TITLE: How do I solve this in an understandable and direct way?
QUESTION [3 upvotes]: For each $i \in \Bbb N$, let $f_i: \Bbb N \mapsto \{0, 1\}$.
Let $A = \{f_i : i \in \Bbb N\}$ and $E = \{n \in \Bbb N : f_n(n) = 0\}$.
Does there exist a $f \in A$ such that $E = \{n \in\Bbb N : f(n) = 1\}$
I think that there does not exist a $f \in A$ such that $E = \{n \in\Bbb N : f(n) = 1\}$. However, I am unsure on how to prove it. How do I show my answer in an understandable way?

REPLY [0 votes]: No.
Let $f=f_k$ for $k\in\mathbb{N}$.
If $k\in E$, where $E = \{n \in \Bbb N : f_n(n) = 0\}$, then $f_k(k) = 0$. Assume $E=E_2$, where $E_2 = \{n \in\Bbb N : f(n) = 1\}$. Then, $k\in E_2$ as well and $f_k(k)=1$, which contradicts the premise $f_k(k)=0$. It follows, $E\neq E_2$, which contradicts what we wanted to show.
If $k\notin E$, then $f_k(k)=1$. Assume $E=E_2$. Then, $k\notin E_2$ as well and $f_k(k)=0$, which contradicts the premise $f_k(k)=1$. It follows $E\neq E_2$, which contradicts what we wanted to show.


TITLE: Resultant force on a pulley doesn't seem logical to me
QUESTION [0 upvotes]: I have a simple mechanics problem (to the left) that can be solved using newton's second law.

Strings are all the same and are light and inextensible and vertical.
When we try finding the tension using these equations:
$3-T=0.3a$ | $T-2=0.2a$
...after solving simultaneously we get that the tension is 2.4 Newtons.
Since the strings are connected to the pulley it's what causes the tension so according to newton's third law it has equal and opposite reaction to the tensions on it. The sum of the tensions is equal to 4.8 and so is the reaction force from the wall.
I'm having trouble with that because I imagine it like the diagram on the right as a combined body having a combined weight and then the tension would be equal to the weight and so...
Can any one help me fix my idea?
Thanks :)

REPLY [1 votes]: On the left, the weights are accelerating.
If you take the asymmetry to its limit, with weights of 5kg and 0kg instead of 3kg and 2kg, then the 5kg weight simply falls with acceleration $g$, and the tension in the string is 0.


TITLE: Limit of zero sets of harmonic functions
QUESTION [6 upvotes]: Let $u_n : \mathbb{R}^n \to \mathbb{R}$ be a sequence of harmonic functions which converge uniformly on compact subsets. The limit function $u$ (which we assume to be not identically $0$) is clearly harmonic (via mean value property). Suppose we denote by $Z_{f}$ the set of zeros of a function $f$.
My question is, is there a convergence of $Z_{u_n}$ to $Z_u$ in any reasonable sense (maybe when restricted to compact regions)? This is probably well-known, and in that case, this is mainly a reference request. Thanks in advance!

REPLY [4 votes]: Yes.
With an appropriate topology, the function mapping a non-zero harmonic function to its zero set is continuous. The result actually applies to open mappings from a locally compact locally connected space $X$ to $\mathbb{R}$, and harmonic functions are just a particular example of such open mappings.
If $X$ is a topological space, then let $H(X)$ denote the collection of all closed subsets of $X$. The set $H(X)$
If $X$ is a compact Hausdorff space and $U,V_1,\dots,V_n$ are non-empty open subsets of $X$, then let $[U;V_1,\dots,V_n]$ be the collection of all closed subsets $C\subseteq X$ where $C\subseteq U$ and where $C\cap V_1\neq\emptyset,\dots,C\cap V_n\neq\emptyset$. Then the Vietoris topology on $H(X)$ is the topology with basis consisting of all sets of the form $[U;V_1,\dots,V_n]$.
Proposition: Let $X$ be a locally compact locally connected Hausdorff space. Let $H$ be a collection of continuous open mappings $f:X\rightarrow\mathbb{R}$. Give $H$ the topology of uniform convergence on compact sets. Define a mapping $T:H\rightarrow X\cup\{\infty\}$ by letting $T(h)=Z(h)\cup\{\infty\}$. Then the function $T$ is continuous.
Proof: Suppose that $h\in H$. Let $z\in Z(h)$. Let $U$ be a neighborhood of $z$. Then there is some connected open set $V$ with $z\in V\subseteq\overline{V}\subseteq U$ and where $V$ is compact. In this case, we have $h(r)>0,h(s)<0$ for some $r,s\in V$. Let $\delta=\min(h(r),-h(s))$. Therefore, if $g\in H$ and $\|g|_U-h|_U\|<\delta$, then $g(r)>0,g(s)<0$. Since $V$ is connected and $r,s\in V,g(r)>0,g(s)<0$, there is some $t\in V$ with $g(t)=0$ (this is the reasoning that Giorgio Metafune has made in the comments).
We conclude that if $Z(h)\cap U\neq\emptyset$, then there is some $\delta>0$ where if $\|(g-h)|_U\|<\delta$, then $Z(g)\cap U\neq\emptyset$ as well.
Suppose now that $h\in H$, and $T(h)\in[U;V_1,\dots,V_n]$ for open subsets $U,V_1,\dots,V_n\subseteq X\cup\{\infty\}$. Then $Z(h)\cup\{\infty\}\subseteq U$ and $Z(h)\cup\{\infty\}\cap V_j\neq\emptyset$ for $1\leq j\leq n$. Without loss of generality, assume that $1\leq m\leq n$ and $\infty\not\in V_j$ for $1\leq j\leq m$ where $\infty\in V_j$ for $m<j\leq n$. Then $V_j$ is an open subset of $X$ with $Z(h)\cap V_j\neq\emptyset$ for $1\leq j\leq m$.
Therefore, for $1\leq j\leq m$ there is a relatively compact open $W_j\subseteq V_j$ and some $\delta_j$ where if $\|(g-h)|_{W_j}\|<\delta_j$, then $Z(g)\cap V_j\neq\emptyset$. Now, set $W=W_1\cup\dots\cup W_m$ and $\delta=\min(\delta_1,\dots,\delta_m)$. Then whenever $\|(g-h)|_W\|<\delta$, we have $Z(g)\cap V_j\neq\emptyset$ for $1\leq j\leq m$.
Now let $C=X\setminus U$. Then $C$ is a compact subset of $X$ with $0\not\in g[C]$. Therefore, there is some $\epsilon>0$ where $|g(c)|\geq\epsilon$ for each $c\in C$. Therefore, if $\|(h-g)|_C\|<\epsilon$, then $g(c)\neq 0$ whenever $c\in C$, so $Z(g)\subseteq U$.
Therefore if $g\in H$ and $\|(h-g)|_{C\cup W}\|<\min(\delta,\epsilon)$, then
$g\in[U;V_1,\dots,V_n]$ as well. We therefore conclude that the function $T$ is continuous at the point $h$.
Q.E.D.


TITLE: Steepest descent method analytically
QUESTION [0 upvotes]: I want to use steepest descent method to minimize the function below and show the first method of iteration analytically with starting point xt[1,1]
f(x) = x^3 + xy + x^2y^2 – 3x
I have gone to the point where I know the next point x1 is found by considering (1,1) + r(3,3) which is (1+3r), (1+3r). However I am not sure how to proceed from there.
I know I need to expand this and find a maximum point but I am not sure.

REPLY [0 votes]: For $f(\bar{x}) = x^3 + xy + x^2y^2 – 3x$ the gradient is
$$\left\{3x^2+2xy^2+y-3, x(2xy+1)\right\}$$
For starting point $\{1,1\}$ the gradient is $\{3,3\}$. Therefore, the first step strives to minimize $f(\bar{x})$ for $\bar{x} = \{1+3r, 1+3r\}$. Typically, this is done with a one-dimensional search procedure like the golden section search.
WolframAlpha finds the minimum is at $r = -0.138246$.
Provided, the one-dimensional search is successful, you get a new starting point for the next round: Compute the gradient for this point, use the line search to obtain the next point, check if the termination criterion (accuracy or number of steps) for your search is reached, and move on until you either succeeded or failed to reach an optimum.


TITLE: Sufficient statistics and the Rayleigh distribution
QUESTION [1 upvotes]: Suppose $Y_1, Y_2, Y_3$ are iid random variables from a Rayleigh distribution with parameter $\theta$. The pdf of $Y$ is:
$$f(y; \theta ) = \begin{cases} \frac{2y}{\theta}e^{-y^2/\theta} & , & 0 < y < \infty
\\
0 & & \text{otherwise}
\end{cases}$$
Show that $\sum\limits_{i=0}^n y^2$ is sufficient for $\theta$.
I know that I need to use the likelihood function, and that will be:
$$\frac{2^n}{\theta^n}e^{-(\sum_{i=0}^n y^2)/\theta}$$
But, I don't know how to factor this correctly to give me my $g(u)$ and $h(y_1\text{ to }y_n)$.  At first, I split the exponential part, but that put my sufficient statistic in the $h(u)$, whereas I need it in the $g(u)$.

REPLY [0 votes]: I suspect your likelihood function should be $$\frac{2^n \prod_i y_i}{\theta^n}e^{-(\sum_i y_i^2)/\theta}$$
You want to split this into something looking like $h(\mathbf{Y})\,g(\theta, T(\mathbf{Y}))$, so you could try for example $h=2^n \prod_i y_i$ and $g=\frac{1}{\theta^n}e^{-(\sum_i y_i^2)/\theta}$


TITLE: How to calculate Rotation Matrix in android from accelerometer and magnetometer sensor
QUESTION [4 upvotes]: I found the rotation matrix returned by SensorManager.getRotationMatrix from link: What's the best 3D angular co-ordinate system for working with smartphone apps
The rotation matrix is:

But I cannot find the steps to reconstruct this matrix when I used the rotation matrix for each axis as specified in link: http://www.freescale.com/files/sensors/doc/app_note/AN3461.pdf
Rotation matrix for each axis

I really appreciate if anyone can show me the steps so that I can construct the above rotation matrix because I cannot find any document that shows me the steps to create this matrix.
And at the above link, there is also a matrix as following:

How can I calculate the matrix Matrix
.What does it mean?

REPLY [1 votes]: I have the same problem with you at first ,but after i compute the rotation
matrix multiplication in the order y, x, z axis, the result turns right. The
formula in the pdf is wrong. Another hint is that the in Android, the angle
around -X axis is pitch(value[1]) and the angle around -Z is azimuth(value[0]).
So the angle around X axis is -pitch(-value[1]) and the angle around Z is
-azimuth(-value[0]).
    Finally, multiply all the rotation matrix together u can get the right result.


TITLE: Capacitated multiple vehicle routing problem with handovers
QUESTION [7 upvotes]: I'm looking for literature about a variant of the capacitated vehicle/fleet routing problem (a.k.a. VRP, CVRP, etc.) that takes into account the possibility of handovers between multiple vehicles, i.e. the ability to drop off an item from a vehicle and let another one pick it up.
Put otherwise, we could say we are allowing an item to be transported (sequentially) by more than one vehicle, e.g.:

at time $T^u$ vehicle $V_1$ takes the item from the pick-up location $P^u$
at time $T^d_1 \geq T^u$ vehicle $V_1$ drops it off at an intermediate point $P_1$
at time $T^u_1 \geq T^d_1$ vehicle $V_2$ picks it up from the intermediate point $P_1$
at time $T^d_2 \geq T^u_1$ vehicle $V_2$ drops it off at the intermediate point $P_2$
...
at time $T^u_{n-1} \geq T^d_{n-1}$ vehicle $V_n$ picks it up from intermediate point $P_{n-1}$
at time $T^d \geq T^u_{n-1}$ vehicle $V_n$ delivers it to the final destination $P^d$.

In the VRP overviews I found this variant is not mentioned so I was wondering if anybody knew if it has been investigated at all.

note: the term handover is something I came up with to describe the problem: it may not be the one commonly used to denote this kind of variant. The intended meaning, w.r.t. the example above is that e.g. when vehicle $V_1$ drops off the object $O_1$ at $P_1$ and $V_2$ picks it up we could say that "$V_1$ hands over object $O_1$ to $V_2$ in $P_1$".

update: Reworded to clarify that I'm talking about multiple vehicles (it was just kind-of implicit in the original wording).

REPLY [6 votes]: These problems are studied but with different terminology such as drop-off. See below and references therein.
The Finite Capacity Dial-A-Ride Problem,
M. Charikar and B. Raghavachari,
in Proceedings of the 39th Annual IEEE Conference on Foundations of Computer Science (1998)
Algorithms for Capacitated Vehicle Routing
M. Charikar, S. Khuller and B. Raghavachari,
in Proceedings of the 30th Annual ACM Symposium on Theory of Computing (1998).
Dial a Ride from k-forest
ACM Transactions on Algorithms, 6(2):2010
Anupam Gupta, MohammadTaghi Hajiaghayi, Viswanath Nagarajan, and R. Ravi


text:
\begin{document}

\begin{frontmatter}

\title{Limit Theory for Moderate Deviation from Integrated GARCH Processes}

\author{Yubo Tao\fnref{myft1}}
\address{90 Stamford Rd, Singapore Management University}
\fntext[myft1]{I would like to thank the co-editor, an associate editor and the referees for helping improve the paper. All possible errors are mine. Yubo Tao, School of Economics, Singapore Management University, 90 Stamford Road, Singapore 178903. Email: yubo.tao.2014@phdecons.smu.edu.sg.}

\begin{abstract}
	This paper develops the limit theory of the GARCH(1,1) process that moderately deviates from IGARCH process towards both stationary and explosive regimes. The GARCH(1,1) process is defined by equations $u_t = \sigma_t \varepsilon_t$, $\sigma_t^2 = \omega + \alpha_n u_{t-1}^2 + \beta_n\sigma_{t-1}^2$ and $\alpha_n + \beta_n$ approaches to unity as sample size goes to infinity. The asymptotic theory extends \cite{BerkesHorvathKokoszka2005} by allowing the parameters to have a slower rate of convergence. The results can be applied to unit root test for processes with mildly-integrated GARCH innovations (e.g. \cite{Boswijk2001}, \cite{Cavaliere2007, Cavaliere2009}) and deriving limit theory of estimators for models involving mildly-integrated GARCH processes (e.g. \cite{JensenRahbek2004}, \cite{FrancqZakoian2012,FrancqZakoian2013}).
\end{abstract}

\begin{keyword}
	Central Limit Theorem \sep Limiting Process \sep Localization \sep Explosive GARCH \sep Volatility Process
\MSC[2010] 62M10 \sep 91B84
\end{keyword}

\end{frontmatter}

\section{Introduction} \label{secIntro}

The model considered in this paper is a GARCH(1,1) process:
\begin{align*}
&\text{(Return Process)} \quad && u_t = \sigma_t \varepsilon_t, \\
&\text{(Volatility Process)} \quad && \sigma_t^2 = \omega + \alpha_n u_{t-1}^2 + \beta_n\sigma_{t-1}^2,\quad
\omega > 0\text{, }\alpha_n \geq 0\text{, and }\beta_n \geq 0,
\end{align*}
where $\lbrace\varepsilon_t\rbrace_{t=0}^{n}$ is a sequence of independent identically distributed (i.i.d) variables such that $E\varepsilon_0 = 0$ and $E\varepsilon_0^2 = 1$.

Unlike conventional GARCH(1,1) process, the innovation process considered in this paper is a mildly-integrated GARCH process whose key parameters, $\alpha_n$ and $\beta_n$, are changing with the sample size, viz.
\begin{equation*}
	\alpha_n = O(n^{-p}), \quad \beta_n = 1+ O(n^{-q}), \text{ where } p, q \in (0, 1),
\end{equation*}
and
\begin{equation*}
	\gamma_n = \alpha_n + \beta_n - 1 = O(n^{-\kappa}), \quad \kappa = \min\lbrace p, q\rbrace.
\end{equation*}
The limiting process of this GARCH process is first derived in \cite{BerkesHorvathKokoszka2005} by imposing the assumption $\kappa \in (1/2,1)$. Extending their results, we obtain the limiting process that applies to parameter values that covers the whole range of $(0,1)$. This is a non-trivial extension because when the process deviates further from the integrated GARCH process, the approximation errors in \cite{BerkesHorvathKokoszka2005} diverges and thus a different normalization is needed.

\section{Main Results}
The main results are summarized in the following one proposition and three theorems. The first proposition modifies the additive representation for $\sigma_t^2$ in \cite{BerkesHorvathKokoszka2005} to accommodate $\kappa \in (0,1)$. Based on the proposition, we establish three theorems to describe the asymptotic behaviours of $\sigma_t^2$ and $u_t$ under the cases $\gamma_n \lesseqqgtr 0$ respectively.

To establish the additive representation of $\sigma_t^2$, we make the following assumptions on the distribution of the innovations $\{\varepsilon_t\}_{t=0}^n$ and the convergence rate of the GARCH coefficients, $\alpha_n$ and $\beta_n$.
\begin{assumption} \label{ass01}
	$\lbrace \varepsilon_t\rbrace_{t=0}^n$ is an i.i.d sequence with $E\varepsilon_0^2 = 1$ and $E|\varepsilon_0|^{4+\delta} < \infty$, for some $\delta > 0$.
\end{assumption}
\begin{assumption} \label{ass02}
	$\alpha_n\log\log n \rightarrow 0$, $n\alpha_n \rightarrow \infty$ and $\beta_n \rightarrow 1$.
\end{assumption}
Assumption \ref{ass01} imposes a non-degeneracy condition on the distribution of $\varepsilon_t^2$ and thus ensures its applicability to the central limit theorem. Assumption \ref{ass02} bounds the convergence rate of $\alpha_n$ so that the normalized sequence could converge to a proper limit. Based on these assumptions, we obtain a modified additive representation for $\sigma_t^2$ in Proposition \ref{propAddRep} on the top of \cite{BerkesHorvathKokoszka2005}.
\begin{proposition} [Additive Representation] \label{propAddRep}
	Under Assumption \ref{ass01} and \ref{ass02}, we have the additive representation for $\sigma_t^2$ as
	\begin{align*}
	\sigma_t^2 &= \sigma_0^2t^{t/2}e^{\sqrt{t}\gamma_n}\left(1+\dfrac{\alpha_n}{\sqrt{t}}\sum_{j=1}^{t}\xi_{t-j} + R_{t}^{(1)}\right) + \omega\left[1+\sum_{j=1}^{t}t^{j/2}e^{\frac{j\gamma_n}{\sqrt{t}}}\left(1+\dfrac{\alpha_n}{\sqrt{t}}\sum_{i=1}^{j}\xi_{t-i} + R_{t,j}^{(2)}\right)\left(1+ R_{t,j}^{(3)}\right)\right]
	\end{align*}
	where $\xi_t = \varepsilon_t^{2}-1$ and the remainder terms satisfy
	\begin{align*}
	\left\vert R_{t}^{(1)} \right\vert = O_p\left(\alpha_n^2 + \gamma_n^{2}\right),  &\quad \max\limits_{1 \leq j \leq t}\left\vert R_{t,j}^{(2)} \right\vert = O_p\left(\alpha_n^2\right) \\
	\max\limits_{1 \leq j \leq t}\dfrac{1}{j\log\log j}\left\vert R_{t,j}^{(2)} \right\vert = O_p\left(\dfrac{\alpha_n^2}{t}\right), &\quad \max\limits_{1 \leq j \leq t}\dfrac{1}{j}\left\vert R_{t,j}^{(3)} \right\vert = O_p\left(\dfrac{\alpha_n^2 + \gamma_n^{2}}{t}\right)
	\end{align*}
\end{proposition}

\begin{remark}
	The key difference between our results and \cite{BerkesHorvathKokoszka2005} is the convergence rate of the approximation errors. In \cite{BerkesHorvathKokoszka2005}, the approximation error $\vert R_{t}^{(p)} \vert$, $\forall p = \{1, 2, 3\}$ is of order $t(\alpha_n^2 + \gamma_n^{2})$ or $t\alpha_n^2$ asymptotically. Hence, these errors are negligible only when $\kappa \in (1/2,1)$. We relax this restrictive assumption by normalizing the original terms with $\sqrt{t}$. Under this new normalization, all the approximation errors remains negligible when $\kappa \in (0,1)$.
\end{remark}

To formulate the theorems below, I introduce the following notations. For $0 < t_1 < t_2 < \cdots < t_N <1$ define $k(m) = \lfloor nt_m \rfloor$, $1 \leq m \leq N$. Further, we need the assumptions for relative convergence rate between $\alpha_n$ and $\gamma_n$ to regulate the asymptotic behaviours of returns and volatilities for near-stationary case.

\begin{assumption} \label{ass03}
	$\dfrac{\sqrt{\lvert \gamma_n \rvert}}{\alpha_n n^{1/4}} \rightarrow \infty$, while $\dfrac{\sqrt{\lvert \gamma_n \rvert^{3}}}{\alpha_n n^{1/4}} \rightarrow 0$, as $n \rightarrow \infty$.
\end{assumption}

Assumption \ref{ass03} imposes a rate condition on the localized parameters $\alpha_n$ and $\gamma_n$. This condition is less restrictive than that in \cite{BerkesHorvathKokoszka2005} in the sense that instead of requiring $\lvert \gamma_n \rvert^{3/2}/\alpha_n$ to converge to 0, we allow it to diverge slowly at a rate of $n^{1/4}$. The relaxation of the assumption also attributes to the change of the normalization.

\begin{theorem} [Near-stationary Case] \label{thmNS}
	Suppose $\gamma_n < 0$, then under Assumption \ref{ass01}-\ref{ass03}, the random variables
	\begin{equation*}
	\dfrac{\sqrt{2\lvert \gamma_n \rvert^{3}}}{\alpha_nk(m)^{1/4}} \dfrac{1}{\sqrt{E\xi_0^2}} \left(\dfrac{\sigma_{k(m)}^2}{\omega {k(m)}^{k(m)/2}} -  \sum_{j=1}^{k(m)-1}e^{\frac{j\gamma_n}{\sqrt{k(m)}}}\right) \xrightarrow{d} \mathcal{N}(0,1).
	\end{equation*}

	In addition, the random variables
	\begin{equation*}
	\left(\dfrac{\lvert\gamma_n\rvert }{\omega {k(m)}^{(k(m)+1)/2}} \right)^{1/2} u_{k(m)}
	\end{equation*}
	are asymptotically independent, each with the asymptotic distribution equals to that of $\varepsilon_0$.
\end{theorem}

\begin{theorem} [Integrate Case] \label{thmIN}
	Suppose $\gamma_n = 0$, then under Assumption \ref{ass01} and \ref{ass02}, the volatility has the  asymptotic distribution
	\begin{equation*}
	\dfrac{k(m)^{1/2}}{n^{3/2}\alpha_n}\dfrac{1}{\sqrt{E\xi_0^{2}}}\left(\dfrac{\sigma_{k(m)}^2}{\omega {k(m)}^{k(m)/2}} - k(m)\right)  \xrightarrow{d} \int_{0}^{t_m}xdW(x).
	\end{equation*}

	In addition, the random variables
	\begin{equation*}
	\left(\omega {k(m)}^{k(m)/2 + 1}\right)^{-1/2}u_{k(m)}
	\end{equation*}
	are asymptotically independent, each with the asymptotic distribution equals to that of $\varepsilon_0$.
\end{theorem}

Similar to the near-stationary case, we have to impose additional assumption on the relative speed of converging to zero between $\alpha_n$ and $\gamma_n$.
\begin{assumption} \label{ass04}
	$ \gamma_n / \alpha_n \rightarrow 0$, as $n \rightarrow \infty$.
\end{assumption}
\begin{theorem} [Near-explosive Case] \label{thmNE}
	Suppose $\gamma_n > 0$, then under Assumption \ref{ass01}, \ref{ass02} and \ref{ass04}, the volatility has the asymptotic distribution
	\begin{equation*}
	\dfrac{\gamma_n e^{-\sqrt{k(m)}\gamma_n}}{\alpha_n\sqrt{k(m)}} \dfrac{1}{\sqrt{E\xi_0^2}} \left(\dfrac{\sigma_{k(m)}^2}{\omega k(m)^{k(m)/2}} - \sum_{j=1}^{k(m)-1}e^{\frac{j\gamma_n}{\sqrt{k(m)}}}\right) \Rightarrow W(t_m).
	\end{equation*}

	In addition, the random variables
	\begin{equation*}
	\left(\dfrac{\gamma_n e^{-\sqrt{k(m)}\gamma_n}}{\omega k(m)^{(k(m)+1)/2}}\right)^{1/2}u_{k(m)}
	\end{equation*}
	are asymptotically independent, each with the asymptotic distribution equals to that of $\varepsilon_0$.
\end{theorem}

\begin{remark}
	As one may notice, the rate of convergence for both volatility process and return process in all three cases decreases to 0 asymptotically. These seemingly awkward results are reasonable in the sense that the convergence rate is a part of the normalization which reflects the order of the process. In other words, when we compute a partial sum of $X$s in form of $\sum_{i=1}^{n} a_i X_i$, the normalization just plays the role of $a_i$ which is usually required to decrease to 0 for applying a central limit theorem.
\end{remark}

\section{Proofs}
In this section, I present detailed proofs for all the propositions and the theorems listed in the previous section. For readers' convenience, I provide a roadmap for understanding the proofs of the theorems. In general, the proofs are done in three steps:

\textit{Step 1:} We decompose the volatility process into 4 components, $\sigma_{k,s}^2$, $s = 1, \cdots, 4$, by expanding the multiplicative form provided in Proposition \ref{propAddRep}.

\textit{Step 2:} We show the first 3 volatility components are negligible after normalization, and the last term converges to a proper limit by using Cramer-Wold device and Liapounov central limit theorem or Donsker's theorem.

\textit{Step 3:} We figure out a normalization to make the normalized volatility converges to 1. Then, applying this normalization to the return process, we complete the proof.

\begin{proof}[\textbf{Proof of Proposition \ref{propAddRep}}]
	First, note the GARCH(1,1) model can be written into the following multiplicative form:
	\begin{align*}
		\sigma_t^2 &= \sigma_0^2\prod_{i=1}^{t}\left(\beta_n + \alpha_n\varepsilon_{t-i}^2\right) + \omega\left[1 + \sum_{j=1}^{t-1}\prod_{i=1}^{j}\left(\beta_n + \alpha_n\varepsilon_{t-i}^{2}\right)\right] \\
		&= \sigma_0^2 t^{t/2} \prod_{i=1}^{t}\dfrac{\left(\beta_n + \alpha_n\varepsilon_{t-i}^2\right)}{\sqrt{t}} + \omega\left[1 + t^{t/2}\sum_{j=1}^{t-1}\prod_{i=1}^{j}\dfrac{\left(\beta_n + \alpha_n\varepsilon_{t-i}^{2}\right)}{\sqrt{t}}\right].
	\end{align*}

	Note that
	\begin{equation*}
		\max\limits_{1 \leq i \leq t} \dfrac{\left\vert \beta_n + \alpha_n\varepsilon_{t-i}^2 - 1 \right\vert}{\sqrt{t}} \leq \dfrac{\vert \gamma_n \vert}{\sqrt{t}} + \alpha_n\max\limits_{1 \leq i \leq t}\dfrac{\vert \varepsilon_{t-i}^2 - 1 \vert}{\sqrt{t}} =  \dfrac{\vert \gamma_n \vert}{\sqrt{t}} + \alpha_n\max\limits_{1 \leq i \leq t-1}\dfrac{\vert \varepsilon_{i}^2 - 1 \vert}{\sqrt{t}}.
	\end{equation*}
	Then by Assumption \ref{ass01} and \cite{ChowTeicher2012}, we have the almost sure convergence of
	\begin{equation*}
		\max_{1 \leq j \leq t-1}\vert \varepsilon_{i}^2 - 1 \vert = O(\sqrt{t}).
	\end{equation*}
	Therefore, the term above is
	\begin{equation*}
		\max\limits_{1 \leq i \leq t} \dfrac{\left\vert \beta_n + \alpha_n\varepsilon_{t-i}^2 - 1 \right\vert}{\sqrt{t}} = o_p(1).
	\end{equation*}

	Now consider the sequence of events
	$$ A_n = \left\lbrace \max\limits_{1 \leq i \leq t} \dfrac{\lvert\beta_n + \alpha_n\varepsilon_{t-i}^2 - 1 \rvert}{\sqrt{t}} \leq \dfrac{1}{2}\right\rbrace. $$
	From the previous result we know $\lim\limits_{n \rightarrow \infty}P(A_n) = 1$. Then by Taylor expansion, $\lvert \log(1+x) - x \rvert \leq 2x^2$, $\lvert x\rvert \leq 1/2$ on the event $A_n$, which implies
	\begin{align*}
		\left\vert R_{t,j}^{(3)} \right\vert &= \left\vert \sum_{i=1}^{j} \log\dfrac{\left(\beta_n + \alpha_n\varepsilon_{t-i}^2\right)}{\sqrt{t}} - \sum_{i=1}^{j}\dfrac{\left(\gamma_n + \alpha_n\xi_{t-i}\right)}{\sqrt{t}} \right\vert \\
		&= \left\vert \sum_{i=1}^{j} \log\dfrac{\left(\gamma_n + \alpha_n\xi_{t-i} + 1\right)}{\sqrt{t}} - \sum_{i=1}^{j}\dfrac{\left(\gamma_n + \alpha_n\xi_{t-i}\right)}{\sqrt{t}} \right\vert \\
		&\leq \sum_{i=1}^{j} \left\vert \log\left(\dfrac{\gamma_n + \alpha_n\xi_{t-i}}{\sqrt{t}} + 1 \right)- \dfrac{\left(\gamma_n + \alpha_n\xi_{t-i}\right)}{\sqrt{t}} \right\vert \\
		&\leq 2 \sum_{i=1}^{j} \dfrac{\left(\gamma_n + \alpha_n\xi_{t-i}\right)^2}{t} \leq \dfrac{4j\gamma_n^2}{t} + \dfrac{4\alpha_n^2\sum_{i=1}^{j}\xi_{t-i}^2}{t}.
	\end{align*}

	By Assumption \ref{ass01} and law of large numbers (LLN), we know
	\begin{equation*}
		\max\limits_{1 \leq j \leq t} \dfrac{1}{j} \left\vert \sum_{i=1}^{j} \xi_{t-i}^2\right\vert \sim \max\limits_{1 \leq j \leq t} \dfrac{1}{j} \left\vert \sum_{i=1}^{j} \xi_{i}^2\right\vert = O_p(1).
	\end{equation*}
	Then by the equation above, we have
	\begin{equation*}
		\max\limits_{1 \leq i \leq j} \dfrac{1}{j}\lvert R_{t,j}^{(3)} \rvert = O_p\left(\dfrac{\gamma_n^2 + \alpha_n^2}{t}\right).
	\end{equation*}

	Now by direct plugging into the key multiplicative term we care about, we have
	\begin{align*}
		\prod_{i=1}^{j}\dfrac{\left(\beta_n + \alpha_n\varepsilon_{t-i}^2\right)}{\sqrt{t}} &= \exp\left\lbrace \sum_{i=1}^{j} \log\left(\dfrac{\beta_n + \alpha_n\varepsilon_{t-i}^2}{\sqrt{t}}\right)\right\rbrace \\
		&= \exp\left\lbrace \dfrac{j\gamma_n}{\sqrt{t}}\right\rbrace \exp\left\lbrace\dfrac{\alpha_n\sum_{i=1}^{j}\xi_{t-i}}{\sqrt{t}} \right\rbrace \exp\left\lbrace R_{t,j}^{(3)}\right\rbrace \\
		&= e^{\frac{j\gamma_n}{\sqrt{t}}}\exp\left\lbrace\dfrac{\alpha_n\sum_{i=1}^{j}\xi_{t-i}}{\sqrt{t}} \right\rbrace\left(1+ R_{t,j}^{(3)}\right).
	\end{align*}

	Further, note $\lbrace \xi_t\rbrace_{t=1}^{n}$ is an i.i.d sequence with $E\xi_0^{2} < \infty$, then we know
	\begin{equation*}
		\max\limits_{1 \leq j \leq t} \left\vert \sum_{i=1}^{j} \xi_{t-i} \right\vert = O_p(\sqrt{t}),
	\end{equation*}
	which implies
	\begin{equation*}
		\max\limits_{1 \leq j \leq t} \left\vert \dfrac{\alpha_n}{\sqrt{t}}\sum_{i=1}^{j} \xi_{t-i} \right\vert = O_p(\alpha_n) = o_p(1).
	\end{equation*}

	Similarly, we define the sequence of events
	\begin{equation*}
		B_n = \left\lbrace \max\limits_{1 \leq j \leq t}\left\vert \dfrac{\alpha_n}{\sqrt{t}}\sum_{i=1}^{j} \xi_{t-i} \right\vert \leq \dfrac{1}{2} \right\rbrace,
	\end{equation*}
	which is known to have the property $\lim\limits_{n \rightarrow \infty} P(B_n) = 1$. Then by Taylor expansion, $\lvert \exp(x) - (1+x) \rvert \leq \sqrt{e}x^2/2 $ when $\lvert x \rvert \leq 1/2$, on the event $B_n$
	\begin{equation*}
		\left\vert R_{t,j}^{(2)}\right\vert = \left\vert\exp\left\lbrace \dfrac{\alpha_n}{\sqrt{t}}\sum_{i=1}^{j} \xi_{t-i} \right\rbrace - \left(1 + \dfrac{\alpha_n}{\sqrt{t}}\sum_{i=1}^{j} \xi_{t-i}\right)\right\vert \leq \dfrac{\sqrt{e}}{2}\left(\dfrac{\alpha_n}{\sqrt{t}}\sum_{i=1}^{j} \xi_{t-i}\right)^2 = O_p\left(\alpha_n^2\right),
	\end{equation*}
	and by law of iterated logarithm, we know
	\begin{equation*}
		\max\limits_{1 \leq j \leq t}\dfrac{1}{j \log\log j}\left(\dfrac{\alpha_n}{\sqrt{t}}\sum_{i=1}^{j} \xi_{t-i}\right)^2 = O_p\left(\dfrac{\alpha_n^2}{t}\right).
	\end{equation*}

	Combining the results above, we have thus showed that
	\begin{equation*}
		\prod_{i=1}^{j}\left(\dfrac{\beta_n + \alpha_n\varepsilon_{t-i}^2}{\sqrt{t}}\right) = e^{\frac{j\gamma_n}{\sqrt{t}}}\left(1 + \dfrac{\alpha_n}{\sqrt{t}}\sum_{i=1}^{j}\xi_{t-i} + R_{t,j}^{(2)}\right)\left(1 + R_{t,j}^{(3)}\right).
	\end{equation*}

	Lastly, by the equation above, we know
	\begin{align*}
		\prod_{i=1}^{t}\left(\dfrac{\beta_n + \alpha_n\varepsilon_{t-i}^2}{\sqrt{t}}\right) &= e^{\frac{t\gamma_n}{\sqrt{t}}}\left(1 + \dfrac{\alpha_n}{\sqrt{t}}\sum_{i=1}^{t}\xi_{t-i} + O_p(\alpha_n^2)\right)\left(1 + O_p(\gamma_n^2 + \alpha_n^2)\right) \\
		&= e^{\sqrt{t}\gamma_n}\left(1 + \dfrac{\alpha_n}{\sqrt{t}}\sum_{i=1}^{t}\xi_{t-i} + O_p(\gamma_n^2 + \alpha_n^2)\right),
	\end{align*}
	and this establishes $R_{t}^{(1)}$.
\end{proof}

\begin{proof}[\textbf{Proof of Theorem \ref{thmNS}}]
	First, we focus on the volatilities. Denote $k = \lfloor nt \rfloor$, $0 < t \leq 1$,
	\begin{align*}
		\sigma_k^2 &= \omega + \sigma_0^2 k^{k/2} e^{\sqrt{k}\gamma_n}\left(1+ \dfrac{\alpha_n}{\sqrt{k}}\sum_{j=1}^{k}\xi_{k-j} + R_k^{(1)}\right)  + \omega k^{k/2}\sum_{j=1}^{k-1}e^{\frac{j\gamma_n}{\sqrt{k}}}\left(1+\dfrac{\alpha_n}{\sqrt{k}}\sum_{i=1}^{j}\xi_{k-i} + R_{k,j}^{(2)}\right)R_{k,j}^{(3)} \\
		&\ \ \ + \omega k^{k/2} \sum_{j=1}^{k-1}e^{\frac{j\gamma_n}{\sqrt{k}}} R_{k,j}^{(2)} + \omega k^{k/2} \sum_{j=1}^{k-1}e^{\frac{j\gamma_n}{\sqrt{k}}}\left(1+\dfrac{\alpha_n}{\sqrt{k}}\sum_{i=1}^{j}\xi_{k-i}\right) \\
		&= \omega + \sigma_{k,1}^{2} + \sigma_{k,2}^{2} + \sigma_{k,3}^{2} + \sigma_{k,4}^{2}.
	\end{align*}

	For $\sigma_{k,1}^{2}$, note $k^{-1/2}\sum_{j=1}^{k}\xi_{k-j}$ is asymptotically normal, then by Proposition \ref{propAddRep},
	\begin{equation*}
		\dfrac{\alpha_n}{\sqrt{k}}\sum_{j=1}^{k}\xi_{k-j} + R_{k}^{(1)} = o_p(1),
	\end{equation*}
	and this implies
	\begin{align*}
		\left\lvert \sigma_{k,1}^2 \right\rvert &= O_p\left(k^{k/2}e^{\sqrt{k}\gamma_n}\right).
	\end{align*}

	For $\sigma_{k,2}^{2}$, note by Lemma 4.1 in \cite{BerkesHorvathKokoszka2005}, we have
	\begin{equation} \label{eqApp01}
		\sum_{j=1}^{k}je^{\frac{j\gamma_n}{\sqrt{k}}} \sim \dfrac{k}{\lvert \gamma_n \rvert^2}\Gamma(2),
	\end{equation}
	and note that
	\begin{equation} \label{eqApp02}
		\max\limits_{1\leq j \leq k-1} \left\lvert \dfrac{\alpha_n}{\sqrt{k}}\sum_{i=1}^{j}\xi_{k-i} + R_{k,j}^{(2)} \right\rvert = o_p(1).
	\end{equation}
	Then by equation (\ref{eqApp01}), (\ref{eqApp02}) and Proposition \ref{propAddRep} we have
	\begin{align*}
		\left\lvert \sigma_{k,2}^{2} \right\rvert &= \left\lvert \omega k^{k/2}\sum_{j=1}^{k-1}je^{\frac{j\gamma_n}{\sqrt{k}}}\left(1 + \dfrac{\alpha_n}{\sqrt{k}}\sum_{i=1}^{j}\xi_{k-i} + R_{k,j}^{(2)}\right)\dfrac{1}{j}R_{k,j}^{(3)} \right\rvert \\
		&= O_p(1)\omega k^{k/2} \dfrac{\alpha_n^2 + \gamma_n^2}{k} \dfrac{k}{\lvert \gamma_n \rvert^2} \\
		&= O_p\left(\dfrac{k^{k/2}\left(\alpha_n^2 + \gamma_n^2\right)}{\gamma_n^2}\right).
	\end{align*}

	For $\sigma_{k,3}^{2}$, similarly, by Proposition \ref{propAddRep} and Lemma 4.1 in \cite{BerkesHorvathKokoszka2005}, we have
	\begin{align*}
		\left\lvert \sigma_{k,3}^{2} \right\rvert &= \left\lvert \omega k^{k/2}\sum_{j=1}^{k-1}e^{\frac{j\gamma_n}{\sqrt{k}}}R_{k,j}^{(2)} \right\rvert \\
		&= O_p(1)\omega k^{k/2}\dfrac{\alpha_n^2}{k}\sum_{j=1}^{k-1}je^{\frac{j\gamma}{\sqrt{k}}}\log\log j \\
		&= O_p\left(\dfrac{k^{k/2} \left(\alpha_n^{2}\log\log k\right) }{\gamma_n^2}\right).
	\end{align*}

	Lastly, for $\sigma_{k,4}^{2}$, by Lemma 4.1 in \ref{propAddRep} we have
	\begin{align*}
		\sigma_{k,4}^{2} &= \omega k^{k/2} \sum_{j=1}^{k-1}e^{\frac{j\gamma_n}{\sqrt{k}}} + \omega k^{k/2} \dfrac{\alpha_n}{\sqrt{k}} \sum_{j=1}^{k-1}e^{\frac{j\gamma_n}{\sqrt{k}}} \sum_{i=1}^{j}\xi_{k-i} \\
		&= O_p\left(\dfrac{k^{k/2}k^{1/2}}{\lvert \gamma_n \rvert}\right) + \omega k^{k/2} \dfrac{\alpha_n}{\sqrt{k}} \sum_{j=1}^{k-1}e^{\frac{j\gamma_n}{\sqrt{k}}} \sum_{i=1}^{j}\xi_{k-i}.
	\end{align*}

	Therefore, we only have to consider the last term in the above equation. Define
	\begin{equation*}
		\tau_m = k(m)^{-1/4}\sum_{j=1}^{k(m)-1}e^{\frac{j\gamma_n}{\sqrt{k(m)}}} \xi_{k(m)-j}, \quad 1 \leq m \leq N,
	\end{equation*}
	and
	\begin{equation*}
		\tau_m^* = k(m)^{-1/2}\sum_{j=1}^{k(m)-1}e^{\frac{j\gamma_n}{\sqrt{k(m)}}} \sum_{i=1}^{j}\xi_{k(m)-i}, \quad 1 \leq m \leq N.
	\end{equation*}

	Then by Cramer-Wold device (Theorem 29.4 of \cite{Billingsley1995}), we have
	\begin{align*}
		\sum_{m=1}^{N}\mu_m\tau_m &= \sum_{i=1}^{k(1)-1}\sum_{m=1}^{N}\dfrac{\mu_m}{k(m)^{1/4}} e^{\frac{(k(m)-i)\gamma_n}{\sqrt{k(m)}}} + \sum_{i=k(1)}^{k(2)-1}\sum_{m=2}^{N}\dfrac{\mu_m}{k(m)^{1/4}} e^{\frac{(k(m)-i)\gamma_n}{\sqrt{k(m)}}}  \\
		&+ \cdots + \sum_{i=k(N-1)}^{k(N)-1}\dfrac{\mu_N}{k(N)^{1/4}} e^{\frac{(k(N)-i)\gamma_n}{\sqrt{k(N)}}} \\
		&= S_1 + S_2 + \cdots + S_N.
	\end{align*}

	Observe that
	\begin{align*}
		ES_1^2 &= E\xi_0^2\left( \sum_{i=1}^{k(1)-1}\sum_{m=1}^{N}k(m)^{-1/4}\mu_m e^{\frac{(k(m)-i)\gamma_n}{\sqrt{k(m)}}}\right)^2 \\
		&= E\xi_0^2 \sum_{m=1}^{N}\dfrac{\mu_m^2}{\sqrt{k(m)}}\sum_{i=1}^{k(1)-1} e^{\frac{2(k(m)-i)\gamma_n}{\sqrt{k(m)}}} + E\xi_0^2 \sum_{1 \leq m \neq l \leq N}\left(k(m)k(l)\right)^{-1/4}\mu_m\mu_l\sum_{i=1}^{k(1)-1} e^{\frac{(k(m)-i)\gamma_n}{\sqrt{k(m)}}+\frac{(k(l)-i)\gamma_n}{\sqrt{k(l)}}} \\
		&= E\xi_0^2 \dfrac{\mu_1^2}{\sqrt{k(1)}}\sum_{i=1}^{k(1)-1} e^{\frac{2(k(1)-i)\gamma_n}{\sqrt{k(1)}}} + E\xi_0^2 \sum_{m=2}^{N}\dfrac{\mu_m^2}{\sqrt{k(m)}}\sum_{i=1}^{k(1)-1} e^{\frac{2(k(m)-i)\gamma_n}{\sqrt{k(m)}}} \\
		&\ \ \ + E\xi_0^2 \sum_{1 \leq m \neq l \leq N}\left(k(m)k(l)\right)^{-1/4}\mu_m\mu_l\sum_{i=1}^{k(1)-1} e^{\frac{(k(m)-i)\gamma_n}{\sqrt{k(m)}}+\frac{(k(l)-i)\gamma_n}{\sqrt{k(l)}}} \\
		&= E\xi_0^2 \dfrac{\mu_1^2}{\sqrt{k(1)}}\sum_{i=1}^{k(1)-1} e^{\frac{2i\gamma_n}{\sqrt{k(1)}}} + E\xi_0^2 \sum_{m=2}^{N}\dfrac{\mu_m^2}{\sqrt{k(m)}}e^{\frac{2(k(m)-k(1))\gamma_n}{\sqrt{k(m)}}}\sum_{i=1}^{k(1)-1} e^{\frac{2i\gamma_n}{\sqrt{k(m)}}} \\
		&\ \ \ + E\xi_0^2 \sum_{1 \leq m \neq l \leq N}(k(m)k(l))^{-1/4}\mu_m\mu_le^{\frac{(k(m)-k(1))\gamma_n}{\sqrt{k(m)}}+\frac{(k(l)-k(1))\gamma_n}{\sqrt{k(l)}}}\sum_{i=1}^{k(1)-1} e^{\frac{i\gamma_n}{\sqrt{k(m)}}+\frac{i\gamma_n}{\sqrt{k(l)}}} \\
		&\sim E\xi_0^2 \mu_1^2\dfrac{1}{2\lvert \gamma_n\rvert} + E\xi_0^2 \sum_{m=2}^{N}\mu_m^2e^{\frac{2(k(m)-k(1))\gamma_n}{\sqrt{k(m)}}} \dfrac{1}{2\lvert \gamma_n \rvert} \\
		&\ \ \ + E\xi_0^2 \sum_{1 \leq m \neq l \leq N}\dfrac{\mu_m\mu_l}{(\sqrt{k(m)} +\sqrt{k(l)}) \lvert\gamma_n\rvert} e^{\frac{(k(m)-k(1))\gamma_n}{\sqrt{k(m)}} + \frac{(k(l)-k(1))\gamma_n}{\sqrt{k(l)}}}  \\
		&= E\xi_0^2 \mu_1^2\dfrac{1}{2\lvert \gamma_n\rvert} + o\left(\dfrac{1}{\lvert \gamma_n \rvert}\right),
	\end{align*}
	we then have
	\begin{align*}
		E\left(\sum_{m=1}^{N} \mu_m \tau_m\right)^2 &= \left(\sum_{m=1}^{N} \mu_m^2\right) E\xi_0\dfrac{1}{2\lvert \gamma_n \rvert} + o\left(\dfrac{1}{\lvert \gamma_n \rvert}\right).
	\end{align*}

	Observe also that, for some $c_i$, $1 \leq i \leq k(N)-1$, we have
	\begin{equation*}
		\sum_{m=1}^{N} \mu_m \tau_m = \sum_{i=1}^{k(N)-1} c_i \xi_i,
	\end{equation*}
	and by Jensen's inequality, we know for some $\delta > 0$,
	\begin{align*}
		\lvert c_i \rvert^{2+\delta} &= \left\vert k(1)^{-1/4}\mu_1 e^{\frac{(k(1)-i) \gamma_n}{\sqrt{k(1)}}} + k(2)^{-1/4}\mu_1 e^{\frac{(k(2)-i) \gamma_n}{\sqrt{k(2)}}}+ \cdots +k(N)^{-1/4}\mu_1 e^{\frac{(k(N)-i) \gamma_n}{\sqrt{k(N)}}} \right\vert^{2+\delta} \\
		&\leq C_1(N)\left[\dfrac{\lvert \mu_1\rvert^{2+\delta}}{k(1)^{1/2+\delta/4}}e^{\frac{(k(1) - i)(2+\delta)\gamma_n}{\sqrt{k(1)}}} + \cdots + \dfrac{\lvert \mu_N\rvert^{2+\delta}}{k(N)^{1/2+\delta/4}}e^{\frac{(k(N) - i)(2+\delta)\gamma_n}{\sqrt{k(N)}}}\right].
	\end{align*}
	This implies that
	\begin{equation*}
		\sum_{i=1}^{k(N)-1} \lvert c_i \rvert^{2+\delta} \sim C_1(N)\lvert \mu_1\rvert^{2+\delta} \dfrac{1}{k(1)^{\delta/4}(2+\delta)\lvert \gamma_n \rvert} + O\left(\dfrac{1}{k(2)^{\delta/4}\lvert \gamma_n\rvert}\right) = o\left(\dfrac{1}{\lvert \gamma_n \rvert}\right).
	\end{equation*}

	Now we can easily check the Liapounov's condition, where
	\begin{equation*}
		\dfrac{\left(\sum_{i=1}^{k(N)-1}\lvert c_i \rvert^{2+\delta} E\lvert \xi_i\rvert^{2+\delta}\right)^{1/(2+\delta)}}{\left(\sum_{i=1}^{k(N)-1}c_i^2 E\xi_i^{2}\right)^{1/2}} = o\left(\lvert \gamma_n \rvert^{1/2 - 1/(2+\delta)}\right) = o_p(1).
	\end{equation*}
	Then by Liapounov central limit theorem (Theorem 27.3, p.362 of \cite{Billingsley1995}), we have
	\begin{equation*}
		\sqrt{2\lvert \gamma_n \rvert}\left[\tau_1, \tau_2, \cdots, \tau_N\right] \xrightarrow{d} \sqrt{E\xi_{0}^2}\left[\eta_1, \eta_2, \cdots, \eta_N\right],
	\end{equation*}
	where $\eta_1, \eta_2, \cdots, \eta_N$ are independent standard normal random variables.

	Now we have to check the relationship between $\tau_m$ and $\tau_m^*$. Note by $k^{-1/2}\left(e^{\frac{\gamma_n}{\sqrt{k}}}-1\right)^{-1} = \left(\gamma_n + o(1)\right)^{-1}$, we have
	\begin{align*}
		\dfrac{1}{\sqrt{k}}\sum_{j=i}^{k-1}e^{\frac{j\gamma_n}{\sqrt{k}}} - \lvert \gamma_n\rvert^{-1}e^{\frac{i\gamma_n}{\sqrt{k}}} &= \dfrac{1}{\sqrt{k}}\dfrac{e^{\frac{k\gamma_n}{\sqrt{k}}} - e^{\frac{i\gamma_n}{\sqrt{k}}}}{e^{\frac{\gamma_n}{\sqrt{k}}}-1} - \lvert \gamma_n\rvert^{-1}e^{\frac{i\gamma_n}{\sqrt{k}}} \\
		&= \left(\gamma_n + o(1)\right)^{-1}\left(e^{\frac{k\gamma_n}{\sqrt{k}}} - e^{\frac{i\gamma_n}{\sqrt{k}}}\right) - \lvert \gamma_n \rvert^{-1} e^{\frac{i\gamma_n}{\sqrt{k}}} \\
		&= \left(\gamma_n^{-1} + O(1)\right) e^{\frac{k\gamma_n}{\sqrt{k}}} - e^{\frac{i\gamma_n}{\sqrt{k}}}O(1).
	\end{align*}

	Then, we know
	\begin{align*}
		E\left[\sqrt{2\lvert \gamma_n\rvert^3}\tau_m^* - \sqrt{2\lvert \gamma_n\rvert}\tau_m\right]^2 &= \dfrac{2\lvert \gamma_n \rvert^3}{\sqrt{k}} E\left[\dfrac{1}{\sqrt{k}}\sum_{i=1}^{k-1}\left(\sum_{j=i}^{k-1}e^{\frac{j\gamma}{\sqrt{k}}}\right)\xi_{k-i} - \lvert \gamma_n\rvert^{-1}\sum_{i=1}^{k-1}e^{\frac{i\gamma_n}{\sqrt{k}}}\xi_{k-i} \right]^{2} \\
		&= \dfrac{2\lvert \gamma_n \rvert^3}{\sqrt{k}} E\xi_0^2\sum_{i=1}^{k-1}\left(\dfrac{1}{\sqrt{k}}\sum_{j=i}^{k-1}e^{\frac{j\gamma_n}{\sqrt{k}}} - \lvert \gamma_n\rvert^{-1}e^{\frac{i\gamma_n}{\sqrt{k}}}\right)^2 \\
		&\sim \dfrac{2\lvert \gamma_n \rvert^3}{\sqrt{k}} E\xi_0^2\left(k\gamma_n^{-2}e^{\sqrt{k}\gamma_n} + \dfrac{\sqrt{k}}{2\lvert\gamma\rvert} - 2\gamma_n^{-1}e^{\sqrt{k}\gamma_n}\dfrac{\sqrt{k}}{\lvert\gamma\rvert}\right) \\
		&= 2E\xi_0^2 O\left(\sqrt{k}\lvert\gamma_n\rvert e^{\sqrt{k}\gamma_n}\right) + o_p(1) \\
		&= o_p(1),
	\end{align*}
	where the last equality comes from the well known limits of $xe^{-x}$,
	\begin{equation*}
		\lim\limits_{x \rightarrow \infty} \dfrac{x}{e^{x}} = \lim\limits_{x \rightarrow \infty}\dfrac{1}{e^x} = 0 \quad \text{and} \quad \lim\limits_{x \rightarrow 0} \dfrac{x}{e^{x}} = 0.
	\end{equation*}

	Therefore, we have
	\begin{equation*}
		\sqrt{2\lvert \gamma_n \rvert^{3}}\left[\tau_1^*, \tau_2^*, \cdots, \tau_N^* \right] \xrightarrow{d} \sqrt{E\xi_0^2}\left[\eta_1, \eta_2, \cdots, \eta_N\right],
	\end{equation*}

	Now combine the results above, we have, for each $k = {\lfloor nt_m \rfloor}$, $m = 1, \cdots, N$
	\begin{align*}
		\dfrac{\sqrt{2\lvert \gamma_n \rvert^{3}}}{\alpha_n k^{1/4}} \dfrac{1}{\sqrt{E\xi_0^2}} \left(\dfrac{\sigma_{k}^2}{\omega {k}^{k/2}} - \sum_{j=1}^{k-1}e^{\frac{j\gamma_n}{\sqrt{k}}}\right) \xrightarrow{d} \mathcal{N}(0,1).
	\end{align*}

	Now, for returns, we know from the above result that
	\begin{align*}
		\dfrac{\lvert\gamma_n\rvert\sigma_{k}^2}{\omega {k}^{(k+1)/2}} - 1 = O_p\left(\dfrac{\alpha_n n^{1/4}}{\sqrt{\lvert \gamma_n \rvert}}\right) = o_p(1).
	\end{align*}
	Therefore, by the return equation, we have
	\begin{equation*}
		\left(\dfrac{\lvert\gamma_n\rvert }{\omega {k}^{(k+1)/2}} \right)^{1/2} u_k =  \left(\dfrac{\lvert\gamma_n\rvert\sigma_{k}^2}{\omega {k}^{(k+1)/2}}\right)^{1/2} \varepsilon_k \sim \varepsilon_k.
	\end{equation*}
\end{proof}

\begin{proof}[\textbf{Proof of Theorem \ref{thmIN}}]
	Similar to Theorem \ref{thmNS}, when $\gamma_n = 0$, the volatility admits the decomposition. Then, for $\sigma_{k,1}^{2}$, by central limit theorem, we know
	\begin{equation*}
		\dfrac{\alpha_n}{\sqrt{k}}\sum_{j=1}^{k}\xi_{k-j} = O_p(\alpha_n) = o_p(1)
	\end{equation*}
	which, combining with Proposition \ref{propAddRep}, implies that
	\begin{equation*}
		\left\lvert \sigma_{k,1}^2 \right\rvert = O_p\left(k^{k/2}\right)
	\end{equation*}

	For $\sigma_{k,2}^{2}$, note that we have established equation (\ref{eqApp02}), then by Proposition \ref{propAddRep}, we have
	\begin{equation*}
		\left\lvert \sigma_{k,2}^{2} \right\rvert = O_p\left(k^{k/2}\alpha_n^{2}\right).
	\end{equation*}

	For $\sigma_{k,3}^{2}$, by Proposition \ref{propAddRep} we have
	\begin{equation*}
		\left\lvert \sigma_{k,3}^{2} \right\rvert = O_p(k^{k/2}\alpha_n^{2}).
	\end{equation*}

	Lastly, for $\sigma_{k,4}^2$, note by Lemma 5.1 in \cite{BerkesHorvathKokoszka2005}, for $k = \lfloor nt \rfloor$, $t \in (0, 1)$ , we have
	\begin{equation*}
		\dfrac{1}{n^{3/2}}\sum_{j=1}^{\lfloor nt \rfloor - 1}\sum_{i=1}^{j}\xi_{k-i} \xrightarrow{d} \sqrt{E\xi_{0}^2}\int_{0}^{t}xdW(x),
	\end{equation*}
	where $W(x)$ is a Wiener process.

	Therefore, for $k(m) = \lfloor nt_m \rfloor$, $m = 1, \cdots, N$, we have
	\begin{align*}
		\dfrac{k(m)^{1/2}}{n^{3/2}\alpha_n}\left(\dfrac{\sigma_{k(m)}^2}{\omega {k(m)}^{k(m)/2}} - k(m)\right) = \dfrac{1}{n^{3/2}}\sum_{j=1}^{\lfloor nt_m \rfloor - 1}\sum_{i=1}^{j}\xi_{k(m)-i} + o_p(1) \xrightarrow{d} \sqrt{E\xi_0^{2}}\int_{0}^{t_m}xdW(x).
	\end{align*}

	Further, note the results above implies that
	\begin{equation*}
		\dfrac{\sigma_{k(m)}^2}{\omega {k(m)}^{k(m)/2 + 1}} - 1 = O_p\left(\left(\dfrac{n}{k}\right)^{3/2}\alpha_n\right) = o_p(1).
	\end{equation*}
	Hence, by return equation, we obtain
	\begin{equation*}
		\left(\dfrac{1}{\omega {k(m)}^{k(m)/2 + 1}}\right)^{1/2}u_{k(m)} = \left(\dfrac{\sigma_{k(m)}^2}{\omega {k(m)}^{k(m)/2 + 1}}\right)^{1/2}\varepsilon_{k(m)} \xrightarrow{d} \varepsilon_{k(m)}.
	\end{equation*}
\end{proof}

\begin{proof}[\textbf{Proof of Theorem \ref{thmNE}}]
	Similar to proof of Theorem \ref{thmNS}, when $\gamma_n > 0$, the volatility admits the additive representation. For $\sigma_{k,1}^{2}$, similar to that in Theorem \ref{thmNS},
	\begin{equation*}
	\left\lvert \sigma_{k,1}^2 \right\rvert = O_p\left(k^{k/2}e^{\sqrt{k}\gamma_n}\right).
	\end{equation*}

	For $\sigma_{k,2}^2$, by Proposition \ref{propAddRep} and equation (\ref{eqApp02}), we have the relation
	\begin{align*}
	\left\lvert \sigma_{k,2}^{2} \right\rvert &= \left\lvert \omega k^{k/2}\sum_{j=1}^{k-1} je^{\frac{j\gamma_n}{\sqrt{k}}}(1+o_p(1))\dfrac{1}{j}R_{k,j}^{(3)} \right\rvert \\
	&=O_p(1)\omega k^{k/2}(\alpha_n^{2} + \gamma_n^{2})\dfrac{e^{\frac{k\gamma_n}{\sqrt{k}}} - e^{\frac{\gamma_n}{\sqrt{k}}}}{e^{\frac{\gamma_n}{\sqrt{k}}}-1} \\
	&< O_p\left(k^{k/2}\left(\alpha_n^2 + \gamma_n^2\right) \dfrac{\sqrt{k}e^{\sqrt{k}\gamma_n}}{\gamma_n}\right),
	\end{align*}
	where the last inequality comes from the fact that
	\begin{equation*}
	\dfrac{e^{\frac{k\gamma_n}{\sqrt{k}}} - e^{\frac{\gamma_n}{\sqrt{k}}}}{e^{\frac{\gamma_n}{\sqrt{k}}}-1} < \dfrac{e^{\sqrt{k}\gamma_n}}{\gamma_n/\sqrt{k}}.
	\end{equation*}

	For $\sigma_{k,3}^2$, by Proposition \ref{propAddRep}, we have
	\begin{align*}
	\left\lvert \sigma_{k,3}^2 \right\rvert &= \left\lvert \omega k^{k/2} \sum_{j=1}^{k-1}e^{\frac{j\gamma_n}{\sqrt{k}}} \left(j\log\log j\right) \dfrac{1}{j\log\log j} R_{k,j}^{(2)}\right\rvert \\
	&= O_p(1)\omega k^{k/2} \left(k\log\log k\right) \dfrac{\alpha_n^{2}}{k}\dfrac{e^{\frac{k\gamma_n}{\sqrt{k}}} - e^{\frac{\gamma_n}{\sqrt{k}}}}{e^{\frac{\gamma_n}{\sqrt{k}}}-1} \\
	&< O_p\left(k^{k/2}\left(\alpha_n^2\log\log k\right)\dfrac{\sqrt{k}e^{\sqrt{k}\gamma_n}}{\gamma_n} \right).
	\end{align*}

	Lastly, for $\sigma_{k,4}^{2}$, we have
	\begin{equation*}
	\sigma_{k,4}^2 = \omega k^{k/2} \sum_{j=1}^{k-1}e^{\frac{j\gamma_n}{\sqrt{k}}} + \omega k^{k/2} \dfrac{\alpha_n}{\sqrt{k}} \sum_{j=1}^{k-1}e^{\frac{j\gamma_n}{\sqrt{k}}} \sum_{i=1}^{j}\xi_{k-i}.
	\end{equation*}

	Now, we introduce the following lemma to assist the proof.
	\begin{lemma} \label{lemApp}
		If Assumption \ref{ass01} and \ref{ass02} hold, then
		\begin{equation*}
		\dfrac{\gamma_n^2}{k}e^{-2\sqrt{k}\gamma_n}E\left(\dfrac{1}{\sqrt{k}}\sum_{j=1}^{k-1}e^{\frac{j\gamma_n}{\sqrt{k}}}\sum_{i=1}^{j}\xi_{k-i} - \dfrac{e^{\sqrt{k}\gamma_n}}{\gamma_n}\sum_{i=1}^{k-1}\xi_i\right)^2 \rightarrow 0.
		\end{equation*}
	\end{lemma}

	Then by Lemma \ref{lemApp}, we have
	\begin{align*}
	\dfrac{\gamma_n e^{-\sqrt{k}\gamma_n}}{\sqrt{k}\alpha_n} \left(\dfrac{\sigma_{k,4}^2}{\omega k^{k/2}} - \sum_{j=1}^{k-1}e^{\frac{j\gamma_n}{\sqrt{k}}}\right) &= \dfrac{\gamma_n e^{-\sqrt{k}\gamma_n}}{\sqrt{k}}\dfrac{1}{\sqrt{k}}\sum_{j=1}^{k-1}e^{\frac{j\gamma_n}{\sqrt{k}}}\sum_{i=1}^{j}\xi_{k-i} + o_p(1) = \dfrac{1}{\sqrt{k}}\sum_{i=1}^{k-1}\xi_i + o_p(1).
	\end{align*}

	Therefore, by Donsker's theorem, we obtain that, for $k(m) = \lfloor nt_m \rfloor$, $t_m \in (0, 1)$ and $m = 1, 2, \cdots, N$,
	\begin{equation*}
	\dfrac{\gamma_n e^{-\sqrt{k(m)}\gamma_n}}{\sqrt{k(m)}\alpha_n} \dfrac{1}{\sqrt{E\xi_0^2}} \left(\dfrac{\sigma_{k(m)}^2}{\omega k(m)^{k(m)/2}} - \sum_{j=1}^{k(m)-1}e^{\frac{j\gamma_n}{\sqrt{k(m)}}}\right) \Rightarrow W(t_m),
	\end{equation*}
	where $W(t)$ is a finite dimensional Wiener process.

	Further, note that
	\begin{equation*}
	\dfrac{\gamma_n}{\sqrt{k}}e^{-\sqrt{k}\gamma_n}\left(\sum_{j=1}^{k-1}e^{\frac{j\gamma_n}{\sqrt{k}}} - \dfrac{\sqrt{k}e^{\sqrt{k}\gamma_n}}{\gamma_n}\right) = o(1),
	\end{equation*}
	then by the result above we know
	\begin{equation*}
	\dfrac{\gamma_n e^{-\sqrt{k(m)}\gamma_n}}{\sqrt{k(m)}}  \left(\dfrac{\sigma_{k(m)}^2}{\omega k(m)^{k(m)/2}} - \sum_{j=1}^{k(m)-1}e^{\frac{j\gamma_n}{\sqrt{k(m)}}}\right) = O_p(\alpha_n) = o_p(1).
	\end{equation*}

	Hence, by return equantion, we derive
	\begin{equation*}
	\left(\dfrac{\gamma_n e^{-\sqrt{k}\gamma_n}}{\omega k^{(k+1)/2}}\right)^{1/2}u_{k} = \left(\dfrac{\gamma_n e^{-\sqrt{k}\gamma_n}}{\omega k^{(k+1)/2}}\sigma_k^2\right)^{1/2}\varepsilon_{k} \sim \varepsilon_k.
	\end{equation*}
\end{proof}

\textbf{Proof of Lemma \ref{lemApp}.} Note that
\begin{equation*}
\dfrac{1}{\sqrt{k}}\sum_{j=1}^{k-1}e^{\frac{j\gamma_n}{\sqrt{k}}}\sum_{i=1}^{j}\xi_{k-i} = \dfrac{1}{\sqrt{k}}\sum_{i=1}^{k-1}\left(\sum_{j=i}^{k-1}e^{\frac{j\gamma_n}{\sqrt{k}}}\right)\xi_{k-i} \quad \text{and} \quad  \sum_{i=1}^{k-1}\xi_i = \sum_{i=1}^{k-1}\xi_{k-i},
\end{equation*}
Then,
\begin{align*}
E\left(\dfrac{1}{\sqrt{k}}\sum_{j=1}^{k-1}e^{\frac{j\gamma_n}{\sqrt{k}}}\sum_{i=1}^{j}\xi_{k-i} - \dfrac{e^{\sqrt{k}\gamma_n}}{\gamma_n}\sum_{i=1}^{k-1}\xi_i\right)^2 &= E\xi_0^{2}\sum_{i=1}^{k-1}\left(\dfrac{1}{\sqrt{k}}\sum_{j=i}^{k-1}e^{\frac{j\gamma_n}{\sqrt{k}}} - \dfrac{\sqrt{k}e^{\sqrt{k}\gamma_n}}{\gamma_n} \right)^2 \\
&= \dfrac{E\xi_0^{2}}{k}\sum_{i=1}^{k-1}\left(\dfrac{e^{\frac{k\gamma_n}{\sqrt{k}}} - e^{\frac{i\gamma_n}{\sqrt{k}}}}{e^{\frac{\gamma_n}{\sqrt{k}}}-1} - \dfrac{\sqrt{k}e^{\sqrt{k}\gamma_n}}{\gamma_n} \right)^2.
\end{align*}

Note by Taylor expansion,
\begin{equation*}
\left\lvert \dfrac{e^{\frac{k\gamma_n}{\sqrt{k}}} - e^{\frac{i\gamma_n}{\sqrt{k}}}}{e^{\frac{\gamma_n}{\sqrt{k}}}-1} - \dfrac{\sqrt{k}e^{\sqrt{k}\gamma_n}}{\gamma_n} \right\rvert \leq C_1\left(\dfrac{\sqrt{k}e^{\frac{i\gamma_n}{\sqrt{k}}}}{\gamma_n} + e^{\sqrt{k}\gamma_n}\right),
\end{equation*}
which implies that
\begin{align*}
\sum_{i=1}^{k-1}\left(\dfrac{e^{\frac{k\gamma_n}{\sqrt{k}}} - e^{\frac{i\gamma_n}{\sqrt{k}}}}{e^{\frac{\gamma_n}{\sqrt{k}}}-1} - \dfrac{\sqrt{k}e^{\sqrt{k}\gamma_n}}{\gamma_n}\right)^2 &\leq 2C_1^2\left(\sum_{i=1}^{k-1}\dfrac{ke^{\frac{2i\gamma_n}{\sqrt{k}}}}{\gamma_n^2} + ke^{2\sqrt{k}\gamma_n} \right) \\
&= O(1)\left(\dfrac{k}{\gamma_n^2}\dfrac{e^{\frac{2k\gamma_n}{\sqrt{k}}} - e^{\frac{2\gamma_n}{\sqrt{k}}}}{e^{\frac{2\gamma_n}{\sqrt{k}}} - 1} + ke^{2\sqrt{k}\gamma_n} \right) \\
&= O(1)\left(\dfrac{k}{\gamma_n^3}e^{2\sqrt{k}\gamma_n} + ke^{2\sqrt{k}\gamma_n}\right).
\end{align*}

Now we can see that
\begin{align*}
\dfrac{\gamma_n^2}{k}e^{-2\sqrt{k}\gamma_n}E\left(\dfrac{1}{\sqrt{k}}\sum_{j=1}^{k-1}e^{\frac{j\gamma_n}{\sqrt{k}}}\sum_{i=1}^{j}\xi_{k-i} - \dfrac{e^{\sqrt{k}\gamma_n}}{\gamma_n}\sum_{i=1}^{k-1}\xi_i\right)^2 &= O(1)\dfrac{\gamma_n^2}{k}e^{-2\sqrt{k}\gamma_n}\dfrac{E\xi_0^2}{k}\left(\dfrac{k}{\gamma_n^3}e^{2\sqrt{k}\gamma_n} + ke^{2\sqrt{k}\gamma_n}\right) = o_p(1).
\end{align*}

\nocite{*}
\bibliography{ExGARCHBib}

\end{document}

config: arxiv


TITLE: Prove that $\{(x,y)\in \mathbb{Z}\times\mathbb{Z}: x^2\equiv y^2 \!\!\!\mod \!\!4\}$ is an equivalence relation
QUESTION [0 upvotes]: I'm trying to prove that the relation R on $\mathbb{Z}$ is an equivalence relation, where R is defined by:
$$\{(x,y)\in \mathbb{Z}\times\mathbb{Z}: x^2\equiv y^2 \!\!\!\mod \!\!4\}$$
I know I need to show that it is reflexive, symmetric, and transitive, but I'm not sure how to do that when working with congruence. Thank you.
I answered my own question below but I'm not sure if I did it correctly. Any feedback would be really helpful. Thanks.

REPLY [1 votes]: Reflexive: Let $x\in\mathbb{Z}$. Then $x^2\equiv x^2\, mod\, 4$ for all $x\in R$. Thus, $(a,a)\in R$, so R is reflexive.
Symmetric: Let $(a,b)\in R$ and let $a^2\equiv b^2\, mod\, 4$. This means that $4\mid a^2-b^2$. Then, it is obvious that $4\mid -(a^2-b^2)$, so $4\mid b^2-a^2$. Therefore, $b^2\equiv a^2\, mod\, 4$. Thus, $(b,a)\in R$, so R is symmetric.
Transitive: Let $(a,b)\in R$ and $(b,c)\in R$. This means that $4\mid a^2-b^2$ and $4\mid b^2-c^2$. Then, there exists integers $l,k$ such that $4l=a^2-b^2$ and $4k=b^2-c^2$. The second expression can be written as $b^2=4k+c^2$. Then, by substitution, we have $4l=a^2-(4k+c^2)=a^2-4k-c^2$. Adding $4k$ to both sides, we get $4l+4k=a^2-c^2$. Factoring out $4$, we get $4(l+k)=a^2-c^2$. Since $l+k$ is an integer, we know that $4\mid a^2-c^2$. Therefore, $a^2\equiv b^2\, mod\, 4$. Thus, $(a,c)\in R$, so R is transitive.
Thus, R is an equivalence relation.
This is what I have for my proof so far. What do you guys think? Any feedback is appreciated. Thank you


TITLE: If $\|A(x)\|_Y \geq c \|x\|_X$ and $\dim X = + \infty$, can $A \in \mathcal{L}(X,Y)$ be a compact operator?
QUESTION [3 upvotes]: Exercise :

Let $X,Y$ be Banach spaces with $\dim X = + \infty$ and $A \in \mathcal{L}(X,Y)$ such that :
  $$\|A(x)\|_Y \geq c\|x\|_X \; \; \forall x \in X \; \text{and} \; c>0$$
  Can the operator $A$ be compact ?

Attempt :
First of all, we know that an operator is compact if it transfers bounded sets to relatively compact sets (which means that they have a compact closure).
Let's assume now that $A$ is compact. Since $X$ is infinite dimensional, the unit ball $B_1^X$ is bounded (but not totally bounded). That would mean that $\overline{A(B_1^X)}$ should be compact. But, from the inequality relation gives, it would be :
$$\|A(B_1^X)\|_Y \geq c\|B_1^X\|_X \implies \|A(B_1^X)\|_Y \geq c' >0$$
But $c'$ could be as large as we'd like and thus the quantity $\|A(B_1^X)\|_Y$ is not bounded, which also means that the $\|\overline{A(B_1^X)}\|_Y$ is also not bounded, thus \overline{A(B_1^X)} is not compact (???).
Question : Is my intuition and especially my final argument correct ? If not, what other way could I approach that specific problem ?

REPLY [4 votes]: You can do this by definition. Since $X$ is infinite-dimensional, there exists a sequence $\{x_n\}$ in the unit ball $B_X$ of $X$ that has no convergent subsequence; thus there exists $\delta>0$ such that $\|x_k-x_j\|\geq\delta$ if $k\ne j$. Then
$$
\|Ax_k-Ax_j\|=\|A(x_k-x_j)\|\geq\,c\,\|x_k-x_j\|\geq\delta c.
$$
So the sequence $\{Ax_j\}$ admits no convergent subsquence, and thus $A(B_X)$ is not precompact, and $A$ is not compact.


TITLE: Is the image of the map $A \to \bigwedge^{k}A $ from matrices of a given rank closed?
QUESTION [0 upvotes]: $\newcommand{\Cof}{\operatorname{cof}}
\newcommand{\id}{\operatorname{Id}}$
Let $V$ be a real $d$-dimensional vector space ($d>2$). Let $2 \le k \le d-1$ be fixed, and let $r>k$.
Define $H_r=\{ A \in \text{End}(V) \mid \operatorname{rank}(A) = r
\}$. $H_r$ is a submanifold of $ \text{End}(V)$.
Consider the map
$$\psi:H_r \to \text{End}(\bigwedge^{k}V) \, \,, \, \, \psi(A)=\bigwedge^{k}A,$$
where $\bigwedge^{k} V$ is the $k$-th exterior power of $V$.
Denote $\tilde H_s=\{ B \in \text{End}(\bigwedge^kV) \mid \operatorname{rank}(B) = s \}$, and note that for $A \in \text{End}(V)$,
 $$\operatorname{rank}(\bigwedge^kA) = \binom {\operatorname{rank}(A)}{k} ,$$  that is $\psi(H_r) \subseteq \tilde H_{\binom {r}{k}}$

Is $\psi(H_r)$ a closed subset of $ \tilde H_{\binom {r}{k}}$? (The topology on $ \tilde H_{\binom {r}{k}}$ is the subspace topology induced by $\text{End}(\bigwedge^kV)$).

Comment: In the special case where $r=d$ the answer is positive; we have $H_d=\text{GL}(V)$ and $ \tilde H_{\binom {d}{k}}=\text{GL}(\bigwedge^{k}V)$, and
$\psi(\text{GL}(V))$ is closed in $\text{GL}(\bigwedge^{k}V)$.

REPLY [0 votes]: The answer is positive, since $\psi:H_r \to \tilde H_{\binom {r}{k}}$ is proper, and every proper map is closed.
Here is a proof $\psi$ is proper:
Let $K \subseteq \tilde H_{\binom {r}{k}}$ be compact, and let $A_n \in \psi^{-1}(K)$. We shall prove $A_n$ has a convergent subsequence in $\psi^{-1}(K)$. It suffices to prove $A_n$ converges in $\text{End}(V)$; indeed, if $A_n \to A$, then $\bigwedge^k A_n \to \bigwedge^k A$, and the limit $\bigwedge^k A$ must be in $K$. In particular, $\binom {r}{k}=\operatorname{rank}(\bigwedge^kA) = \binom {\operatorname{rank}(A)}{k}$, so $\operatorname{rank}(A)=r$, that is $A \in H_r$.
By using SVD, we can assume $A_n=\text{diag}(\sigma_1^n,\dots,\sigma_r^n,0,\dots,0)$  is diagonal, where the first $r$ diagonal elements are non-zero, and the last $d-r$ elements are zero. (Since the orthogonal group is compact, the isometric components surely converge after passing to a subsequence).
$\bigwedge^k A_n$ is diagonal, and its first $\binom {r}{k}$ elements are of the form $\Pi_{s=1}^k \sigma_{i_s}^n$, where all the $1 \le i_s \le r$ are distinct. So, every such product converges when $n \to \infty$ to a positive number. Indeed, $\psi(A_n)=\bigwedge^k A_n \in K \subseteq \tilde H_{\binom {r}{k}}$, so it converges (after passing to a subsequence) to an element $D \in K$. Since $\text{rank}(D)=\binom {r}{k}$, it follows that the products $\Pi_{s=1}^k \sigma_{i_s}^n$ must converge to positive numbers. (If even one of them converges to zero instead, the rank of the limit $D$ would be too low, which is a contradiction).
Now, let $1\le i \neq j \le r$. Since $r \ge k+1$, we can choose some $1 \le i_1,\dots,i_{k-1} \le r$ all different from $i,j$. Since both products $$(\Pi_{s=1}^{k-1} \sigma_{i_s}^n)\sigma_{i}^n,(\Pi_{s=1}^{k-1} \sigma_{i_s}^n)\sigma_{j}^n$$ converge to positive numbers, so does their ratio $C_{ij}^n=\frac{\sigma_i^n}{\sigma_j^n}$.
We know that $$\Pi_{s=1}^k \sigma_{s}^n=\Pi_{s=1}^k \sigma_{1}^n\frac{\sigma_s^n}{\sigma_1^n}=\Pi_{s=1}^k \sigma_{1}^nC_{s1}^n=(\sigma_{1}^n)^k  \Pi_{s=1}^k C_{s1}^n$$ converges to a positive number. Since all the $C_{s1}^n$ converge to positive numbers, we deduce $\sigma_1^n$ converges. W.L.O.G the same holds for every $\sigma_i^n$, so $A_n$ indeed converges. (and we know the limit must have the right rank).


text:
\section{Experiments} \label{sec:exp}

\subsection{Synthetic experiment -- simulation of FO-BLO divergence}

Theorem \ref{th:counter} is proven by explicitly constructing the following counterexample: $\Omega_\mathcal{T} = \{ \mathcal{T}_1, \mathcal{T}_2 \}$ where both tasks $\mathcal{T}_1, \mathcal{T}_2$ are equiprobable under $p(\mathcal{T})$ and functions $\mathcal{L}^{in} (\phi, \mathcal{T}_i) = \mathcal{L}^{out} (\phi, \mathcal{T}_i)$ are piecewise-polynomials of $\phi$ for $i \in \{ 1, 2 \}$ (see Appendix \ref{sec:proofs} for details). Figure \ref{fig:synth} is a simulation of this example for a case with a single parameter ($p = 1$) and inner-GD length of $r = 10$. More details and additional parameters of the simulation can be found in Appendix \ref{sec:synth}.

\subsection{Few-shot classification}

We compare UFO-BLO with other algorithms on Omniglot \cite{omniglot} and Mini-ImageNet \cite{imagenet}, popular few-shot classification benchmarks. We use MAML formulation of the few-shot meta-learning problem (Section \ref{sec:fewshot}). Both datasets consist of many classes with a few images for each class. We take train and test splits as in \cite{maml,reptile}. To sample from $p(\mathcal{T})$ in the $K$-shot $m$-way setting, $m$ classes are chosen randomly and $K + 1$ examples are drawn from each class randomly: $K$ examples for training and $1$ for testing, i.e. $s = mK, t = m$. We reuse convolutional architectures for $g(\phi, X)$ from \cite{maml} and set UFO-BLO inner-loop length to $r = 10$, as used by \cite{reptile}. In addition to exact BLO and FO-BLO, we compare with Reptile \cite{reptile} -- a modification of MAML which, similarly to FO-BLO, does not require storing inner-loop states in memory. Table \ref{tab:acc}  presents experimental results. On Omniglot, exact BLO shows the best performance on a range of setups, but is memory-inefficient (See Table \ref{tab:comp}). Out of all memory-efficient approaches (FO-BLO, Reptile \cite{reptile}, UFO-BLO), UFO-BLO with $q = 0.2$ shows the best performance in all Omniglot setups. Similarly, on Mini-ImageNet, UFO-BLO with $q = 0.2$ outperforms FO-BLO but performs slightly worse than the memory-inefficient exact BLO. More experimental details and extensions can be found in Appendix \ref{sec:expdet}.

\begin{table}[t]
  \caption{1-shot accuracy ($\%$) on Omniglot and Mini-ImageNet. The first three rows of results for the 20-way column are taken from
  \cite{reptile}.}
  \label{tab:acc}
  \centering
  \begin{tabular}{llllll}
    \toprule
    \, & \multicolumn{4}{c}{Omniglot} & Mini-ImageNet \\
    \midrule
    Algorithm & 20-way & 30-way & 40-way & 50-way & 10-way \\
    \midrule
    Exact BLO & $95.8 \pm 0.3$ & $90.8 \pm 0.2$ & $89.0 \pm 0.7$ & $87.9 \pm 0.4$ & $29.5 \pm 0.4$ \\
    \midrule
    Reptile \cite{reptile} & $89.4 \pm 0.1$ & $85.4 \pm 0.3$ & $82.5 \pm 0.3$ & $79.5 \pm 0.3$ & $31.7 \pm 0.2$ \\
    FO-BLO & $89.4 \pm 0.5$ & $81.1 \pm 1.2$ & $71.6 \pm 1.5$ & $64.4 \pm 2.1$ & $27.1 \pm 0.1$ \\
    UFO ($q = 0.1$) & $88.0 \pm 0.7$ & $84.6 \pm 0.7$ & $83.8 \pm 0.6$ & $81.6 \pm 0.7$ & $27.1 \pm 0.1$ \\
    UFO ($q = 0.2$) & $\mathbf{92.2 \pm 0.4}$ & $\mathbf{88.7 \pm 0.9}$ & $\mathbf{88.3 \pm 0.1}$ & $\mathbf{87.5 \pm 0.6}$ & $28.8 \pm 0.5$ \\
    \bottomrule
  \end{tabular}
\end{table}