text:
\section{Proofs}\label{sectionproof}
\subsection{Proofs of Theorem \ref{consistency} and Corollary \ref{corconvsn}}

\begin{lem}\label{superlem}
Assuming \textbf{(H1)} and \textbf{(H2)} and taking $c_{\beta}> 0$, it comes
\[
\lambda_{\max} \left( S_{n}^{-1} \right) = \mathcal{O} \left( n^{\beta-1} \right) \quad a.s \quad \text{and} \quad \lambda_{\max} \left( S_{n} \right) =\mathcal{O}(n)\quad  a.s.
\]
\end{lem}

\begin{proof}[Proof of Lemma \ref{superlem}]
First, note that
\[
\lambda_{\min}\left( S_{n} \right) \geq \lambda_{\min} \left(\sum_{i=1}^{n} \frac{c_{\beta}}{i^{\beta}} Z_i Z_i^{T} \right),
\]
and one can check that
\begin{equation}
\label{vitzi}
\frac{1}{\sum_{i=1}^{n}\frac{c_{\beta}}{i^{\beta}}} \sum_{i=1}^{n} \frac{c_{\beta}}{i^{\beta}}Z_{i}Z_{i}^{T} \xrightarrow[n\to + \infty]{a.s} I_{q} .
\end{equation}
Since
\[\sum_{i=1}^{n} \frac{c_{\beta}}{i^{\beta}} \sim \frac{c_{\beta}}{1-\beta}n^{1-\beta},\] it comes
\[
\lambda_{\max} \left( S_{n}^{-1} \right) = \mathcal{O} \left( n^{\beta -1 } \right) \quad a.s.
\]
Let us now give a bound of $\lambda_{\max}\left( S_{n} \right)$. First, remark that $S_{n}$ can be written as
\begin{equation}
\label{decsn} S_{n} = S_{0} + \sum_{i=1}^{n} \mathbb{E}\left[ \overline{\Phi}_{i}\overline{\Phi}_{i}^{T} |\mathcal{F}_{i-1} \right] + \sum_{i=1} \Xi_{i} + \sum_{i=1}^{n} \frac{c_{\beta}}{i^{\beta}}Z_{i}Z_{i}^{T},
\end{equation}
where $\Xi_{i} := \overline{\Phi}_{i}\overline{\Phi}_{i}^{T} - \mathbb{E}\left[ \overline{\Phi}_{i}\overline{\Phi}_{i}^{T} |\mathcal{F}_{i-1} \right]$ and $\left( \mathcal{F}_{i} \right)$ is the $\sigma$-algebra generated by the sample, i.e $\mathcal{F}_{i} := \sigma \left( \left( X_{1}, Y_{1} \right) , \ldots ,\left( X_{i} , Y_{i} \right) \right)$. Thanks to assumption \textbf{(H1b)}, $\mathbb{E}\left[ \left\| \overline{\Phi}_{i}\overline{\Phi}_{i}^{T} \right\|_{F}^{2} |\mathcal{F}_{i-1} \right] \leq C''$, and applying Theorem~\ref{theomartmoy}, it comes for all positive constant $\delta > 0$,
\begin{equation}\label{vitXi}
\left\| \sum_{i=1}^{n} \Xi_{i} \right\|_{F}^{2} = o \left( n (\ln n)^{1+\delta} \right) \quad a.s,
\end{equation}
where $\left\| . \right\|_{F}$ is the Frobenius norm. Then,
\[
\left\| S_{n} \right\|_{F} = \mathcal{O} \left( \max \left( \left\| S_{0} \right\|_{F}, C''n , \sqrt{n (\ln n)^{1+\delta}} , n^{1-\beta} \right)\right) \quad a.s,
\]
which concludes the proof.
\end{proof}

\begin{proof}[Proof of Theorem \ref{consistency}]
 With the help of a Taylor's decomposition, it comes
\[
G \left( \theta_{n+1} \right) = G \left( \theta_{n} \right) + \nabla G \left( \theta_{n} \right)^{T}\left( \theta_{n+1} - \theta_{n} \right) + \frac{1}{2}\left( \theta_{n+1} - \theta_{n} \right)^{T} \int_{0}^{1} \nabla^{2} G \left( \theta_{n+1} + t \left( \theta_{n+1} - \theta_{n} \right) \right) dt \left( \theta_{n+1} - \theta_{n} \right) .
\]
Then, assumption \textbf{(H2)} yields
\[
G \left( \theta_{n+1} \right) \leq G \left( \theta_{n} \right) + \nabla G \left( \theta_{n} \right)^{T}\left( \theta_{n+1} - \theta_{n} \right) + \frac{C'}{2}\left\| \theta_{n+1} - \theta_{n} \right\|^{2}.
\]
Replacing $\theta_{n+1}$,
\begin{align*}
G & \left( \theta_{n+1} \right)\leq G \left( \theta_{n} \right) - \gamma_{n+1}\nabla G \left( \theta_{n} \right)^{T}\overline{S}_{n}^{-1}\nabla_{h} g \left( X_{n+1} , Y_{n+1} , \theta_{n} \right) + \frac{C'}{2}\gamma_{n+1}^{2} \left\| \overline{S}_{n}^{-1} \nabla_{h} g \left( X_{n+1} , Y_{n+1} ,\theta_{n} \right) \right\|^{2}  \\
& \leq G \left( \theta_{n} \right) - \gamma_{n+1} \nabla G \left( \theta_{n} \right)^{T} \overline{S}_{n}^{-1} \nabla_{h} g \left( X_{n+1} , Y_{n+1} , \theta_{n} \right) + \frac{C'}{2}\gamma_{n+1}^{2} \left( \lambda_{\max}  \left( \overline{S}_{n}^{-1} \right) \right)^{2} \left\|  \nabla_{h} g \left( X_{n+1} , Y_{n+1} ,\theta_{n} \right) \right\|^{2}.
\end{align*}
Assumption \textbf{(H1a)} leads to
\begin{align*}
\mathbb{E}\left[ G \left( \theta_{n+1} \right) |\mathcal{F}_{n} \right] & \leq G \left( \theta_{n} \right) - \gamma_{n+1} \nabla G \left( \theta_{n} \right)^{T} \overline{S}_{n}^{-1} \nabla G \left( \theta_{n} \right)  + \frac{CC'}{2}\gamma_{n+1}^{2} \left( \lambda_{\max}  \left( \overline{S}_{n}^{-1} \right) \right)^{2} \\
& \leq G \left( \theta_{n} \right) - \gamma_{n+1}\lambda_{\min} \left( \overline{S}_{n}^{-1} \right) \left\| \nabla G \left( \theta_{n} \right) \right\|^{2} + \frac{CC'}{2}\gamma_{n+1}^{2}\left( \lambda_{\max}  \left( \overline{S}_{n}^{-1} \right) \right)^{2}.
\end{align*}
Thanks to Lemma \ref{superlem}, and since $\beta < \alpha - 1/2$,
\[
\sum_{n\geq 1} \gamma_{n+1}^{2}\left(  \lambda_{\max} \left (\overline{S}_{n}^{-1} \right) \right)^{2} < + \infty \quad a.s,
\]
so that, applying Robbins-Siegmund Theorem (see \cite{Duf97} for instance), $G \left( \theta_{n} \right)$ converges almost surely to a finite random variable and
\[
\sum_{n\geq 1} \gamma_{n+1} \lambda_{\min} \left( \overline{S}_{n}^{-1} \right)\left\| \nabla G \left( \theta_{n} \right) \right\|^{2} = \sum_{n\geq 1 } \gamma_{n+1} \lambda_{\max} \left( \overline{S}_{n} \right)^{-1} \left\| \nabla G \left( \theta_{n} \right) \right\|^{2} < + \infty \quad a.s.
\]
Lemma~\ref{superlem} implies $\sum_{n\geq 1} \gamma_{n+1} \lambda_{\max} \left( \overline{S}_{n}\right)^{-1} = + \infty$ almost surely, so that there is alsmot surely a subsequence $\theta_{\varphi_{n}}$ such that $\left\| \nabla G \left( \theta_{\varphi_{n}} \right) \right\|^{2} $ converges to $0$.

\end{proof}

\begin{proof}[Proof of Corollary \ref{corconvsn}] Let us give the convergence of each term in decomposition (\ref{decsn}). Since $\beta > 0$ and applying (\ref{vitXi}), it comes
\[
\frac{1}{n}\left\| \sum_{i=1}^{n}\Xi_{i} \right\|_{F} \xrightarrow[n\to + \infty]{a.s} 0 \quad a.s  \quad \text{and} \quad \frac{1}{n}\left\| \sum_{i=1}^{n} \frac{c_{\beta}}{i^{\beta}}Z_{i}Z_{i}^{T} \right\|_{F} \xrightarrow[n\to + \infty]{a.s} 0 \quad a.s.
\]
Finally, since $\overline{\theta}_{n}$ converges almost surely to $\theta$, assumption \textbf{(H3)} together with Toeplitz lemma yield
\[
\frac{1}{n}\sum_{i=1}^{n} \mathbb{E}\left[ \overline{\Phi}_{i}\overline{\Phi}_{i}^{T} |\mathcal{F}_{i-1} \right] \xrightarrow[n\to + \infty]{a.s} \mathbb{E}\left[ \nabla_{h}f \left( X, \theta \right) \nabla_{h} f\left( X , \theta \right)^{T} \right] = L(\theta).
\]
\end{proof}

\subsection{Proof of Theorem \ref{ratetheta}}
First, $\theta_{n+1}$ can be written as
\begin{equation}
\label{decxi} \theta_{n+1} - \theta = \theta_{n} - \theta -\gamma_{n+1}\overline{S}_{n}^{-1} \nabla G \left( \theta_{n} \right) + \gamma_{n+1}\overline{S}_{n}^{-1}\xi_{n+1},
\end{equation}
with $\xi_{n+1} :=\nabla G\left( \theta_{n} \right) - \nabla_{h} g \left( X_{n+1} , Y_{n+1} , \theta_{n} \right)$. Remark that $\left( \xi_{n} \right)$ is a sequence of martingale differences adapted to the filtration $\left( \mathcal{F}_{n} \right)$. Moreover, linearizing the gradient and noting $H = \nabla^{2}G(\theta)$, it comes
\begin{align}
\label{decareutiliser} \theta_{n+1} - \theta & = \theta_{n} - \theta - \gamma_{n+1}\overline{S}_{n}^{-1} H \left( \theta_{n} - \theta \right) + \gamma_{n+1}\overline{S}_{n}^{-1}\xi_{n+1} - \gamma_{n+1}\overline{S}_{n}^{-1}\delta_{n} \\
\notag & = \left( 1-\gamma_{n+1} \right)\left( \theta_{n} - \theta \right) + \gamma_{n+1}\left( H^{-1} -\overline{S}_{n}^{-1} \right) H \left( \theta_{n} - \theta \right)\\
\label{decdelta} & + \gamma_{n+1}\overline{S}_{n}^{-1}\xi_{n+1} -\gamma_{n+1}\overline{S}_{n}^{-1}\delta_{n},
\end{align}
where $\delta_{n} = \nabla G \left( \theta_{n} \right) - H \left( \theta_{n} - \theta \right) $ is the remainder term in the Taylor's decomposition of the gradient. By induction, one can check that for all $n \geq 1$,
\begin{align}
\notag \theta_{n} - \theta & = \beta_{n,0} \left( \theta_{0} - \theta \right) +  \sum_{k=0}^{n-1} \beta_{n,k+1}\gamma_{k+1} \left( H^{-1} - \overline{S}_{k}^{-1}\right) H \left( \theta_{k} - \theta \right) -  \sum_{k=0}^{n-1} \beta_{n,k+1}\gamma_{k+1}\overline{S}_{k}^{-1}\delta_{k} \\
\label{decbeta} &  +  \sum_{k=0}^{n-1} \beta_{n,k+1} \gamma_{k+1}\overline{S}_{k}^{-1}\xi_{k+1} ,
\end{align}
with, for all $k,n \geq 0$ and $k \leq n$, \[\beta_{n,k} = \prod_{j=k+1}^{n} \left( 1-\gamma_{j} \right) \quad \mbox{and} \quad \beta_{n,n}=1.\]
Applying Theorem~\ref{theomartbeta},
\begin{equation}
\label{lembeta}\left\| \sum_{k=0}^{n-1}\beta_{n,k+1}\gamma_{k+1}\overline{S}_{k}^{-1} \xi_{k+1} \right\|^{2} = \mathcal{O} \left( \frac{\ln n}{n^{\alpha}} \right) \quad a.s.
\end{equation}
We can now prove Theorem~\ref{ratetheta}.
\begin{proof}[Proof of equation~\ref{ratethetaas}]
The aim is to give the rate of convergence of each term of decomposition (\ref{decbeta}). The rate of the martingale term is given  by equation \eqref{lembeta}.  Remark that there is a rank $n_{\alpha}$ such that for all $n \geq n_{\alpha}$, we have $\gamma_{n}\leq 1$, so that, for all $n \geq n_{\alpha}$,
\begin{align*}
\left\| \beta_{n,0}\left( \theta_{0} - \theta \right) \right\| & \leq \left\| \theta_{0} - \theta \right\| \prod_{i=1}^{n_{\alpha}-1} \left| 1- \gamma_{i+1} \right| \prod_{i=n_{\alpha}} \left( 1-\gamma_{i+1} \right) \\
& \leq \left\| \theta_{0} - \theta \right\| \prod_{i=1}^{n_{\alpha}-1} \left| 1- \gamma_{i+1} \right| \exp \left( -\sum_{i=n_{\alpha}}^{n}\gamma_{i+1} \right) .
\end{align*}
Since $\alpha < 1$, this term converges at an exponential rate, and more precisely
\begin{equation}\label{vitexp}
\left\| \beta_{n,0}\left( \theta_{0} - \theta \right) \right\| = \mathcal{O} \left( \exp \left( - \frac{c_{\alpha}}{1-\alpha}n^{1-\alpha} \right) \right)  \quad a.s.
\end{equation}
Let us denote
\[
\Delta_{n} :=  \sum_{k=0}^{n-1}\beta_{n,k+1}\gamma_{k+1}\left( H^{-1} - \overline{S}_{k}^{-1} \right) H \left( \theta_{k} - \theta \right) -  \sum_{k=0}^{n} \beta_{n,k+1}\gamma_{k+1} \overline{S}_{k}^{-1}\delta_{k} .
\]
The aim is so to prove that this term is negligible. First, remark that the Taylor's decomposition of the gradient yields
\[
\delta_{n} = \int_{0}^{1} \left( \nabla^{2}G \left( \theta + t \left( \theta_{n} - \theta \right) \right) - H \right) \left( \theta_{n} - \theta \right) dt ,
\]
and thanks to assumption \textbf{(H4a)}, since $\theta_{n}$ converges almost surely to $\theta$ and by dominated convergence,
\[
\left\| \delta_{n} \right\| \leq \left\| \theta_{n} - \theta \right\| \int_{0}^{1} \left\| \nabla^{2}G \left( \theta + t \left( \theta_{n} - \theta \right) \right) - H \right\|_{op} dt = o \left( \left\| \theta_{n} - \theta \right\| \right) \quad a.s
\]
In a similar way, since $\overline{S}_{n}^{-1}$ converges almost surely to $H^{-1}$,
\[
\left\| \left( H^{-1} - \overline{S}_{n}^{-1} \right) H \left( \theta_{n} - \theta \right) \right\| = o \left( \left\| \theta_{n} - \theta \right\| \right) \quad a.s.
\]
Moreover, since
\[
\Delta_{n+1} = \left( 1-\gamma_{n+1} \right) \Delta_{n} + \gamma_{n+1} \left( H^{-1} - \overline{S}_{n}^{-1} \right) H \left( \theta_{n} - \theta \right) - \gamma_{n+1}\overline{S}_{n}^{-1}\delta_{n},
\]
we have,
\begin{align*}
\left\| \Delta_{n+1} \right\| & \leq \left| 1- \gamma_{n+1} \right| \left\| \Delta_{n} \right\| + o \left( n^{-\alpha} \left\| \theta_{n} - \theta \right\| \right) \quad a.s \\
& \leq \left| 1- \gamma_{n+1} \right| \left\| \Delta_{n} \right\| + o \left( n^{-\alpha} \left\| \Delta_{n} + \sum_{k=0}^{n-1} \beta_{n,k+1} \gamma_{k+1} \overline{S}_{k}^{-1} \xi_{k+1} + \beta_{n,0}\left( \theta_{0} - \theta \right) \right\| \right) \quad a.s.
\end{align*}
Then, since $\beta_{n,0}\left( \theta_{0} - \theta \right)$ converges at an exponential rate to $0$ and thanks to equation \eqref{lembeta},
\[
\left\| \Delta_{n+1} \right\| \leq \left( 1- \gamma_{n+1} \right) \left\| \Delta_{n} \right\| + o \left( \gamma_{n+1} \sqrt{\frac{\ln n}{n^{\alpha}}} + \gamma_{n+1}\left\| \Delta_{n} \right\| \right) \quad a.s ,
\]
and applying a stabilization Lemma (see \cite{Duf97} for instance),
\[
\left\| \Delta_{n} \right\| = \mathcal{O} \left( \sqrt{\frac{\ln n}{n^{\alpha}}}\right) \quad a.s
\]
which concludes the proof.
\end{proof}

\begin{proof}[Proof of equation \eqref{tlc}]
In order to get the asymptotic normality, we will apply the Central Limit Theorem in \cite{Jak88} to the martingale term in decomposition (\ref{decbeta}) and prove that other terms of this decomposition are negligible. The rate of convergence of the first term on the right-hand side of equality (\ref{decbeta}) is given by equality (\ref{vitexp}). Moreover, applying Theorem \ref{ratetheta} and Remark \ref{rempourletrucpourri} as well as Lemma E.2 in \cite{CG2015}, one can check that for all $\delta > 0$,
\[
\left\| \sum_{k=0}^{n-1}\beta_{n,k+1}\gamma_{k+1}\left( H^{-1} - \overline{S}_{k}^{-1} \right) H \left( \theta_{k} - \theta \right) \right\|^{2} = o \left( \frac{(\ln n)^{1+\delta}}{n^{ \alpha + \beta}} \right) \quad a.s,
\]
and this term is so negligible. In the same way,
\[
\left\| \sum_{k=0}^{n-1}\beta_{n,k+1}\gamma_{k+1}\overline{S}_{k}^{-1} \delta_{k} \right\|^{2} = o \left( \frac{(\ln n)^{2+ \delta}}{n^{ 2\alpha }} \right) \quad a.s.
\]
Note that the martingale term can be written as
\[
\sum_{k=0}^{n-1} \beta_{n,k+1}\gamma_{k+1}\overline{S}_{k}^{-1}\xi_{k+1}= \sum_{k=0}^{n-1} \beta_{n,k+1} c_{\alpha}\gamma_{k+1} \left(  \overline{S}_{k}^{-1} - H^{-1} \right) \xi_{k+1} + \sum_{k=0}^{n-1} \beta_{n,k+1}\gamma_{k+1} H^{-1}\xi_{k+1} .
\]
Applying Theorem \ref{theomartbeta} and Remark \ref{rempourletrucpourri},
\[
\left\| \sum_{k=0}^{n} \beta_{n+1,k+1} \gamma_{k+1} \left(  \overline{S}_{k}^{-1} - H^{-1} \right) \xi_{k+1} \right\|^{2} = O \left( \frac{\ln n}{n^{\alpha + 2\beta}} \right) \quad a.s.
\]
Finally, let us now prove that the martingale term verifies assumption in \cite{Jak88}, i.e that we have
\begin{equation}
\label{cond1} \forall \nu > 0 , \quad \lim_{n\to + \infty} \mathbb{P}\left[ \sup_{0\leq k \leq n} \sqrt{\frac{n^{\alpha}}{c_{\alpha}}} \left\| \beta_{n+1,k+1} c_{\alpha}(k+1)^{-\alpha}H^{-1} \xi_{k+1} \right\| > \nu \right] = 0 .
\end{equation}
\begin{equation}
\label{cond2} \frac{n^{\alpha}}{c_{\alpha}} \sum_{k=0}^{n} \beta_{n+1,k+1}^{2} \gamma_{k+1}^{2} H^{-1} \xi_{k+1} \xi_{k+1}^{T} H^{-1} \xrightarrow[n\to + \infty]{a.s} \frac{1}{2}\sigma^{2} H^{-1} .
\end{equation}
\medskip

\noindent\textbf{Proof of equation (\ref{cond1}):}  Thanks to assumption \textbf{(H5)}, one can check that for all $n\geq 1$,
\begin{equation}
\label{majxipourri}\mathbb{E}\left[ \left\| \xi_{n+1} \right\|^{2+2\eta} |\mathcal{F}_{k} \right] \leq 2^{2+2\eta}C_{\eta}
\end{equation}
Then, applying Markov's inequality,
\begin{align*}
\mathbb{P}\left[ \sup_{0\leq k \leq n} \sqrt{\frac{n^{\alpha}}{c_{\alpha}}} \left\| \beta_{n+1,k+1} \gamma_{k+1}H^{-1} \xi_{k+1} \right\| > \nu \right] & \leq \sum_{k=0}^{n} \mathbb{P}\left[  \sqrt{\frac{n^{\alpha}}{c_{\alpha}}} \left\| \beta_{n+1,k+1} \gamma_{k+1}H^{-1} \xi_{k+1} \right\| > \nu \right] \\
& \leq \sum_{k=0}^{n} \mathbb{P}\left[ \left\| \xi_{k+1} \right\| > \frac{\sqrt{c_{\alpha}}\nu}{n^{\alpha/2}\left| \beta_{n+1,k+1}\right|\gamma_{k+1}\left\| H^{-1}\right\|_{op}} \right] \\
& \leq \frac{c_{\alpha}^{1+\eta}\left\| H^{-1} \right\|_{op}^{2+2\eta}}{\nu^{2+2\eta}} 2^{2+2\eta}C_{\eta}n^{\alpha\left( 1+ \eta \right)}\sum_{k=0}^{n}\left| \beta_{n+1,k+1} \right|^{2+2\eta} k^{-2\alpha\left( 1+ \eta \right)}.
\end{align*}
Moreover, note that there is a rank $n_{\alpha}$ such that for all $j \geq n_{\alpha}$ we have $c_{\alpha} j^{-\alpha} \leq 1$. For the sake of readibility of the proof (in other way, one can split the sum into two parts as in the proof of Lemma 3.1 in \cite{CCG2015}), we consider from now on that $n_{\alpha} =1$. Then
\begin{align*}
\left| \beta_{n+1,k+1} \right| & =  \prod_{i=k+2}^{n+1} \left( 1-c_{\alpha}i^{-\alpha} \right)  \leq  \exp \left( - c_{\alpha}\sum_{i=k+2}^{n+1} i^{-\alpha} \right).
\end{align*}
Applying Lemma E.2 in \cite{CG2015}, it comes
\[
 \sum_{k=0}^{n}\left| \beta_{n+1,k+1} \right|^{2+2\eta} k^{-2\alpha\left( 1+ \eta \right)} =\mathcal{O} \left( n^{-\alpha -2 \alpha \eta} \right) ,
\]
and so
\[
\lim_{n\to + \infty} \frac{c_{\alpha}^{ 1+ \eta}\left\| H^{-1} \right\|_{op}^{2+2\eta}}{\nu^{2 +2 \eta}} 2^{2+2\eta}C_{\eta}n^{\alpha (1+\eta)}\sum_{k=0}^{n}\left| \beta_{n+1,k+1} \right| k^{-2\alpha( 1+ \eta)} = 0 .
\]

\medskip

\noindent\textbf{Proof of equation (\ref{cond2}): } First, note that
\begin{align*}
\sum_{k=0}^{n} \beta_{n+1,k+1}^{2} \gamma_{k+1}^{2} H^{-1} \xi_{k+1}\xi_{k+1}^{T}H^{-1}& = \sum_{k=0}^{n} \beta_{n+1,k+1}^{2} \gamma_{k+1}^{2} H^{-1} \mathbb{E}\left[ \xi_{k+1}\xi_{k+1}^{T}|\mathcal{F}_{k} \right] H^{-1}\\ &+\sum_{k=0}^{n} \beta_{n+1,k+1}^{2} \gamma_{k+1}^{2} H^{-1} \Xi_{k+1}H^{-1},
\end{align*}
with $\Xi_{k+1} = \xi_{k+1}\xi_{k+1}^{T} - \mathbb{E}\left[ \xi_{k+1}\xi_{k+1}^{T} |\mathcal{F}_{k} \right]$. $\left( \Xi_{k} \right)$ is a sequence of martingale differences adapted to the filtration $\left( \mathcal{F}_{k} \right)$. Applying Theorem~\ref{theomartbeta}, it comes
\[
\left\| \sum_{k=0}^{n}\beta_{n+1,k+1}^{2}\gamma_{k+1}^{2}H^{-1} \Xi_{k+1} H^{-1} \right\|^{2} = O \left( \frac{\ln n}{n^{3\alpha}}\right) \quad a.s.
\]
Moreover, note that
\begin{align*}
\sum_{k=0}^{n} & \beta_{n+1,k+1}^{2} \gamma_{k+1}^{2} H^{-1} \mathbb{E}\left[ \xi_{k+1}\xi_{k+1}^{T}|\mathcal{F}_{k} \right] H^{-1} =  -\sum_{k=0}^{n} \beta_{n+1,k+1}^{2} \gamma_{k+1}^{2} H^{-1} \nabla G\left(\theta_{k} \right) \nabla G\left( \theta_{k} \right)^{T} H^{-1} \\
& + \sum_{k=0}^{n} \beta_{n+1,k+1}^{2} \gamma_{k+1}^{2} H^{-1} \mathbb{E}\left[ \nabla_{h} g \left( X_{k+1},Y_{k+1} , \theta_{k} \right) \nabla_{h} g \left( X_{k+1} , Y_{k+1} , \theta_{k}\right)^{T}|\mathcal{F}_{k} \right] H^{-1}.
\end{align*}
Applying Theorem \ref{ratetheta}, equality (\ref{vitexp}) and Lemma E.2 in \cite{CG2015}, since the gradient of $G$ is Lipschitz, one can check that for all $\delta > 0$,
\[
\left\| \sum_{k=0}^{n} \beta_{n+1,k+1}^{2} \gamma_{k+1}^{2} H^{-1} \nabla G\left(\theta_{k} \right) \nabla G\left( \theta_{k} \right)^{T} H^{-1} \right\|_{F}^{2} = o \left( \frac{(\ln n)^{1+\delta}}{n^{2\alpha}} \right) \quad a.s.
\]
Moreover, noting
\[
R_{k}=\mathbb{E}\left[ \nabla_{h} g \left( X_{k+1},Y_{k+1} , \theta_{k} \right) \nabla_{h} g \left( X_{k+1} , Y_{k+1} , \theta_{k}\right)^{T}|\mathcal{F}_{k} \right] - \mathbb{E}\left[ \nabla_{h} g \left( X_{k+1},Y_{k+1} , \theta \right)\nabla_{h} g \left( X_{k+1},Y_{k+1} , \theta \right)^{T} |\mathcal{F}_{k} \right] ,
\]
\begin{align*}
& \sum_{k=0}^{n}  \beta_{n+1,k+1}^{2} \gamma_{k+1}^{2} H^{-1} \mathbb{E}\left[ \nabla_{h} g \left( X_{k+1},Y_{k+1} , \theta_{k} \right) \nabla_{h} g \left( X_{k+1} , Y_{k+1} , \theta_{k}\right)^{T}|\mathcal{F}_{k} \right] H^{-1} \\
&  = \sum_{k=0}^{n} \beta_{n+1,k+1}^{2} \gamma_{k+1}^{2} H^{-1} R_{k} H^{-1} + \sum_{k=1}^{n}  \beta_{n+1,k+1}^{2} \gamma_{k+1}^{2} H^{-1} \mathbb{E}\left[ \nabla_{h} g \left( X_{k+1},Y_{k+1} , \theta \right) \nabla_{h} g \left( X_{k+1} , Y_{k+1} , \theta\right)^{T}|\mathcal{F}_{k} \right] H^{-1}
\end{align*}
Moreover, applying Theorem \ref{ratetheta}, Lemma E.2 in \cite{CG2015}, one chan check that for all $\delta > 0$,
\[
\left\| \sum_{k=0}^{n} \beta_{n+1,k+1}^{2} \gamma_{k+1}^{2} H^{-1} R_{k} H^{-1} \right\|_{F} = o \left( \frac{(\ln n)^{1+\delta}}{n^{2\alpha}} \right) \quad a.s.
\]
Furthermore, since $\mathbb{E}\left[ \nabla_{h} g \left( X_{k+1},Y_{k+1} , \theta \right) \nabla_{h} g \left( X_{k+1} , Y_{k+1} , \theta\right)^{T}|\mathcal{F}_{k} \right] = \sigma^{2}H$, applying Lemma A.1 in \cite{GB2017},
\[
\lim_{n\to \infty}\left\| H^{-1} \left( \frac{n^{\alpha}}{c_{\alpha}}\sum_{k=0}^{n}  \beta_{n+1,k+1}^{2} \gamma_{k+1}^{2}  \mathbb{E}\left[ \nabla_{h} g \left( X_{k+1},Y_{k+1} , \theta \right) \nabla_{h} g \left( X_{k+1} , Y_{k+1} , \theta\right)^{T}|\mathcal{F}_{k} \right] - \frac{1}{2}\sigma^{2}H \right) H^{-1} \right\|_{F} =0,
\]
which concludes the proof.
\end{proof}

\subsection{Proof of Theorem \ref{ratethetabar}}
Let us first give some decompositions of the averaged estimates. First, note that
\[
\overline{\theta}_{n} = \frac{1}{n+1}\sum_{k=0}^{n} \theta_{k}.
\]
Second, one can write decomposition (\ref{decareutiliser}) as
\begin{equation}
\label{equationcle} \theta_{n} - \theta = H^{-1} \overline{S}_{n}\frac{\left( \theta_{n} - \theta \right) - \left( \theta_{n+1} - \theta \right)}{\gamma_{n+1}} + H^{-1}\xi_{n+1} - H^{-1} \delta_{n} .
\end{equation}
Then, summing these equalities and dividing by $n+1$, it comes
\begin{equation}
\label{decmoy} \overline{\theta}_{n} - \theta = H^{-1}\frac{1}{n+1}\sum_{k=0}^{n} \overline{S}_{k}\frac{\left( \theta_{k} - \theta \right) - \left( \theta_{k+1} - \theta \right)}{\gamma_{k+1}} + H^{-1} \frac{1}{n+1} \sum_{k=0}^{n} \xi_{k+1} - H^{-1} \frac{1}{n+1}\sum_{k=0}^{n} \delta_{k} .
\end{equation}

\begin{proof}[Proof of Theorem \ref{ratethetabar}]
Let us give the rate of convergence of each term on the right-hand side of equality (\ref{decmoy}). First, in order to apply a Law of Large Numbers and a Central Limit Theorem for martingales, let us calculate
\[
\lim_{n\to \infty}\frac{1}{n+1}\sum_{k=0}^{n} \mathbb{E}\left[ \xi_{k+1}\xi_{k+1}^{T} |\mathcal{F}_{k} \right].
\]
By definition of $\left( \xi_{n} \right)$, it comes
\begin{align*}
\frac{1}{n+1}\sum_{k=0}^{n} \mathbb{E}\left[ \xi_{k+1}\xi_{k+1}^{T} |\mathcal{F}_{k} \right]&  = \frac{1}{n+1}\sum_{k=0}^{n} \mathbb{E}\left[ \nabla_{h} g \left( X_{k+1} ,Y_{k+1} , \theta_{k} \right)\nabla_{h} g \left( X_{k+1} ,Y_{k+1} , \theta_{k} \right)^{T} |\mathcal{F}_{k} \right] \\
&  - \frac{1}{n+1}\sum_{k=0}^{n} \nabla G\left( \theta_{k} \right) \nabla G \left( \theta_{k} \right)^{T} .
\end{align*}
By continuity and since $\theta_{k}$ converges amost surely to $\theta$, Toeplitz lemma implies
\[
\frac{1}{n+1}\sum_{k=0}^{n} \mathbb{E}\left[ \xi_{k+1}\xi_{k+1}^{T} |\mathcal{F}_{k} \right] \xrightarrow[n\to + \infty]{a.s} \mathbb{E}\left[ \nabla_{h} g \left( X,Y , \theta \right)\nabla_{h} g \left( X,Y , \theta \right)^{T} \right] .
\]
Furthermore,
\[
\mathbb{E}\left[ \nabla_{h} g \left( X,Y , \theta \right)\nabla_{h} g \left( X,Y , \theta \right)^{T} \right] = \mathbb{E}\left[ \epsilon^{2} \nabla_{h}f\left( X,Y,\theta \right) \nabla_{h}f \left( X,Y , \theta \right)^{T} \right] = \sigma^{2} H .
\]
Finally, since $\mathbb{E}\left[ \left\| \xi_{n+1} \right\|^{2+2\eta} |\mathcal{F}_{n} \right] \leq 2^{2+2\eta}C_{\eta}$,  applying a Law of Large Numbers for martingales (see \cite{Duf97}),
\[
\frac{1}{(n+1)^{2}} \left\| H^{-1} \sum_{k=0}^{n} \xi_{k+1} \right\|^{2} = \mathcal{O}\left( \frac{\ln n}{n} \right) \quad a.s.
\]
Moreover, applying a Central Limit Theorem for martingales (see \cite{Duf97}),
\[
\frac{1}{\sqrt{n}}H^{-1} \sum_{k=0}^{n} \xi_{k+1} \xrightarrow[n\to + \infty]{\mathcal{L}} \mathcal{N}\left( 0 , \sigma^{2}H^{-1} \right) .
\]
Let us now prove that other terms on the right-hand side of equality (\ref{decmoy}) are negligible. Thanks to assumption \textbf{(H4b)} and since $\theta_{n}$ converges almost surely to $\theta$,
\[
\left\| \delta_{n} \right\| \leq \left\| \theta_{n} - \theta \right\| \int_{0}^{1} \left\| \nabla^{2}G \left( \theta + t \left( \theta_{n} - \theta \right) \right) - H \right\|_{op} dt = O \left( \left\| \theta_{n} - \theta \right\|^{2} \right) \quad a.s
\]
Then applying Theorem \ref{ratetheta}, for all $\delta > 0$,
\[
\frac{1}{n+1}\left\| H^{-1}\sum_{k=0}^{n} \delta_{k} \right\| \leq \frac{\left\| H^{-1} \right\|_{op}}{n+1} \sum_{k=1}^{n} \left\| \delta_{k} \right\| = o \left( \frac{(\ln n)^{1+\delta}}{n^{\alpha}} \right) \quad a.s
\]
and since $\alpha > 1/2$, this term is negligible. Finally, as in \cite{Pel00}, using an Abel's transform,
\begin{align}
\notag \frac{1}{n+1}\sum_{k=0}^{n} \overline{S}_{k}\frac{\left( \theta_{k} - \theta \right) - \left( \theta_{k+1} - \theta \right)}{c_{\alpha}(k+1)^{-\alpha}} & = - \frac{\overline{S}_{n+1}\left( \theta_{n+1} - \theta\right) }{(n+1)\gamma_{n+2}} + \frac{\overline{S}_{0}\left( \theta_{0}-\theta \right)}{(n+1)\gamma_{1}} \\
\label{termedemerde}&  - \frac{1}{n+1}\sum_{k=1}^{n+1} \left( \gamma_{k}^{-1}\overline{S}_{k-1} - \gamma_{k+1}^{-1}\overline{S}_{k}  \right) \left( \theta_{k} - \theta \right) .
\end{align}
Let us now give the rates of convergence of each term on the right-hand side of equality (\ref{termedemerde}). First, applying Theorem \ref{ratetheta} and since $\overline{S}_{n}$ converges almost surely to $H$,
\[
\left\| \frac{\overline{S}_{n+1}\left( \theta_{n+1} - \theta\right) }{\gamma_{n+2}(n+1)} \right\|^{2} = O \left( \frac{\ln n}{n^{2 - \alpha}} \right) \quad a.s \quad \quad \quad \text{and} \quad \quad \quad \left\| \frac{\overline{S}_{0}\left( \theta_{0}-\theta \right)}{(n+1)\gamma_{1}} \right\|^{2} = O \left( \frac{1}{n^{2}} \right) \quad a.s
\]
and these terms are negligible since $\alpha < 1$. Furthermore, since
\[
\overline{S}_{k}  =\overline{S}_{k-1} - \frac{1}{k}\overline{S}_{k-1} + \frac{1}{k} \left( \overline{\Phi}_{k} \overline{\Phi}_ {k}^{T} + \frac{c_{\beta}}{k^{\beta}} Z_{k}Z_{k}^{T} \right),
\]
it comes
\begin{align*}
 (*) :& =\frac{1}{n+1}\sum_{k=1}^{n+1} \left( \gamma_{k}\overline{S}_{k-1} - \gamma_{k+1}\overline{S}_{k}  \right) \left( \theta_{k} - \theta \right) \\
& =  \frac{1}{n+1}\sum_{k=1}^{n+1} \left( \gamma_{k}^{-1} - \gamma_{k+1}^{-1} \right) \overline{S}_{k-1} \left( \theta_{k} - \theta \right) - \frac{1}{n+1}\sum_{k=1}^{n+1}\frac{\gamma_{k+1}^{-1}}{k}\overline{S}_{k-1}\left( \theta_{k} - \theta \right)  \\
&  + \underbrace{\frac{1}{n+1}\sum_{k=1}^{n+1}\frac{\gamma_{k+1}^{-1}}{k } \left(   \overline{\Phi}_{k} \overline{\Phi}_{k}^{T} + \frac{c_{\beta}}{k^{\beta}} Z_{k}Z_{k}^{T} \right) \left( \theta_{k} - \theta \right)}_{:=(**)} .
\end{align*}
For the first term on the right-hand side of previous equality, since $\left| \gamma_{k}^{-1} - \gamma_{k+1}^{-1} \right| \leq  \alpha c_{\alpha}^{-1} k^{\alpha -1}$ and since $\overline{S}_{k-1}$ converges almost surely to $H$, applying Theorem \ref{ratetheta}, for all $\delta > 0$,
\[
\frac{1}{(n+1)^{2}}\left\| \sum_{k=1}^{n+1} \left( \gamma_{k}^{-1} - \gamma_{k+1}^{-1} \right) \overline{S}_{k-1} \left( \theta_{k} - \theta \right) \right\|^{2} = o \left( \frac{(\ln n)^{1+\delta}}{n^{2-\alpha} }\right) \quad a.s
\]
which is negligible since $\alpha < 1$. Furthermore, since $\overline{S}_{k-1}$ converges almost surely to $H$, one can check that
\[
\frac{1}{(n+1)^{2}} \left\|  \sum_{k=1}^{n} \frac{\gamma_{k+1}^{-1}}{k} \overline{S}_{k-1}\left( \theta_{k} - \theta \right) \right\|^{2} = o \left( \frac{(\ln n)^{1+\delta}}{n^{2- \alpha}} \right) \quad a.s.
\]
Let us now give the rate of convergence of $(**)$. In this aim, let us consider $\delta > 0$ and introduce the events $\Omega_{k}= \left\lbrace \left\| \theta_{k}   - \theta \right\| \leq \frac{(\ln k)^{1/2 + \delta}}{k^{\alpha/2 }} \right\rbrace$. Since $\delta > 0$, and thanks to Theorem \ref{ratetheta}, $\frac{\left\| \theta_{n} - \theta \right\|^{2}n^{\alpha}}{(\ln n)^{1+\delta}}$ converges almost surely to $0$, so that $\mathbf{1}_{\Omega_{k}^{C}}$ converges almost surely to $0$. Furthermore,
\begin{align*}
(**) &= \underbrace{\frac{1}{n+1}\sum_{k=1}^{n+1}\frac{\gamma_{k+1}^{-1}}{k } \left(   \overline{\Phi}_{k} \overline{\Phi}_{k}^{T} + \frac{c_{\beta}}{k^{\beta}} Z_{k}Z_{k}^{T} \right) \left( \theta_{k} - \theta \right)\mathbf{1}_{\Omega_{k}}}_{:= (***)} \\&+ \frac{1}{n+1}\sum_{k=1}^{n+1}\frac{\gamma_{k+1}^{-1}}{k } \left(   \overline{\Phi}_{k} \overline{\Phi}_{k}^{T} + \frac{c_{\beta}}{k^{\beta}} Z_{k}Z_{k}^{T} \right) \left( \theta_{k} - \theta \right) \mathbf{1}_{\Omega_{k}^{C}}
\end{align*}
Since $\mathbf{1}_{\Omega_{k}^{C}}$ converges almost surely to $0$,
\[
\sum_{k\geq 1}\frac{\gamma_{k+1}^{-1}}{k } \left\|   \overline{\Phi}_{k} \overline{\Phi}_{k}^{T} + \frac{c_{\beta}}{k^{\beta}} Z_{k}Z_{k}^{T} \right\| \left\| \theta_{k} - \theta \right\|_{op} \mathbf{1}_{\Omega_{k}^{C}} < + \infty \quad a.s
\]
so that
\[
\frac{1}{n+1}\left\| \sum_{k=1}^{n+1}\frac{\gamma_{k+1}^{-1}}{k } \left(   \overline{\Phi}_{k} \overline{\Phi}_{k}^{T} + \frac{c_{\beta}}{k^{\beta}} Z_{k}Z_{k}^{T} \right) \left( \theta_{k} - \theta \right) \mathbf{1}_{\Omega_{k}^{C}} \right\| = O \left( \frac{1}{n}\right) \quad a.s.
\]
Moreover,
\begin{align*}
(***) \leq \frac{1}{n+1}\sum_{k=1}^{n+1} \frac{\gamma_{k+1}^{-1}}{k}\left\| \overline{\Phi}_{k} \overline{\Phi}_{k}^{T} + \frac{c_{\beta}}{k^{\beta}}Z_{k}Z_{k}^{T} \right\|_{op} \frac{(\ln k)^{1/2 + \delta}}{k^{\alpha /2}} .
\end{align*}
One can consider the sequence of martigales differences $\left( \Xi_{k} \right)$ adapted to the filtration $\left( \mathcal{F}_{k} \right)$ and defined for all $k \geq 1$ by
\[
\Xi_{k} = \left\| \overline{\Phi}_{k} \overline{\Phi}_{k}^{T} + \frac{c_{\beta}}{k^{\beta}}Z_{k}Z_{k}^{T} \right\|_{op} - \mathbb{E}\left[ \left\| \overline{\Phi}_{k} \overline{\Phi}_{k}^{T} + \frac{c_{\beta}}{k^{\beta}}Z_{k}Z_{k}^{T} \right\|_{op}  |\mathcal{F}_{k-1} \right] .
\]
Then,
\[
(***) \leq \frac{1}{n+1}\sum_{k=1}^{n+1} \frac{\gamma_{k+1}(\ln k)^{1/2 + \delta}}{k^{\alpha/2 +1}} \mathbb{E}\left[ \left\| \overline{\Phi}_{k} \overline{\Phi}_{k}^{T} + \frac{c_{\beta}}{k^{\beta}}Z_{k}Z_{k}^{T} \right\|_{op} |\mathcal{F}_{k-1} \right] + \frac{1}{n+1}\sum_{k=1}^{n+1} \frac{\gamma_{k+1}(\ln k)^{1/2 + \delta}}{k^{\alpha/2 +1}} \Xi_{k} .
\]
\end{proof}
Then, thanks to Assumption \textbf{(H1b)} and Toeplitz lemma,
\[
\frac{1}{n+1}\sum_{k=1}^{n+1} \frac{\gamma_{k+1}(\ln k)^{1/2 + \delta}}{k^{\alpha/2 +1}} \mathbb{E}\left[ \left\| \overline{\Phi}_{k} \overline{\Phi}_{k}^{T} + \frac{c_{\beta}}{k^{\beta}}Z_{k}Z_{k}^{T} \right\|_{op} |\mathcal{F}_{k-1} \right] = o \left( \frac{(\ln n)^{1/2 + \delta}}{n^{1- \alpha/2}} \right) \quad a.s .
\]
Furthermore, since $\alpha < 1$, with assumption \textbf{(H1b)}, Theorem~\ref{theomartmoy} leads  to
\[
\left\| \frac{1}{n+1}\sum_{k=1}^{n+1} \frac{\gamma_{k+1}(\ln k)^{1/2 + \delta}}{k^{\alpha/2 +1}} \Xi_{k} \right\|^{2} = O \left( \frac{1}{n^{2}} \right) \quad a.s.
\]

\subsection{Proof of Corollary \ref{ratesn}}
\begin{proof}
The aim is to give the rate of convergence of each term of decomposition (\ref{decsn}). Note that the rate of the martingale term is given by (\ref{vitXi}), while, thanks to equation (\ref{vitzi}), we have
\[
\frac{1}{n^{2}}\left\| \sum_{k=1}^{n} \frac{c_{\beta}}{k^{\beta}} Z_{k+1}Z_{k+1}^{T} \right\|_{F}^{2} = O \left( \frac{1}{n^{2\beta}} \right) \quad a.s.
\]
Finally, since the functional $h \longmapsto \mathbb{E}\left[ \nabla_{h}f \left( X , h \right) \nabla_{h} f ( X,h )^{T} \right]$ is $C_{f}$-Lipschitz on a neighborhood of $\theta$ and since $\overline{\theta}_{n}$ converges almost surely to $\theta$, applying Theorem \ref{ratethetabar} as well as Toeplitz lemma, for all $\delta > 0$,
\begin{align*}
\frac{1}{n^{2}}\left\| \sum_{k=1}^{n} \left( \mathbb{E}\left[ \overline{\Phi}_{k+1}\overline{\Phi}_{k+1}^{T} |\mathcal{F}_{k} \right] - H \right) \right\|_{F}^{2} & = \mathcal{O} \left(  \frac{C_{f}^{2}}{n^{2}} \left( \sum_{k=1}^{n} \left\| \overline{\theta}_{k} - \theta \right\| \right)^{2} \right) \quad a.s  \\
& = o \left( \frac{(\ln n)^{1+\delta}}{n} \right) \quad a.s,
\end{align*}
which concludes the proof.

\begin{rmq}\label{rempourletrucpourri}
Note that to prove equality \eqref{tlc} without knowing the rate of convergence of $\overline{\theta}_{n}$, it is necessary to have a first rate of convergence of $\overline{S}_{n}$. For that purpose, we study the asymptotic behaviour of
\[
\frac{1}{n^{2}}\left\| \sum_{k=1}^{n} \left( \mathbb{E}\left[ \overline{\Phi}_{k+1}\overline{\Phi}_{k+1}^{T} |\mathcal{F}_{k} \right] - H \right) \right\|_{F}^{2}.
\]
Equality \eqref{ratethetaas} yields for all $\delta > 0$,
\[
\left\| \overline{\theta}_{n} - \theta \right\| \leq \frac{1}{n+1}\sum_{k=0}^{n} \left\| \theta_{k} - \theta \right\| = O \left( \frac{(\ln n)^{1 /2 + \delta /2}}{n^{\alpha /2}} \right) ,
\]
so that, since the functional $h \longmapsto \mathbb{E}\left[ \nabla_{h}f \left( X , h \right) \nabla_{h} f ( X,h )^{T} \right]$ is $C_{f}$-Lipschitz on a neighborhood of $\theta$,
\begin{align*}
\frac{1}{n^{2}}\left\| \sum_{k=1}^{n} \left( \mathbb{E}\left[ \overline{\Phi}_{k+1}\overline{\Phi}_{k+1}^{T} |\mathcal{F}_{k} \right] - H \right) \right\|_{F}^{2} & = \mathcal{O} \left(  \frac{C_{f}^{2}}{n^{2}} \left( \sum_{k=1}^{n} \left\| \overline{\theta}_{k} - \theta \right\| \right)^{2} \right) \quad a.s  \\
& = o \left( \frac{(\ln n)^{1+\delta}}{n^{\alpha}} \right) \quad a.s,
\end{align*}
and then, since $\beta < \alpha  -1/2$,
\[
\left\| \overline{S}_{n} - H \right\|_{F} = O  \left( \max \left\lbrace  \frac{c_{\beta}}{n^{\beta}} , \sqrt{\frac{(\ln n)^{1+\delta}}{n^{\alpha}}} \right\rbrace  \right) \quad a.s
\]
\end{rmq}
\end{proof}

\subsection{Proof of Theorem \ref{theothetatilde}\label{prooftilde}}
Only the main lines of the proof are given since it is a mix between the proof of Theorem~\ref{ratethetabar} and the ones in \cite{BGBP2019}.

\begin{proof}[Proof of THeorem \ref{theothetatilde}]
Let us denote $\overline{H}_{n}^{-1} = (n+1) \widetilde{H}_{n}^{-1}$. First, remark that as in the proof of Corollary \ref{ratesn}, one can check that that on $\widetilde{\Gamma}_{\theta}$,
\[
\overline{H}_{n} \xrightarrow[n\to + \infty]{a.s} H \quad \quad \text{and} \quad \quad \overline{H}_{n}^{-1} \xrightarrow[n\to + \infty]{a.s} H^{-1} .
\]
Furthermore, decomposition \eqref{decxi} can be rewritten as
\begin{equation}
\label{decxitilde} \widetilde{\theta}_{n+1} = \widetilde{\theta}_{n} - \theta - \frac{1}{n+1} \overline{S}_{n}^{-1} \nabla G \left( \widetilde{\theta}_{n} \right) + \frac{1}{n+1}\overline{H}_{n}^{-1} \widetilde{\xi}_{n+1},
\end{equation}
where $\widetilde{\xi}_{n+1} := \nabla G \left( \widetilde{\theta}_{n} \right) - \nabla_{h} g \left( X_{n+1} , Y_{n+1} , \widetilde{\theta}_{n} \right)$. Then, $\left( \widetilde{\xi}_{n} \right)$ is a sequence of martingale differences adapted to the filtration $\left( \mathcal{F}_{n} \right)$. Linearizing the gradient, decomposition \eqref{decdelta} can be rewritten as
\begin{align}
\notag \widetilde{\theta}_{n+1} - \theta & = \frac{n}{n+1} \left( \widetilde{\theta}_{n} - \theta \right) + \frac{1}{n+1}\left( H^{-1} - \overline{H}_{n}^{-1} \right)H \left( \widetilde{\theta}_{n} - \theta \right) \\
\label{decdeltatilde} & + \frac{1}{n+1}\overline{H}_{n}^{-1}\widetilde{\xi}_{n+1} - \frac{1}{n+1}\overline{H}_{n}^{-1} \widetilde{\delta}_{n}
\end{align}
where $\widetilde{\delta}_{n} = \nabla G \left( \widetilde{\theta}_{n} \right) - H \left( \widetilde{\theta}_{n} - \theta \right)$ is the reminder term in the Taylor's decomposition of the gradient. Then, by induction, it comes
\begin{align}
\label{dectot} \widetilde{\theta}_{n} - \theta = \frac{1}{n}\left( \widetilde{\theta}_{0} - \theta \right)+ \frac{1}{n}\sum_{k=0}^{n-1} \left( H^{-1} - \overline{H}_{k}^{-1} \right) H \left( \widetilde{\theta}_{k} - \theta \right) - \frac{1}{n}\sum_{k=0}^{n-1} \overline{H}_{k}^{-1} \widetilde{\delta}_{k} + \frac{1}{n}\sum_{k=0}^{n-1}\overline{H}_{k}^{-1}\widetilde{\xi}_{k+1} .
\end{align}
Since $\overline{H}_{k}^{-1}$ converges almost surely to $H^{-1}$, on $\widetilde{\Gamma}_{\theta}$, one can check that
\[
\frac{1}{n}\sum_{k=0}^{n-1} \overline{H}_{k}^{-1} \widetilde{\xi}_{k+1}\widetilde{\xi}_{k+1}^{T}\overline{H}_{k}^{-1} \xrightarrow[n\to + \infty]{a.s} \sigma^{2}H^{-1}.
\]
Thanks to assumption \textbf{(H5)}, applying a law of large numbers for martingales, one can check that
\begin{equation}\label{vitxitilde}
\left\|\frac{1}{n}\sum_{k=0}^{n-1} \overline{H}_{k}^{-1} \widetilde{\xi}_{k+1} \right\|^{2} = \mathcal{O} \left( \frac{\ln n}{n} \right) \quad a.s
\end{equation}
In the same way, applying  a central limit theorem, it comes
\begin{equation}\label{tlctilde}
\frac{1}{\sqrt{n}} \sum_{k=0}^{n-1} \overline{H}_{k}^{-1} \widetilde{\xi}_{k+1} \xrightarrow[n\to + \infty]{\mathcal{L}} \mathcal{N}\left( 0 , \sigma^{2}H^{-1} \right) .
\end{equation}
Let us now prove that the other terms in decomposition \eqref{dectot} are negligible. First, clearly
\begin{equation}
\label{vitdebile}\frac{1}{n}\left\| \widetilde{\theta}_{0} - \theta \right\| = \mathcal{O} \left( \frac{1}{n} \right) \quad a.s.
\end{equation}
Furthermore, let us denote
\[
\widetilde{\Delta}_{n} = \frac{1}{n}\sum_{k=0}^{n-1} \left( H^{-1} - \overline{H}_{k}^{-1} \right) H \left( \widetilde{\theta}_{k} - \theta \right) - \frac{1}{n}\sum_{k=0}^{n-1} \overline{H}_{k}^{-1} \widetilde{\delta}_{k} .
\]
Furthermore, as in the proof of Theorem \ref{ratetheta}, one can verify
\begin{align*}
& \left\| \left( H^{-1} - \overline{H}_{n}^{-1} \right) H \left( \widetilde{\theta}_{n} - \theta \right) \right\| = o \left( \left\| \widetilde{\theta}_{n} - \theta \right\| \right) \quad a.s
& \left\| \widetilde{\delta}_{n} \right\| = o \left( \left\| \widetilde{\theta}_{n} - \theta \right\| \right) \quad a.s
\end{align*}
Then,
\begin{equation}\label{deltatilde}
\left\| \widetilde{\Delta}_{n+1} \right\| = \frac{n}{n+1} \left\| \widetilde{\Delta}_{n} \right\| + o \left( \widetilde{\theta}_{n} - \theta \right) \quad a.s.
\end{equation}
As in the proof of Theorem 6.2 in \cite{BGBP2019} (see equations (6.24) to (6.32)), it comes
\begin{equation}
\label{vitdeltatilde}\left\| \widetilde{\Delta}_{n} \right\|^{2} = \mathcal{O} \left( \frac{\ln n}{n} \right) \quad a.s.
\end{equation}
Then, thanks to equalities \eqref{vitxitilde},\eqref{vitdebile} and \eqref{vitdeltatilde}, it comes
\begin{equation}\label{vitthetatilde}
\left\|\widetilde{\theta}_{n} - \theta \right\|^{2} = \mathcal{O} \left( \frac{\ln n}{n} \right) \quad a.s.
\end{equation}
In order to get the asymptotic normality of $\widetilde{\theta}_{n}$, let us now give the rate of convergence of each term on the right-hand side of decomposition \eqref{dectot}. First, since
\[
\widetilde{\delta}_{n} = \mathcal{O} \left( \left\| \widetilde{\theta}_{n} - \theta \right\|^{2} \right) \quad a.s,
\]
and since $\overline{H}_{n}^{-1}$ converges almost surely to $H^{-1}$, thanks to equality \eqref{vitthetatilde}, for all $\delta > 0$
\[
\frac{1}{n}\left\| \sum_{k=0}^{n-1} \overline{H}_{k}^{-1}\widetilde{\delta}_{k} \right\| = o \left( \frac{(\ln n)^{2+\delta}}{n} \right) \quad a.s
\]
which is so negligible. Furthermore, as in the proof of Corollary \ref{ratesn}, one can check that for all $\delta > 0$,
\[
\left\| \overline{H}_{n}^{-1} - H^{-1} \right\|^{2} = \mathcal{O} \left( \max \left\lbrace \frac{c_{\beta}^{2}}{n^{2\beta}} , \frac{(\ln n)^{1+\delta}}{n} \right\rbrace \right) \quad a.s .
\]
Then, for all $\delta > 0$,
\[
\frac{1}{n}\left\|\sum_{k=1}^{n} \left( \overline{H}_{k}^{-1} - H^{-1} \right) H \left( \widetilde{\theta}_{k} - \theta \right) \right\| = o \left( \max \left\lbrace \frac{c_{\beta}(\ln n)^{1/2+\delta}}{n^{\beta + 1/2}} , \frac{(\ln n)^{2+\delta}}{n} \right\rbrace \right) \quad a.s
\]
and this term is no negligible.
\end{proof}

\subsection{Proof of Corollary \ref{estsigma}}
First, note that
\[
\left( \hat{Y}_{k} - Y_{k} \right)^{2} - \sigma^{2} = \left( f \left( X_{k} , \overline{\theta}_{k-1} \right) - f \left( X_{k} , \theta \right) \right)^{2} -2 \epsilon_{k} \left( f \left( X_{k} , \overline{\theta}_{k-1} \right) - f \left( X_{k} , \theta \right) \right) + \left( \epsilon_{k}^{2} - \sigma^{2} \right) .
\]
Thanks to a Taylor's decomposition, there are $U_{0},\ldots,U_{n-1} \in \mathbb{R}^{q}$ such that
\[
R_{1} := \frac{1}{n}\sum_{k=1}^{n} \left( f \left( X_{k} , \overline{\theta}_{k-1} \right) - f \left( X_{k} , \theta \right) \right)^{2} \leq \frac{1}{n}\sum_{k=1}^{n} \left\| \nabla_{h} f \left( X_{k} , U_{k-1} \right) \right\|^{2} \left\| \overline{\theta}_{k-1} - \theta \right\|^{2}
\]
Let us consider the filtration $\left( \mathcal{F}^{(1)}_{n} \right)$ defined by $\mathcal{F}_{n}^{(1)} = \sigma \left( \left( X_{k} , \epsilon_{k-1}\right), k=1,\ldots ,n  \right)$. Then, considering $\widetilde{\xi}_{k} = \mathbb{E}\left[ \left\| \nabla_{h} f \left( X_{k} , U_{k-1} \right) \right\|^{2} |\mathcal{F}^{(1)}_{k-1} \right] - \left\| \nabla_{h} f \left( X_{k} , U_{k-1} \right) \right\|^{2}$, it comes thanks to assumption \textbf{(H1b)},
\[
R_{1} \leq \frac{1}{n}\sum_{k=1}^{n}\sqrt{C''} \left\| \overline{\theta}_{k-1} - \theta \right\|^{2} -  \frac{1}{n}\sum_{k=1}^{n} \widetilde{\xi}_{k} \left\| \overline{\theta}_{k-1} - \theta \right\|^{2}
\]
Then, applying Toeplitz Lemma and Theorem \ref{ratethetabar} to the term on the left hand-side of previous inequality, one can check that for all $\delta > 0$,
\[
\frac{1}{n}\sum_{k=1}^{n}\sqrt{C''} \left\| \overline{\theta}_{k-1} - \theta \right\|^{2}  = o \left( \frac{(\ln n)^{2+ \delta}}{n}\right) \quad a.s.
\]
Furthermore, since $\left( \widetilde{\xi}_{n} \right)$ is a sequence of martingale differences with bounded squared moments, with the help of Theorems \ref{ratethetabar} and \ref{theomartmoy}, it comes
\[
\frac{1}{n}\sum_{k=1}^{n} \widetilde{\xi}_{k} \left\| \overline{\theta}_{k-1} - \theta \right\|^{2} = \mathcal{O} \left( \frac{1}{n} \right) \quad a.s
\]
so that
\[
R_{1} = o \left( \frac{(\ln n)^{2+\delta}}{n}\right) \quad a.s.
\]
Furthermore, let us consider
\[
R_{2} = \frac{1}{n}\sum_{k=1}^{n}\epsilon_{k} \left( f \left( X_{k} , \overline{\theta}_{k-1} \right) - f \left( X_{k} , \theta \right) \right) .
\]
We have a sum of martingale differences, and thanks to Theorems \ref{ratethetabar} and \ref{theomartmoy}, one can check that
\[
\left\| R_{2} \right\|^{2} = o \left( \frac{(\ln n)^{2+\delta}}{n^{2}} \right) \quad a.s.
\]
Then, considering that
\[
\epsilon_{n}^{2} - \sigma^{2} = \epsilon_{n}^{2} - \mathbb{E}\left[ \epsilon_{n}^{2} |\mathcal{F}^{(1)}_{n-1} \right]
\]
one can apply a Central Limit Theorem for martingale to get the asymptotic normality, and a strong law of large numbers for martingales to get the almost sure rate of convergence.


TITLE: Calculus of Variations-question on rotating curve of max volume
QUESTION [1 upvotes]: My calc of variations is still rusty.  I'm assuming implementation of arclength revolution formula is necessary, but how to find y(1/2a)?

REPLY [1 votes]: We can decrease the problem to find the largest area in 2D because it will give the largest volume in rotating the curve.
The are to be maximized can be defined as
$$A[y]=\int_{x_1}^{x_2}y\ dx$$
subject to
$$2a=\int_{x_1}^{x_2}\sqrt{1+y'\ ^2} dx$$
We can use Lagrange multiplier such as
$$H=y+\lambda \sqrt{1+y'\ ^2}$$
and
$$\frac{\partial H}{\partial y}-\frac{d}{dx}\bigg(\frac{\partial H}{\partial y'}\bigg)=0$$
$$1-\frac{d}{dx}\bigg(\frac{\lambda y'}{\sqrt{1+y'\ ^2}}\bigg)=0$$
$$\Rightarrow \frac{x-\alpha}{\lambda}=\frac{y'}{\sqrt{1+y'\ ^2}}$$
If $\bar x=\frac{x-\alpha}{\lambda}$
$$\bar x=\frac{y'}{\sqrt{1+y'\ ^2}}\Rightarrow y'=\frac{dy}{dx}=\frac{\bar x}{\sqrt{1-\bar x^2}}$$
which can be solved for $y(x)$ such that
$$y(x)=-\sqrt{1-\bar x^2}+\beta=\beta-\sqrt{1-\bigg(\frac{x-\alpha}{\lambda}\bigg)^2}$$
The equation must satisf boundary conditions
$$(0,0)\Rightarrow 0=\beta-\sqrt{1-\bigg(\frac{-\alpha}{\lambda}\bigg)^2}$$
$$(a,0)\Rightarrow 0=\beta-\sqrt{1-\bigg(\frac{a-\alpha}{\lambda}\bigg)^2}$$
$$2a=\int_{0}^{a}\sqrt{1+y'\ ^2} dx\qquad y'=\frac{x-\alpha}{\lambda^2 \sqrt{1-\frac{\big(\alpha-x\big)^2}{\lambda^2}}}$$


TITLE: Is closurness a necessary condition in cantor's interesction theorem?
QUESTION [1 upvotes]: Cantor's Intersection Theorem:
Let $X$ be a complete metric space and let $\{F_n\}$ be a sequence of decreasing non-empty closed subsets of $X$. If $d(F_n)\rightarrow 0$ then $F=\bigcap_{n=1}^\infty F_n$ contains exactly a unique point of $X$
Where $d(F_n)$ means the diameter of the set $F_n$ and $d(F_n)\rightarrow 0$ means that the sequence of the diameters converge to $0$.
I found that I can find an example where all the conditions except the closureness of the subsets hold but we still have the intersection a single point.
The example is, Consider $X$ to be the complex plane.For $n= 1,2,3,4,5,...$, define
$F_n=\{(x,y):\sqrt{x^2 + y^2} < \frac{1}{n}\}$. So $F_n$ is disc of radius $\frac{1}{n}$.
It's clear that the sequence $\{F_n\}$ is a decreasing one. every $F_n$ is an open (but not closed) sphere of $X$ the sequence of diameters converges to $0$ but we still have:
$\bigcap_{n=1}^{\infty} F_n=\{(0,0)\}$.
So, it seems that this condition is sufficient but not necessary, Am I right?
What are the necessary conditions of this theorem?
Added:
I wonder if there is an example where only a one subset is not closed but the other conditions hold and we have an empty intersection?

REPLY [2 votes]: Yes, sufficient but not necessary. But to show one needs the condition consider examples like $F_n=(0,1/n)$ in the space of reals, where this time the intersection is empty (here the $F_n$ are not closed). To drop the other condition, just consider a constant sequence $F_n=[0,1]$ where this time the $F_n$ are closed but the lengths $d(F_n)$ fail to approach zero, and there are more than one point in the intersection (in fact the entire interval $[0,1]$ is the intersection).
Added: The OP has asked about what if only one of the $F_n$ is not assumed to be closed. Suppose that not necessarily closed set happens when the index $n$ is some particular natural number $n=n_1.$ All the sets $F_k$ where $k\ge n_1+1$ are then assumed to be closed. Their intersection $G$ is then nonempty by Cantor, and $G$ is a subset of $F_{n_1+1}$ because the sets are a decreasing sequence. Thus there is a point $p$ in $G$, which then lies in every $F_j$ for which $j \ge n_1+1.$  Now again from the fact that the whole sequence of sets is decreasing, $F_{n_1+1}$ is a subset of each $F_j$ for $1 \le j \le F_{n_1},$  so that the point $p$ also lies in each set $F_j$ where $1 \le j \le n_1.$ this shows that for every index $j\ge 1$ we have $p$ in $F_j,$ which is to say that $p$ is in the intersection of all the $F_j$'s.


TITLE: Recognizing Permutation of Group with different Label
QUESTION [0 upvotes]: Problem description:
Assume, a group, $G \leq S_{26}$ , $S_{26}$ is a symmetric group.
Each permutation of $G$ is labeled using $1,2,....26$ as usual.
Suppose, $f$ is a function  that changes label of permutations.
Instead of $\{1,2,....26\}$, $f$ uses  $\{a,b,....z \} $ .
now, take   each element $\pi \in G$,  take  $f(\pi)$ and from set  $H$.
Question:
$G, H$ are given and $f$ is not.
For which group, we can find the bijection $f$ between $G$ and $H$?
A toy example:if $G$ has  permutation $(123)$ and $H$ has  permutation (d  e  f) we can say, $f(1)=d, f(2)=e, f(3)=f$.

REPLY [1 votes]: Your questions are very difficult to understand. You have not really said what $H$ is for example. Is $H = \{ f(\pi) : \pi \in G \}$?
If I am understanding the question correctly, then $f$ is uniquely determined if and only if $C_{S_n}(G)=1$.
When $G$ is transitive, this is the case if and only if $G$ is nonabelian and primitive.


TITLE: About the polynomials satisfying $P\left((x-y)^2\right)+P\left((y-z)^2\right)+P\left((z-x)^2\right)=18P\left(\left(\frac{(x+y+z)^2}3\right)^2\right)$
QUESTION [0 upvotes]: Find all polynomials $P(x)$ that satisfy
$$P\left((x - y)^2\right) + P\left((y - z)^2\right) + P\left((z - x)^2\right) = 18P\left(\left(\frac{(x + y + z)^2}{3}\right)^2\right)$$
when $x,y,z \in \mathbb R$ and $xy + yz + zx = 0$.

This is from POSN Camp 2 Thailand.
I've done it and got $P(x) = 0$ or $P(x) = ax^2$ for some $a \in \mathbb R$. I checked my answer and found that $P(x) = 0$ only. But, I found out after discussing with my friend who did it later that $P(x) = ax^2$ are solutions, too, and that I had checked it wrongly. What it the solution of this problem? I am confused. Am I correct or wrong?

REPLY [1 votes]: Note: I believe the RHS should be $18 P \left ( \frac{ (x+y+z)^2 } { 3} \right ) $. If not, ignore this.

Set $ x= y = z = 0$, so $P(0) = 0$.
Set $ x = y = 0 $, so $ P \left ( z^2 \right ) = 9 P \left ( \frac{z^2}{3} \right )$ holds for all $z$. (Do some work here.) Hence, the solutions are a subset of $P(x) = ax^2 $.
Verify that $P(x) = ax^2$ indeed satisfies the equation. Hence, these are all of the solutions.


TITLE: Existance of a simple function
QUESTION [3 upvotes]: Let $f : [a, b] \to \mathbb{R}$ be a measurable function. Suppose that $\varepsilon, M > 0$ are given. Show that there is some simple function $\varphi : [a, b] \to \mathbb{R}$ such that $|f(x) - \varphi(x)| < \varepsilon$ for all $x$ except where $|f(x)| > M$.
I was wondering if I could get a hint.

REPLY [1 votes]: Hint: a bounded measurable function is an uniform limit of a sequence of simple functions. Use this with $f\chi_{\{|f|\leqslant M\}}$.


TITLE: 'Linux' math program with interactive terminal?
QUESTION [43 upvotes]: Are there any open source math programs out there that have an interactive terminal and that work on linux?
So for example you could enter two matrices and specify an operation such as multiply and it would then return the answer or a error message specifying why an answer can't be computed? I am just looking for something that can perform basic matrix operations and modular arithmetic.

REPLY [2 votes]: freemat
FreeMat is a free environment for rapid engineering and scientific prototyping and data processing. It is similar to commercial systems


TITLE: An equation in the convolution measure algebra on reals
QUESTION [0 upvotes]: Let us consider the measure algebra $M(\mathbb{R})$ consisting of all Radon measures on reals.
Let $\mu$ be a Radon measure in  $M(\mathbb{R})$ and $\delta_0$ be the point mass measure concentrated on 0,  which is also the multiplicative identity of $M(\mathbb{R})$.

Q. Does there exist any strictly positive real number $r>0$ and Radon measures  $\mu_1,\cdots \mu_n$ satisfying the following idenity? $$\mu^**\mu=r\delta_0+\sum_1^n\mu_i^**\mu_i.$$

Remark. $M(\mathbb{R})=C_0(\mathbb{R})^*$ is a unital dual Banach $*$-algebra. The involution on $M(\mathbb{R})$ is defined as follow
$$\langle \mu^*,f\rangle=\int f(t)\overline{d\mu(-t)}.$$

REPLY [4 votes]: The answer is no in general. Indeed, let $\hat\mu$ denote the Fourier transform of $\mu$, so that $\hat\mu(t)=\int e^{itx}\mu(dx)$ for real $t$. Then the equality
$$\mu^**\mu=r\delta_0+\sum_1^n\mu_i^**\mu_i$$
would imply
$$|\hat\mu|^2=r+\sum_1^n|\hat\mu_i|^2\ge r>0.$$
Taking now any $\mu$ with $\hat\mu(t)\to0$ as $t\to\infty$ gives a contradiction.


text:
\section{Background and Motivation}\label{sec:background}
\subsection{Notation}
We denote by $\mathbb{R}$, $\mathbb{R}^{n}$ and $\mathbb{R}^{n \times m}$ the real numbers, $n$-dimensional real vectors and $n \times m$ dimensional real matrices, respectively.
The interior of a set  is denoted by $\text{int}(\cdot)$, its cardinality by $|\cdot|$ and $\left \lfloor \cdot \right \rfloor$ denotes the \emph{floor} operation, i.e. the rounding to the nearest lower integer.
The probability of an event is denoted by $P(\cdot)$ and the composition of two functions $f$ and $g$ by $g\circ f(\cdot) = g(f(\cdot))$.

\subsection{Explicit MPC}
Model predictive control (MPC) is an optimal control scheme that uses a system model to predict the future evolution of a system.
We consider discrete linear time-invariant (LTI) systems:
\begin{align}\label{eq:LTI}
x_{k+1} = Ax_k + Bu_k,
\end{align}
where $x \in \mathbb{R}^{n_x}$ is the state vector, $u \in \mathbb{R}^{n_u}$ is the control input, $A \in \mathbb{R}^{n_x \times n_x}$ is the system matrix, $B \in \mathbb{R}^{n_x \times n_u}$ is the input matrix and the pair $(A,B)$ is controllable.

Using a standard quadratic cost function, the following constrained finite time optimal control problem with a horizon of $N$ steps should be solved at each sampling time to obtain the MPC feedback law:

\begin{subequations}\label{eq:FTOCP}
	\begin{align}
		&\underset{\tilde u}{\text{minimize}} && x_N^T P x_N + \sum_{k=0}^{N-1}{x_k^T Q x_k + u_k^T R u_k} \span \span \span \\
		&\text{subject to} &&x_{k+1} = Ax_k + Bu_k, \\
		&            &&C_x x_k \leq c_x,\, C_f x_N \leq c_f, \label{eq:state_terminal_cons}\\
		&            &&C_u u_k \leq c_u, \label{eq:input_cons}\\

		&			 &&x_0 = x_{\text{init}}, \\
		& && \forall \,\, k = 0,\dots, N-1,
	\end{align}
\end{subequations}

where $\tilde u = [u_0, \dots, u_{N-1}]^T$ is a vector that contains the sequence of control inputs and $P \in \mathbb{R}^{n_x \times n_x}$, $Q \in \mathbb{R}^{n_x \times n_x}$ and $R \in \mathbb{R}^{n_u \times n_u}$ are the weighting matrices. The weighting matrices are chosen such that $P \succeq 0$ and $Q \succeq 0$ are positive semidefinite, and $R \succ 0$ is positive definite. The state, terminal and input constraints are bounded polytopic sets $\mathcal X$, $\mathcal X_f$ and $\mathcal U$ defined by the matrices $C_x\in \mathbb R^{n_{\text{cx}}\times n_x}$, $C_f\in \mathbb R^{n_{\text{cf}}\times n_x}$, $C_u\in \mathbb R^{n_{\text{cu}}\times n_u}$ and the vectors $c_x\in \mathbb R^{n_{\text{cx}}}$, $c_f\in \mathbb R^{n_{\text{cf}}}$, $c_u\in \mathbb R^{n_{\text{cu}}}$.
The terminal cost defined by $P$ as well as the terminal set $\mathcal X_f$ are chosen in such a way that stability of the closed-loop system and recursive feasibility of the optimization problem are guaranteed~\cite{mayne2000}. The set of initial states $x_{\text{init}}$ for which~\eqref{eq:FTOCP} has a feasible solution depending on the prediction horizon $N$ is called feasibility region and is denoted by $\mathcal X_N$.

The optimization problem \eqref{eq:FTOCP} can be reformulated as a multi-parametric problem \cite{bemporad2002} that only depends on the current system state $x_{\text{init}}$:
\begin{subequations}\label{eq:condensed}
	\begin{align}
		&\underset{\tilde u}{\text{minimize}}&& \tilde u^T F \tilde u + x_{\text{init}}^T G \tilde u + x_{\text{init}}^T H x_{\text{init}} \span \\
		&\text{subject to} && C_c \tilde u \leq T x_{\text{init}} + c_c,
	\end{align}
\end{subequations}
where $F \in \mathbb{R}^{N n_u \times N n_u}$, $G \in \mathbb{R}^{n_x \times N n_u}$, $H \in \mathbb{R}^{n_x \times n_x}$, $C_c \in \mathbb{R}^{N n_{\text{ineq}} \times N n_u}$, $T \in \mathbb{R}^{N n_{\text{ineq}} \times n_x}$, $c_c \in \mathbb{R}^{N n_{\text{ineq}}}$ and $n_{\text{ineq}}$ is the total number of inequalities in~\eqref{eq:FTOCP}.

The solution of the multi-parametric quadratic programming problem~\eqref{eq:condensed} is a piecewise affine (PWA) function of the form \cite{bemporad2002}:
\begin{align}
\mathcal{K}(x_{\text{init}}) =
	\begin{cases}
		K_1 x_{\text{init}}+g_1 & \text{if} \quad x_{\text{init}} \in \mathcal{R}_1, \\
		& \vdots \\
		K_{n_{\text{r}}} x_{\text{init}}+g_{n_{\text{r}}} & \text{if} \quad x_{\text{init}} \in \mathcal{R}_{n_{\text{r}}},
	\end{cases}
\label{eq:exp_mpc}
\end{align}
with $n_{\text{r}}$ regions, $K_i \in \mathbb{R}^{Nn_u \times n_x}$ and $g_i \in \mathbb{R}^{Nn_u}$.
Each region $\mathcal R_i$ is described by a polyhedron
\begin{align}
\mathcal{R}_i = \{x \in \mathbb{R}^{n_x} \mid Z_i x \leq z_i\} \quad \forall i = 1,\dots,n_r,
\end{align}
where $Z_i \in \mathbb{R}^{c_i \times n_x}$, $z_i \in \mathbb{R}^{c_i}$ describe the $c_i$ hyperplanes $a_{i,j} x_{\text{init}} \leq b_{i,j}$ of the $i$-th region with $j = 1,\dots,c_i$, $a_{i,j} \in \mathbb{R}^{1 \times n_x}$ and $b_i \in \mathbb{R}$.
The formulation \eqref{eq:exp_mpc} is defined on the bounded polytopic partition $\mathcal{R}_\Omega = \cup_{i=1}^{n_{\text{r}}} \mathcal{R}_i$ with $\text{int}(\mathcal{R}_i) \cap \text{int}(\mathcal{R}_j) = \emptyset$ for all $i \neq j$.

Most hyperplanes are shared by neighbouring regions and the feedback law can be identical for two or more regions.
Hence, the memory needed to store the explicit MPC controller~\eqref{eq:exp_mpc} can be approximated as
\begin{align}\label{eq:memory_empc}
\Gamma_{\mathcal{K}} = \alpha_{\text{bit}} \left( \left( n_{\text{h}} \left( n_x + 1 \right) \right) + n_{\text{f}} \left(n_x n_u + n_u \right) \right),
\end{align}
where $n_{\text{h}}$ is the number of unique hyperplanes, $n_{\text{f}}$ is the number of unique feedback laws and $\alpha_{\text{bit}}$  is the memory necessary to store a real number. Since for the actual implementation of the explicit MPC law only the input of the first time step is needed, only the  first $n_u$ rows of $K_j$ and $g_j$ for $j=1,\dots,n_{\text{f}}$ have to be stored which equals $n_x n_u+n_u$ numbers per unique feedback law.

\subsection{Artificial Neural Networks}
This subsection shortly recaps the fundamental concepts of artificial neural networks.
A feed-forward neural networks is defined as a sequence of layers of neurons which determines a function $\mathcal{N}:\mathbb{R}^{n_x} \rightarrow \mathbb{R}^{n_u}$ of the form
\begin{equation}\label{eq:neural_network}
\begin{split}
\mathcal{N}(x;\theta,M,L) = \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad\\
\quad \bigg \{
\begin{array}{lll}
		f_{L+1} \circ g_L \circ f_L \circ \dots \circ g_1 \circ f_1(x) & \text{for} & L \geq 2, \\
		f_{L+1} \circ g_1 \circ f_1(x), & \text{for} & L = 1,
\end{array}
\end{split}
\end{equation}
where the input of the network is $x \in \mathbb{R}^{n_x}$ and the output of the network is $ u \in \mathbb{R}^{n_u}$.
$M$ is the number of neurons in each hidden layer and $L$ is the number of hidden layers.
If $L\geq2$, $\mathcal N$ is described as a \emph{deep} neural network and if $L=1$ as a \emph{shallow} neural network.
Each hidden layer consists of an affine function:
\begin{align}\label{eq:affine_function}
f_l(\xi_{l-1}) = W_l\xi_{l-1}+b_l,
\end{align}
where $\xi_{l-1} \in \mathbb{R}^M$ is the output of the previous layer with $\xi_0 = x$. The second element of the neural network is a nonlinear activation function $g_l$.
In this paper, exclusively rectifier linear units (ReLU) are considered as activation function, which compute the element-wise maximum between zero and the affine function of the current layer $l$:
\begin{align}\label{eq:ReLU}
g_{l}(f_{l}) = \max(0,f_{l}).
\end{align}
The parameter $\theta = \{\theta_1, \dots, \theta_{L+1}\}$ contains all the weights and biases of the affine functions of each layer
\begin{align}
\theta_l = \{W_l,b_l\} \quad \forall l = 1, \dots, L+1,
\end{align}
where the weights are
\begin{align}
W_l \in
	\begin{cases}
		\mathbb{R}^{M \times n_x} & \text{if} \quad l=1, \\
		\mathbb{R}^{M \times M} & \text{if} \quad l=2,\dots,L, \\
		\mathbb{R}^{n_u \times M} & \text{if} \quad l=L+1,
	\end{cases}
\end{align}
and the biases are
\begin{align}
b_l \in
	\begin{cases}
		\mathbb{R}^M & \text{if} \quad l=1,\dots,L, \\
		\mathbb{R}^{n_u} & \text{if} \quad l = L+1.
	\end{cases}
\end{align}

\subsection{Motivation}

Implementing the explicit model predictive controller defined by \eqref{eq:exp_mpc} requires storing the unique sets of hyperplanes and feedback laws that define the regions and the affine controllers in each region~\eqref{eq:memory_empc}.
The number of regions can grow exponentially with respect to the horizon and the number of constraints which leads to large memory requirements that can prohibit the application of explicit MPC for larger case studies.

The main motivation of this work is to find an efficient representation and approximation of the MPC control law, which can significantly reduce the memory requirements for the representation of the exact controller as well for achieving a high-quality approximation that outperforms other approximate explicit MPC techniques.

The basic idea that motivates this work is described in the following Lemma.
\begin{lem}{\cite{montufar2014}}\label{lemma:expo}
Every neural network $\mathcal N(x;\theta, M, L)$ with input $x\in \mathbb R^{n_x}$ defined as in \eqref{eq:neural_network} with ReLUs as activation functions and $M \geq n_x$ represents a piecewise affine function. In addition, a lower bound on the maximal number of affine regions that the neural network represents is given by the following expression:
\begin{align*}
\left( \prod \limits_{l=1}^{L-1} \left \lfloor \frac{M}{n_x}\right \rfloor^{n_x} \right) \sum \limits_{j=0}^{n_x} \binom{L}{j}.
\end{align*}
\end{lem}
\begin{pf}

The neural network $\mathcal N(x;\theta, M, L)$ is a piecewise affine function because it only contains compositions of affine transformations with a piecewise affine function (ReLUs).
For the derivation of the maximal number of regions, see \cite{montufar2014}. \QEDA
\end{pf}

Lemma~\ref{lemma:expo} gives clear insights about why deep networks, as often observed in practice, obtain better performance to approximate complex functions when compared to shallow networks. In particular, Lemma~\ref{lemma:expo} implies that the number of affine regions that a neural network can represent grows exponentially with the number of layers $L$ as long as the width of the network $M$ is not smaller than the number of inputs $n_x$. The bound of Lemma~\ref{lemma:expo} can be slightly improved if $M\geq 3n_x$ as recently shown in \cite{serra2018}.

At the same time, the number of parameters contained in $\theta$ that are necessary to fully describe the neural network $\mathcal N(x;\theta, M, L)$ are determined by the dimensions of the weights and biases at each layer. Assuming that storing each number requires $\alpha_{\text{bit}}$ bits, the total amount of memory necessary to store the neural network $\mathcal N(x;\theta, M, L)$ can be computed as:
\begin{align}\label{eq:mem_deep}
\Gamma_{\mathcal{N}} =\alpha_{\text{bit}}(&(n_x+1)M + (L-1)(M+1)M\notag \\& + (M+1)n_u).
\end{align}
Since $\Gamma_{\mathcal{N}}$ only grows linearly with respect to the number of layers $L$, deep ReLU networks can represent exponentially many more linear regions than shallow ones for a fixed amount of memory. This fact can be clearly seen in Fig.~\ref{fig:regions_weights}.
\begin{figure}[t]
\begin{center}
\includegraphics[width=0.9\columnwidth]{figures/regions_weights}
\caption{Number of regions with respect to the number of weights a neural network can represent. The parameters for this plot were chosen to $n_x = 2$, $M = 10$, $L = 1, \dots, 50$ and $n_u = 4$.}
\label{fig:regions_weights}
\end{center}
\end{figure}

We believe that this observation, while somewhat obvious, is a very powerful result with important implications in control theory and constitutes the main motivation for this paper.

config: arxiv


TITLE: A problem on limits
QUESTION [0 upvotes]: When is it the case that $\limsup _{n}(\alpha a_{n})=\alpha \limsup _{n} a_{n} ?$ When does it not hold?
I believe that this holds when $\alpha \geq$ 0, then the equality can be obtained. Are there any other conditions?

REPLY [3 votes]: This holds only if $\alpha \geq 0$, as the hint above mentions.
For $\alpha < 0$ this becomes false. For example,
$$\limsup_{n \to \infty} (-1) (-1)^n = 1$$
$$-\limsup_{n \to \infty} (-1)^n = -1$$


TITLE: Problems with solving a 2 variable limit
QUESTION [3 upvotes]: There is this exercise I'm trying to solve but can't seem to get it, it states the following:
Study the continuity in the origin (depending of $\alpha \in \mathbb{R}$) of $$f_\alpha(x,y) = \frac{x^6y^3}{(x^8+y^6)^\alpha}$$ when $(x,y) \neq (0,0)$ and $$f_\alpha(x,y) = 0$$ when $(x,y) = (0,0)$
I have tried various methods, passing to polar coordinates and so on but can't really seem to get anywhere, I know the only possible candidate for the limit is $0$ and thats the result I got several times, so I'm guessing it is continous at $0$ (maybe I'm wrong) and I would have to prove this using the definition but don't know how to do it.
Any hints are appreciated and sorry for my broken English.

REPLY [1 votes]: By polar coordinates for $\alpha<\frac32$ the limit is clearly equal to zero.
For $\alpha\ge \frac32$ by $x=t$ and $y=t$ with $t\to0$
$$ \frac{x^6y^3}{(x^8+y^6)^\alpha}= \frac{t^9}{(t^8+t^6)^\alpha}= \frac{1}{(t^{8-\frac9{\alpha}}+t^{6-\frac9{\alpha}})^\alpha}$$
which tends to $1$, when $\alpha=3/2$ or $\infty$, when $\alpha>3/2$.


TITLE: Proving that the least upper bound of  is less than or equal to the greatest lower bound of B
QUESTION [0 upvotes]: I'm failing to understand the basics concepts of upper/lower bounds and I was hoping I could get some guidance through a simple question.
Suppose that $A$ and $B$ are two non-empty bounded sets of real numbers, such that $x \leq y$ for every $x \in A$ and $y \in B$. Show $lubA \leq glbB$
I have no idea how to even start this proof. I obviously need to somehow relate $x \leq y$ to the boundaries of their sets, that is $lubA \leq glbB$. Although I'm not even sure if my idea is correct. Any help?

REPLY [0 votes]: The least upper bound of a set is called its supremum ($\text{sup}$) and the greatest lower bound of a set is called its infimum ($\text{inf}$).
Firstly, we have,
$\forall x\in A\text{ and } y\in B, x\leq y
\implies y\text{ is an upper bound for every }x
\implies\text{sup}A\leq y,\text{ }\forall y\in B$ as $\text{sup}A$ is the least upper bound of the set $A$.
Now we have,
$\forall y\in B, \text{ sup}A\leq y\implies\text{sup}A\leq\text{inf}B$ as $\text{inf} B$ is the greatest lower bound of the set $B$.
Thus, proved.


TITLE: Doubt in Proof of Dirichlet Asymptotic formula for partial sum of divisor function $d(n)$
QUESTION [2 upvotes]: I am studying Analytic number theory from Tom M Apostol Introduction to analytic number theory and I  have doubt in the proof given in the book.
Dirichlet gives two proof of formula
1 st has error term $O(x)$  and I don't have any doubt in it.

I have doubt in his 2nd proof where he deduces error term to be $(2C-1) x +O(\sqrt{x})$. He writes before giving the argument of Lattice Points that we take advantage of symmetry of region about $q=d$. The total number of lattice points in region is equal to twice the number below line q=d plus the number of bisecting line segments.
  I don't understand how apostol derives the formula  which I am mentioning in the image along with the diagram.

Can somebody please tell how to derive this formula.

REPLY [1 votes]: $$\sum_{n\le x} d(n)= \sum_{n\le x}\sum_{md=n}1=\sum_{md\le x}1= 2\sum_{md\le x,d<m} 1+ \sum_{d^2\le x}  1=2\sum_{d<\sqrt{x}}(\lfloor x/d\rfloor -d) + \lfloor \sqrt{x}\rfloor$$
In the figure :
$ \lfloor \sqrt{x}\rfloor$ is the number of lattice point on the diagonal,
$\lfloor x/d\rfloor -d$ is the number of lattice points with abscissa $d$ and on the right side of the diagonal (the horizontal segment plotted in the black region)
The asymptotic is immediate from there since $\lfloor x/d\rfloor = x/d+O(1)$


TITLE: Do Significant Figures apply with numbers without units?
QUESTION [1 upvotes]: Say I need to divide a number with a unit by a number without a unit, e.g.: $\frac{16.4\: \text{meters}}{2}$
Following the rules on significant figures, will the correct answer be $8.2$m (since $2$ has no unit), or $8$ m (since $2$ has only one significant figure)?

REPLY [0 votes]: Think of it in terms of percentage error.
If a reading is $100 \pm 2$ metres this is $100 \pm 2\%$.
Now divide the 100 metres by 5 to give 20 metres.
The percentage error does not change so the value is $20 \pm 2\% = 20.0 \pm 0.4$ metres.
If you are not given a percentage error then you would have to judge it from the given value.
110 might have the second 1 as significant so approximately $10\%$?  Whereas 111 might indicate a $1\%$ error?


TITLE: An infinite subset of a countable set is countable
QUESTION [8 upvotes]: In my book, it proves that an infinite subset of a coutnable set is countable. But not all the details are filled in, and I've tried to fill in all the details below. Could someone tell me if what I wrote below is valid?
Let $S$ be an infinite subset of a countable set $T$. Since $T$ is countable, there exists a bijection $f: \mathbb{N} \rightarrow T$. And $S \subseteq T = \{ f(n) \ | \ n \in \mathbb{N} \}$. Let $n_{1}$ be the smallest positive integer such that $f(n_{1}) \in S$. And continue where $n_{k}$ is the smallest positive integer greater than $n_{k-1}$ such that $f(n_{k}) \in S$. And because $S$ is infinite, we continue forever. Now consider the function $\beta : \mathbb{N} \rightarrow S$ which sends $k \rightarrow f(n_{k})$.
So in order for $S$ to be countable, $\beta$ would have to be a bijection. $\beta$ is injective since if $f(n_{k}) = f(n_{j})$, then $n_{k} = n_{j}$ because $f$ is injective. We also can conclude $k = j$ because the various $n_{i}$ chosen were strictly increasing.
Edit: $\beta$ also needs to be surjective. So for every $r \in S$ there exists a $q \in \mathbb{N}$ such that $\beta(q) = f(n_{q}) = r$. We know that $f^{-1}(r) \in \{n_{1}, n_{2}, ..., n_{k} ... \}$, so we can let $f^{-1}(r) = n_{d}$. Thus $\beta(d) = f(n_{d}) = r$.

REPLY [5 votes]: The set $A$ is countable if there is $f:A\to\mathbb N$ which is injective.
Suppose $B\subseteq A$ then $f|_B:B\to\mathbb N$ is also injective, simply because you restrict an injective function you cannot counterexample the injectivity.

Suppose that $f:A\to\mathbb N$ is a bijection and $B\subseteq A$ is infinite. Let $N=\{f(b)\mid b\in B\}$, the image of $B$.
Since $B$ is infinite, we have that $N$ is infinite. Let us show that an infinite subset of $\mathbb N$ is countable:
We define inductively a function $g:\mathbb N\to N$, $g(0)=\min N$ the minimal number in $N$. Suppose $g(n)$ was defined, let $g(n+1)$ be the least number in $N$ which is larger than $g(n)$.
It is obvious that $g$ is a bijection between $\mathbb N$ and $N$. Now we have that $g^{-1}\circ f|_B:B\to\mathbb N$ is a bijection, as wanted.

REPLY [4 votes]: To fix the last statement:
Suppose that $q=f^{-1}(r)$. Then $\{ 0,1,2,\dots,q\}$ is a finite set. Intersect this with $f^{-1}(S) = \{ n \in \mathbb{N} \;|\; f(n) \in S\}$ and get a finite set. By your definition of the $n_k$'s this set is $\{n_0,n_1,\dots,n_\ell\}$ for some $\ell$. Thus $q=n_\ell$. Therefore, $\beta(\ell)=f(n_\ell)=f(q)=r$. Patched.


TITLE: Proof Fragment and Questions
QUESTION [0 upvotes]: Consider the following proof fragment.
There exists an integer k such that $n=3k+1$. Then $n^2=(3k+1)^2=9k^2+6k+1=3(3k^2+2k)+1$
For each of the statements (a), (b), (c) below, answer the following. If yes, explain why. If no, explain why not. The letter $n$ denotes an integer.
(a) If $n$ is odd, then $n^2$ is odd.
(b) If $n^2$ is divisible by $3$, then $n$ is divisible by $3$.
(c) If $n$ leaves remainder $1$ on division by $3$, then so does $n^2$.
Now what I did was computed a few values from $-4$ to $5$.
(a) for example, holds because squaring an odd will get an odd
(b) does not hold cause we are always taking one more than a multiple of 3; therefore, $n^2$ cannot be divisible by $3$.
(c) holds because in essence, we are just minusing $1$ and $3k+1-1=3k$ and $3k$ is a multiple of 3 which would make its square a multiple of 3.
Fairly certain by logic is sound. But question: would giving examples be proof enough? Or would I need to generalize it? If I need to generalize it, how? Thank you in advance.

REPLY [0 votes]: You could try and formalize part (b) a bit:
Since $n^2=(3k+1)^2=9k^2+6k+1=3(3k^2+2k)+1$, we know that $3$ cannot divide $n^2$, since it divides $n^2-1$.
Since $p \implies q$ and $\neg p$  what can you say about the truth value of this type of statement, more generally? The "if" part of your statement is always false.
for a quick proof of the my statement:
 Suppose $p$ is prime and that $p \mid k$. Then $k=pr$ for some $r \in \mathbb{Z}$. Suppose $p \mid k+1$, then $k+1=pl$ for some $l \in \mathbb{Z}$. But then $1=k+1-k=pl-pr=p(l-r)$ so $p \mid 1$. Since $3 \neq 1$, my first statement stands.

For part (a), your logic is not sound. If $n=3k+1$, we could easily let $k=1$, meaning that $n=3+1=4$ which is even. You should think a lit bit more about why (a) could hold (as in, do the algebra)

For part (c), yes it holds kind of for the reason you mention. If $n=3k+1$, it will always leave a remainder of one upon division by $3$, and the algebra you did shows that $n^2$ will as well.

Rule of thumb: more or less, the only time a "proof by example" works is when it is a counter-example. If you wish to prove a statement false, one example will suffice.
For example, the statement (c) is saying that $\forall n \in \mathbb{Z}\, s.t \, n=3k+1$ we know that $n^2=3m+1$ for some $m \in \mathbb{Z}$.
Of course, to show something is true for all anything, examples will not suffice.


TITLE: Check if in each case borrow and overflow are generated?
QUESTION [0 upvotes]: Note: Just tell me if my reasoning or logic is wrong.
So I have two numbers 01001 and 1110. And the question asks to subtract them as both signed and unsigned system.
Unsigned System
For this I can use, subtraction via addition (2's compliment system).
Question: 01001 - 1110
1110 -> 01110 // Sign extension
01110 -> 10001 // Inverting Bits
10001 + 1 -> 10010 // Adding one

Now,

   01001
   10010
 + -----
   11011

From my book it says, "No final carry in step 2 indicates a final borrow, or a greater “2nd number” than the “1st number”, hence an impossible subtraction." So I concluded Overflow generated and borrow occurred as the second number is greater than first number.
Signed System
Question: 01001 - 1110
1110 -> 11110 // Sign extension

Now,
Carry:1 1 000
        0 1001
        1 1110
      + -------
        0 0111

The book says, "In 2’s complement subtraction modeled by Equation (7), and also in 2’s complement addition, overflow occurs if and only if the carry bit arriving at the sign column is different from the carry bit leaving this column." Hence, No overflow and borrow occurred.
Is there anything I left or didn't consider? For some reason I feel like I messed up signed part even though the result makes sense. 9 -2 = 7

REPLY [0 votes]: Lets say the question is 01001 - 1110
Now we have two options. Calculate is using Unsigned System and then do it using signed system.
Unsigned System
01001 is five bits while 1110 is four bytes.
The first step is to sign extend and make sure they both are of equal length.
For unsigned, we sign extend by adding "0s" on left most side.
Hence,
1110 -> 01110

Now the subtraction.
In term of decimal the question looks like this: 9-14. Which is -5
// Taking 2s compliment of 01110
01110 -> 10001 // Inverting Bits
10001 + 1 -> 10010 // Adding one

Now,

   01001  // 9
   10010  // -14
 + -----
   11011  // -5 if we consider it signed system, otherwise 27

"No final carry in step 2 indicates a final borrow, or a greater “2nd number” than the “1st number”, hence an impossible subtraction." In unsigned, -5 can't be represented anyways. Hence Overflow occured.
Signed System
For signed System the procedure is more or less the same. The major difference would be the sign extension.
For signed numbers, to do signed extension, we just repeat whatever the MSB is.
For 1110, MSB is 1
Hence,
1110 -> 11110

Now the subtraction.
In term of decimal the question looks like this: 9-(-2). Which is 11.
// Taking 2s compliment of 01110
11110 -> 00001 // Inverting Bits
00001 + 1 -> 00010 // Adding one

Now,

   01001  // 9
   00010  // -2
 - -----
   01011  // 11, as expected

Since the carry bit arriving arriving at the sign column is same from the carry bit leaving this column. We have no overflow.


TITLE: Finding measure of a set
QUESTION [0 upvotes]: Let $(X,\mathcal A, \mu)$ be a measure space and let $f\in L^p(\mu)$ for $1\leq p<\infty$. Let $E_n=\{x\in X: \frac{1}{n}\leq |f(x)|\leq n\}$ for all $n\in \mathbb N$. I have to show that $\mu(E_n)<\infty$. Can you please give a hint?

REPLY [2 votes]: Hint:
$\int|f|d\mu\geq\int_{E_n}|f|d\mu\geq\frac1{n}\mu(E_n)$


TITLE: Inclusion Exclusion principles in combinatorics
QUESTION [3 upvotes]: I am looking to get some help with the following problem, I am not sure if my workings are correct but I have given it a good shot, thanks!

How many ways are there for a child to take $11$ pieces of candy with $4$ types of candy if they child does not take exactly two pieces of candy of any type?

My idea was to see it up and think of it in the way that there are 11 pieces balls, that have to go into 4 bins, but you cant take from the same bin twice.
So I would use the negation $N(a_1', a_2', a_3', a_4')$
Then the total number of ways would be $N(u) = \binom{11+4-1}{11}$
Then for each next term take $5$ off so
$$N(a_1', a_2', a_3', a_4') = \binom{11+4-1}{11}-\binom{4}{1}\binom{6+4-1}{6} + \binom{4}{2}\binom{1+4-1}{1}-outofballs$$

REPLY [2 votes]: Assuming I understand your notation correctly, you incorrectly have $N(a_1)=\binom{6+4-1}6$, when it should be $$N(a_1)=\binom{9+3-1}{9}.$$ If the child took $2$ of the first type of candy, there are $\binom{9+3-1}{9}$ ways to choose $9$ candies from the other $3$ types of candy.
Similarly, $N(a_1,a_2) = \binom{7+2-1}7$ and $N(a_1,a_2,a_3) = \binom{5+1-1}5$, while $N(a_1,a_2,a_3,a_4) = 0$.


TITLE: Proving that specific ring is isomorphic to quotient ring
QUESTION [1 upvotes]: let's consider the ring $\mathbb Z[\sqrt 2] := {a+b\sqrt{2} \mid a,b\in \mathbb Z}$. I now want to prove that $\mathbb Z[\sqrt 2]  \cong \mathbb Z [X]/(X^2-2)\mathbb Z[X]$.
What I found on the Internet is how to prove that $\mathbb C \cong \mathbb R[X]/(X^2+1)\mathbb R[X]$: http://people.virginia.edu/~mve2x/3354_Fall2010/lecture26.pdf (page 3)
I really like their approach there because it is clear that they construct an explicit ring isomorphism. However, I cannot take their ring isomorphism because they explicitly exploit that $\mathbb R$ is a field ...
Kind regards,
MathIsFun

REPLY [2 votes]: Take $\varphi :\mathbb Z[X]\to \mathbb Z[\sqrt 2]$ defined by $$\varphi (p(X))=p(\sqrt 2).$$
This is obviously surjective. Moreover $$\varphi (p(X))=0\iff X^2-2\mid p(X).$$
I let you conclude.


TITLE: Complete is equivalent to Misner-complete for Riemannian manifolds.
QUESTION [0 upvotes]: I'm trying to prove that, if $M$ is a smooth Riemannian manifold, then completeness of $M$ is equivalent to Misner-completeness.

A pesudo-Riemannian (or semi-Riemannian) manifold $M$ is Misner-complete if and only if every geodesic $\gamma: [0,b) \to M,$ $b < \infty,$ lies in a compact subset of $M.$

Now one way is simple, I think. If $M$ is complete, then for a geodesic $\gamma: [0,b) \to M,$ let $p = \gamma(0), v = \gamma'(0).$ Since $M$ is complete, there is the maximal geodesic starting at $p$ with initial velocity $v,$ $\gamma_v : \mathbb R \to M$ with $\gamma_v |_{[0,b)} = \gamma.$ But then $\gamma_v([0,b])$ is compact and contains $\gamma([0,b)).$
Now for the other way I am completely lost. It would seen some of the equivalences for a Riemann manifold to be complete via Hopf-Rinow theorem would make things easier, but I've had no success so far.
Any hints are appreciated, thanks.
(for reference, this is problem 13, chapter 5 of O'Neill, Semi-Riemannian Geometry, also problem 5.9.6 of Riemannian Geometry by Petersen, third edition)

REPLY [1 votes]: Here is a solution, mostly constructed by my Riemann geometry professor, for anyone seeking further questions regarding similar problems. Any remarks or corrections are welcome:
Let $M$ be Riemannian and Misner-complete. Take $\gamma : [0,b) \to M$ a geodesic, which can be reparameterized as to be an unit speed geodesic. Then $\gamma([0,b)) \subseteq K$ a compact subset of $M.$ Also, since $b< \infty,$ the length of $\gamma$ is finite. For a sequence $\{t_n\} \in [0,b)$ converging to $b,$ since $K$ is compact, the sequence $\gamma(t_n)$ has a converging subsequence, so we can as well assume $\{t_n\}$ is such that $\gamma(t_n) \to p \in K.$ We now prove this $p$ is the extension of $\gamma.$ Since it is a geodesic, extension as a continuous curve is equivalent to extension as a geodesic, so by proving this we are done.
Assume there is another sequence $\{s_n\}$ in $[0,b)$ converging to $b$ with $\gamma(s_n)$ converging to $q\neq p,$ so we have $d(p,q) > 0$ (where $d$ is the Riemann distance). Take $\epsilon > 0$ such that $ 0 < \epsilon < \frac{1}{3}d(p,q).$ For $n \in \mathbb N$ large enough, $\gamma(t_n) \in B(p,\epsilon )$ and $\gamma(s_n) \in B(q,\epsilon) $ (the open balls for Riemann distance). By the triangle inequality,
$$ d(p,q) \leq d(p, \gamma(t_n)) +d(\gamma(t_n),\gamma(s_n))  + d(q, \gamma(s_n)) < 2\epsilon +d(\gamma(t_n),\gamma(s_n))  $$
which gives $\epsilon < d(\gamma(t_n),\gamma(s_n)),$ for $n$ large enough.
By a standard process we can construct two increasing subsequences of $\{t_n\}$ and $\{s_n\}$ such that $t_{n_k} < s_{n_k},$ for all $k \in \mathbb{N}.$
To finish, take an $l \in \mathbb N$ with $\epsilon < d(\gamma(t_{n_k}),\gamma(s_{n_k}))$ for $k > l$ and such that $b < l\cdot \epsilon.$ We have
$$\epsilon < d(\gamma(t_{n_k}), \gamma(s_{n_k})) \leq L(\gamma|_{[t_{n_k}, s_{n_k}]}),$$
where $L(\gamma|_{[t_{n_k}, s_{n_k}]})$ is the length of $\gamma$ over the interval ${[t_{n_k}, s_{n_k}]}.$ Summing over $k,$
$$b < l\cdot \epsilon < \sum_{k=1}^l L(\gamma|_{[t_{n_k}, s_{n_k}]}) \leq b,$$
a contradiction.


TITLE: Law of inertia, Frames of reference
QUESTION [0 upvotes]: Ok, we all know that an Inertial frame of reference is defined like this:
"An inertial frame of reference is one frame where Newton's First Law holds, therefore, a body has a constant velocity or velocity equal zero. And, the sum of all forces equals zero, there is no aceleration etc..."
But, I always read that in an inertial frame of reference holds $F=ma$, the second law....
It's confuses me... why we can assume $F=ma$ in an inertial frame of reference?

REPLY [0 votes]: Ok, we all know that an Inertial frame of reference is defined like this:
"An inertial frame of reference is one frame where Newton's First Law holds, therefore, a body has a constant velocity or velocity equal zero. And, the sum of all forces equals zero, there is no aceleration etc..."

No. An inertial frame is one in which all of Newton's Laws hold. And the first law has content, it tells us that a particle on the top of Norton's Dome that is at rest stays there at rest forever unless disturbed. It's not a practical result since the top of Norton's Dome is unstable anyway.
But it isn't enough to have the first law hold since there are examples of systems where the first law holds and the second doesn't (replace the second law with anything that gives zero acceleration when net force is zero) and examples where the second holds but the first doesn't (e.g. where a particle sits on top of Norton's Dome for an hour and then spontaneously falls down in some direction).

But, I always read that in an inertial frame of reference

People get the first and second law confused and think all kinds of weird things. If you read the originals, the first one tells you about things staying in uniform motion when in uniform motion and having no net force. The second tells you what happens when there is a net force, and they are both talking about kinematics described in an inertial frame. And the third describes the forces themselves and again is about the descriptions in an inertial frame.


TITLE: the image of $\text{Div}(X)\to \text{Pic}(X)$ is generated by those line bundles with $H^{0}(X,L) \ne 0$
QUESTION [0 upvotes]: Let $X$ be complex manifold, the image of $\text{Div}(X)\to
 \text{Pic}(X)$ is generated by those line bundles with $H^{0}(X,L) \ne
 0$ ( i.e. it admits a global non trivial holomorphic section).

The proof for the image of a divisor $D$, gives a non trivial global section goes as follows:

Proof: Any divisor $D=\sum a_i\left[Y_i^i\right] \in \operatorname{Div}(X)$ can be written as $D=\sum a_i^{+}\left[Y_i^{+}\right]-\sum a_j^{-}\left[Y_j^{-}\right]$with $a_k^{\pm} \geq 0$. Hence, $\mathcal{O}(D) \cong \mathcal{O}\left(\sum a_i^{+}\left[Y_i\right]\right) \otimes$ $\mathcal{O}\left(\sum a_j^{-}\left[Y_j\right]\right)^*$ and the line bundles $\mathcal{O}\left(\sum a_i^{+}\left[Y_i\right]\right)$ and $\mathcal{O}\left(\sum a_j^{-}\left[Y_j\right]\right)$ are both associated to effective divisors and, therefore, admit non-trivial global sections.

My question is we have proved that $\mathcal{O}\left(\sum a_j^{-}\left[Y_j\right]\right)$ admits non trivial global section however $\mathcal{O}(D)$ tensors a dual bundle of it, it's not clear that the dual bundle admits non trivial global section?

REPLY [2 votes]: It is enough to consider effective divisors.
Write a divisor $D$ as $D_1-D_2$, with $D_1,D_2$ both effective, and let $L_1:=\mathcal O(D_1)$, $L_2:=\mathcal O(D_2)$ be the associated bundles. Then
$\mathcal O(D)=\mathcal O(D_1)\otimes \mathcal O(-D_2)= L_1\otimes L_2^* \in \langle L_1,L_2\rangle$, with $H^0(X,L_1)\neq 0$ and $H^0(X,L_2)\neq 0$, because  $D_1,D_2$ are effective.
Therefore, the image of the map $\mathrm{Div}(X)\to \mathrm{Pic}(X)$ is generated by line bundle having global sections, i.e. arising from effective diviors.
For the question in the comment, it is also correct: if both $L=\mathcal O(D)$ and $L^*= \mathcal O(-D)$ have non trivial global section, there exist divisor $A\in |D|$ and $B\in |-D|$. But $A+B$ is a principal divisor, and  $A$ and $B$ are bot effective, whence $A=B=0$.


text:
\subsection{Theoretical Properties} \label{sec:theory}

We now present certain theoretical analysis for the proposed learning
algorithm.
The following result provides a bound between the solution of the
modified system \eqref{eq:ode_mod} and the original system
\eqref{eq:ode}. The difference between the two systems is due to the
use of the parameterized external input $\tgamma(t)$ \eqref{global_gamma} in
the modified system \eqref{eq:ode_mod}, as opposed to the original external input
$\gamma(t)$ in the original system \eqref{eq:ode}. Again, we emphasize that in many
practical situations when the external input is already known in a
parametric form, the modified system \eqref{eq:ode_mod} is equivalent
to the original system \eqref{eq:ode}.

\begin{proposition}
  \label{thm:error_mod}
  Consider the original system \eqref{eq:ode} with input $\gamma(t)$
  and the modified system \eqref{eq:ode_mod} with input $\tgamma(t)$
  \eqref{global_gamma}, and
assume the function $\f(\x,\gamma)$ is Lipschitz
continuous with respect to both $\x$ and $\gamma$, with Lipschitz
constants $L_1$ and $L_2$, respectively.
  If the difference in the inputs is bounded by
  	$$\|\gamma(t)-\tgamma(t)\|_{L^\infty([0, T])}\leq \eta, $$
where $T>0$ is a finite time horizon. Then,
	\begin{equation*}
	|\x(t)-\tx(t)|\leq L_2\, \eta\, t\, e^{L_1 t}, \quad \forall t\in [0, T].
	\end{equation*}
\end{proposition}
\begin{proof}
	For any $t\in [0, T]$,
	\begin{align*}
		\x(t)&=\x(0)+\int_0^t \f(\x(s), \gamma(s))\,ds,\\
		\tx(t)&=\x(0)+\int_0^t \f(\tx(s), \tgamma(s))\, ds.
	\end{align*}
	We then have
	\begin{align*}
		|\x(t)-\tx(t)|&\leq \int^t_0 |\f(\x(s), \gamma(s))-\f(\tx(s), \tgamma(s))| \, ds\\

		&\leq \int_0^t \abs{\f(\x(s), \gamma(s))-\f(\x(s), \tgamma(s))}\,ds+\int_0^t \abs{\f(\x(s), \tgamma(s))-\f(\tx(s), \tgamma(s))}\,ds\\
		&\leq L_2\int_0^t \abs{\gamma(s)-\tgamma(s)}\,ds + L_1\int_0^t \abs{\x(s)-\tx(s)}\,ds\\
		&\leq L_2\,\eta\,t +L_1  \int_0^t \abs{\x(s)-\tx(s)}\,ds.
	\end{align*}

        By using Gronwall's inequality, we obtain
        $$
        |\x(t)-\tx(t)| \leq L_2\,\eta\, t\, e^{L_1 t}.
        $$

\end{proof}

We now recall the celebrated universal approximation property of
neural networks.

\begin{proposition}[\cite{pinkus1999}]
	For any function $F\in C(\mathbb{R}^n)$ and a positive real
        number $\varepsilon>0$, there exists a single-hidden-layer neural
        network $N(\cdot\,; \Theta)$ with parameter $\Theta$ such that
	\begin{equation*}
	\max_{\y\in D} |F(\y)-N(\y\,;\Theta)| \leq \varepsilon,
	\end{equation*}
	for any compact set $D\in \mathbb{R}^n$, if and only if the activation functions are continuous and are not polynomials.
\end{proposition}

Relying on this result, we assume the trained neural network model
\eqref{trained} has sufficient accuracy, which is equivalent
to assuming accuracy in the trained FNN operator $\N$ of
\eqref{evo_mod_new} to the
one-step evolution operator $\tpphi$ in \eqref{evo_mod}.
More specifically, let $\mathcal{D}$ be the convex hull of the
training data set $\mathcal{S}$, defined \eqref{training_set}. We then assume
\begin{equation}
\label{eq:err_NN}
\left\|\N(\cdot; \Theta^*)-\tpphi(\cdot)\right\|_{L^\infty(\mathcal{D})}<\mathcal{E},
\end{equation}
where $\mathcal{E}\geq 0$ is a sufficiently small real number.
\begin{proposition}
  \label{prop:NN_error}
  Consider the modified system \eqref{evo_mod} and the trained network
  model \eqref{eq:prediction} over the time instances
  \eqref{tline}. Assume the exact evolution operator \eqref{evo_mod}
  is Lipschitz continuous with respect to $\x$, with Lipschitz
  constant $L_\phi$. If the network training is sufficiently
  accurate such that \eqref{eq:err_NN} holds, then
	\begin{equation}
	\label{eq:approx_err}
	\|\hx (t_n)-\tx(t_n)\| \leq \frac{1-L_\phi^n}{1-L_\phi}
        \mathcal{E}, \qquad n=0,\dots, N.
	\end{equation}
\end{proposition}

  \begin{proof}
    Let $\PPhi = \hI + \tpphi$, where $\hI$ is defined in
    \eqref{hatI}, we can rewrite the one-step evolution
    \eqref{evo_mod} as
    $$
    \tx(t_{n+1})=[\PPhi(\cdot, \ggamma_n, \delta_n)](\tx(t_n)),
    $$
    Meanwhile, the learned model \eqref{eq:prediction} satisfies, by
    using \eqref{trained},
    $$
    \hx(t_{n+1}) = [\tN(\cdot; \Theta^*)](\hx(t_n)).
    $$
    Let $e_n = \|\hx(t_n) - \tx(t_n)\|$, we then have
    \begin{equation*}
      \begin{split}
    e_n = & \left\|[\tN(\cdot; \Theta^*)](\hx(t_{n-1})) -[\PPhi(\cdot, \ggamma_{n-1}, \delta_{n-1})](\tx(t_{n-1}))\right\|
   \\
    \leq &\left\|[\tN(\cdot; \Theta^*) - \PPhi(\cdot, \ggamma_{n-1},
      \delta_{n-1}) ](\hx(t_{n-1}))\right\| + \\
    &
    \left\|\left[\PPhi(\hx(t_{n-1}), \ggamma_{n-1}, \delta_{n-1})\right] -
      \left[\PPhi(\tx(t_{n-1}), \ggamma_{n-1}, \delta_{n-1})\right]\right\|\\
     \leq &~\mathcal{E} + L_\phi \left\|\hx(t_{n-1}) - \tx(t_{n-1})\right\|
    \end{split}
  \end{equation*}
  This gives
  $$
  e_n \leq \mathcal{E} + L_\phi e_{n-1}.
  $$
  Repeated use of this relation and with $e_0=0$ immediately gives the conclusion.
  \end{proof}

Note that the assumption of Lipschitz continuity on the evolution
operator in \eqref{evo_mod} is equivalent to assuming Lipschitz
continuity on the right-hand-side of the original system
\eqref{eq:ode}. This is a very mild condition, commonly assumed for the well-posedness
of the original problem \eqref{eq:ode}.

Upon combining the results from above and using triangular inequality, we immediately obtain the following.
\begin{theorem}
	Under the assumptions of Proposition \ref{thm:error_mod} and
        \ref{prop:NN_error}, the solution of the trained network
  model \eqref{eq:prediction} and the true solution of the original
  system \eqref{eq:ode} over the time instances satisfies
  \eqref{tline} satisfy
	\begin{equation}
	\label{eq:err_est}
	\left\|\hx(t_n)-\x(t_n)\right\| \leq L_2\,\eta\, t_n\, e^{L_1 t_n}
        + \frac{1-L_\phi^n}{1-L_\phi}
        \mathcal{E}, \qquad n=0,\dots, N.
	\end{equation}
\end{theorem}

\begin{remark} \label{remark1}
It is worth noting that the DNN structure employed here is
to accomplish the approximation \eqref{eq:err_NN}. Such an
approximation can be conducted by any other proper approximation
techniques using, for example, (orthogonal) polynomials,
Gaussian process, radial basis, etc. The target function is
the one-step evolution operator $\tpphi$ in \eqref{evo_mod}.
Since for many problems of practical interest,
$\tpphi:\R^{d+n_b+1}\to\R^d$ often resides in high dimensions and is
highly nonlinear, DNN represents a more flexible and practical choice
and is the focus of this paper.
\end{remark}


TITLE: Under what conditions can a body can be approximated as a black body?
QUESTION [1 upvotes]: I have read this post and from it was my understanding that the definition of a blackbody is:

Black body means a body which ABSORBS all wavelengths completely.

After reading this  answer on another post
I gained the understanding that a $\color{green}{\text{black body doesn't have to be in thermal equilibrium.}}$
Wikipedia agrees with both the statements above.
So all is well, but then I read this (from the Imperial College London, Department of Physics):

Black Body Radiation
Stars have high densities -> frequent collisions that can lead to thermodynamic equilibrium where all particles (electrons, ions, photons) have a single temperature
Can often approximate stars as black bodies!

This has got to be a joke right? So does it seem that any object in thermal equilibrium with its surroundings will emit as a black body? I don't see how thermal equilibrium implies a black body (even if it is an approximation).
This also contradicts the line in green above.

REPLY [3 votes]: No, thermal equilibrium does not mean something is a blackbody. A red apple can be in thermal equilibrium with its surroundings. It is manifestly not a blackbody radiator.
The outer parts of a star, from whence the radiation we see, escapes, is approximately in thermal equilibrium at a certain temperature. A star absorbs nearly all radiation incident upon it. A star is approximately a blackbody radiator.


TITLE: Relation between probabilities such as $P(AB>a) > P(Ab>a)$
QUESTION [1 upvotes]: Let $X=AB$,
$A$ and $B$ are random variables which are NOT independent and I know that $A>0$, $B\geq b >0$ with $b$ is a deterministic constant.
Then for any constant $a$, probability $P(X>a) > P(Ab>a)$ because $p=P(X>Ab)=1$.
Now if I know $p=P(B>b')$ with $b'>b$,
is there any relation (equality or inequality) among $P(X>a)$, $P(Ab'>a)$ and $p$ ?

REPLY [0 votes]: Not quite an answer, but too long for a comment.
I think we need to be careful about equality.  In the setup you said $A>0, B\ge b> 0$.  In that case you can only conclude $X = AB \ge Ab$, but you cannot conclude $P(X > Ab)=1$.  In fact, $P(X>Ab) = P(B>b)$ which can be any value from $0$ to $1$.
Also, you said $P(X > a) > P(Ab > a)$, but that is again untrue.  What you can conclude is $P(X > a) \ge P(Ab > a)$, first because it is possible $B \equiv b$ in which case $X=Ab$ and the two probabilities are equal, and second because it is possible that for certain values of $a$ (e.g. $a < 0$) we have $P(Ab> a) =1$ already and $P(X > a)$ cannot possibly be $>1$.

Anyway on to your actual question: In general, $P(Ab' > a)$ only has to do with the marginal distribution of $A$, and $P(B > b')$ only has to do with the marginal distribution of $B$, but $P(X > a)$ has to do with the joint distribution. If $A$ and $B$ are dependent, you can make the $3$ distributions look like a lot of different things, so I don't think you can prove anything in general.
E.g. one natural way to proceed is:
$$
\begin{align}
P(X > a) &= P(X > a | B > b') P(B > b') + P(X > a | B \le b') P(B \le b') \\
 & \ge P(Ab' > a | B > b') P(B > b') + 0
\end{align}$$
So this has $2$ of the $3$ quantities you want.  However, since $A$ and $B$ are dependent, we cannot even say $P(Ab' > a | B> b') = P(Ab' > a)$, so I don't see a way to proceed further.  I think I can make up examples where $P(Ab' > a | B>b')$ is $>$ or is $< P(Ab' > a)$.