a_{j_k} be the even and odd terms respectively, ordered in increasing order of magnitude. We will define four sequences; b_n (indexed by \mathbb Z_+) which will be the desired rearranged sequence, a -1, 1 valued sequence s_n (indexed by \mathbb N), which will be a “state/control” variable, and two ... 3 For small r the curvature of the surface \partial D can be neglected, so D \cap B(x,r) is half the d-dimensional ball with radius r. Choosing the origin of the coordinate system at position x and orienting the x_1-axis along the inward normal, the integral is given by the vector v with components$$v_p=\frac{2}{V_{d}(r)}\int\cdots\int_{-\... 6 The right hand side evaluates to $$\sum_{m=0}^{\infty} m^{k-1} x^m,$$ and so it remains to verify that the coefficient of$x^m$in the l.h.s. is also$m^{k-1}$. By differentiating the l.h.s., we have \begin{split} [x^m]\ \bigg(-\sum_{n\geq 1} \frac{J_k(n)}{n} \ln(1-x^n)\bigg) &= \frac1m [x^{m-1}]\ \sum_{n\geq 1} J_k(n)\frac{x^{n-1}}{1-x^n}\\ &=\... 4 As you said, the sum is$\Pr[X \leq \alpha n]$where$X$is drawn from a Binomial distribution with$n$trials having$p$probability of success. Bounds on this sum (for$\alpha < p$) are called "tail bounds", "concentration inequalities", etc. These bounds are proven for many settings, especially sums of independent random variables, ... 7 Yes, if$\alpha<p$(if$\alpha>p$, the sum is almost 1). To see this, write $$\sum_{\ell = 0}^{\alpha n} \binom{n}{\ell}p^\ell(1-p)^{n-\ell}\leqslant t^{-\alpha n}(pt+(1-p))^n$$ for every$t\in (0,1]$. Choose a positive$t=t_0$for which RHS is minimal possible, taking the logarithmic derivative equal to 0 we get$-\alpha/t_0+p/(pt_0+1-p)=0$,$-\...

9

By the Chernoff–Hoeffding theorem, the sum in question is $\le\exp(-nD(a||p))$ for $a\le p$, where $a:=\alpha$ and $$D(a||p):=a\ln\frac ap+(1-a)\ln\frac{1-a}{1-p},$$ the Kullback–Leibler divergence between the distributions of Bernoulli random variables with parameters $a$ and $p$. In particular, this gives your bound for $p=1/2$. On the other hand, if \$a>...

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