&\tD^{-2}\trace \Big[\Big( \bK\frac{\partial \vec(\bG) }{\partial \vec(\bE)} \Big)^2\Big]\\
=~& \tD^{-2}\trace \Big( \bK (\be_{t_0}\be_{t_0'}^\top \otimes \bI_n) \frac{\partial \vec(\bF) }{\partial \vec(\bE)} \bK (\be_{t_0}\be_{t_0'}^\top \otimes \bI_n)\frac{\partial \vec(\bF) }{\partial \vec(\bE)}\Big)\\
\le~& \tD^{-2}\fnorm{(\be_{t_0'}^\top \otimes\bI_n)\frac{\partial \vec(\bF) }{\partial \vec(\bE)} \bK (\be_{t_0}\otimes \bI_n)}^2.
%=~& \fnorm*{\frac{\partial \vec(\bF) }{\partial \vec(\bE)} \bK}^2\\
%\le~& n\fnorm{\bS}^2,
\end{align*}

For the second term in (<ref>), recall $\Rem = \bu\bv^\top$, and $\bu = \vec(\bG)$, $\bv^\top = \frac{\partial \tD^{-1}}{\partial \vec(\bE) }$ from (<ref>), then $\trace[(\bK\Rem)^2] = \trace( \bK \bu\bv^\top\bK \bu\bv^\top)= (\bv^\top\bK \bu)^2$, thus,

\begin{align*}
&\trace[(\bK\Rem)^2] \\
=~& \Big[ \frac{\partial \tD^{-1}}{\partial \vec(\bE) } \bK \vec(\bG)\Big]^2\\
=~& \tD^{-6}  \Big[ \vec(\bF)^\top\frac{\partial \vec(\bF)}{\partial \vec(\bE)} \bK (\be_{t_0}\be_{t_0'}^\top \otimes \bI_n) \vec(\bF) \Big]^2\\
\le~ & \tD^{-6} \norm{\vec(\bF)^\top\frac{\partial \vec(\bF)}{\partial \vec(\bE)} \bK (\be_{t_0} \otimes \bI_n)}^2 \norm{(\be_{t_0'}^\top \otimes \bI_n) \vec(\bF)}^2.
\end{align*}

where the first inequality uses $
\frac{\partial \tD^{-1}}{\partial \vec(\bE)} = -\frac12\tD^{-3}\frac{\partial \fnorm{\bF}^2}{\partial \vec(\bE)} = -\tD^{-3}\vec(\bF)^\top\frac{\partial \vec(\bF)}{\partial \vec(\bE)}$ by chain rule, and the inequality uses Cauchy-Schwarz inequality.

For the third term in (<ref>), recall $\Rem = \bu\bv^\top$, and $\bu = \vec(\bG)$, $\bv^\top = \frac{\partial \tD^{-1}}{\partial \vec(\bE) }$, then
$\trace\Big[ \bK\frac{\partial \vec(\bG) }{\partial \vec(\bE) } \bK \Rem\Big] =
\bv^\top\bK\frac{\partial \vec(\bG) }{\partial \vec(\bE) } \bK  \bu$, hence

\begin{align*}
&2\tD^{-1}\trace\Big[ \bK\frac{\partial \vec(\bG) }{\partial \vec(\bE) } \bK \Rem\Big] \\
=~& 2\tD^{-1}  \frac{\partial \tD^{-1}}{\partial \vec(\bE) }\bK \frac{\partial \vec(\bG) }{\partial \vec(\bE) }\bK \vec(\bG)\\
=~& -2\tD^{-4} \vec(\bF)^\top \frac{\partial \vec(\bF) }{\partial \vec(\bE)}\bK
(\be_{t_0}\be_{t_0'}^\top \otimes \bI_n)\frac{\partial \vec(\bF) }{\partial \vec(\bE)} \bK(\be_{t_0}\be_{t_0'}^\top \otimes \bI_n) \vec(\bF)\\
\le~&2 \tD^{-4}  \fnorm{(\be_{t_0'}^\top \otimes \bI_n) \vec(\bF) \vec(\bF)^\top \frac{\partial \vec(\bF) }{\partial \vec(\bE)}\bK (\be_{t_0} \otimes \bI_n)} \\
& \fnorm{(\be_{t_0'}^\top \otimes \bI_n) \frac{\partial \vec(\bF) }{\partial \vec(\bE)} \bK(\be_{t_0}\otimes \bI_n)},
\end{align*}

where the last inequality uses Cauchy-Schwarz inequality.

Summing over all $(t_0, t_0')\in [T]\times [T]$ for these three terms in (<ref>), using $\opnorm{ \frac{\partial \vec(\bF) }{\partial \vec(\bE)} } \le 1$, $\bK = \bS \otimes \bI_n$, and $\fnorm{\bS} \le \fnorm{\bS^{\frac12}}^2= \trace(\bS)$, we obtain

\begin{align}
&\sum_{t_0, t_0'} \trace\big[\big(\bK \nabla \bff(\bz)\big)^2\big] \nonumber\\
\le~& \tD^{-2}\fnorm{\bK}^2 + \tD^{-6} \fnorm{\bF}^2 \fnorm{\bK}^2 \fnorm{\bF}^2 + 2 \tD^{-4} \fnorm{\bF}^2 \fnorm{\bK}^2\nonumber\\
=~ & \tD^{-2}n\fnorm{\bS}^2 + \tD^{-6} \fnorm{\bF}^4 n\fnorm{\bS}^2  + 2 \tD^{-4} \fnorm{\bF}^2 n\fnorm{\bS}^2\nonumber\\
\le~& \big[\tD^{-2}n\trace(\bS) + \tD^{-6} \fnorm{\bF}^4 n\trace(\bS)  + 2 \tD^{-4} \fnorm{\bF}^2 n\trace(\bS)\big] \trace(\bS).\label{eq: stein-RHS2}
\end{align}

Second term in (<ref>).
Recall that $\Rem = \bu\bv^\top$, hence $[\trace(\bK\Rem)]^2 =\trace[(\bK\Rem)^2]$.
By calculation of second term in  (<ref>), we obtain

\begin{align}\label{eq: LHS2}
\sum_{t_0,t_0'} [\trace(\bK\Rem)]^2
=\sum_{t_0, t_0'}  \trace[(\bK\Rem)^2]
\le  \tD^{-6} \fnorm{\bF}^4 n  \trace(\bS)^2.
\end{align}

Combining the results (<ref>), (<ref>), (<ref>), (<ref>),  (<ref>), we obtain

\begin{align*}
\fnorm{\bE^\top \bF/\tD - \bS (n\bI_T - \hbA )/\tD}^2\Bigr]\\
\le~&  2\big[\tD^{-2} \fnorm{\bF}^2+\tD^{-2}n\trace(\bS) + 2\tD^{-6} \fnorm{\bF}^4 n\trace(\bS)  + 2 \tD^{-4} \fnorm{\bF}^2 n\trace(\bS)\big] \trace(\bS)\\
\le~& 4 \trace(\bS),
\end{align*}

thanks to $\tD^{2} = \fnorm{\bF}^2 + n\trace(\bS)$.

Apply <cit.> with $\brho = \bU\be_t, \bfeta = \bV\be_{t'}$, we obtain

\begin{align*}
&\E\Big[\norm[\Big]{\bU^\top \bZ \bV -
\sum_{j=1}^p\sum_{i=1}^n
\frac{\partial}{\partial z_{ij} }\Bigl(\bU^\top \be_i \be_j^\top \bV \Bigr)
}_{\rm F}^2\Big]\\
=~ &\sum_{t,t'=1}^T\E \big(\be_t^\top\bU^\top \bZ\bV\be_{t'} - \sum_{j=1}^p\sum_{i=1}^n
\frac{\partial }{\partial z_{ij}} \be_{t}\bU^\top \be_i \be_j^\top \bV\be_{t'} \big)^2\\
\le~ & \sum_{t,t'=1}^T \left[\E \big[\norm*{\bU\be_t}^2 \norm*{\bV\be_{t'}}^2\big] + \E
\sum_{ij} \Big[2\norm*{\bV\be_{t'}}^2 \fnorm*{\frac{\partial \bU \be_t}{\partial z_{ij}}} + 2\norm*{\bU\be_{t}}^2 \fnorm*{\frac{\partial \bV \be_{t'}}{\partial z_{ij}}}\Big] \right]\\
%\sum_{ij,lk}\Big( \frac{\partial}{\partial x_{lk}}\be_{t}\bU^\top \be_i \be_j^\top \bV\be_{t'} \Big)^2\\
=~ & \E \fnorm*{\bU}^2 \fnorm*{\bV}^2+ \E \sum_{ij}\Big[
2\fnorm*{\bV}^2\fnorm*{ \frac{\partial \bU}{\partial z_{ij}} }^2
+ 2\fnorm*{\bU}^2\fnorm*{ \frac{\partial \bV}{\partial z_{ij}} }^2\Big].
\end{align*}

By Kirszbraun's theorem, there exists an $L_1$-Lipschitz function $\bar\bU: \R^{n\times p} \to \R^{n\times T}$ such that $\bar\bU = \bU$ on $\Omega$, and an $L_2$-Lipschitz function $\bar\bV: \R^{n\times p} \to \R^{n\times T}$ such that $\bar\bV = \bV$ on $\Omega$. By projecting $\bar\bU$ and $\bar\bV$ onto the Euclidean ball of radius 1 and $K$ if necessary, we assume without loss of generality that $\fnorm{\bar\bU}\le 1$ and $\fnorm{\bar\bV}\le K$.

\begin{align*}
&\E\Big[I(\Omega) \fnorm{\bU^\top \bZ \bV -
\sum_{j=1}^p\sum_{i=1}^n
\frac{\partial}{\partial z_{ij} }\Bigl(\bU^\top \be_i \be_j^\top \bV \Bigr)
=~&\E\Big[I(\Omega) \fnorm{\bar\bU^\top \bZ\bar \bV -
\sum_{j=1}^p\sum_{i=1}^n
\frac{\partial}{\partial z_{ij} }\Bigl(\bar\bU^\top \be_i \be_j^\top \bar\bV \Bigr)
\le~&\E\Big[ \fnorm{\bar\bU^\top \bZ\bar \bV -
\sum_{j=1}^p\sum_{i=1}^n
\frac{\partial}{\partial z_{ij} }\Bigl(\bar\bU^\top \be_i \be_j^\top \bar\bV \Bigr)
\le~& \E \Big[\big(\fnorm{\bar\bU}^2 \fnorm{\bar\bV}^2\big) +
\fnorm{\bar\bV}^2\fnorm*{ \frac{\partial \bar\bU}{\partial z_{ij}} }^2
+ \fnorm{\bar\bU}^2\fnorm*{ \frac{\partial \bar\bV}{\partial z_{ij}}
\le~& K^2 + 2\E \Big[
\sum_{ij}\Big( K^2\fnorm*{\frac{\partial \bar\bU}{\partial z_{ij}} }^2 + \fnorm*{\frac{\partial \bar\bV}{\partial z_{ij}} }^2\Big) \Big]\\
=~& K^2 + 2\E \Big[I(\Omega)
\sum_{ij}\Big( K^2\fnorm*{\frac{\partial \bar\bU}{\partial z_{ij}} }^2 + \fnorm*{\frac{\partial \bar\bV}{\partial z_{ij}} }^2\Big) + I(\Omega^c)
\sum_{ij}\Big( K^2\fnorm*{\frac{\partial \bar\bU}{\partial z_{ij}} }^2 + \fnorm*{\frac{\partial \bar\bV}{\partial z_{ij}} }^2\Big) \Big] \\
=~& K^2 + 2\E \Big[I(\Omega)
\sum_{ij}\Big( K^2\fnorm*{\frac{\partial \bU}{\partial z_{ij}} }^2 + \fnorm*{\frac{\partial \bV}{\partial z_{ij}} }^2\Big) + I(\Omega^c)
\sum_{ij}\Big( K^2\fnorm*{\frac{\partial \bar\bU}{\partial z_{ij}} }^2 + \fnorm*{\frac{\partial \bar\bV}{\partial z_{ij}} }^2\Big) \Big] \\
\le~& K^2 + 2\E \Big[I(\Omega)
\sum_{ij}\Big( K^2\fnorm*{\frac{\partial \bU}{\partial z_{ij}} }^2 + \fnorm*{\frac{\partial \bV}{\partial z_{ij}} }^2\Big) \Big]
+ 2C( K^2 L_1^2  + L_2^2 ),
\end{align*}

where the last inequality uses $\sum_{ij}\fnorm{\frac{\partial \bar\bU}{\partial z_{ij}} }^2\le nT (n^{-1/2}L_1)^2 = TL_1^2$, $\sum_{ij}\fnorm{\frac{\partial \bar\bV}{\partial z_{ij}} }^2\le TL_2^2$ by Lipschitz properties of $\bar\bU$, $\bar\bV$, and $P(\Omega^c)\le C/T$.

For each $j\in[p]$, let $\E_j (\cdot) $ denote the conditional expectation $\E[\cdot |\{\bZ e_k, k\ne j\}]$. The left-hand side of the desired inequality can be rewritten as

\begin{align*}
\fnorm[\big]{
p \bU^\top \bV  - \sum_{j=1}^p
(\E_j\bU^\top \bZ - \bL^\top) \be_j
\be_j^\top (\bZ^\top \E_j\bV - \hat \bL)
\Big]
\end{align*}

with $\bL\in\R^{p\times T}$ defined by
$\bL^\top \be_j = \E_j\bU^\top \bZ \be_j - \bU^\top \bZ \be_j + \sum_{i=1}^n \partial_{ij} \bU^\top \be_i$
and $\hat \bL$ defined similarly with $\bU$ replaced by $\bV $.
We develop the terms in the sum over $j$ as follows:

\begin{align}
\nonumber
&p \bU^\top \bV  - \sum_j
(\E_j\bU^\top \bZ - \bL^\top) \be_j
\be_j^\top (\bZ^\top \E_j\bV - \hat \bL)
\\=\quad&
\sum_j
\Bigl(\bU^\top \bV  -  \E_j[\bU^\top]\E_j\bV\Bigr)
\label{eq:first-term-swap-trick}
\\&  + \sum_j
\Bigl(
\E_j[\bU^\top]\E_j\bV
\E_j\bU^\top \bZ \be_j \be_j^\top \bZ^\top\E_j\bV
\Bigr)
\label{eq:difficult-term}
\\ & - \bL^\top \hat \bL \label{eq:Xi-hat-Xi-term}
\\ & + \sum_j \Bigl(\E_j\bU^\top \bZ \be_j \be_j^\top \hat \bL\Bigr)
\Bigl(\bL^\top \be_j \be_j^\top \bZ^\top \E_j[\bU]\Bigr).
\label{eq:last-two-terms}
\end{align}

First, for (<ref>),
by the Cauchy-Schwarz inequality
$\E\bigl[ \|\bL^\top \hat \bL\|_{\rm F}\bigr]
\le \E\bigl[\|\bL\|_{\rm F}^2\bigr]^{\frac 12}\E\bigl[\|\hat\bL\|_{\rm F}^2\bigr]^{\frac 12}$.
For a fixed $j\in[p]$ and $t\in [T]$,

\begin{align*}
\E[(\be_j^\top \bL e_t)^2]
\sum_{i=1}^n
\E[(\be_i^\top(\E_j[\bU] - \bU)e_t )^2]
\E
\sum_{i=1}^n
\sum_{l=1}^n
\Bigl(
\frac{\be_i^\top \partial \bU e_t }{\partial z_{lj}}
\Bigr)^2\\
\E
\sum_{i=1}^n
\sum_{l=1}^n
\Bigl(
\frac{\be_i^\top \partial \bU e_t }{\partial z_{lj}}
\Bigr)^2,
\end{align*}

where the two inequalities are due to the second-order stein inequality in <Ref>, and Gaussian-Poincaré inequality in <Ref>, respectively.
Summing over $j\in[p]$ and $t\in[T]$ we obtain
$\E[\|\bL\|_{\rm F}^2] \le 2 \E \sum_{lj} \| \partial_{lj} \bU \|_{\rm F}^2
= 2 \|\bU\|_\partial^2$.
Combined with the same bound for $\hat \bL$, we obtain
$\E[\|\eqref{eq:Xi-hat-Xi-term}\|_{\rm F}^2]
\le
2 \|\bU\|_\partial\|\bV \|_\partial$.
We now turn to the two terms in (<ref>).
By the triangle inequality for the Frobenius norm,

\begin{align*}
\E\Bigl[\|\sum_j \E_j\bU^\top \bZ \be_j \be_j^\top \hat \bL\|_{\rm F}\Bigr]
\sum_j \E\Bigl[\|\E_j\bU^\top \bZ \be_j\|_2 \|\be_j^\top \hat \bL\|_2\Bigr]
\\ &\le \E[
\sum_j \| \E_j\bU^\top \bZ \be_j\|_2^2
]^{\frac 12}
\E[ \sum_j \|\be_j^\top \hat \bL\|_2^2]^{\frac 12}
\\&\le (p \E[\fnorm{\bU}^2])^{\frac 12}
\E[ \|\hat \bL\|_{\rm F}^2]^{\frac 12}
\end{align*}

where we used that $\| \ba \bb^\top \|_{\rm F} = \|\ba\|_2\|\bb\|_2$ for two vectors $\ba,\bb$,
the Cauchy-Schwarz inequality, $\E[\|\bA \bz_j\|_2^2|\bA] = \|\bA\|_{\rm F}^2$
if matrix $\bA$ is independent of $\bz_j\sim \calN(0,I_n)$ (set $\bz_j=\bZ \be_j$),
and Jensen's inequality.

Next, we decompose (<ref>) as
$\sum_j \bU^\top(\bV  - \E_j\bV )
\sum_j(\bU - \E_j\bU)^\top\E_j\bV
We have by the submultiplicativity of the Frobenius norm
and the Cauchy-Schwarz inequality

\begin{align*}
\E[\|\bU^\top\sum_{j}(\bV  - \E_j\bV)\|_{\rm F}]
& \le \E[\sum_j\fnorm{\bU} \| \bV  - \E_j\bV \|_{\rm F}]
\\&\le \E[p \fnorm{\bU}^2]^{\frac 12}
\E[\sum_j\| \bV  - \E_j\bV \|_{\rm F}^2]^{\frac 12}.
\end{align*}

By the Gaussian Poincaré inequality applied $p$ times,
$\E[\sum_j\| \bV  - \E_j\bV \|_{\rm F}^2] \le \|\bV  \|_{\partial}^2$,
so that the previous display is bounded from above by $\sqrt{p}\|\bV  \|_{\partial}$.
Similarly, $\E[\|\sum_j(\E_j[\bU] - \bU)^\top \E_j\bV \|_{\rm F}]
\le \sqrt{p} \| \bU \|_{\partial}$
and $\E[\|\eqref{eq:first-term-swap-trick}\|_{\rm F}] \le \sqrt p (\|\bU\|_{\partial}
+ \|\bV \|_{\partial})$.

For the last remaining term, (<ref>),
we first use
$\E[\| \eqref{eq:difficult-term} \|_{\rm F}]
\le
\E[\| \eqref{eq:difficult-term} \|_{\rm F}^2]^{\frac 12}
$ by Jensen's inequality and now proceed to bound
$\| \eqref{eq:difficult-term} \|_{\rm F}^2$. We have

\| \eqref{eq:difficult-term} \|_{\rm F}^2
\|\sum_j \E_j\bU^\top \E_j\bV - \E_j\bU^\top X \be_j \be_j^\top \bZ^\top \E_j\bV\|_{\rm F}^2
\sum_{j,k} \trace[\bM_j^\top \bM_k],

where $\bM_j = \E_j\bU^\top \E_j\bV - \E_j\bU^\top \bZ \be_j \be_j^\top \bZ^\top \E_j\bV$.
We first bound $\sum_j \|\bM_j\|_{\rm F}^2$.
Since the variance of $\ba^\top\bb - \ba^\top \bg \bg^\top \bb$ for standard normal $\bg\sim \calN(0,I_p)$
is $2\|(\ba \bb^\top + \bb \ba^\top)/2\|_{\rm F}^2\le 2 \|\ba\|_2^2 \|\bb\|_2^2$,
applying this variance bound on each pair of coordinates $(t,t')\in[T]\times [T]$
gives $\sum_j \|\bM_j\|_{\rm F}^2 \le
\sum_j 2 \|\E_j[\bU]\|_{\rm F}^2\|\E_j\bV\|_{\rm F}^2
\le 2p$.

We now bound $\sum_{j\ne k} \trace[\bM_j^\top\bM_k]$.
Setting $\bz_j = \bZ \be_j \sim \calN(0,I_n)$ for every $j\in[p]$,
we will use many times
the identity

\begin{equation}
\label{stein}
\E[(\bz_j^\top f(\bZ)  - \sum_i \partial_{ij} f(\bZ)^\top \be_i )g(\bZ)
\E[
\sum_{i}
f(\bZ)^\top \be_i \partial_{ij} g(\bZ)
\end{equation}

which follows from Stein's formula
for $f:\R^{n\times p}\to \R^n$ and $g:\R^{n\times p}\to \R$.
With $f^{tt'}(\bZ) =(\bz_j^\top \E_j[\bU] \be_{t'}) \E_j\bV e_t$
and $g^{tt'}(\bZ) = \be_{t'}^\top \bM_k e_t$,
we find

\begin{align*}
&\E\trace[\bM_j^\top \bM_k]
=\E\trace [\bM_j^\top \sum_t \be_{t'} \be_{t'}^\top \bM_k]
=\E[\sum_{tt'} e_t^\top  \bM_j^\top \be_{t'} \be_{t'}^\top \bM_k e_t]
\\&=\E
\sum_{tt'}
\Bigl(\bz_j^\top f^{tt'}(\bZ) - \sum_i \be_i^\top \partial_{ij} f^{tt'}(\bZ)\Bigr)
\\&
\sum_{tt'}\sum_i
\be_i^\top f^{tt'}(\bZ) \partial_{ij} g^{tt'}(\bZ).
\end{align*}

where $g_{tt'}(\bZ) =
e_t^\top \E_k \bV  ^\top \E_k \bV  e_t'
e_t^\top \E_k   \bU^\top \bz_k \bz_k^\top \E_k \bV  e_t'
)$ and

$$\partial_{ij} g_{tt'} =
\partial_{ij}
\be_{t'}^\top \bM_k e_t
\be_{t'}^\top \partial_{ij}[\E_k \bU ^\top \E_k \bV ] e_t
\bz_k^\top\partial_{ij}[\E_k \bU \be_{t'} \be_{t}^\top \E_k  \bU^\top] \bz_k

Now define   $\tilde f^{tt'}(\bZ) =
\partial_{ij}[\E_k \bU \be_{t'} \be_{t}^\top \E_k   \bU^\top] \bz_k$
and $\tilde g^{tt'}(\bZ) = \sum_i \be_i^\top f^{tt'}(\bZ)$.
By definition of $\tilde f^{tt'}(\bZ)$,
the previous display is equal to
$\bz_k^\top\tilde f^{tt'}(\bZ) - \sum_l \partial_{lk} \be_l^\top\tilde f^{tt'}(\bZ)$.
We apply (<ref>) again with respect to $\bz_k$, so that

\begin{align*}
\E \trace[\bM_j^\top \bM_k]
\sum_{il, tt'}
\be_i^\top \partial_{lk}[f^{tt'}(\bZ)]
\be_l^\top \tilde f^{tt'}(\bZ)
\\&=
\sum_{il, tt'}
\Bigl(\be_i^\top \partial_{lk}\Bigl[\E_j\bV e_t \be_{t'}^\top \E_j\bU^\top\Bigr]\bz_j\Bigr)
\Bigl(\be_l^\top \partial_{ij}\Bigl[\E_k[\bU]      \be_{t'} e_t^\top \E_k[\bV ]^\top\Bigr] \bz_k\Bigr).
\end{align*}

To remove the indices $t,t'$, we rewrite the above using $\sum_t \be_t \be_t^\top=I_T$
and $\sum_{t'} \be_{t'} \be_{t'}^\top = I_T$ so that it equals

\E\sum_{il}\trace
\Bigl\{
\partial_{lk}\Bigl[
\E_j\bU^\top \bz_j \be_i^\top \E_j\bV
\Bigr]
\partial_{ij}\Bigl[
\E_k[\bV ]^\top \bz_k \be_l^\top \E_k[\bU]
\Bigr]
\Bigr\}

Summing over $j,k$,
using $\trace [\bA^\top \bB ] \le \|\bA\|_{\rm F}\|\bB\|_{\rm F})$
and the Cauchy-Schwarz inequality,
the above is bounded from above by

\begin{align*}
\Bigl\{
\E\sum_{jk, il}
\Big\|
\partial_{lk}
\Bigl[
\E_j
\bU^\top
\bz_j
\be_i^\top
\E_j
[\bV ]
\Bigr]
\Big\|_{\rm F}^2
\Bigl\}^{\frac 12}
\Bigl\{
\E
\sum_{jk, il}
\Big\|
\partial_{ij}\Bigl[
\E_k[\bV ]^\top \bz_k
\be_l^\top
\E_k[\bU]
\Bigr]
\Big\|_{\rm F}^2
\Bigr\}^{\frac 12}.
\end{align*}

At this point the two factors are symmetric,
with $(\bV , \bU)$ in the left factor
replaced with $(\bU,\bV )$ on the right factor.
We focus on the left factor; similar bound will
apply to the right one by exchanging the roles of $\bV $ and $\bU$.
If $\bz_j$ is independent of matrices $A^{(q)}$
$\E_j[\|\sum_{q=1}^n (\be_q^\top\bz_j) A^{(q)}\|_{\rm F}^2
\sum_{q=1}^n
\|A^{(q)}\|_{\rm F}^2
$ so that
with $A^{(q)} =\partial_{lk}[\E_j \bU^\top \be_q \be_i^\top \E_j \bU]$, the
first factor in the above display is equal to

\begin{align*}
\E\sum_{jk, ilq}
\Big\|
\partial_{lk}
\Bigl(
\E_j
\bU^\top
\be_q
\be_i^\top
\E_j\bV
\Bigr)
\Big\|_{\rm F}^2
\Bigl\}^{\frac 12}
\\&\stackrel{\text{(i)}}{=}
\Bigl\{
\E\sum_{jk, ilq}
\Big\|
\partial_{lk}
\Bigl(
\E_j
\bU^\top
\Bigr)
\be_q
\be_i^\top
\E_j
[\bV ]
\E_j
[\bU] ^\top
\be_q
\be_i^\top
\partial_{lk}
\Bigl(
\E_j
[\bV ]
\Bigr)
\Big\|_{\rm F}^2
\Bigl\}^{\frac 12}
\\&\stackrel{\text{(ii)}}{\le}
\Bigl\{
\E\sum_{jk, ilq}
\Big\|
\partial_{lk}
\Bigl(
\E_j
\Bigr)
\be_q
\be_i^\top
\E_j
[\bV ]
\Big\|_{\rm F}^2
\Bigl\}^{\frac 12}
\Bigl\{
\E\sum_{jk, ilq}
\Big\|
\E_j
\bU^\top
\be_q
\be_i^\top
\partial_{lk}
\Bigl(
\E_j
[\bV ]
\Bigr)
\Big\|_{\rm F}^2
\Bigl\}^{\frac 12}
\\&\stackrel{\text{(iii)}}{=}
\Bigl\{
\E\sum_{jk, ilq}
\Big\|
\E_j [ \partial_{lk} \bU]^\top
\be_q
\Big\|_2^2
\Big\|
\be_i^\top
\E_j\bV
\Big\|_2^2
\Bigl\}^{\frac 12}
\Bigl\{
\E\sum_{jk, ilq}
\Big\|
\E_j
\bU^\top
\be_q
\Big\|_2^2
\Big\|
\be_i^\top
\E_j
[ \partial_{lk} \bV ]
\Big\|_2^2
\Bigl\}^{\frac 12}
\\&\stackrel{\text{(iv)}}{=}
\Bigl\{
\E\sum_{jk, l}
\Big\|
\E_j [ \partial_{lk} \bU]^\top
\Big\|_{\rm F}^2
\Big\|
\E_j\bV
\Big\|_{\rm F}^2
\Bigl\}^{\frac 12}
\Bigl\{
\E\sum_{jk, l}
\Big\|
\E_j
\bU^\top
\Big\|_{\rm F}^2
\Big\|
\E_j
[ \partial_{lk} \bV ]
\Big\|_{\rm F}^2
\Bigl\}^{\frac 12}
\end{align*}

where (i) is the chain rule,
(ii) the triangle inequality,
(iii) holds
provided that the order of the derivation $\partial_{lk}$
and the expectation sign $\E_j$ can be switched and using
$\|\ba \bb^\top\|_{\rm F}^2 = \|\ba\|_2^2 \|\bb\|_2^2$ for vectors $\ba, \bb$,
and (iv) holds using
$\sum_i \|\bA \be_i\|_2^2 = \|\bA\|_{\rm F}^2 = \sum_q \|\bA \be_q\|_2^2$ for a matrix $\bA$ with $n$ columns.
Finally, by Jensen's inequality, the above display is bounded by

\begin{equation*}
\Bigl\{
\E\sum_{k, l}
\Big\|
\partial_{lk}
\bU
\Big\|_{\rm F}^2
\sum_j
\Big\|
\E_j
[\bV ]
\Big\|_{\rm F}^2
\Bigl\}^{\frac 12}
\Bigl\{
\E\sum_{k, l}
\Big\|
\partial_{lk}
\bV
\Big\|_{\rm F}^2
\sum_j
\Big\|
\E_j
\Big\|_{\rm F}^2
\Bigl\}^{\frac 12}
\end{equation*}

Since $\fnorm{\bU}\vee \|\bV \|_{\rm F} \le 1$ almost surely,
the previous display is bounded by
$\sqrt{p} (\|\bU\|_\partial + \|\bV \|_{\partial})$.
In summary,

\E[
\|\eqref{eq:difficult-term}
\|_{\rm F}^2
]^{\frac 12}
\le(2p + [2\sqrt{p}(\|\bU\|_{\partial} + \|\bV \|_{\partial})]^2)^{\frac 12}
\le \sqrt{2p} + 2\sqrt{p}(\|\bU\|_{\partial} + \|\bV \|_{\partial}).

Combining the bounds on the terms
with the triangle inequality completes the proof.

Note that $\bQ_1 = \frac{\bE^\top\bF/\tD  - \bS(n\bI_T - \hbA)/\tD}{\fnorm{\bS^{\frac 12}}}$, where $\tD = \big(\fnorm*{\bF}^2 + n \trace(\bS)\big)^{\frac 12}$ is defined in <Ref>.
Now, Apply <Ref>, we obtain

\begin{align*}
\E \| \bQ_1\|^2_{\rm F}
= \E \big[\| \bE^\top \bF/\tD - \bS (n\bI_T - \hbA )/\tD\|_{\rm F}^2\big] \frac{1}{\trace(\bS)}
\le 4.
\end{align*}

Before proving Proposition <ref>], we state the following  <Ref>, whose proof is deferred to the end of this section.

We have

\begin{align*}
\sum_{ij}\bF^\top\bZ\be_j\be_i^\top\frac{\partial \bF}{\partial z_{ij}}
&= \bJ_1 - \bF^\top\bZ\bH (n\bI_T -\hbA),\\
\sum_{ij}\Big(\frac{\partial \bF}{\partial z_{ij} }\Big)^\top
\bZ \be_j \be_i^\top \bF
&= \bJ_2 -  \hbA \bF^\top\bF.
\end{align*}

where $\bJ_1 = \sum_{ij}\bF^\top\bZ\be_j \sum_t \Delta_{ij}^{it} \be_t^\top$ with $\fnorm*{\bJ_1}\le n^{\frac12} \fnorm*{\bF}^2$, and
$\bJ_2=\sum_{ijlt}\be_t D_{ij}^{lt}\be_l^\top\bZ\be_j \be_i^\top\bF$ with $\fnorm*{\bJ_2}\le n^{\frac 12}\opnorm*{\bZ}\fnorm{\bH} \fnorm*{\bF}$.

We first apply <Ref>. To be more specific, let $\bU = n^{-\frac 12}\bF/D$ and $\bV =n^{-\frac 12} \bZ^\top\bU$ with $D = (\fnorm*{\bF}^2/n + \fnorm{\bH}^2)^{\frac 12}$, then $\fnorm*{\bU}\le 1$, $\fnorm*{\bV}\le n^{-\frac12}\opnorm*{\bZ}$. <Ref> yields

\begin{align}\label{eq: UZV}
&\E \Big(\Big\|\bU^\top \bZ \bV -
\sum_{j=1}^p\sum_{i=1}^n
\frac{\partial}{\partial z_{ij} }\Bigl(\bU^\top \be_i \be_j^\top \bV \Bigr)\Big\|_{\rm F}^2\Big)\\
\le~ & \E [\fnorm*{\bU}^2 \fnorm*{\bV}^2]+ \E \sum_{ij}\Big[
2\fnorm*{\bV}^2\fnorm*{ \frac{\partial \bU}{\partial z_{ij}} }^2
+ 2\fnorm*{\bU}^2\fnorm*{ \frac{\partial \bV}{\partial z_{ij}} }^2\Big]\notag\\
\le~&\E \big(n^{-1}\opnorm*{\bZ}^2\big) +
2\E \Big(n^{-1}\opnorm*{\bZ}^2\sum_{ij}
\fnorm*{ \frac{\partial \bU}{\partial z_{ij}} }^2\Big)
+ 2 \E \Big(\sum_{ij} \fnorm*{ \frac{\partial \bV}{\partial z_{ij}} }^2 \Big)\\ \notag
\le~&\frac{4p}{n}
+  \E \big(\frac 1n\opnorm{\bZ}^2\big)
+ 6\E \Big(n^{-1}\opnorm*{\bZ}^2\sum_{ij}
\fnorm*{ \frac{\partial \bU}{\partial z_{ij}} }^2\Big),
\end{align}

where the last inequality uses the following bound derived using $\bV = n^{-\frac 12} \bZ^\top\bU$, and $\fnorm{\bU}\le1$,

\begin{align}
\sum_{ij}
\fnorm*{ \frac{\partial \bV}{\partial z_{ij}} }^2\notag
=~& \sum_{ij}
n^{-1}\fnorm*{\frac{\partial \bZ^\top}{\partial z_{ij}} \bU + \bZ^\top \frac{\partial \bU^\top}{\partial z_{ij}} }^2\notag\\
\le~& 2{n}^{-1} \Big(p\fnorm*{\bU}^2 + \opnorm*{\bZ}^2 \sum_{ij} \fnorm*{ \frac{\partial \bU}{\partial z_{ij}} }^2\Big)\notag\\
\le~& \frac{2p}{n} + 2n^{-1}\opnorm*{\bZ}^2 \sum_{ij} \fnorm*{ \frac{\partial \bU}{\partial z_{ij}} }^2. \label{eq: Vzij}
\end{align}

Note that $\E [(n^{-\frac12}\opnorm*{\bZ})^2] \le (1 + \sqrt{p/n})^2 + 1/n$ by (<ref>).
Now we establish the connection between $\bQ_2$ and the term inside Frobenius norm in (<ref>). By definitions of $\bU$ and $\bV$,

\begin{equation}\label{eq: uxv}
\bU^\top \bZ \bV = n^{-\frac 32}D^{-2}\bF^\top \bZ \bZ^\top\bF.
\end{equation}

Next, by product rule,

\begin{align*}
&\sum_{j=1}^p\sum_{i=1}^n\frac{\partial}{\partial z_{ij} }\Bigl(\bU^\top \be_i \be_j^\top \bV \Bigr)\\
=~& n^{-\frac 32}\sum_{j=1}^p\sum_{i=1}^n\frac{\partial}{\partial z_{ij} }\Bigl(\bF^\top \be_i \be_j^\top \bZ^\top\bF D^{-2} \Bigr)\\
=~& n^{-\frac 32}\sum_{j=1}^p\sum_{i=1}^n\Big( \underbrace{\frac{\partial \bF^\top}{\partial z_{ij}} \be_i \be_j^\top \bZ^\top\bF D^{-2}}_{(i)} +\underbrace{\bF^\top\be_i \be_j^\top\frac{\partial \bZ^\top\bF}{\partial z_{ij}}  D^{-2}}_{(ii)} +  \underbrace{\bF^\top\be_i \be_j^\top\bZ^\top\bF\frac{\partial D^{-2}}{\partial z_{ij}}}_{(iii)} \Big).
\end{align*}

We now rewrite the above three terms $(i), (ii)$ and $(iii)$.

(i) For term ($i$), by <Ref>,

\begin{align*}
&n^{-\frac 32}D^{-2}\sum_{j=1}^p\sum_{i=1}^n\frac{\partial \bF^\top}{\partial z_{ij}} \be_i \be_j^\top \bZ^\top\bF\\
=~& n^{-\frac 32}D^{-2} \bigl[ \bJ_1 - \bF^\top\bZ\bH(n\bI_T -\hbA) \bigr]^\top\\
=~& n^{-\frac 32}D^{-2} \bigl[ \bJ_1^\top - (n\bI_T -\hbA) \bH^\top \bZ^\top \bF \bigr].
\end{align*}

(ii) For term ($ii$), by product rule and <Ref>,

\begin{align*}
&n^{-\frac 32}D^{-2}\sum_{j=1}^p\sum_{i=1}^n \bF^\top\be_i \be_j^\top\frac{\partial \bZ^\top\bF}{\partial z_{ij}} \\
=~&n^{-\frac 32}D^{-2}\Big( p\bF^\top\bF +
\sum_{j=1}^p\sum_{i=1}^n \bF^\top\be_i \be_j^\top\bZ^\top\frac{\partial \bF}{\partial z_{ij}} \Big)\\
=~&n^{-\frac 32}D^{-2}\Big( p\bF^\top\bF +
(\bJ_2 - \hbA\bF^\top\bF)^\top \Big)\\
=~&n^{-\frac 32}D^{-2}\big( p\bF^\top\bF - \bF^\top\bF\hbA + \bJ_2^\top \big).
\end{align*}

(iii) For term ($iii$), by chain rule,

\begin{align*}
&n^{-\frac 32}\sum_{j=1}^p\sum_{i=1}^n \bF^\top\be_i \be_j^\top\bZ^\top\bF\frac{\partial D^{-2}}{\partial z_{ij}}\\
=~& -2n^{-\frac 32}D^{-3}\sum_{j=1}^p\sum_{i=1}^n \bF^\top\be_i \be_j^\top\bZ^\top\bF\frac{\partial D}{\partial z_{ij}}\\
=~& -n^{-\frac 32}D^{-2} \underbrace{\Big(2D^{-1}\sum_{j=1}^p\sum_{i=1}^n \bF^\top\be_i \be_j^\top\bZ^\top\bF\frac{\partial D}{\partial z_{ij}}\Big)}_{\bJ_3}
\end{align*}

Combining (<ref>) and the above three expressions for (i)-(ii)-(iii),

\begin{align*}
&\bU^\top \bZ \bV - \sum_{j=1}^p\sum_{i=1}^n \frac{\partial}{\partial z_{ij} }\Bigl(\bU^\top \be_i \be_j^\top \bV \Bigr)\\
=~& n^{-\frac 32}D^{-2}\Bigl[\bF^\top \bZ \bZ^\top\bF  + (n\bI_T -\hbA) \bH^\top \bZ^\top \bF - p\bF^\top\bF + \bF^\top\bF\hbA\\
& -\bJ_1^\top - \bJ_2^\top + \bJ_3\Bigr] \\
=~& \bQ_2 - n^{-\frac 32}D^{-2} (\bJ_1^\top + \bJ_2^\top - \bJ_3).
\end{align*}

That is,

\begin{equation}\label{eq: V6=A + B}
\bQ_2 =  \bU^\top \bZ \bV - \sum_{j=1}^p\sum_{i=1}^n \frac{\partial}{\partial z_{ij} }\Bigl(\bU^\top \be_i \be_j^\top \bV \Bigr) + n^{-\frac 32}D^{-2} (\bJ_1^\top + \bJ_2^\top - \bJ_3).
\end{equation}

Note that <Ref> implies that

\begin{align}\label{eq: steinX_4}
n^{-\frac 32}D^{-2} \fnorm*{\bJ_1} \le \frac{\fnorm*{\bF}^2/n}{\fnorm*{\bF}^2/n + \fnorm{\bH}^2} \le 1,
\end{align}

\begin{align}\label{eq: steinX_3}
n^{-\frac 32}D^{-2} \fnorm*{\bJ_2} \le \Big[n^{-\frac 12} \opnorm*{\bZ} \frac{\fnorm*{\bF}n^{-\frac 12} \fnorm{\bH}}{\fnorm*{\bF}^2/n + \fnorm{\bH}^2} \Big]
\le \frac12 (n^{-\frac 12} \opnorm*{\bZ}).
\end{align}

Since $\bJ_3=2D^{-1}\sum_{j=1}^p\sum_{i=1}^n \bF^\top\be_i \be_j^\top\bZ^\top\bF\frac{\partial D}{\partial z_{ij}}$, by Cauchy-Schwarz inequality

\begin{align*}
n^{-3} D^{-4}\fnorm{\bJ_3}^2 =
&\sum_{t,t'} ([\bJ_3]_{t,t'})^2\\
&= 4 n^{-3} D^{-6}\sum_{t,t'} \big[\sum_{i,j} \be_t^\top\bF^\top\be_i \be_j^\top\bZ^\top\bF\be_{t'}\frac{\partial D}{\partial z_{ij}}\big]^2\\
&\le 4 n^{-3} D^{-6} \sum_{t,t'}  \Big[\sum_{i,j} [\be_t^\top\bF^\top\be_i \be_j^\top\bZ^\top\bF\be_{t'}]^2 \sum_{i,j}\big(\frac{\partial D}{\partial z_{ij}}\big)^2\Big]\\
&= 4 n^{-3} D^{-6}
\fnorm{\bF}^2
\fnorm{\bZ^\top\bF}^2  \sum_{i,j}\big(\frac{\partial D}{\partial z_{ij}}\big)^2\\
&= 4 n^{-3} D^{-6}
\fnorm{\bF}^4 \opnorm{\bZ}^2 \sum_{i,j}\big(\frac{\partial D}{\partial z_{ij}}\big)^2.
\end{align*}

(1) Under <Ref>(i) that $\tau>0$. By <Ref>, we have $\sum_{i,j}\big(\frac{\partial D}{\partial z_{ij}}\big)^2\le C(\tau')n^{-1}D^2$. Then,

\begin{equation}\label{eq: steinX_5}
n^{-3} D^{-4}\fnorm{\bJ_3}^2
\le C(\tau')n^{-4}D^2 \fnorm{\bF}^4 \opnorm{\bZ}^2
\le C(\tau')n^{-2} \opnorm{\bZ}^2,
\end{equation}

by $\fnorm*{\bF}^2/(nD^2) \le 1$.

By (<ref>) and triangular inequality, we have

\begin{equation}\label{eq: steinX_2}
\begin{aligned}
\E[\fnorm{\bQ_2}^2]
\le~& 2 \E \Big[ \fnorm{\bU^\top \bZ \bV - \sum_{j=1}^p\sum_{i=1}^n \frac{\partial}{\partial z_{ij} }\Bigl(\bU^\top \be_i \be_j^\top \bV \Bigr)}^2 \Big]\\
&+ 2\E [\fnorm{n^{-\frac 32}D^{-2} (\bJ_1^\top + \bJ_2^\top- \bJ_3)}^2].
\end{aligned}
\end{equation}

By (<ref>), (<ref>), (<ref>), the second term in (<ref>) can be upper bounded by

\begin{equation}
6\E [( 1 + \frac 14 n^{-1}\opnorm{\bZ}^2) + C(\tau') n^{-2}\opnorm{\bZ}^2) ] \le C(\tau') (1 + \frac pn),
\end{equation}

where the last inequality uses (<ref>).

For the first term in (<ref>), since $\sum_{ij} \fnorm{ \frac{\partial \bU}{\partial z_{ij}} }^2 \le C(\tau') (T\wedge \frac pn)$ by <Ref>, we have

\begin{align*}%\label{eq: steinX_1}
&\E \Big(\Big\|\bU^\top \bZ \bV -
\sum_{j=1}^p\sum_{i=1}^n
\frac{\partial}{\partial z_{ij} }\Bigl(\bU^\top \be_i \be_j^\top \bV \Bigr)\Big\|_{\rm F}^2\Big) \\
\le~&\frac{4p}{n}
+  \E \big(n^{-1}\opnorm{\bZ}^2\big)
+ 6\E \Big(n^{-1}\opnorm{\bZ}^2\sum_{ij}
\fnorm*{ \frac{\partial \bU}{\partial z_{ij}} }^2\Big)\\
\le~& \frac{4p}{n}  + [1 + C(\tau') (T\wedge \frac pn) ]\E \big(n^{-1}\opnorm{\bZ}^2\big) \\
\le~& \frac{4p}{n}  + [1 + C(\tau') (T\wedge \frac pn) ] [(1 + \sqrt{p/n})^2 + 1/n]\\
\le~& C(\tau')(T \wedge (1 + \frac pn))(1 + \frac pn).
\end{align*}

Therefore, under <Ref>(i), we obtain

\begin{align*}
\E [\fnorm{\bQ_2}^2] \le C(\tau')(T \wedge (1 + \frac pn))(1 + \frac pn).
\end{align*}

(2) Under <Ref>(ii), let $\Omega = U_1\cap U_2\cap U_3$, then we have
$\P(\Omega^c) \le C(\gamma, c) \frac 1T$ by  (<ref>).
<Ref>, on $\Omega$, we have
(i) The map $\bZ\mapsto \bU$ is $n^{-1/2} L_1$-Lipschitz, where $L_1=8\max(1, (2\eta)^{-1})$, and $\fnorm{\bU}\le 1$,
(ii) The map $\bZ \mapsto \bV$ is $n^{-1/2} L_2$-Lipschitz, where $L_2 = (1 + (2 +\sqrt{p/n})L_1)$, and $\fnorm{\bV}\le (2 + \sqrt{p/n})$.
Applying <Ref> with $K = 2 + \sqrt{p/n}$ yields

\begin{align*}
&\E\Big[I(\Omega) \fnorm{\bU^\top \bZ \bV -
\sum_{j=1}^p\sum_{i=1}^n
\frac{\partial}{\partial z_{ij} }\Bigl(\bU^\top \be_i \be_j^\top \bV \Bigr)
\le~& K^2 + C(\gamma, c)( K^2 L_1^2  + L_2^2 ) + 2\E \Big[I(\Omega)
\sum_{ij}\Big( K^2\fnorm*{\frac{\partial \bU}{\partial z_{ij}} }^2 + \fnorm*{\frac{\partial \bV}{\partial z_{ij}} }^2\Big) \Big]\\
\le~& C(\gamma, c),
\end{align*}

where the last inequality holds because $K \le C(\gamma), L_1 = C(\gamma, c), L_2 = C(\gamma, c)$, and on $\Omega$, $\sum_{ij} \fnorm{\frac{\partial \bU}{\partial z_{ij}} }^2\le C(\gamma, c)$ from <Ref>, and $\E [\fnorm{\frac{\partial \bV}{\partial z_{ij}} }^2 ]\le C(\gamma, c)$ by product rule. Therefore, under <Ref>(i), we obtain

\begin{align*}
\E [I(\Omega)\fnorm{\bQ_2}^2]
\le~& 2 \E \Big[ \fnorm{\bU^\top \bZ \bV - \sum_{j=1}^p\sum_{i=1}^n \frac{\partial}{\partial z_{ij} }\Bigl(\bU^\top \be_i \be_j^\top \bV \Bigr)}^2 \Big]\\
&+ 2\E [\fnorm{n^{-\frac 32}D^{-2} (\bJ_1^\top + \bJ_2^\top- \bJ_3)}^2]\\
\le~& C(\gamma, c),
\end{align*}

where the last inequality used (<ref>), (<ref>), and that
$n^{-3} \E[D^{-4}\fnorm{\bJ_3}^2] \le  C(\gamma, c)$ in analogy to  (<ref>).

We will apply <Ref>.
Let $\bU = \bV = n^{-\frac12} D^{-1} \bF$ with $D = (\fnorm*{\bF}^2/n + \fnorm{\bH}^2)^{\frac 12}$, then $\fnorm*{\bU}=\fnorm*{\bV}\le 1$. Let $\bW_0 = p \bU^\top \bU - \sum_{j=1}^p \big(\sum_{i=1}^n \frac{\partial \bU^\top\be_i}{\partial z_{ij}} - \bU^\top \bZ\be_j\big) \big(\sum_{i=1}^n \frac{\partial \bU^\top\be_i}{\partial z_{ij}} - \bU^\top \bZ\be_j\big)^\top$,
then <Ref> gives

\begin{align*}
\E [\fnorm*{\bW_0}] &\le 2 \|\bU\|_\partial \|\bV\|_\partial
+ \sqrt p \bigl( \sqrt 2 + (3+\sqrt{2})(\|\bU\|_{\partial} + \|\bV\|_{\partial}) \bigr) \\
&= 2\|\bU\|_\partial^2 + \sqrt{p} (\sqrt 2 + 2(3+\sqrt{2}) \|\bU\|_{\partial}).
\end{align*}

We will prove under <Ref>(i), and the proof under <Ref>(ii) on set $\Omega = U_1\cap U_2\cap U_3$ follows from almost similar arguments with $\tau'$ replaced by $\eta$, which is a constant that depends only on $\gamma, c$.

By definition of $\norm*{\bU}_{\partial}$ and <Ref>,
$\norm{\bU}_{\partial}^2= \sum_{ij}\E  \fnorm{\frac{\partial \bU}{\partial z_{ij}} }^2 \le C(\gamma, \tau')$. Thus $\E [\fnorm*{\bW_0}] \le C(\tau', \gamma) \sqrt{p}$.

Now we establish the connection between $\bW_0$ and $\bQ_3$. Since $\bU = n^{-\frac12} D^{-1} \bF$, by product rule,

\begin{align*}
&\sum_{i=1}^n \frac{\partial \bU^\top\be_i}{\partial z_{ij}} - \bU^\top \bZ\be_j \\
=~& n^{-\frac12} \Big(\sum_{i=1}^n \frac{\partial D^{-1}\bF^\top \be_i}{\partial z_{ij}} - D^{-1}\bF^\top \bZ\be_j\Big)\\
=~& n^{-\frac12}\Big(\sum_{i=1}^n D^{-1}\frac{\partial \bF^\top \be_i}{\partial z_{ij}} - D^{-1}\bF^\top \bZ\be_j\Big) + n^{-\frac12}\sum_{i=1}^n \bF^\top \be_i\frac{\partial D^{-1}}{\partial z_{ij}} \\
=~& n^{-\frac12}D^{-1}\Big(\sum_{i=1}^n \frac{\partial \bF^\top \be_i}{\partial z_{ij}} - \bF^\top \bZ\be_j \Big) - n^{-\frac12}D^{-2}\sum_{i=1}^n\bF^\top \be_i\frac{\partial D}{\partial z_{ij}}.
\end{align*}

For the first term in the last display, we have by <Ref>

\begin{align*}
\sum_{i=1}^n \frac{\partial \be_i^\top \bF}{\partial z_{ij}} = \sum_{i=1}^n \sum_{t=1}^T \frac{\partial \be_i^\top \bF\be_t}{\partial z_{ij}}\be_t^\top = \sum_{it} (D_{ij}^{it} + \Delta_{ij}^{it})\be_t^\top = -\be_j^\top \bH (n\bI_T - \hbA) +
\sum_{it} \Delta_{ij}^{it}\be_t^\top.
\end{align*}

\begin{align}
& \sum_{i=1}^n \frac{\partial \bU^\top\be_i}{\partial z_{ij}} - \bU^\top \bZ\be_j \nonumber\\
=~& n^{-\frac12}D^{-1}\big[ -(n\bI_T - \hbA) \bH^\top \be_j + \sum_{it} \Delta_{ij}^{it}\be_t - \bF^\top\bZ\be_j \big] - n^{-\frac12}D^{-2}\sum_{i=1}^n\bF^\top \be_i\frac{\partial D}{\partial z_{ij}}\nonumber\\
=~& -n^{-\frac12}D^{-1}\big[ (n\bI_T - \hbA) \bH^\top + \bF^\top\bZ\big]\be_j + n^{-\frac12}D^{-1} \sum_{it} \Delta_{ij}^{it}\be_t - n^{-\frac12}D^{-2}\sum_{i=1}^n\bF^\top \be_i\frac{\partial D}{\partial z_{ij}}.\label{eq: V8_proof}
\end{align}

Let $\bW_1 = -n^{-\frac12}D^{-1}\big[ (n\bI_T - \hbA) \bH^\top + \bF^\top\bZ\big]$ be the first term in (<ref>).
For the second term in (<ref>), recall
$\Delta_{ij}^{lt} = -(\be_t^\top \otimes \be_l^\top)(\bI_T\otimes \bX)
\bM^\dagger (\bI_T\otimes \bSigma^{\frac12})
\bigl(\bF^\top \otimes \bI_{p}\bigr)(\be_i \otimes\be_j)$ in <Ref>,

\begin{align*}
& n^{-\frac12}D^{-1} \sum_{it} \Delta_{ij}^{it}\be_t \\
=~& - n^{-\frac12}D^{-1}\sum_{it} \be_t (\be_t^\top \otimes \be_i^\top)
(\bI_T\otimes \bX)
\bM^\dagger (\bI_T\otimes \bSigma^{\frac12}) \bigl(\bF^\top\be_i \otimes \be_j \bigr)\\
=~& - n^{-\frac12}D^{-1}\sum_{i} (\bI_T \otimes \be_i^\top\bX)
\bM^\dagger (\bI_T\otimes \bSigma^{\frac12}) \bigl(\bF^\top\be_i \otimes  \bI_p\bigr)\be_j\\
=~& -n^{-\frac12}D^{-1} \sum_{i=1}^n(\bI_T \otimes \be_i^\top\bX) \bM^\dagger (\bI_T\otimes \bSigma^{\frac12}) (\bF^\top \be_i\otimes \bI_p) \be_j\\
=~& \bW_2 \be_j,
\end{align*}

where $\bW_2 =  -n^{-\frac12}D^{-1} \sum_{i=1}^n(\bI_T \otimes \be_i^\top\bX) \bM^\dagger (\bI_T\otimes \bSigma^{\frac12}) (\bF^\top \be_i\otimes \bI_p)$.

For the third term in (<ref>),

$$- n^{-\frac12}D^{-2}\bF^\top\sum_{i=1}^n \be_i\frac{\partial D}{\partial z_{ij}} = \bW_3 \be_j,$$

where $\bW_3 = - n^{-\frac12}D^{-2} \bF^\top \frac{\partial D}{\partial \bZ}$, here we slightly abuse the notation and let $\frac{\partial D}{\partial \bZ}$ denote the $n\times p$ matrix with $(i,j)$-th entry being $\frac{\partial D}{\partial z_{ij}}$.
Therefore, (<ref>) can be simplified as

\begin{align*}
\sum_{i=1}^n \frac{\partial \bU^\top\be_i}{\partial z_{ij}} - \bU^\top \bZ\be_j = \big[\bW_1 + \bW_2 + \bW_3 \big]\be_j.
\end{align*}

\begin{align*}
\bW_0 &= p \bU^\top \bU - \sum_{j=1}^p\Big(\sum_{i=1}^n \frac{\partial \bU^\top\be_i}{\partial z_{ij}} - \bU^\top \bZ\be_j\Big) \Big(\sum_{i=1}^n \frac{\partial \bU^\top\be_i}{\partial z_{ij}} - \bU^\top \bZ\be_j\Big)^\top\\
&= p \bU^\top \bU - [\bW_1 + \bW_2 + \bW_3][\bW_1 + \bW_2 + \bW_3]^\top\\
&= n^{\frac 12} \bQ_3 - \bW_1(\bW_2+\bW_3)^\top -(\bW_2+\bW_3)\bW_1^\top - (\bW_2+\bW_3) (\bW_2+\bW_3)^\top,
\end{align*}

where the last equality is due to

\begin{align*}
&p \bU^\top \bU - \bW_1\bW_1^\top \\
=~& n^{-1} D^{-2} \Big[p\bF^\top\bF -\bF^\top \bZ \bSigma^{-1}\bZ^\top \bF  - (n\bI_T -\hbA)\bH^\top\bH (n\bI_T -\hbA) \\
& - (n\bI_T -\hbA)\bH^\top\bZ^\top\bF - \bF^\top\bZ\bH(n\bI_T -\hbA)\Big]\\
=~& n^{\frac12} \bQ_3.
\end{align*}

\begin{align}\label{eq: V8-W}
\bQ_3 = n^{-\frac 12} \Big[ \bW_0 +
\bW_1(\bW_2+\bW_3)^\top + (\bW_2+\bW_3)\bW_1^\top + (\bW_2+\bW_3) (\bW_2+\bW_3)^\top
\Big].
\end{align}

We then bound the norms of $\bW_1, \bW_2, \bW_3$. For $\bW_1$,

\begin{align*}
\fnorm*{\bW_1} &= n^{-\frac12}D^{-1} \fnorm{(n\bI_T - \hbA) \bH^\top + \bF^\top\bZ} \\
&\le n^{\frac12} \big(D^{-1} \fnorm{\bH} + n^{-1}D^{-1} \fnorm*{\bF}\opnorm*{\bZ} \big)\\
&\le n^{\frac12} + n^{\frac12}  \big( D^{-1} \fnorm*{\bF}/\sqrt{n}\opnorm*{\bZ/\sqrt{n}}\big)\\
&\le n^{\frac12} ( 1 + \opnorm{\bZ/\sqrt{n}}),
\end{align*}

where we used $\opnorm*{\bI_T - \hbA/n}\le 1$ by <Ref> , $D^{-1} \fnorm{\bH} \le 1$, and $D^{-1} \fnorm*{\bF}/\sqrt{n} \le 1$.

For $\bW_2$,

\begin{align*}
\opnorm*{\bW_2}
&= n^{-\frac12} D^{-1}\opnorm[\Big]{\sum_{i=1}^n(\bI_T \otimes \be_i^\top\bX) \bM^\dagger (\bI_T\otimes \bSigma^{\frac12}) (\bF^\top \be_i\otimes \bI_p)}\\
&\le n^{-\frac12} D^{-1}\sum_{i=1}^n\opnorm{(\bI_T \otimes \be_i^\top\bX) \bM^\dagger (\bI_T\otimes \bSigma^{\frac12}) (\bF^\top \be_i\otimes \bI_p)}\\
&\le n^{-\frac12} D^{-1} \sum_{i=1}^n  \opnorm{(\bI_T \otimes \be_i^\top\bX) \bM^\dagger (\bI_T\otimes \bSigma^{\frac12}) }
\opnorm{(\bF^\top \be_i\otimes \bI_p)}\\
&\le n^{-\frac12} \opnorm{\bSigma}^{\frac12}D^{-1} \sum_{i=1}^n  \opnorm{(\bI_T \otimes \be_i^\top\bX) (\bI_T \otimes (\bX_{\hat{\mathscr{S}}}^\top\bX_{\hat{\mathscr{S}}} + n\tau\bP_{\hat{\mathscr{S}}})^{\dagger})}
\opnorm{(\bF^\top \be_i\otimes \bI_p)}\\
&= n^{-\frac12} \opnorm{\bSigma}^{\frac12}D^{-1} \sum_{i=1}^n  \opnorm{\bI_T \otimes [\be_i^\top\bX_{\hat{\mathscr{S}}}(\bX_{\hat{\mathscr{S}}}^\top\bX_{\hat{\mathscr{S}}} + n\tau\bP_{\hat{\mathscr{S}}})^{\dagger}]}
\norm{\bF^\top \be_i}\\
&\le n^{-\frac12} \opnorm{\bSigma}^{\frac12}D^{-1} \opnorm{\bX_{\hat{\mathscr{S}}}(\bX_{\hat{\mathscr{S}}}^\top\bX_{\hat{\mathscr{S}}} + n\tau\bP_{\hat{\mathscr{S}}})^{\dagger}}
\sum_{i=1}^n \norm{\bF^\top \be_i}\\
&\le n^{-\frac12}D^{-1} n^{-\frac12} (\tau/\opnorm{\bSigma})^{-\frac12}
n^{\frac 12} \fnorm{\bF}\\
&\le (\tau')^{-\frac12},
\end{align*}

where the third inequality uses $\bM^\dagger \preceq \bI_T \otimes (\bX_{\hat{\mathscr{S}}}^\top\bX_{\hat{\mathscr{S}}} + n\tau\bP_{\hat{\mathscr{S}}})^{\dagger}$,
the fourth inequality uses the result that $\opnorm{\be_i^\top\bA} \le \opnorm{\bA}$,
the penultimate inequality uses $\opnorm{\bX_{\hat{\mathscr{S}}}(\bX_{\hat{\mathscr{S}}}^\top\bX_{\hat{\mathscr{S}}} + n\tau\bP_{\hat{\mathscr{S}}})^{\dagger}} \le (n\tau)^{-1/2}$ and the Cauchy-Schwarz inequality,
the last inequality follows from $n^{-1/2}D^{-1}\fnorm{\bF}\le 1$.
It immediately follows that $\fnorm{\bW_2} \le \sqrt{T}C(\tau')$ since the rank of $\bW_2$ is at most $T$.

For $\bW_3$, using $n^{-\frac12} D^{-1} \fnorm*{\bF}\le 1$, and $\fnorm*{ \frac{\partial D}{\partial \bZ}}\le n^{-\frac12} D C(\tau')$ from <Ref>, we obtain

\begin{align*}
\fnorm*{\bW_3}
= n^{-\frac12} D^{-2} \fnorm{\bF^\top  \frac{\partial D}{\partial \bZ}} \le n^{-\frac12} D^{-2} \fnorm*{\bF}\fnorm*{ \frac{\partial D}{\partial \bZ}}
\le D^{-1}\fnorm*{ \frac{\partial D}{\partial \bZ}}
\le n^{-1/2} C(\tau'),
\end{align*}

The desired inequality follows by combining (<ref>) and the bounds for $\bW_0, \bW_1, \bW_2, \bW_3$.

By <Ref>, $\frac{\partial F_{lt}}{\partial z_{ij}} = D_{ij}^{lt} + \Delta_{ij}^{lt}$, where

\begin{align}
D_{ij}^{lt} &= -(\be_j^\top\bH \otimes \be_i^\top)(\bI_{nT} - \bN)(\be_t\otimes \be_l),\label{eq: D_ijlt}\\
%		\Delta_{ij}^{lt} &=-(\be_t^\top \otimes \be_l^\top)(\bI_T\otimes \bZ)
%		\tilde\bM^\dagger \bigl(\bF^\top \otimes \bI_{p}\bigr) (\be_i \otimes\be_j).
\Delta_{ij}^{lt} &= -(\be_t^\top \otimes \be_l^\top)(\bI_T\otimes \bX)
\bM^\dagger (\bI_T\otimes \bSigma^{\frac12})
\bigl(\bF^\top \otimes \bI_{p}\bigr)(\be_i \otimes\be_j). \label{eq: Delta_ijlt}
\end{align}

For the first equality, since $\be_i^\top \frac{\partial \bF}{\partial z_{ij} }=\sum_t (D_{ij}^{it} + \Delta_{ij}^{it}) \be_t^\top$, we have

\begin{align*}
&\sum_{i=1}^n \sum_{j=1}^p\bF^\top\bZ\be_j\be_i^\top\frac{\partial \bF}{\partial z_{ij}}\\
=~& \sum_{ij}\bF^\top\bZ\be_j \sum_{t=1}^T (D_{ij}^{it} + \Delta_{ij}^{it}) \be_t^\top\\
=~& \sum_{ij}
\bF^\top \bZ \be_j \sum_{t} D_{ij}^{it} \be_t^\top
+ \underbrace{\sum_{ij} \bF^\top \bZ \be_j \sum_{t} p \Delta_{ij}^{it} \be_t^\top}_{\bJ_1},
\end{align*}

where the first term can be simplified as below

\begin{align*}
\bF^\top \bZ \be_j \sum_{t=1}^T D_{ij}^{it} \be_t^\top\\
\bF^\top \bZ \be_j \sum_{t=1}^T(\be_j^\top\bH \otimes \be_i^\top)(\bI_{nT} - \bN)(\be_t\otimes \be_i)\be_t^\top\\
=~& -
\bF^\top \bZ \bH
\Bigl[
\sum_i
\bI_T \otimes \be_i^\top
(\bI_{nT} - \bN)
(\bI_T \otimes \be_i)
\Bigr]\\
=~& -
\bF^\top \bZ\bH(n\bI_T - \hbA).
\end{align*}

For the second equality, since $\frac{\partial \bF}{\partial z_{ij} }
=\sum_{lt} \be_l (D_{ij}^{lt} + \Delta_{ij}^{lt}) \be_t^\top$,

\begin{align*}
\sum_{ij}\Big(\frac{\partial \bF}{\partial z_{ij} }\Big)^\top
\bZ \be_j \be_i^\top \bF
= \underbrace{\sum_{ijtl} \be_t D_{ij}^{lt} \be_l^\top\bZ\be_j \be_i^\top \bF}_{\bJ_2} +\sum_{ijtl} \be_t \Delta_{ij}^{lt} \be_l^\top\bZ\be_j \be_i^\top \bF,
\end{align*}

where the second term can be simplified as below,

\begin{align*}
&\sum_{ijtl} \be_t \Delta_{ij}^{lt} \be_l^\top\bZ\be_j \be_i^\top \bF\\
~=& \sum_{ijtl} \be_t\Delta_{ij}^{lt}(\be_i\otimes \be_j)^\top (\bF \otimes \bZ^\top\be_l)\\
~=& -\sum_{ijtl} \be_t(\be_t^\top \otimes \be_l^\top)(\bI_T\otimes \bX)
\bM^\dagger (\bI_T\otimes \bSigma^{\frac12})\bigl(\bF^\top \otimes \bI_{p}\bigr) (\be_i \otimes\be_j)(\be_i\otimes \be_j)^\top (\bF \otimes \bZ^\top\be_l)\\
~=& -\sum_{l} (\bI_T \otimes \be_l^\top)(\bI_T\otimes \bX)
\bM^\dagger (\bI_T\otimes \bSigma^{\frac12}) (\bF^\top\otimes \bI_p) (\bF \otimes \bZ^\top\be_l)\\
~=& -\sum_{l} (\bI_T \otimes \be_l^\top)(\bI_T\otimes \bX)
\bM^\dagger \bigl(\bF^\top\bF \otimes \bX^\top\be_l\bigr) \\
~=& -\sum_{l} (\bI_T \otimes \be_l^\top)(\bI_T\otimes \bX)
\bM^\dagger \bigl(\bI_T \otimes \bX^\top\be_l\bigr) \bF^\top\bF\\
~=& - \hbA \bF^\top \bF,
\end{align*}

where the last line uses the expression of $\hbA$ in (<ref>).

It remains to bound the norm of $\bJ_1$ and $\bJ_2$.
To bound $\fnorm{\bJ_2}$, recall the definition of $\bJ_2$,

\begin{align*}
\bJ_2 &= \sum_{ijlt} \be_t D_{ij}^{lt} \be_l^\top\bZ\be_j \be_i^\top\bF\\
&= - \sum_{ijlt} \be_t (\be_j^\top\bH \otimes \be_i^\top) (\bI_{nT} - \bN ) (\be_t\otimes \be_l) \be_l^\top\bZ\be_j \be_i^\top\bF\\
&= - \sum_{ijlt} \be_t (\be_t^\top\otimes \be_l^\top) (\bI_{nT} - \bN) (\bH^\top\be_j \otimes \be_i) \be_l^\top\bZ\be_j \be_i^\top\bF\\
&= - \sum_{ijl} (\bI_T\otimes \be_l^\top) (\bI_{nT} - \bN) (\bH^\top\be_j \otimes \be_i) \be_l^\top\bZ\be_j \be_i^\top\bF\\
&= - \sum_{l} (\bI_T\otimes \be_l^\top) (\bI_{nT} - \bN) (\bH^\top\bZ^\top\be_l \otimes \bF),
\end{align*}

Since $\bN$ is non-negative definite with $\opnorm*{\bN} \le 1$, $ \opnorm*{\bI_{nT} - \bN} \le 1$,

\begin{align*}
\fnorm*{\bJ_2} &\le \sum_{l} \fnorm*{(\bI_T\otimes \be_l^\top) (\bI_{nT} - \bN) (\bH^\top\bZ^\top\be_l \otimes \bF)}\\
&\le \sum_{l} \opnorm*{(\bI_T\otimes \be_l^\top) (\bI_{nT} - \bN)}\fnorm*{(\bH^\top\bZ^\top\be_l \otimes \bF)}\\
&\le \sum_{l} \fnorm*{(\bH^\top\bZ^\top\be_l \otimes \bF)}\\
&= \sum_{l} \fnorm*{\bH^\top\bZ^\top\be_l} \fnorm*{\bF}\\
&\le n^{\frac 12} \fnorm{\bZ\bH} \fnorm{\bF}\\
&\le n^{\frac 12} \opnorm*{\bZ}\fnorm{\bH} \fnorm*{\bF}.
\end{align*}

To bound $\fnorm{\bJ_1}$, recall the definition of $\bJ_1$,

$$\bJ_1 = \sum_{ij}\bF^\top\bZ\be_j \sum_t \Delta_{ij}^{it} \be_t^\top.$$

For each $t, t'\in [T]$,

\begin{align*}
\be_{t'}^\top\bJ_1 \be_{t} & = \be_{t'}^\top \big[\sum_{ij}\bF^\top\bZ\be_j \sum_t \Delta_{ij}^{it} \be_t^\top\big]\be_t\\
\be_{t'}^\top \bF^\top \bZ \be_j  \Delta_{ij}^{it}
\\&
(\be_i^\top\bF \otimes \be_{t'}^\top \bF^\top)
\bN
(\be_t \otimes \be_i)\\
(\be_i^\top\bF \otimes \bF^\top)
\bN
( \bI_T\otimes \be_i)\be_t,
\end{align*}

\bJ_1 = \sum_i
(\be_i^\top\bF \otimes \bF^\top)
\bN
( \bI_T\otimes \be_i),
and hence

\begin{align*}
\fnorm{\bJ_1} &= \fnorm{\sum_i(\be_i^\top\bF \otimes \bF^\top)\bN(\bI_T\otimes \be_i)}\\
&\le \sum_i\fnorm{(\be_i^\top\bF \otimes \bF^\top)\bN(\bI_T\otimes \be_i)}\\
&\le \sum_i\fnorm{(\be_i^\top\bF \otimes \bF^\top)}\opnorm{\bN(\bI_T\otimes \be_i)}\\
&\le \sum_i\norm{\be_i^\top\bF} \fnorm{\bF}\\
&\le n^{\frac 12}\fnorm{\bF}^2,
\end{align*}

where the first inequality is by sub-additivity of Frobenius norm, the second inequality uses $\fnorm{\bA_1\bA_2}\le \fnorm{\bA_1}\opnorm{\bA_2}$ for any matrices $\bA_1,\bA_2$ with appropriate dimensions, the third inequality is by $\opnorm{\bN}\le1$ from <Ref>, the last inequality is by Cauchy-Schwarz inequality.


feasible computability is desirable. Some algorithms interpreted in continuous
structures (like the real or complex numbers) become even incorrect (see
examples in Section 8); it seems desirable to add corresponding comments
(which mention the computable analysis approach) in the next editions of such
textbooks.

Acknowledgements. The authors thank Pavel Alaev, Sergey Goncharov, Valentina
Harizanov, Peter Hertling, Iskander Kalimullin, Julia Knight, Russell Miller,
Andrey Morozov, and Xizhong Zheng for useful discussions and bibliographical
hints.

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