text:
\section{Existence and Uniqueness of Solutions to the Mean-field system and exponential convergence to the invariant measure}
\label{sec ex uniq and conve}

\begin{lemma}[A priori estimate]
\label{lem apriori}
Let Assumptions~\ref{ass exist and uniq} and \ref{ass coercivity} hold.
Fix $p'\geq 2$ and let $\mathbb E\int_0^T |\theta_{0,t}|^{p'} \,dt < \infty$.
If $\lambda := \frac{p'}{2}\left(\frac{\sigma^2\kappa}4 + \eta_1\right)>0$
and
$(\theta_{s,t})_{s\geq 0, t\in [0,T]}$ is a solution to~\eqref{eq mfsgd}-\eqref{eq mfsgd 2} then
\begin{equation}
\label{eq a priori estimate}
\int_{0}^{T}\mathbb E [|\theta_{s,t}|^{p'}]\,dt
\leq e^{-\lambda s}\int_0^T \mathbb E [|\theta_{0,t}|^{p'}]\,dt  + c_{T,p,\sigma,\kappa,\eta_1,\bar \eta}\int_0^s e^{-\lambda (s-v)} \,dv\,.
\end{equation}
\end{lemma}
\begin{proof}
Note that Assumption~\ref{ass exist and uniq} together with the Young's inequality implies that : $\forall a,y\in \mathbb R^p$ $|a y|\leq \frac{\kappa}{2} |a|^2 + (2 \kappa)^{-1}|y|^2$ and so
\begin{equation}
\label{eq coercivity of nabla U}
 (\nabla_a U)(a)\,a \geq \frac{\kappa}{2} |a|^2  - \frac{1}{2\kappa}| (\nabla_a U)(0)|^2\,.
\end{equation}
From~\eqref{eq coercivity of nabla U} and Assumptions~\ref{ass coercivity}  we get that for all $(t, a,\mu)\in [0,T] \times \mathbb R^p \times \mathcal V^W_2$
we have
\begin{equation}
\label{eq coercivity}
\frac{\sigma^2}2\nabla_a U(a)\cdot a + (\nabla_a \frac{\delta \mathbf H_t^{0}}{\delta m})(a,\mu) \cdot a
\geq \bigg(\frac{\sigma^2 \kappa}{4} + \eta_1\bigg) |a|^2 - \eta_2 \mathcal E(\mu,\delta_0)^2 - \bar \eta_t\,.
\end{equation}
Let $\mu_{s,t} := \mathcal L(\theta_{s,t}|\mathcal F^W_t)$.
Applying It\^o's formula in~\eqref{eq mfsgd}, first with $t\in [0,T]$ fixed, then integrating over $[0,T]$, we have
\begin{align*}
& e^{\lambda s} \int_{0}^{T}|\theta_{s,t}|^{p'}\,dt
=\int_0^T |\theta_{0,t}|^{p'} \,dt  + \lambda \int_{0}^{s} e^{\lambda v} \int_{0}^{T} |\theta_{v,t}|^{p'}\,dt\,dv+\frac{\sigma^2}2 p' (p'-1)\int_{0}^{s}e^{\lambda v} \int_0^T |\theta_{v,t}|^{p'-2}  \,dt \,dv\\
&\quad -p'\int_{0}^{s}e^{\lambda v} \int_{0}^{T} \bigg(\frac{\sigma^2}{2}\nabla_a U(\theta_{v,t})
+(\nabla_a \frac{\delta \mathbf H_t^{0}}{\delta m})(\theta_{v,t},\mu_{v,\cdot})\bigg) \theta_{v,t} |\theta_{v,t}|^{p'-2}\,dt\,dv \\
&\quad + \sigma p' \int_0^T \int_0^s e^{\lambda v}  |\theta_{v,t}|^{p'-2}\theta_{v,s}\,  dB_v\,dt\,.\\
\end{align*}
Applying~\eqref{eq coercivity} and taking expectation we thus get
\begin{equation}
\label{eq pre p moment est}
\begin{split}
 e^{\lambda s} \int_{0}^{T}\mathbb E [|\theta_{s,t}|^{p'}]\,dt
\leq & \int_0^T \mathbb E [|\theta_{0,t}|^{p'}]\,dt  + \left(\lambda -p'\bigg(\frac{\sigma^2\kappa}4 + \eta_1\bigg) \right)\int_{0}^{s}e^{\lambda v} \int_{0}^{T} \mathbb E \left[|\theta_{v,t}|^{p'}\right]\,dt\,dv \\
& + p' \int_0^s e^{\lambda v} \int_0^T \mathbb E \bigg[\bigg(\frac{\sigma^2}2 p' (p'-1) + \eta_2 \mathcal E(\mu,\delta_0)^2  + \bar \eta_t  \bigg)|\theta_{v,t}|^{p'-2}\bigg]\,dt\,dv\,.
\end{split}
\end{equation}
From Young's inequality we get $\mathcal E(\mu,\delta_0)^2 |\theta_{v,t}|^{p'-2} \leq \frac2{p'}\mathcal E(\mu,\delta_0)^{p'} + \frac{p'-2}{p'} |\theta_{v,t}|^{p'}$.
Then we see that
\[
\mathbb E \int_0^T \bigg[ \frac2{p'}\mathcal E_t(\mu,\delta_0)^{p'} + \frac{p'-2}{p'} |\theta_{v,t}|^{p'} \bigg]\,dt
\leq \mathbb E \int_0^T \bigg[ \frac2{p'}\mathcal W_{p'}(\mu_t,\delta_0)^{p'} + \frac{p'-2}{p'} |\theta_{v,t}|^{p'} \bigg]\,dt
\leq \mathbb E \int_0^T |\theta_{v,t}|^{p'}\,dt\,.
\]
Let $\varepsilon > 0$.
From Young's inequality we know there is $c_{\varepsilon,p'}$ such that
\[
\bigg(\frac{\sigma^2}2 p' (p'-1) + \bar \eta_t  \bigg)|\theta_{v,t}|^{p'-2} \leq \varepsilon|\theta_{v,t}|^{p'} + c_{\varepsilon,p'}\bigg(\frac{\sigma^2}2 {p'} (p'-1) + \bar \eta_t  \bigg)^{p'/2}
\]
Then from~\eqref{eq pre p moment est} we have
\[
\begin{split}
 e^{\lambda s} \int_{0}^{T}\mathbb E [|\theta_{s,t}|^{p'}]\,dt
\leq & \int_0^T \mathbb E [|\theta_{0,t}|^{p'}]\,dt  + \left(\lambda -p'\bigg(\frac{\sigma^2\kappa}4 + \eta_1 - \eta_2\bigg) + p'\varepsilon\right)\int_{0}^{s}e^{\lambda v} \int_{0}^{T} \mathbb E \left[|\theta_{v,t}|^{p'}\right]\,dt\,dv \\
& + p' c_{\varepsilon,p'} \int_0^s e^{\lambda v} \int_0^T \mathbb E \bigg[\bigg(\frac{\sigma^2}2 p' (p'-1) +  \bar \eta_t  \bigg)^{p'/2}\bigg]\,dt\,dv\,.
\end{split}
\]
Fix $\varepsilon = \frac{1}{2}\left(\frac{\sigma^2\kappa}4 + \eta_1 - \eta_2\right)$.
Then from~\eqref{eq pre p moment est} with $\lambda = \frac{p'}{2}\left(\frac{\sigma^2\kappa}4 + \eta_1 - \eta_2\right)>0$ we have
\begin{equation*}
 e^{\lambda s} \int_{0}^{T}\mathbb E [|\theta_{s,t}|^{p'}]\,dt
\leq \int_0^T \mathbb E [|\theta_{0,t}|^{p'}]\,dt  + \int_0^s e^{\lambda v} c_{T,p,\sigma,\kappa,\eta_1,\eta_2,\bar \eta}\,dv\,.
\end{equation*}
This concludes the proof.
\end{proof}

\begin{proof}[Proof of Lemma~\ref{thm:WellposednessMeanField}]
Let
\[
C(I; \mathcal V_{q}^W) := \left\{ \nu = (\nu_{s,\cdot})_{s\in I} : \nu_{s,\cdot} \in  \mathcal V_{q}^W\,\,\text{and}\,\,\lim_{s'\to s} \rho_q(\nu_{s',\cdot},\nu_{s,\cdot}) = 0 \,\,\,\forall s\in I \right\}\,.
\]
Consider $\mu \in C(I,\mathcal V_{q}^W)$.
For each $\mu_{s,\cdot} \in \mathcal V_q^W$, $s\geq 0$ we obtain unique solution to~\eqref{eq process}-\eqref{eq adjoint proc} which we denote $(X_{s,\cdot}, Y_{s,\cdot},Z_{s,\cdot})$.
Moreover, for each $t\in [0,T]$ the SDE
\begin{equation}
\label{eq lang in proof}
d\theta_{s,t}(\mu) = - \bigg((\nabla_a \frac{\delta \mathbf H_t^{0}}{\delta m})(\theta_{s,t}(\mu), \mu_{s,t}) + \frac{\sigma^2}{2}(\nabla_a U)(\theta_{s,t}(\mu))\bigg)\,ds + \sigma\,dB_s
\end{equation}
has unique strong solution and for each $s\in I$, see e.g.~\cite[Theorem 3.1]{krylov1981stochastic}.
We denote the measure in
$\mathcal V_q^W$ induced by
$\theta_{s,t}$ conditioned on $\mathcal F_t^W$ for each $t\in[0,T]$ as $\mathcal L(\theta(\mu)_{s,\cdot}|W)$.
Consider now the map $\Psi$ given by $\mathcal C(I,\mathcal V_{q}^W) \ni \mu \mapsto  \{ \mathcal L(\theta_{s,\cdot}(\mu)\,|\,W(\omega^W)) : \omega^W \in \Omega^W, s\in I\}$.

\paragraph{Step 1.}
We need to show that $ \{ \mathcal L(\theta(\mu)_{s,\cdot}|W(\omega^W)) : \omega^W \in \Omega^W, s\in I\} \in \mathcal C(I, \mathcal V_{q}^W)$.
This amounts to showing that we have the appropriate integrability and continuity.
Integrability follows from the same argument as in the proof Lemma~\ref{lem apriori} with $p'=q$ and with $\theta_{s,t}$ replaced by $\theta_{s,t}(\mu)$.
To establish the continuity property note that for $s'\geq s$ we have
\[
|\theta_{s',t}(\mu) - \theta_{s,t}(\mu)|\leq c \int_s^{s'}\left( 1
  +  ( |\theta_{r,t}(\mu)| + \int_0^T\int |a|\mu_{r,t}(da)\,dt\right)\,dr+\sigma|B_s-B_{s'}|\,.
\]
This, together with Lebesgue's theorem, leads to
\[
\lim_{s'\rightarrow s} \int_0^T \mathbb E[|\theta_{s',t}(\mu) - \theta_{s,t}(\mu)|^q ]dt = 0\,.
\]
which establishes the required continuity property.

\paragraph{Step 2.}
Fix $\lambda\in \mathbb R$.
Fix $t\in[0,T]$.
From It\^o's formula we get
\[
\begin{split}
d\Big(e^{\lambda s} & |\theta_{s,t}(\mu) - \theta_{s,t}(\mu')|^{q}  \Big)
=  \,  e^{\lambda s}
\biggl[ \lambda |\theta_{s,t}(\mu) - \theta_{s,t}(\mu')|	^q  \\
& -  \frac{q}{2}(\theta_{s,t}(\mu) - \theta_{s,t}(\mu')) \bigg( \sigma^2 \Big[ (\nabla_a U)(\theta_{s,t}(\mu)) -  (\nabla_a U)(\theta_{s,t}(\mu')) \Big] \\
& \qquad	+
  2 \bigg[ \nabla_a \frac{\delta \mathbf H_t^{0}}{\delta m})(\theta_{s,t}(\mu),\mu_{s,\cdot}) - (\nabla_a \frac{\delta \mathbf H_t^{0}}{\delta m})(\theta_{s,t}(\mu'),\mu'_{s,\cdot})\bigg]\bigg)|\theta_{s,t}(\mu) - \theta_{s,t}(\mu')|^{q-2}\bigg] ds \,.
\end{split}
\]
Note that from Assumptions~\ref{ass exist and uniq} and~\ref{ass monotonicity of DmH0} we get that for any $a,a'\in \mathbb R^p$ and for any $\mu,\mu' \in \mathcal V_2^W$ we have
\[
\begin{split}
 \bigg(\sigma^2 \left(\nabla_aU(a')-\nabla_aU(a)\right)
+ & 2\Big[
(\nabla_a \frac{\delta \mathbf H_t^{0}}{\delta m})(a',\mu') - (\nabla_a \frac{\delta \mathbf H_t^{0}}{\delta m})(a,\mu)
\Big] \bigg)  \cdot (a'-a) \\
& \geq (\sigma^2\kappa + \eta_1)|a'-a|^2 - \eta_2 \mathcal E_t(\mu,\mu')^2	\,.
\end{split}
\]
Hence we get
\begin{equation*}
\begin{split}
	  d(e^{\lambda s} &  |\theta_{s,t}(\mu)  - \theta_{s,t}(\mu')|^q  )\\

& \leq  e^{\lambda s}\left[-\Big(\lambda - \frac{q}2(\sigma^2\kappa +  \eta_1)\Big)  |\theta_{s,t}(\mu) - \theta_{s,t}(\mu')|^q +  \frac{q}2\eta_{2} \mathcal E_t(\mu_{s,t},\mu'_{s,t})^2|\theta_{s,t}(\mu) - \theta_{s,t}(\mu')|^{q-2}\right]  ds	\,.

\end{split}
\end{equation*}
Fix $\lambda > \frac{q}2(\sigma^2\kappa +  \eta_1)>0$.
From Young's inequality we get for $\varepsilon > 0$ that
\[
\eta_{2} \mathcal E_t(\mu_{s,t},\mu'_{s,t})^2|\theta_{s,t}(\mu) - \theta_{s,t}(\mu')|^{q-2}
\leq \frac{2}{q\varepsilon^{q/2}}\eta_2^{q/2} \mathcal E_t(\mu_{s,t},\mu'_{s,t})^q + \frac{\varepsilon^{q/(q-2)}}{q}|\theta_{s,t}(\mu) - \theta_{s,t}(\mu')|^{q}\,.
\]
Taking $\varepsilon > 0$ s.t. $2\varepsilon^{q/(q-2)} < \frac12\left(\lambda - \frac{q}{2}(\sigma^2\kappa + \eta_1)\right)$ we get that
\[
d(e^{\lambda s}|\theta_{s,t}(\mu)  - \theta_{s,t}(\mu')|^q  )
\leq e^{\lambda s} c_{q,\kappa,\sigma,\eta_1,\eta_2} \mathcal E_t(\mu_{s,t},\mu'_{s,t})^q\,.
\]
Recall that for any $s\in I$ we have
\begin{equation}
\label{eq wass est}
\begin{split}
\rho_q(\Psi(\mu_{s,\cdot}),\Psi(\mu'_{s,\cdot}))^q & = \mathbb E^W\Big[\mathcal W^T_q\big(\mathcal L( \theta_{s,\cdot}(\mu)\mid W),(\mathcal L( \theta_{s,\cdot}(\mu')\mid W) \big)^q  \Big]\\
& \leq\int_0^T \mathbb E\left[  |\theta_{s,t}(\mu)-\theta_{s,t}(\mu')|^q  \right] dt\,.
\end{split}
\end{equation}

Hence
\[
d(e^{\lambda s}\rho_q(\Psi(\mu)_{s,\cdot},\Psi(\mu')_{s,\cdot})^q) \leq  e^{\lambda s}c_{q,\kappa,\sigma,\eta_1,\eta_2} \rho_q(\mu_{s,\cdot},\mu'_{s,\cdot})^q\,.
\]
Recall that for all $t\in[0,T]$, $\theta_{0,t}(\mu)=\theta_{0,t}(\mu')$.
Fix $S>0$ and note that
\begin{equation} \label{eq theta inequlity}
\begin{split}
&  e^{\lambda S} \rho_q(\Psi(\mu)_s,\Psi(\mu')_s)^q
\leq   c_{q,\kappa,\sigma,\eta_1,\eta_2} \int_0^S e^{\lambda s}  \rho_q(\mu_{s,\cdot},\mu'_{s,\cdot})^qds\,.

\end{split}
\end{equation}
\paragraph{Step 3.}
Let $\Psi^k$ denote the $k$-th composition of the mapping $\Psi$ with itself.
Then, for any integer $k > 1$,
\[
\rho_q(\Phi^k(\mu)_{S,\cdot},\Phi^k(\mu')_{S,\cdot})^q \leq
 e^{-\lambda S} c_{q,\kappa,\sigma,\eta_1,\eta_2}^k \frac{S^k}{k!} \sup_{s\in[0,S]} \mathbb\rho_q(\mu_{s,\cdot},\mu_{s,\cdot}')^q\,.
\]
Hence, for any $S\in I$ there is $k$, such that $\Phi^k$ is a contraction and then Banach fixed point theorem gives existence of the unique solution in $C([0,S];\mathcal V_{q}^W)$. The estimate \eqref{eq:UnifBound} follows simply from  Lemma~\ref{lem apriori}.
\end{proof}

Recall that $P_{s,t} \mu^0 := \mathcal L\big( \theta_{s,t} | \mathcal F_t^W \big)$,
where $(\theta_{s,t})_{s\geq 0, t\in [0,T]}$ is the unique solution to the  system~\eqref{eq mfsgd}-\eqref{eq mfsgd 2} started with the initial condition $(\theta^0_t)_{t\in [0,T]}$ such that $\mathcal L(\theta^0_t(\omega^W)) = \mathbb P^B \circ \theta^0_t(\omega^W)^{-1} = \mu^0(\omega^W)$, given by Lemma~\ref{thm:WellposednessMeanField}.
Let $P_s \mu^0 := (P_{s,t} \mu^0)_{t\in [0,T]}$ and note that $P_s \mu^0 \in \mathcal V_{q}^W$ for any $s\geq 0$.
Moreover note that due to uniqueness of solutions to~\eqref{eq mfsgd}-\eqref{eq mfsgd 2} we have $P_{s+s'} \mu^0 = P_s\big(P_{s'}\mu^0\big)$.

\begin{lemma}
\label{lemma exp conv est}
Let Assumptions~\ref{ass exist and uniq} and~\ref{ass monotonicity of DmH0} hold.
If $\lambda = \frac{q}{2}\left(\sigma^2\kappa + \eta_1 - \eta_2\right) \geq 0$ and if $\mu^0, \bar \mu^0 \in \mathcal V_{q}^W$, then for all $s\geq 0$ we have
\begin{equation}
\label{eq inv meas conv rate}
\rho_q(P_s \mu^0, P_s \bar \mu^0)\leq e^{-\frac1q\lambda s}  \rho_q(\mu^0, \bar \mu^0)\,.
\end{equation}
\end{lemma}

\begin{proof}
Let $(\theta_{s,\cdot})_{s\geq 0}$ and $( \theta'_{s,\cdot})_{s\geq 0}$
denote two solutions to~\eqref{eq mfsgd}-\eqref{eq mfsgd 2} with initial conditions
$(\theta^0_t)_{t\in [0,T]}$ such that $\mathcal L(\theta^0_t(\omega^W)) = \mu^0(\omega^W)$
and
$(\theta'^0_t)_{t\in [0,T]}$
such that $\mathcal L(\theta'^0_t(\omega^W)) = \bar \mu^0(\omega^W)$
for all $\omega^W \in \Omega^W$.

For $\lambda = \frac{q}{2}\left(\sigma^2\kappa + \eta_1 - \eta_2\right)  \geq 0$, the same computation as in the  Step 2 of the proof of Lemma~\ref{thm:WellposednessMeanField} gives
\begin{equation*}
\begin{split}
	  d\bigg(e^{\lambda s} &  \mathbb E \int_{0}^T |\theta_{s,t}  - \theta'_{s,t}|^q dt  \bigg)\\
& \leq  e^{\lambda s}\left[\Big(\lambda - \frac{q}{2}\big(\sigma^2\kappa + \eta_1\big)\Big)  \mathbb E \int_{0}^T \Big[ |\theta_{s,t} - \theta'_{s,t}|^q + \frac{q}{2}\eta_2\mathcal E_t(\mathcal L(\theta_{s,t}|W),  \mathcal L(\theta'_{s,t})|W)^2 |\theta_{s,t} - \theta'_{s,t}|^{q-2} \Big] \,dt   \right] ds \,.\\

\end{split}
\end{equation*}
From Young's inequality
\[
\mathcal E_t\big(\mathcal L(\theta_{s,t}|\mathcal F_t^W),  \mathcal L(\theta'_{s,t}|\mathcal F_t^W)\big)^2 |\theta_{s,t} - \theta'_{s,t}|^{q-2} \leq \frac2q \mathcal E_t\big(\mathcal L(\theta_{s,t}|\mathcal F_t^W),  \mathcal L(\theta'_{s,t}|\mathcal F_t^W)\big)^q + \frac{q-2}{q}|\theta_{s,t} - \theta'_{s,t}|^q\,.
\]
Moreover, from Assumption~\ref{ass monotonicity of DmH0} we get
\[
\mathbb E \int_{0}^T \mathcal E_t(\mathcal L(\theta_{s,t}|\mathcal F_t^W),  \mathcal L(\theta'_{s,t}|\mathcal F_t^W))^q \,dt \leq \rho_q(\mathcal L(\theta_{s,\cdot}|W),  \mathcal L(\theta'_{s,\cdot})|W)^q \leq \mathbb E \int_{0}^T |\theta_{s,t} - \theta'_{s,t}|^q\,dt\,.
\]
Hence
\[
\begin{split}
d\Big(e^{\lambda s} \rho_q(P_s \mu^0, P_s \bar \mu^0)^q \Big)
& \leq d\bigg(e^{\lambda s} \mathbb E \int_{0}^T|\theta_{s,t}-\theta'_{s,t}|^q\,dt\bigg) \\
& \leq e^{\lambda s} \left[\Big(\lambda - \frac{q}{2}\big(\sigma^2\kappa + \eta_1 - \eta_2\big)\Big)  \mathbb E \int_{0}^T \Big[ |\theta_{s,t} - \theta'_{s,t}|^q \,dt\right] ds \leq 0 \,ds\,.
\end{split}
\]
Integrating this leads to
\[
\rho_q( P_s \mu^0, P_s \bar \mu^0)^2 \leq e^{-\lambda s}  \rho_q(\mu^0,  \bar \mu^0)^2\,.
\]
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm conv to inv meas rate}]
We follow a similar argument as in the proof of \cite[Corollary 1.8]{majka2017coupling} or in \cite[Section 3]{komorowski2012central}.
Choose $s_0 > 0$ such that $e^{-\frac1q \lambda s_0} < 1$.
Then $P_{s_0} : \mathcal V_q^W \to \mathcal V_q^W$
is a contraction due to Lemma~\ref{lemma exp conv est}.
By Banach's fixed point theorem and Lemma~\ref{lemma L2overmeasurescomplete} there is a (unique) $\tilde \mu \in \mathcal V_q^W$ such that $P_{s_0}\tilde \mu = \tilde \mu$.
Note that $\tilde \mu$ depends on the above choice of $s_0$.

Let $\mu^\ast := \int_0^{s_0} P_s\tilde \mu \,ds$.
Now take an arbitrary $r \leq s_0$.
Then
\[
P_r \mu^\ast = P_r \int_0^{s_0} P_s\tilde \mu\,ds = \int_0^{s_0} P_{r+s} \tilde \mu \,ds 	= \int_r^{r+s_0} P_s\tilde \mu\,ds
= \int_r^{s_0} P_s\tilde \mu\,ds + \int_{s_0}^{r+s_0} P_s\tilde \mu\,ds\,.
\]
Since $\tilde \mu = P_{s_0}\tilde \mu$ we get that
\[
P_r \mu^\ast = \int_r^{s_0} P_s\tilde \mu\,ds + \int_{s_0}^{r+s_0} P_{s+s_0}\tilde \mu\,ds = \int_r^{s_0} P_s\tilde \mu\,ds + \int_{0}^{r} P_{s}\tilde \mu\,ds = \mu^\ast\,.
\]
For $r > s_0$ choose $k\in \mathbb N$ so that $ks_0 < r$ and $(k+1)s_0 \geq r$.
Then $P_r \mu^\ast = (P_{s_0})^k P_{(r-ks_0)} \mu^\ast = \mu^\ast$.
To see that $\mu^\ast$ is unique consider $\nu^\ast \neq \mu^\ast$ such that $P_r \nu^\ast = \nu^\ast$ for any $r\geq 0$. Then from Lemma~\ref{lemma exp conv est} we have, for any $r>s_0$, that
\[
\rho_q(\mu^\ast, \nu^\ast) = \rho_q(P_r \mu^\ast, P_r \nu^\ast) \leq e^{-\frac1q\lambda r} \rho_q(\mu^\ast,\nu^\ast)
\]
which is a contradiction as $e^{-\frac1q \lambda r} < 1$.
From Lemma~\ref{lemma exp conv est} we immediately get~\eqref{eq exp conv to inv meas}.
\end{proof}

\begin{proof}[Proof of Lemma~\ref{lem pde}]
The linear PDE~\eqref{eq fpe} has unique solution $\nu_{\,\cdot,t} \in C^{1,\infty}((0,\infty)\times \mathbb R^p; \mathbb R)$ for each $t\in [0,T]$ and $\omega^W \in \Omega^W$ due to e.g Ladyzenskaja, Solonnikov and Ural'ceva~\cite[Chapeter IV]{LSU68}.
The measurability of $\mathcal F_t^W$-measurability of $\nu_{s,t}(a)$ is a question of measurability of an explicitly defined function and this is proved e.g. in~\cite[Lemma~3.2]{emmrich2017nonlinear}.
Consider the stochastic process
\[
d\theta_{s,t} = - b_{s,t}(\theta_{s,t})\,ds + \sigma \,dB_s\,.
\]
Let $\mu_{s,t}$ denote the law of $\theta_{s,t}$ given $\omega \in \Omega^W$
From Girsanov's theorem we see that the $\mu_{s,t}$ has, for each $s>0$ and $t\in [0,T]$, smooth density and moreover $\mu_{s,t}(a) > 0$.
Applying It\^o's formula to $\varphi \in C^2_b(\mathbb R^p)$, taking expectation (over $\Omega^B$) and using $\mu_{s,t}$ to denote the law of $\theta_{s,t}$ given $\omega \in \Omega^W$ we can check that $\mu_{s,t}$ satisfies
\[
\int \varphi(a) \mu_{s,t}(da) = \int \varphi(a)\mu_{0,t}(da) + \int_0^s \Big[ \int - b_{s,t}(a)\nabla_a \varphi(a)\mu_{r,t}(da) + \int \frac12\sigma^2 \Delta_a \varphi(a) \mu_{r,t}(da)  \Big]\,ds\,.
\]
Integrating by parts we see that is a solution to~\eqref{eq fpe}.
As the solutions are unique we conclude that $\mu_{s,\cdot} = \nu_{s,\cdot}$ and so $\nu_{s,t}(a) > 0$ for all $s>0$, $t\in [0,T]$ and $\Omega^W$-a.s.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm unique minimiser}]

Fix $t\in [0,T]$ and $\omega^W \in \Omega^W$.
Let $b_{s,t}(a) := (\nabla_a \frac{\delta \mathbf H_t^{0}}{\delta m})(a, \mu_s) + \frac{\sigma^2}{2}(\nabla_a U)(a)$
and $\mu_{s,t} = \mathcal L(\theta_{s,t} | \mathcal F_t^W)$.
As in the proof of Lemma~\ref{lem pde} we see that $\mu_{s,t}$ is a solution to
\begin{equation}
\label{eq fpe2}
\partial_s \mu_{s,t} = \nabla_a \cdot \bigg( b_{s,t} \mu_{s,t} + \frac{\sigma^2}{2}\nabla_a \mu_{s,t}\bigg)\,,\,\,\,s\geq0\,, \,\,\, \mu_{s,0} = \mu^0_t := \mathcal L(\theta_{0,t}| \mathcal F_t^W)\,.
\end{equation}
Due to Lemma~\ref{lem pde} we know that the solution is unique and moreover for each $t\in [0,T]$ and $\omega^W \in \Omega^W$ fixed we have $\mu_{\,\cdot,t} \in C^{1,\infty}((0,\infty)\times \mathbb R^p; \mathbb R)$.
Also, $P_s \mu^0 = \mu_{s,\cdot}$ so $P_s$ is the solution operator for~\eqref{eq fpe2}.

Since $\mu^\ast$ is an invariant measure $0 = \rho(P_s \mu^\ast, \mu^\ast) = \big(\mathbb E^W\big[\mathcal W_2^T(P_s \mu^\ast, \mu^\ast)^2\big)^{1/2}$ we get that for almost all  $t\in [0,T]$ and $\omega^W \in \Omega^W$ we have $(P_s \mu^\ast)_t = \mu^\ast_t$ and so $\partial_s \mu^\ast_{s,t} = 0$.
Hence for almost all  $t\in [0,T]$ and $\omega^W \in \Omega^W$ we have that $\mu^\ast_t$ is the solution to the stationary Kolmogorov--Fokker--Planck equation
\begin{equation}
\label{eq stationary fpe}
0 = \nabla_a \cdot \bigg( \Big( (\nabla_a \frac{\delta \mathbf H_t^{0}}{\delta m})(\cdot, \mu^\ast) + \frac{\sigma^2}{2}(\nabla_a U) \Big) \mu^\ast_{t} + \frac{\sigma^2}{2}\nabla_a \mu^\ast_{t}\bigg)\,.
\end{equation}
This implies that $\mu^\ast \in \mathcal I^\sigma$.
Consider now some $\nu \in \mathcal I^\sigma$.
Then from~\eqref{eq foc} we get that
\[
\nu_t(a) = \mathcal Z^{-1} e^{ -\frac{2}{\sigma^2} \frac{\delta \mathbf H_t^{0}}{\delta m}(a,\nu_t(a))}\gamma(a)\,,\quad \mathcal Z:= \int e^{ -\frac{2}{\sigma^2} \frac{\delta \mathbf H_t^{0}}{\delta m}(a,\nu_t(a))}\gamma(a)da\,.
\]
From this and Assumption~\ref{as coefficients new}
we see that almost all $t\in [0,T]$ and $\omega^W \in \Omega^W$ we have that $\nu_t$ solves~\eqref{eq stationary fpe}.
But the solution to~\eqref{eq stationary fpe} is unique and so $\nu = \mu^\ast$.
This proves item i).

We will show by contradiction that $\mu^\ast$ is at least (locally) optimal.
Assume that $\mu^\ast$ is not  the (locally) optimal control for $J^\sigma$ defined in~\eqref{eq objective bar J}.
Then for some $\mu^0 \in \mathcal V_2^W$ it holds that $J^\sigma(\mu^0) < J^\sigma(\mu^\ast)$.
On the other we know from~\eqref{eq exp conv to inv meas} that $\lim_{s\to\infty} P_s \mu^0 = \mu^\ast$.
From this and Theorem~\ref{th with measure flow} and from lower semi-continuity of $J^\sigma$ we get
\[
\begin{split}
& J^\sigma(\mu^\ast) - J^\sigma(\mu^0) \leq \liminf_{s\to \infty} J^\sigma(P_s \mu^0) - J^\sigma(\mu^0) \\
& = - \liminf_{s\to \infty}\int_0^s \mathbb E^W \int_0^T  \left[\int  \left| \left(\nabla_a\frac{\delta \mathbf H^\sigma}{\delta m}\right)(a,(P_s\mu^0)_t) \right|^2 \,(P_s\mu^0)_t(da)   \right]\,dt\,ds\\
& \leq 0	\,.
\end{split}
\]
This is a contradiction and so $\mu^\ast$ must be (locally) optimal.
On the other hand for any other (locally) optimal control $\nu^\ast \in \mathcal V_2^W$ we have for any $\nu \in \mathcal V_2^W$, due to Theorem~\ref{thm necessary cond linear} that
\[
0 \leq \mathbb E^W \bigg[ \int_0^T \int  \frac{\delta \mathbf H^\sigma_t}{\delta m}(a,\nu^\ast)(\nu_t - \nu^\ast_t)(da)\,dt \bigg] \,.
\]
Due to Lemma~\ref{lem constant} this implies that $\nu^\ast \in \mathcal I^\sigma$, where $\mathcal I^\sigma$ is defined in~\eqref{eq foc}.
But we have already shown that $\mathcal I^\sigma = \{\mu^\ast\}$ and so the set of local minimizers is a singleton and thus $\mu^\ast$ is the global minimizer of $J^\sigma$ and item ii) is proved.
\end{proof}


TITLE: A question on Isometry between the orthogonal subspaces of Hilbert spaces
QUESTION [2 upvotes]: I was reviewing my class-notes on Functional analysis and the professor had mentioned that given a closed proper subspace $U$ of an hilbert space $\mathcal{H}$, $\exists  $ a closed subspace  $U^{\perp}$ such that
$U\oplus U^{\perp}=\mathcal{H}$ is an isometry from $\{<z,w>\} \to z+w$ where $z \in U,w\in U^{\perp}$
I know the definition of an isometry(that it is a distance preserving map between two metric spaces) but why is this map an isometry? And why isnt $U\oplus U^{\perp}$ the same as $\mathcal{H}$
I would be grateful if someone could help me out here. Thanks

REPLY [2 votes]: There are two understandings of the notation $U \oplus V$.
The first interpretation means that if you write $U \oplus V$, then $U$ and $V$ are both subspaces of a common inner product space and orthogonal to each other (i.e. $\langle x,y\rangle =0$ for $x\in U,y\in V$ and $U \oplus V=\{x+y\mid x\in U, y \in V\}$. In this interpretation, you have $\mathcal{H}=U \oplus U^\bot$.
In the second interpretation, $U,V$ can be different spaces(i.e. they don't have to be subspaces of a common inner product space) and
$$
U \oplus V :=\{ (u,v)\mid u\in U, v\in V\},
$$
equipped with the usual vector space structure and with the scalar product
$$
\langle (u,v), (a,b)\rangle = \langle u,a\rangle +\langle v,b \rangle.
$$
This is the interpretation used in your statement.
That your map is an isometry is a consequence of Pythagoras theorem, ie that $\Vert u+v\Vert^2= \Vert u\Vert^2 +\Vert v\Vert^2$ if $u,v$ are orthogonal.


TITLE: Gradient of a line
QUESTION [0 upvotes]: I have the question "A(1,3), B(5,7),C(4,8),D(a,b) form a rectangle ABCD. Find a and b."
I have worked out the gradient of AB which is 1 and the gradient of CD which is 2 b/a, however , when I checked the solutions b = 4 and a = 0 and I do not understand how this was achieved.

REPLY [1 votes]: Continuing after your first step, for $CD,AB$ opposite sides have equal slope =
$$\frac{b-8}{a-4}=1$$
in which step you made a mistake.
For $ BC,AD $ slope of opposite sides
$$\frac{b-3}{a-1}=-1$$
the lines are perpendicular , so they must be negative reciprocal of each other. That tallies.
These two equations  solve to $(a,b)= (0,4)$


TITLE: How do you show the Riemann function is Riemann integrable?
QUESTION [0 upvotes]: Consider the so-called Riemann function $f:[0,1]\to\mathbb R$, defined by $$f(x)= \begin{cases}0 & \text{if } x \text{ is irrational or }x=0,\\ \frac1n & \text{if }x=\frac{m}n \text{ where the fraction }\frac{m}n\text{ is irreducible and } n \geq1.\end{cases}$$
Show that the function $f$ is Riemann integrable on $[0,1]$. Show that for given $\epsilon>0$ there exists only finitely many $x\in(0,1]$ such that $x=\frac{m}n$ and $\frac1n\geq\frac{\epsilon}{2(b-a)}$. Use this fact to find $\delta > 0$ such that for every partition P satisfying $\|P\|<\delta$ we have $S(f,P)-s(f,P)<\epsilon$.
Particularly, I wonder how you show there are only finitely many $x$ and how that relates to the condition $\|P\|<\delta$.

REPLY [2 votes]: Naturally, there are only finitely many $n$ such that $\frac{1}{n} \geq \frac{\epsilon}{2(b-a)}$, because $\frac{1}{n}$ is a function that decreases with $n$. Now, since our domain is $(0,1]$, we must have $m \leq n$, which means only finitely many m exist, so only finitely many $x=\frac{m}{n}$ exist. Let the number of such $x$ be $N$.
Take any partition $P$ of $[0,1]$, $ P = x_0 < x_1 < ... < x_k=1$. Note that:
$$
S(f,P) = \sum (x_{i+1} -x_i) M_i, M_i=\displaystyle\sup_{[x_{i+1},x_i]} f(x)
$$
$$
s(f,P) = \sum (x_{i+1} -x_i) m_i, m_i=\displaystyle\inf_{[x_{i+1},x_i]} f(x)
$$
But then $m_i=0$ for all $i$, since in every interval there is some irrational number. So $s(f,P)=0$. Now note that $f(x) \leq 1/2$ except at $x=1$. In order to take care of that, we will consider a partition of [0,1] where $\delta > 2N$. Let it have say T partition points.Now, we have that:
$$
S(f,P) = \sum_{i=0}^{T-1} (x_{i+1} -x_i) M_i, M_i=\displaystyle\sup_{[x_{i+1},x_i]} f(x) \leq \sum_{i=0}^{T-2} (x_{i+1} -x_i) * 1/2 + (1-x_{T-1})
$$
Finish it from here.


TITLE: Complex Values in Second Order Differential Equations
QUESTION [1 upvotes]: Recently I've learned that second order linear homogeneous differential equations can be solved by assuming the function to be something like this.
$$Ay''+By'+Cy=0$$
$$y = e^{St} $$
$$AS^2+BS+C=0$$
When encountering underdamped systems, the value of S would be imaginary, leaving you with Euler's identity.
$$e^{i\alpha}=\cos(\alpha t)+i\sin(\alpha t)$$
When solving for the fundamental solutions our professor disregarded the imaginary coefficient and claimed that the fundamental solutions are the imaginary component and the real component. How is this so??

REPLY [1 votes]: This transformation is functional only for linear differential equations with constant coefficients.
Note that if $i \alpha $ is solution $-i \alpha$ is also solution because the characteristic polynomial has real coefficients.
Now if the solution is
$y = C_1e^{i \alpha t}+C_2 e^{-i \alpha t}$
This is equivalent to
$
y = D_1 \sin(\alpha t)+D_2\cos(\alpha t)
$
with $D_1, D_2$ real constants, because solving for initial conditions
$
y(0) = C_1+C_2 = y_0\\
y'(0)= i\alpha C_1-i\alpha C_2 = y'_0
$
Solving for $C_1,C_2$ gives
$$
C_1 = \frac{\alpha y_0-i y'_0}{2\alpha}\\
C_2 = \frac{\alpha y_0+i y'_0}{2\alpha}
$$
and finally
$$
y = y_0 \cos(\alpha t)+\frac{y'_0}{\alpha}\sin(\alpha t)
$$
hence $D_1 = \frac{y'_0}{\alpha}$ and $D_2 = y_0$


TITLE: Intuitive explanation of Nakayama's Lemma
QUESTION [35 upvotes]: Nakayama's lemma states that given a finitely generated $A$-module $M$, and $J(A)$ the Jacobson radical of $A$, with $I\subseteq J(A)$ some ideal, then if $IM=M$, we have $M=0$.
I've read the proof, and while being relatively simple, it doesn't give much insight on why this lemma should be true, for example - is there some way to see how the fact that $J(A)$ is the intersection of all maximal ideals related to the result?
Any intuition on the conditions and the result would be of great help.

REPLY [4 votes]: There is a more primitive and natural version of Nakayama's lemma:

Let $R$ be a commutative ring, and $M$ be a finitely generated $R$-module and $I\subset R$ be an ideal. If $IM=M$ then $\exists f\in I,\forall m\in M,f\cdot m=m$.

The proof is a bit harder than the above version, but this pretty much says "If $IM=M$ then one of the elements of $I$ stablizes all elements of $M$." And I think this is much more natural than the above version. Assuming this version, then we can see that the condition of "Jacobson ideal" is only used to show that "$1-f$ is a unit" which leads to "$M=0$".
For completion, I give a proof of the above version here:
Proof: Let $m_1,...,m_n$ be a set of generators of $M$. Since $IM=M$, we have a matrix $A'=(a_{ij})\in \textrm{M}_{n\times n}(R)$ such that $A'\underline{m}=\underline{m}$ where $\underline{m}=(m_1,...,m_n)^t$ and $a_{ij}\in I,\forall i,j\leq n$. Now let $A=\textrm{Id}-A'$ so we have $A\underline{m}=\underline{0}$.
Note that $\det A\in R$ is well-defined and there exists a matrix $B\in \textrm{M}_{n\times n}(R)$ s.t. $BA=AB=\det A\cdot \textrm{Id}$, the proof is in Corollary 9.161 of Advanced Modern Algebra by Joseph J. Rotman. Also $\det$ commutes with the quotient map $R\rightarrow R/I$, with $A\equiv \textrm{Id} \ (\textrm{mod  }I)$ we can deduce that $\det A\in 1+I$, so $\exists f\in I$, s.t. $\det A=1-f$.
To show $f$ stablizes $M$ it suffices to show that $f\cdot m_i=m_i,\forall i$.
$$\underline{0}=BA\underline{m}=\det A\cdot \underline{m}=(1-f)\cdot \underline{m}$$
So
$$f\cdot \underline{m}=\underline{m}$$
The result follows. Q.E.D


text:
\section{Conclusions}
We provided the first effective and realizable algorithms for differentially private estimation of mean and covariance in the multivariate setting.
We demonstrated that not only do these algorithms have strong theoretical guarantees (matching the state-of-the-art), but they are also accurate even at relatively low sample sizes and high dimensions.
They significantly outperform all prior methods, possess few hyperparameters, and remain precise even when given minimal prior information about the data.
In addition, we showed that our methods can be used for a private version of PCA, a task which is common in data science and exploratory data analysis.
As we are seeing the rise of a number of new libraries for practical differentially private statistics and data analysis~\cite{OpenDP20,AllenBW20} we believe our results add an important tool to the toolkit for the multivariate setting.

config: arxiv


TITLE: Construct $\frac{1}{z}$ from $z \in \mathbb{C}$
QUESTION [2 upvotes]: Suppose $z \in \mathbb{C}$. How can we construct $\frac{1}{z}$ with tools without calculating? My teacher suggested something with a parallel line, but I couldn't figure it out.
A unit distance is not given!

REPLY [0 votes]: If you prefer a solution with inversion:
For a point $z$ outside the unit circle, intersect the circle with diameter $0z$ with the unit cercle. Intersect the line of the two intersection points with the line $0z$. The intersection point gives $1/\bar z$. Now, reflect with respect to the horizontal axis to find $1/z$.
You can check that this works, because you have lots of similar (rectangular) triangles.
For a point inside, draw the line $0z$, draw the line through $z$ orthogonal to it, intersect it with the unit circle to get the points $A$ and $B$. Draw the line through $A$ that is orthogonal to the radius $0A$, intersect it with $0z$ to get $1/\bar z$. Reflect it with respect to the horizontal axis to find $1/z$.