Let $\omega:[0,\infty)\to\mathbb R$ be bounded above on every finite interval and subadditive. I want to show that $$\omega_0:=\inf_{t>0}\frac{\omega(t)}t=\lim_{t\to\infty}\frac{\omega(t)}t\;.\tag 1$$ Notice that $\omega_0\in[-\infty,\infty)$. I’ve found a proof which starts by choosing some $\gamma>\omega_0$ and stating that there exists a $t_0>0$ with $$\frac{\omega(t_0)}{t_0}<\gamma\tag 2\;.$$ Why does such a $t_0$ exist?

Why is it that if we sandwich a liminf of an expectation between two equal quantities we get that the limit exists? Can we somehow deduce the limsup from that and conclude that it’s the same or am I missing something else. A book I’m reading used a rule called Fatous lemma to give the […]

I’m struggling to understand the beginning of the solution to the following exercise: Let $(X_n)_{n\geq 1}$ and $X$ be random variables. Prove that $X_n \to X$ almost surely if and only if for every $\epsilon>0$ $$P(\limsup\limits_{n\to \infty}\{|X_n-X|\geq \epsilon\})=0$$ Solution: Notice that $\omega \in\limsup\limits_{n\to \infty}\{|X_n-X|\geq \epsilon\}$ iff there exist a subsequence $(n_k)_{k\geq 1}$ such that $|X_{n_k}-X|\geq […]

This originally comes from $f_1(x,y)=\frac{x}{y}$, where $X=\mathbb{R}^{n}, Y=\mathbb{R}^m, x \in X, f: X \rightarrow Y, x \neq 0, f(x) \neq 0$ $$\limsup_{(h_x,h_y)\to(0,0)} \frac{\left|\frac{x+h_x}{y+h_y}-\frac{x}y\right|} {\sqrt{{h_x}^2+{h_y}^2}}$$ If I understand it correctly, $x$ and $y$ can be anything but zero, and $h_x,h_y$ go towards zero. Moreover, both numerator and denominator cannot be negative. But since $x,y$ could be […]

I’m a little confused about the definition of limit supremum; what does it mean that the following limit is finite? $$\limsup _{h\rightarrow \infty}\;\sup_{x\in \mathbb R}\; A(x,h)$$ where $A(x,h)$ is a function of $x$, and $h$.

I have read this solution, but I could not understand it. It has shown $$\sigma_n\leqslant \frac 1n\sum_{j=1}^ks_j+\sup_{l\geqslant k}s_l,$$ but how does it go to $$\sup(\sigma_n)\leqslant \frac 1n\sum_{j=1}^ks_j+\sup_{l\geqslant k}s_l$$ I have thought of regarding the RHS as upper bound of $\sigma_n$, but I am not sure about this, since RHS varies as $n$ changes. The writer […]

Given a sequence $x_{n}$ and initial data that $0< a\leq x_{n}\leq b< \infty $, for $a,b\in \mathbb{R}$. I need to show that: $$\limsup \frac{1}{x_{n}}\cdot \limsup x_{n}\geq 1.$$ I think the the simplest way to do that is to show that $$\liminf \frac{1}{x_{n}}=\frac{1}{\limsup x_{n}}.$$ I started to write some things, but nothing leaded me to the […]

Given $a_{n}=\frac{(-1)^{n}n+1}{n}$, compute $$\lim\limits{\inf(a_{n})}$$ $$\lim\limits{\sup(a_{n})}$$ $$\inf\{a_{n}\}$$ $${\sup(a_{n})}$$ My attempt: I tried taking different values for the sequence and reached the following results: $$\lim\limits{\inf(a_{n})}=1$$ $$\lim\limits{\sup(a_{n})}=-1$$ $$\inf\{a_{n}\}=3/2$$ $${\sup(a_{n})}=-1$$ The teacher told me to do it more formally, anyone can help me please?

The limsup on sequences of extended real numbers is usually taken to be either of these two things, which are equivalent: the sup of all subsequential limits. The limit of the sup of the tail ends of the sequence. For the situation with nets, the same arguments guarantee the existence of the above quantities 1. […]

Let $X$ is a topological space. A function $u:X\rightarrow [-\infty,\infty)$ is upper semicontinuous if the set $\{x\in X:u(x)<\alpha\}$ is open in $X$ for each $\alpha\in \mathbb{R}$. Show that $u$ is upper semicontinuous iff $\limsup_{y\rightarrow x} u(y)\leq u(x)$ for all $x\in X$. My Try: Let $x\in X$. Let $A=\{y\in X:u(y)<u(x)\}$. Then $A$ is open in $X$. […]

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