• a mapping $a$ on $\mathbb{N} \mapsto (0,\infty)$ with $n\mapsto a(n)\equiv a_n$.

  • a mapping $b$ on $\mathbb{N} \mapsto \mathbb{R}$ with $n\mapsto b(n)\equiv b_n$.

Suppose that $\forall t\in \mathbb{R}$ and $\forall x \in \mathbb{R}$ $$ \lim_{n\rightarrow \infty} n \Big(1-G\Big(a_n(t+x)+b_n\Big)\Big)= \exp(-t-x) $$ where $G:\mathbb{R}\rightarrow [0,1]$ is continuous and strictly monotone increasing on $\mathbb{R}$.

Show that convergence is locally uniform with respect to $x$ $\forall t \in \mathbb{R}$.

As hint from the professor I was suggested to use result 0.1 in Resnick's book (Extreme Values, regular Variation, and Point processes) but I have some doubts.

First of all, let $$ U_n(x,t)\equiv n \Big(1-G\Big(a_n(t+x)+b_n\Big)\Big) $$ and $$ U_0(x,t)\equiv \exp(-t-x) $$ I think that local uniformity here intended as $$ \lim_{n\rightarrow \infty} \sup_{x\in [a,b]}|U_n(x,t)-U_0(x,t)|=0 $$ $\forall a\in \mathbb{R}, \forall b\in \mathbb{R}, a<b, \forall t \in \mathbb{R}$. Correct?

Secondly, result 0.1 n Resnick's book gives some sufficient conditions for local uniform convergence: for any $t\in \mathbb{R}$

  • $U_n(\cdot ,t)$ with domain $\mathbb{R}$

  • $U_n(\cdot,t)$ non decreasing on $\mathbb{R}$

  • $U_0(\cdot ,t)$ continuous on $\mathbb{R}$

  • $\lim_{n\rightarrow \infty} U_n(x,t)=U_0(x,t)$ $\forall x \in \mathbb{R}$

My doubt is related to the "non-decreasing part": my function $U_n(\cdot, t)$ is actually strictly decreasing on $\mathbb{R}$ because $G$ is strictly increasing in $x$ and $1-G(\cdot)$ is thus strictly decreasing in $x$. Can we extend Resnick's result to strictly decreasing functions? This and this questions are related (but the answers are given for non-decreasing functions!)

  • $\begingroup$ Can't you apply Resnick's result to $-U$. May be I am missing something. $\endgroup$ – Kavi Rama Murthy Apr 24 '18 at 12:21
  • $\begingroup$ You mean: by assumption $\lim_{n\rightarrow \infty} n\Big(1-G\Big(a_n(t+x)+b_n \Big)\Big)=\exp(-t-x)$ $\Leftrightarrow$ $\lim_{n\rightarrow \infty} n\Big(-1+G\Big(a_n(t+x)+b_n \Big)\Big)=-\exp(-t-x)$; define $\tilde{U}_n(x,t)\equiv n\Big(-1+G\Big(a_n(t+x)+b_n \Big)\Big)$ and $\tilde{U}_0(x,t)\equiv -\exp(-t-x)$; apply Resnick's to $\tilde{U}_n(\cdot,t), \tilde{U}_0(\cdot, t)$ to conclude $\lim_{n\rightarrow \infty}\sup_{x\in [a,b]}|\tilde{U}_{n}(x,t)- \tilde{U}_0(x,t)|=0$ $\forall a<b, t\in \mathbb{R}$. $\endgroup$ – TEX Apr 24 '18 at 12:31
  • $\begingroup$ The final claim is equivalent to $\lim_{n\rightarrow \infty}\sup_{x\in [a,b]}|-U_{n}(x,t)+U_0(x,t)|=0$ $\forall a<b, t\in \mathbb{R}$ which is equivalent to $\lim_{n\rightarrow \infty}\sup_{x\in [a,b]}|U_{n}(x,t)-U_0(x,t)|=0$ $\forall a<b, t\in \mathbb{R}$ $\endgroup$ – TEX Apr 24 '18 at 12:31
  • $\begingroup$ Apply the Professor sugestions to function $-U_n (x,t)$ $\endgroup$ – MotylaNogaTomkaMazura Apr 24 '18 at 13:46

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