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$\newcommand\Om\Omega\newcommand\R{\Bbb R}$The answer is no. Indeed, for real $k>0$, let $$G_k:=\{(x,y)\in\Bbb R^2\colon x>1,|y|<k\sqrt x\}. $$ For $(x,y)\in G_2$, let $$f_0(x,y):=y^2/x-1.$$ For all $ …

$\newcommand\ep\varepsilon$ So the question can be restated as follows: if the Laplace transforms converge pointwise, do the CDFs also converge pointwise? The answer is yes. Indeed, let $P_n$ ($n=1, …

$\newcommand{\ep}{\varepsilon}$Let us write $n$ instead of $N$ and $k$ instead of $n$. For brevity, let us also write $X_i$ and $Y_i$ instead of $X_{n,i}$ and $Y_{n,i}$. For any function $f$ of severa …

$\newcommand\ep\epsilon\newcommand\th\theta\newcommand\de\delta$The answer is yes, even with $C=1$ for all such $f$. Indeed, let $f$ satisfy all your conditions on $f$. Let $g(0):=0$. For real $x>0$, …

$\newcommand\ep\epsilon\newcommand\th\theta$Yes, $|\th_\ep-\th_0|$ can decay arbitrarily slowly compared to $f(\th_\ep,\ep)$. Indeed, let $g\colon\Bbb R\to\Bbb R$ be any smooth function such that $g(0 …

The answer is yes, such an example exists. For instance, let $b_1=0$ and $b_n=(-1)^j h_j(n)$ for $j=1,2,\dots$ and $n\in N_j:=\{2^j,\dots,2^{j+1}-1\}$, where $$h_j(n):=c_j\frac{\min(n-2^j,2^{j+1}-n)}{ …

$\newcommand\de\delta$The answer is no. In fact, $d_H(A,C_n(L_n))$ can be however large or even infinite for all $n$. Indeed, e.g. suppose that $X=[-1,a)$ for some $a\in(0,\infty]$ endowed with the st …

We have $$d(t):=\int dx\Big(\frac{f(x+t)-f(x)}{t}-f'(x)\Big) =\int dx\int_0^1 ds\,(f'(x+st)-f'(x)),$$ whence $$|d(t)|\le\int_0^1 ds\, J_t(s),$$ where $$J_t(s):=\int dx\,|f'(x+st)-f'(x)|\underset{t\to0 …

$\newcommand{\ts}{\tilde s}$Yes, this is true. Indeed, take any real $c\ge1/2$ and let \begin{equation} s_n(y):=\sum_{-n^2\le k\le n^2}\frac1{(ny-k)^2} \end{equation} and \begin{equation} \ts_ …

$\newcommand\de\delta\newcommand\ep\varepsilon$You wrote Could you help me to show that under Ass1 and Ass2 $$d_H(A, A_n)\rightarrow_{a.s.} 0$$ Of course, this is not true in such generality. For in …

$\newcommand{\ep}{\varepsilon}$To begin, note that \begin{equation*} P(B_n)=1-\prod_{k\ge n}P(X_k\le k-n) =p:=1-\prod_{j\ge0}(1-P(X>j)). \tag{1}\label{1} \end{equation*} Next, \begin{equation* …

$\newcommand\ep\varepsilon\newcommand\de\delta$Note that $$\sum_{k=0}^n k^a\binom nk=n^a2^n E\Big(\frac{X_n}n\Big)^a, \tag{1}\label{1}$$ where $X_n$ is a binomial random variable with parameters $n,1/ …

No. Indeed, make the substitution $x=u^{-1/2}$. That is, let $G_n(u):=F_n(u^{-1/2})$ and $G(u):=F(u^{-1/2})$, so that $\int_0^1\frac{F_n(x)}{x^3}\,dx=\frac12\,\int_1^\infty G_n$ and $\int_0^1\frac{F(x …

The following answer is incomplete. Clearly, $0\le f\le1$. Let $h:=1-f$, so that $0\le h\le1$, $h(a,0)=0$, and $h(0,b)=1$ if $b\ge1$. Here and in what follows, $a$ and $b$ are nonnegative integers. We …

$\newcommand{\R}{\mathbb R}\newcommand{\N}{\mathbb N}\newcommand{\ep}{\varepsilon}\newcommand{\si}{\sigma}$This is to complement Nate River's answer by obtaining a simple and natural sufficient condit …