text:
\section{Proofs of technical lemmas}
\label{sec:proof}

\subsection{Proof of Lemma \ref{pert-free}}
\label{sec:proofepsilon-free}

Here we consider the effect of removing the perturbation, so we shall assume $\frac{dx^{(t)}}{d\epsilon} = 0$. From formula \eqref{eq:perturbed_f:derivative1a} proved in section \ref{sec:usefulderivatives} we have
\begin{align}
\frac{d}{d\epsilon} h_{t,s;\epsilon,\delta} = \frac{d\mathbb{E}[\bar{\epsilon}^{(t,s)}]}{d\epsilon} \frac{d}{d\mathbb{E}[\bar{\epsilon}^{(t,s)}]} h_{t,s;\epsilon,\delta}
	= - \frac{\ln 2}{n(1-\epsilon)} \sum_{i=1}^n (1 - \mathbb{E}\<\sigma_i\>_{t,s;\epsilon,\delta})
	= - \frac{\ln 2}{n} \sum_{i=1}^{n}  ( 1 - \mathbb{E} \< \sigma_i \>_{t, s;\epsilon,\delta;\sim H_i} ).
\end{align}
where $\< \sigma_i \>_{t, s;\epsilon,\delta;\sim H_i}$ is the Gibbs expectation with fixed $H_i=0$.
Thus $\vert \frac{d}{d\epsilon} h_{t,s;\epsilon,\delta} \vert \leq \ln 2$.
We also remarked below equation \eqref{eq:perturbed_f:derivative3a} that $\vert \frac{d}{d\delta} h_{t,s;\epsilon,\delta}\vert \leq \ln (2) /n^{\theta}$.
Thus
by the mean value theorem
\begin{align}
 \vert h_{t,s;\epsilon,\delta} - h_{t,s;\epsilon=0,\delta}\vert & \leq \epsilon \ln 2\, , \\
 \vert h_{t,s;\epsilon,\delta} - h_{t,s;\epsilon, \delta=0}\vert & \leq \frac{\ln 2}{n^{\theta}}\,.
\end{align}
By triangle inequality we get \eqref{eq:pert-free:1}.
Note that $\tilde{h}_{\epsilon,\delta}(\underline{\sfx})$ is in the form $h_{T,1;\epsilon,\delta}(\underline{\sfx}) + g(\underline{\sfx})$,
therefore
$$
\tilde{h}_{\epsilon,\delta}(\underline{\sfx}) - \tilde{h}_{\epsilon=0,\delta=0}(\underline{\sfx}) = h_{T,1;\epsilon,\delta}(\underline{\sfx}) - h_{T,1;\epsilon=0,\delta=0}(\underline{\sfx})\,.
$$
Consquently \eqref{eq:pert-free:2} follows immediately from \eqref{eq:pert-free:1}.

\subsection{Proof of Lemma \ref{thm:BEC:overlap}}
\label{sec:proof:BEC:overlap}
The first GKS inequality \eqref{eq:GKS1} implies

\begin{align}
\< \sigma_A \>_{t,s; \epsilon,\delta} ( 1 - \< \sigma_A \>_{t,s; \epsilon;\delta} ) \geq 0. \label{eq:GKS:implication}
\end{align}

Moreover, Nishimori's identity \eqref{eq:Nishimori} implies

\begin{align}
\mathbb{E}  \< \sigma_A \>_{t,s; \epsilon,\delta}  = \mathbb{E} [ \< \sigma_A \>_{t,s; \epsilon,\delta}^2 ] \label{eq:Nishimori:case1}
\end{align}

which can be written as

\begin{align}
\mathbb{E} [ \< \sigma_A \>_{t,s; \epsilon,\delta} ( 1 - \< \sigma_A \>_{t,s; \epsilon,\delta} ) ] = 0. \label{eq:Nishimori:case1:2}
\end{align}

As a result of \eqref{eq:GKS:implication} and \eqref{eq:Nishimori:case1:2}, we have $\< \sigma_A \>_{t,s; \epsilon,\delta}$ equal to either $0$ or $1$.

\subsection{Proof of Lemma~\ref{thm:bd_fixed_t}}
\label{sec:proof:bd_fixed_t}
Using the fundamental theorem of calculus, the desired difference has an integral form
\begin{align}
\mathbb{E} \< Q_1^k \>_{t,s'; \epsilon,\delta}  - &\mathbb{E} \< Q_1^k \>_{t,0; \epsilon,\delta}
	 = \frac{1}{n^k} \int_{0}^{s'} ds \sum_{i_1,\dots,i_k=1}^{n} \frac{d}{ds}\mathbb{E} \< \sigma_{i_1} \cdots \sigma_{i_k} \>_{t,s;\epsilon_n}
	 \nonumber \\
	& = \frac{1}{n^k} \int_{0}^{s'} ds \sum_{i_1,\dots,i_k=1}^{n}\Big\{ \frac{\alpha K}{T} \sum_{j=1}^{n}\big( \mathbb{E}\< \sigma_{i_1} \cdots \sigma_{i_k} \>_{e_{j,s}^{(t)} + 1} - \mathbb{E}\< \sigma_{i_1} \cdots \sigma_{i_k} \>_{e_{j,s}^{(t)}} \big) \nonumber \\
	& - \frac{\alpha n}{T}\big( \mathbb{E}\< \sigma_{i_1} \cdots \sigma_{i_k} \>_{m_s^{(t)} + 1} - \mathbb{E}\< \sigma_{i_1} \cdots \sigma_{i_k} \>_{m_s^{(t)}}\big)\Big\}
	\label{eq:bd_fixed_t:1}
\end{align}
where~\eqref{eq:bd_fixed_t:1} follows from the Poisson property \eqref{eq:dPoissonMean}.
Since $| \< \sigma_{i_1} \cdots \sigma_{i_k} \> |\le 1$ we see that the absolute value of \eqref{eq:bd_fixed_t:1} is bounded by $2 \alpha (K+1) n / T$.

\subsection{Proof of Lemma~\ref{thm:rmv-delta}}
\label{sec:proof:rmv-delta}
As indices $t,s,\epsilon$ are fixed in this proof, we omit them for concision.

Recall that~\eqref{eq:Q1:delta-der} and \eqref{eq:GKS2} imply $\mathbb{E}\<Q_1\>_{\delta}$ is increasing in $\delta$ and therefore $\mathbb{E}[\<Q_1\>_{\delta}] \geq \mathbb{E}[\<Q_1\>_{0}]$. Using also
$|\<Q_1\>| \leq 1$ we obtain the inequality
\begin{align*}
\mathbb{E}[\<Q_1\>_{\delta}]^K - \mathbb{E}[\<Q_1\>_{0}]^K
	= (\mathbb{E}[\<Q_1\>_{\delta}] - \mathbb{E}[\<Q_1\>_{0}]) \sum_{k=0}^{K-1} \mathbb{E}[\<Q_1\>_{\delta}]^{K-k} \mathbb{E}[\<Q_1\>_{\delta}]^{k}
	\leq K(\mathbb{E}[\<Q_1\>_{\delta}] - \mathbb{E}[\<Q_1\>_{0}]).
\end{align*}
This inequality, together with $q_1^{(t)} \equiv \mathbb{E} \tanh V^{(t)} \in [0,1]$ and $\mathbb{E}[\<Q_1\>_{\delta}] \geq \mathbb{E}[\<Q_1\>_{0}]$, gives
\begin{align}
& \int_{\delta_0}^{\delta_1} d\delta \int_{\varepsilon_0}^{\varepsilon_1} d\epsilon \Big | \mathbb{E}[\<Q_1\>_{\delta}]^K - \mathbb{E}[\<Q_1\>_{0}]^K
- K (q_1^{(t)})^{K-1} ( \mathbb{E}[\<Q_1\>_{\delta}] - \mathbb{E}[\<Q_1\>_{0}] ) \Big | \nonumber \\
& \leq 2 K \int_{\delta_0}^{\delta_1} d\delta \int_{\varepsilon_0}^{\varepsilon_1} d\epsilon (\mathbb{E}[\<Q_1\>_{\delta}] - \mathbb{E}[\<Q_1\>_{0}]) \nonumber \\
& = 2 K \int_{\delta_0}^{\delta_1} d\delta \big ( \int_{\varepsilon_0}^{\varepsilon_1} d\epsilon \mathbb{E}[\<Q_1\>_{\delta}] - \int_{\varepsilon_0}^{\varepsilon_1} d\epsilon \mathbb{E}[\<Q_1\>_{0}] \big ).
\label{93}
\end{align}
We use the mean value theorem to upper bound \eqref{93} as
\begin{align*}
2 K \int_{\delta_0}^{\delta_1} d\delta \ \delta \max_{\delta' \in [0,\delta]} \big ( \frac{d}{d\delta} \int_{\varepsilon_0}^{\varepsilon_1} d\epsilon \mathbb{E}[\<Q_1\>_{\delta}] \big ) \Big |_{\delta=\delta'}
& = 2 K \int_{\delta_0}^{\delta_1} d\delta \ \delta \max_{\delta' \in [0,\delta]} \big ( \int_{\varepsilon_0}^{\varepsilon_1} d\epsilon \frac{d}{d\delta} \mathbb{E}[\<Q_1\>_{\delta}] \big ) \Big |_{\delta=\delta'}
\end{align*}
where the equality follows from the fact that $\underline{\sfx}(\epsilon)$ is independent of $\delta$ and therefore we can exchange the order of derivative and integral. Using \eqref{eq:Q1:delta-der} the last equation equals
\begin{align}
& 2 K \int_{\delta_0}^{\delta_1} d\delta \ \delta \max_{\delta' \in [0,\delta]} \big ( \int_{\varepsilon_0}^{\varepsilon_1} d\epsilon \frac{1}{n^{1+\theta}(1-\delta/n^{\theta})} \sum_{i, j=1}^{n} \mathbb{E} [\< \sigma_i \sigma_j \>_{\delta}
- \< \sigma_i \>_{\delta} \< \sigma_j \>_{\delta}] \big ) \Big |_{\delta=\delta'} \nonumber \\
& = 2 Kn^{1-\theta} \int_{\delta_0}^{\delta_1} d\delta \frac{\delta}{1-\delta/n^{\theta}} \max_{\delta' \in [0,\delta]} \big ( \int_{\varepsilon_0}^{\varepsilon_1} d\epsilon \mathbb{E}[\< ( Q_1-\<Q_1\>_{\delta} )^2 \>_{\delta}] \big ) \Big |_{\delta=\delta'} \nonumber \\
& \leq \frac{6K}{n^{\theta}} \int_{\delta_0}^{\delta_1} d\delta \frac{\delta}{1-\delta/n^{\theta}} \label{eq:rmv-delta:3} \\
& \leq \frac{3K(\delta_1^2-\delta_0^2)}{n^{\theta}(1-\delta_1/n^{\theta})} \nonumber
\end{align}
where \eqref{eq:rmv-delta:3} follows from Lemma~\ref{thm:overlap:thermal:concentration}.

\subsection{Proof of Lemma \ref{thm:momentMatching:solution}}
\label{sec:proof:momentMatching:solution}

For each $n$, we seek distributions $\sfx^{(t)}\in \mathcal{B}$ for $V^{(t)}$, $t=1,\dots, T$ which solve equation \eqref{eq:momentMatching}.
By symmetry between vertices
$\mathbb{E}\langle Q_1 \rangle_{t,0; \epsilon,0} = \mathbb{E}\langle \sigma_1 \rangle_{t,0; \epsilon,0}$ so the equation becomes
\begin{align}
 \mathbb{E}\tanh V^{(t)} = \mathbb{E}\langle \sigma_1 \rangle_{t,0; \epsilon,0}.
\end{align}
Recall that in our interpolation scheme the right hand side depends only on  $\{\sfx^{(t')}\}_{t' <t}$ and is thus independent of $\sfx^{(t)}$. Thus it suffices to choose
$V^{(t)}$ for $t=1, \dots, T$ as follows: $V^{(t)} = +\infty$ with probability $\mathbb{E}\langle \sigma_1 \rangle_{t,0; \epsilon,0}$ and $V^{(t)}= 0$ with
probability $1-\mathbb{E}\langle \sigma_1 \rangle_{t,0; \epsilon,0}$. These are the distributions $\hat \sfx^{(t)}_n \in \mathcal{B}$ of the Lemma. It is clear that this solution is unique and $\hat{x}_n^{(t)} = \mathbb{E} \tanh V^{(t)}$.

Finally, we verify $d \hat{x}_n^{(t)} / d\epsilon \geq 0$. To simplify the notation we use $\< - \> \equiv \< - \>_{t,0;\epsilon,0}$ and $\< - \>_{\sim \bar{H}_j^{(t,0)}}$ denotes $\< - \>_{t,0;\epsilon,0}$ with $\bar{H}_j^{(t,0)}$ set to 0.
Recall the definition of $\mathbb{E}[\bar{\epsilon}^{(t,s)}]$ in section \ref{sec:usefulderivatives}. By the chain rule we have
\begin{align}
\frac{d\hat{x}^{(t)}}{d\epsilon}
	= \frac{d\mathbb{E}[\bar{\epsilon}^{(t,0)}]}{d\epsilon} \frac{d\mathbb{E}\<\sigma_1\>_{t,0;\epsilon,0}}{d \mathbb{E}[\bar{\epsilon}^{(t,0)}]}.
\end{align}
To compute the last derivative, first we use the identity $e^{\pm x} = (1 \pm \tanh x)\cosh x $ to write
\begin{align*}
\mathbb{E}\<\sigma_1\>_{t,0;\epsilon,0}
	& = \mathbb{E} \Biggl[ \frac{ \<e^{\bar{H}^{(t,0)}_j (\sigma_j-1)} \sigma_1 \>_{\sim \bar{H}_j^{(t,0)}} }{ \< e^{\bar{H}^{(t,0)}_j (\sigma_j-1)} \>_{\sim \bar{H}_j^{(t,0)}} } \Biggr]
	 = \mathbb{E}\Biggl[ \frac{ \< \sigma_1 \>_{\sim \bar{H}_j^{(t,0)}} + \< \sigma_1\sigma_j \>_{\sim \bar{H}_j^{(t,0)}} \tanh \bar{H}_j^{(t,0)} }{ 1 + \< \sigma_j \>_{\sim \bar{H}_j^{(t,0)}} \tanh \bar{H}_j^{(t,0)}} \Biggr] \\
	& = \mathbb{E}[\bar{\epsilon}_j^{(t,0)}] \cdot \mathbb{E}
	\Biggl[ \frac{ \< \sigma_1 \>_{\sim \bar{H}_j^{(t,0)}} + \< \sigma_1\sigma_j \>_{\sim \bar{H}_j^{(t,0)}} }{ 1 + \< \sigma_j \>_{\sim \bar{H}_j^{(t,0)}} } \Biggr]
	+ (1 - \mathbb{E}[\bar{\epsilon}_j^{(t,0)}]) \mathbb{E} \< \sigma_1 \>_{\sim \bar{H}_j^{(t,0)}}.
\end{align*}
Then it is straightforward to compute
\begin{align}
\frac{d\mathbb{E}\<\sigma_1\>_{t,0;\epsilon,0}}{d \mathbb{E}[\bar{\epsilon}^{(t,0)}]}
	& = \sum_{j=1}^{n} \frac{d\mathbb{E}\<\sigma_1\>_{t,0;\epsilon,0}}{d \mathbb{E}[\bar{\epsilon}_j^{(t,0)}]}
	\- = \sum_{j=1}^{n} \mathbb{E} \Biggl[ \frac{ \< \sigma_1 \sigma_j \>_{\sim \bar{H}_j^{(t,0)}} - \< \sigma_1 \>_{\sim \bar{H}_j^{(t,0)}} \< \sigma_j \>_{\sim \bar{H}_j^{(t,0)}} }{ 1+\< \sigma_j \>_{\sim \bar{H}_j^{(t,0)}} } \Biggr]. \label{eq:der-Gibbs-barEp}
\end{align}
The second GKS inequality \eqref{eq:GKS2} ensures \eqref{eq:der-Gibbs-barEp} non-negative, leaving the sign of $dx^{(t)}/d\epsilon$ determined by $d\mathbb{E}[\bar{\epsilon}^{(t,0)}] / d\epsilon$. Using \eqref{eq:BP-update:BEC} and \eqref{eq:E-epsilon-bar},
\begin{align*}
& \frac{\mathbb{E}[\bar{\epsilon}^{(1,0)}]}{d\epsilon} = 1,
&& \frac{\mathbb{E}[\bar{\epsilon}^{(t,0)}]}{d\epsilon} = \frac{1-\mathbb{E}[\bar{\epsilon}^{(t,0)}]}{1-\epsilon} + (1-\mathbb{E}[\bar{\epsilon}^{(t,0)}]) \frac{\alpha K(K-1)(1-q)}{T} \sum_{t'=1}^{t-1} \hat{x}^{(t') K-2} \frac{d\hat{x}^{(t')}}{d\epsilon}.
\end{align*}
This equation implies that the claim $d \hat{x}^{(t)} / d\epsilon \geq 0$ is true for $t=1$ by direct calculation.  Then we also get the claim for $t \geq 2$ by induction.

\subsection{Proof of Lemma \ref{equivalence}}
\label{proof:h_RS_coupled:equivalence}

We first note that the generalized entropy functionals can easily be shown to be upper bounded and are defined on a closed convex set of probability measures. Hence we can replace the supremum in
the lemma by a maximum. For the BEC $\underline\sfx\in \mathcal{B}^T$ and $\tilde{h}_{\epsilon=0,\delta=0}(\underline{\sfx})$ becomes a function of $\underline x\in [0,1]^T$, therefore the proof of the lemma
can be carried out directly by elementary real analysis calculations.

Here we give an analysis that applies more generally to functionals over $\underline\sfx\in \mathcal{X}^T$ in the general case of symmetric channels.
Let us outline the strategy of the proof: (i) We first show
that the stationarity condition for $\tilde{h}_{\epsilon=0,\delta=0}(\underline{\sfx})$ implies that all $\sfx^{(t)}$ are equal for $t=1, \dots, T$; (ii)
We then show that a maximum of  $\tilde{h}_{\epsilon=0,\delta=0}(\underline{\sfx})$
is necessarily a stationary point.

Before carrying out point (i) it is convenient to
 express $\tilde{h}_{\epsilon=0,\delta=0}(\underline \sfx)$ more explicitly in terms of the distribution $\underline{\sfx}$ thanks to a
formalism from coding theory (see e.g. \cite{KumYMP:2014,MCT:2008}). We define an entropy functional\footnote{The notation $H$ for the entropy should not be confused with the notation $H$ for the
perturbation field in the model.} $H: \mathcal{X} \rightarrow \mathbb{R}$ as
\begin{align}
H(\sfx)\equiv \int \ln (1+e^{-2 a}) \sfx(da) = \ln 2 - \int \ln (1+ \tanh a) \sfx(da). \label{eq:entropy}
\end{align}
The argument $a$ is to be interpreted as a half-log-likelihood ratio.
Two convolution operators $\circledast, \boxast: \sX \times \sX \rightarrow \sX$ are
defined for $a_1 \sim \sfx_1, a_2 \sim \sfx_2$ such that $\sfx_1 \circledast \sfx_2$ is
the distribution of $a_1+a_2$ and $\sfx_1 \boxast \sfx_2$ is the distribution of $\tanh^{-1} (\tanh a_1 \tanh a_2)$.
Therefore, the entropies of convolutions are
\begin{align}
H(\circledast_{i=1}^k \sfx_i ) &  = \int \ln (1+e^{-2 \sum_{i=1}^k a_i}) \prod_{i=1}^k \sfx_i (da_i), \label{eq:entropy-conv1} \\
H(\boxast_{i=1}^k \sfx_i ) & = \ln 2 - \int \ln \Big( 1 + \prod_{i=1}^k \tanh a_i \Big ) \prod_{i=1}^k \sfx_i (da_i). \label{eq:entropy-conv2}
\end{align}
We define $\sfx^{\circledast 0} \equiv \Delta_0$, where $\Delta_0$ is the identity of $\circledast$ and it is a distribution with
solely a point mass at 0. We also define $\Lambda^{\circledast}(\sfx) \equiv \sum_{l=0}^{\infty} \Lambda_l \sfx^{\circledast l}$,
where $\Lambda_l = \frac{(\alpha K)^l}{l!} e^{-\alpha K}$ denotes the probability that a variable node has degree $l$,
and $\lambda^{\circledast}(\sfx) \equiv  \sum_{l=1} \lambda_l \sfx^{\circledast (l-1)}$,
where $\lambda_l = \frac{l \Lambda_l}{\Lambda'(1)} = \frac{(\alpha K)^{l-1}}{(l-1)!} e^{-\alpha K}$ denotes the probability that an edge is connected to a variable node of degree $l\geq 1$.
One can check that (see Appendix~\ref{proof:RSFreeEntropy_coupled2})
\begin{align}
\tilde{h}_{\epsilon=0,\delta=0} \big ( \underline{\sfx} \big )
	& = - \alpha K H \Big( \frac{1}{T} \sum_{t=1}^{T} \sfc \boxast (\sfx^{(t)})^{\boxast (K-1)} \Big )
	+ H \Big ( \Lambda^{\circledast} \Big ( \frac{1}{T} \sum_{t=1}^{T} \sfc \boxast (\sfx^{(t)})^{\boxast (K-1)} \Big ) \Big ) \nonumber \\
	& \hspace{1cm} + \alpha H(\sfc) + \frac{\alpha(K-1)}{T} \sum_{t=1}^{T} H \big (\sfc \boxast (\sfx^{(t)})^{\boxast K} \big ).
\label{eq:RSFreeEntropy_coupled2}
\end{align}
We will need differentiation rules for functionals. The directional (or Gateaux) derivative of a functional\footnote{To have well defined directional derivatives it is
understood that we extend the space $\sX$ to the Banach space of signed probability measures over $\bar{\mathbb{R}}$.}
$F: \sfx\in \mathcal{X}\to \mathbb{R}$ at point $\sfx\in \mathcal{X}$ in
the direction $\eta = \sfx_2 -\sfx_1$
where $\sfx_1,\sfx_2\in \mathcal{X}$ is by definition the following linear functional of $\eta$:

\begin{align*}
dF(\sfx) [\eta] \equiv \lim_{\gamma \rightarrow 0} \frac{F(\sfx + \gamma \eta) - F(\sfx)}{\gamma}.
\end{align*}

We employ the following computational rules that are easily proved for linear functionals $F$:

\begin{lemma}[{\cite[Propositions 14 and 15]{KumYMP:2014}}]
Let $F: \sX \rightarrow \mathbb{R}$ be a linear functional, and $*$ be
either $\circledast$ or $\boxast$. Then for $k\geq 1$ integer, $\sfx, \sfx_1, \sfx_2 \in \sX$, and setting $\eta = \sfx_2 - \sfx_1$, we have
\begin{align*}
d F(\sfx^{* k})[\eta] = k F(\sfx^{* (k-1)} * \eta).
\end{align*}
For any polynomials $p,q$, we have
\begin{align*}
d F \big (p^{\circledast}(q^{\boxast}(\sfx)) \big )[\eta] = F \big ( p'^{\circledast}(q^{\boxast}(\sfx) \circledast (q'^{\boxast}(\sfx) \boxast \eta )\big ).
\end{align*}
where $p'$ and $q'$ are the derivatives of the polynomials.
\label{thm:entropyFunctional:derivative}
\end{lemma}

\begin{lemma}[{\cite[Theorem 4.41]{MCT:2008}}]
For any $\sfx_1, \sfx_2, \sfx_3 \in \sX$, we have
\begin{align*}
H((\sfx_1 - \sfx_2)\circledast \sfx_3) + H((\sfx_1 - \sfx_2) \boxast \sfx_3) = H(\sfx_1 - \sfx_2).
\end{align*}
\label{thm:entropyFunctional:duality}
\end{lemma}

We can now proceed to prove (i).
Fix $t\in \{1, \cdots, T\}$. Consider the functional $\tilde h_{\epsilon=0, \delta=0}(\underline \sfx)$ as functional w.r.t its $t$-th component only. We denote
$d_{t} \tilde h_{\epsilon=0, \delta=0}(\underline\sfx) [\eta^{(t)}]$, the directional derivative of this functional
 at the point $\sfx$ with respect to its $t$-th component (only) in the direction
$\eta^{(t)}$. This a linear functional of $\eta^{(t)}$ and corresponds to a ``partial'' Gateaux derivative as indicated by the notation $d_t$.
Let $$
\sfT(\sfc, \underline{\sfx}) \equiv \lambda^{\circledast} \big( \frac{1}{T} \sum_{t=1}^{T} \sfc \boxast \sfx^{(t)\boxast (K-1)} \big)\,.
$$
Using Lemma~\ref{thm:entropyFunctional:derivative}, $d_{t} \tilde{h}_{\epsilon=0,\delta=0}(\underline{\sfx})[\eta^{(t)}]$ is the sum of the following three terms:
\begin{align}
- \alpha K d_{t}  H \Big ( \frac{1}{T} \sum_{t=1}^{T} \sfc \boxast \sfx^{(t)\boxast (K-1)} \Big) [\eta^{(t)}]&=
- \frac{\alpha K(K-1)}{T} H \big ( \sfc \boxast \sfx^{(t)\boxast (K-2)} \boxast \eta^{(t)} \big ), \label{eq:dir_derivative:term1}\\

	d_{t} H \Big (\Lambda^{\circledast}\Big ( \frac{1}{T} \sum_{t=1}^{T}
	\sfc \boxast \sfx^{(t)\boxast (K-1)} \Big) \Big ) [\eta^{(t)}] &=
	\frac{\alpha K(K-1)}{T} H \big ( \sfT(\sfc, \underline{\sfx})
	\circledast \big ( \sfc \boxast \sfx^{(t)\boxast (K-2)} \boxast \eta^{(t)} \big ) \big ), \label{eq:dir_derivative:term2}\\

- \alpha(K-1)d_{t}   H \Big (\frac{1}{T}\sum_{t=1}^{T}\sfc \boxast (\sfx^{(t)\boxast K} \Big ) [\eta^{(t)}]
	& = - \frac{\alpha K(K-1)}{T} H \big( \sfc \boxast \sfx^{(t)\boxast (K-1)} \boxast \eta^{(t)} \big ). \label{eq:dir_derivative:term3}
\end{align}

In addition, we use Lemma~\ref{thm:entropyFunctional:duality} to rewrite \eqref{eq:dir_derivative:term2} as

\begin{align}
\frac{\alpha K(K-1)}{T} \left \{ H \left ( \sfc \boxast \sfx^{(t)\boxast (K-2)} \boxast \eta^{(t)} \right ) -
H \left ( \sfT(\sfc, \underline{\sfx}) \boxast \big ( \sfc \boxast \sfx^{(t)\boxast (K-2)} \boxast \eta^{(t)} \big ) \right )  \right \}. \label{eq:dir_derivative:term2a}
\end{align}

Putting \eqref{eq:dir_derivative:term1}, \eqref{eq:dir_derivative:term3} and \eqref{eq:dir_derivative:term2a} together, we have

\begin{align}
d_{t} \tilde{h}_{\epsilon=0,\delta=0}(\underline{\sfx})[\eta^{(t)}]
	& = \frac{\alpha K(K-1)}{T} H \left ( \big(\sfx^{(t)} - \sfT(\sfc, \underline{\sfx}) \big) \boxast \big( \sfc \boxast (\sfx^{(t)\boxast (K-2)} \boxast \eta^{(t)} \big) \right ),
	\label{eq:dhcoupled_dxt}
\end{align}
which implies that $\underline \sfx$ is a stationnary point of $\tilde{h}_{\epsilon=0,\delta=0}(\underline{\sfx})[\eta^{(t)}]$ if and only if it satisfies the equation
\begin{align}\label{stat-point-equ}
 \sfx^{(t)} = \sfT(\sfc, \underline{\sfx}), \qquad t=1, \dots, T.
\end{align}
In particular we have $\sfx^{(1)} = \dots = \sfx^{(T)}$ as claimed.

Now we prove (ii). We proceed by contradiction and show that: if $\underline \sfx$ is not a stationary point then it cannot be a maximum.
From the Taylor expansion of the logarithm and \eqref{eq:channelSymmetry} we find for any $\sfx\in \mathcal{X}$
\begin{align}
H(\sfx)
	& = \ln 2 - \sum_{p=1}^{\infty} \frac{(-1)^{p+1}}{p} \int \sfx(da) (\tanh a)^p
	\- = \ln 2 - \sum_{p=1}^{\infty} \frac{1}{2p(2p-1)} \int \sfx(da) (\tanh a)^{2p} \, .\label{eq:entropy:expansion}
\end{align}
Let $\sfx_1, \sfx_2, \sfx_3, \sfx_4 \in \sX$. From \eqref{eq:entropy-conv2} and \eqref{eq:entropy:expansion}
\begin{align}
H((\sfx_1-\sfx_2) \boxast (\sfx_3-\sfx_4)) = - \sum_{p=1}^{\infty} \frac{1}{2p(2p-1)} \bigg\{ \int (\sfx_1-\sfx_2)(da) (\tanh a)^{2p} \bigg\} \bigg\{ \int (\sfx_3-\sfx_4)(da) (\tanh a)^{2p} \bigg\}
\end{align}
which implies that \eqref{eq:dhcoupled_dxt} can be written as
\begin{align}
d_{t} \tilde{h}_{\epsilon=0,\delta=0}(\underline{\sfx})[\eta^{(t)}]
	& = - \frac{\alpha K(K-1)}{T} \sum_{p=1}^{\infty} \frac{1}{2p(2p-1)} \bigg\{ \int ( \sfx^{(t)} - \sfT(\sfc, \underline{\sfx}) )(da) (\tanh a)^{2p} \bigg\} \nonumber \\
	& \hspace{1cm} \bigg\{ \int ( \sfc \boxast \sfx^{(t)\boxast (K-2)} )(da) (\tanh a)^{2p} \bigg\} \bigg\{ \int \eta^{(t)}(da) (\tanh a)^{2p} \bigg\}. \label{eq:dhcoupled_dxt:expansion}
\end{align}
Now, take an $\underline\sfx$ that is not a stationary point. Then there must exist an $t^*$ such that $\sfx^{(t^*)} \neq \sfT(\sfc, \underline{\sfx})$. Hence we can look at the directional derivative in the
non-trivial direction
$\eta^{(t^*)} = \sfx^{(t^*)} - \sfT(\sfc, \underline{\sfx})$. From \eqref{eq:dhcoupled_dxt:expansion} we see that
\begin{align*}
d_{t^{*}} \tilde{h}_{\epsilon=0,\delta=0}(\underline{\sfx})[\eta^{(t^*)}]
	= &\frac{\alpha K(K-1)}{T} \sum_{p=1}^{\infty} \frac{1}{2p(2p-1)} \bigg\{ \int ( \sfc \boxast \sfx^{(t)\boxast (K-2)})(da) (\tanh a)^{2p} \bigg\}
	\nonumber \\ &
	\times
	\bigg\{ \int ( \sfx^{(t^*)} - \sfT(\sfh, \sfc, \underline{\sfx}) )(da) (\tanh a)^{2p} \bigg\}^2
\end{align*}
so the directional derivative is strictly positive. Hence $\underline\sfx$ cannot be a maximum since there exists one direction in which the functional increases.

config: arxiv


TITLE: Prove that if $p(A)=0$ where A is a matrix of a linear operator ($A \in L(V)$), $p(\lambda)=0$ if $\lambda \in \sigma(A)$
QUESTION [1 upvotes]: I think it's all in the title. $p$ is some random polynomial.
I don't know how to approach this one. I've tried taking the roots of $p$, placing them on the diagonal of a new matrix and reasoning that the linear operator of the new matrix has a common specific value (if it's called that in English) with A, but I have no idea how to prove that that specific value is also a root of $p$.

REPLY [1 votes]: Another way to look at this problem is noticing that $p$ has to be a (polynomial) multiple of the minimal polynomial of $A$. Now what happens to the minimal polynomial if you apply it on an Eigenvalue of $A$? What happens consequently to $p$ ?

REPLY [0 votes]: I'll give you an example ($p(x)=x^2+x+1$) and leave the general case to you.
Suppose that $\lambda\in\sigma(A)$ with corresponding eigenvector $u\neq0$. Then
$$
0=p(A)u=A^2u+Au+u=A(\lambda u)+\lambda u+u=
\\
\lambda Au+\lambda u+u=\lambda^2u+\lambda u+u=(\lambda^2+\lambda+1)u=p(\lambda)u.
$$
Since $u\neq0\Rightarrow p(\lambda)=0$.


TITLE: Relation between Work-Energy Theorem and Collisions
QUESTION [1 upvotes]: I am stuck on this idea relating the Work-Energy theorem and collisions. The derivation for the Work-Energy theorem is entirely mathematical, and simply involves
Net work on a system = $$W_{net} = \int_A^B \vec{F_{net}}\cdot d\vec{r} =\ldots = \frac{m}{2}\int_A^B\left(\frac{d}{dt} v^2\right) dt = \Delta K_{sys}$$
My interpretation is that, if this is an equality, then the change in the kinetic energy must be due to the work of some external force on our system. But if we consider a collision involving two balls with no net external forces, by definition (and experimentation and the like), $\Delta K_{sys} = 0 $ for elastic collisions, $\Delta K_{sys} < 0$ for inelastic collisions, and $\Delta K_{sys} > 0$ for superelastic collisions. Textbooks online and Kleppner and Kolenkow attribute the change in/constancy of kinetic energy of the system to interaction forces during the collision and the work they do. But how are these interaction forces between balls of any relevance if they are internal to the system? How can the work of these internal forces change the system's kinetic energy if such a change can only be attributed to the work of external forces?
Similarly, even if we were to try to compute in some way the relevance of these interaction forces, how does them being conservative or non-conservative affect the velocities/kinetic energies of the balls? As far as I know, the conservativity of a force only relates to the work it does being path-independent or equivalently, that it is the gradient of some potential function, which is irrelevant to kinetic energies.
For further confusion, KK differentiates between elastic collisions having conservative interaction forces and inelastic collisions having non-conservative interaction forces, but doesn't say which for superelastic collisions. The example KK uses for superelastic collisions is the collision of two cocked mouse traps which release the energy stored in the springs. But the spring forces there should be conservative. In another example I found online, the instructor gave an example of a superelastic collision where the balls had some explosive chemical that ignited during the collision and increased the kinetic energies post-collision. I am not sure about the conservativity of the forces involved in the chemical explosion, but I would guess they are not conservative. Is there a defined category of interaction forces involved in superelastic collisions?
Thanks!

REPLY [1 votes]: As has been indicated, you have asked multiple questions, whereas you should focus on one question per post. However, I think a major part of the problem you are having is due to the mixing of two different but related concepts:
(1) The change in kinetic energy of a particle or rigid object due to net work done on the particle or object is net work done by all forces, not just external forces, per the work energy theorem
(2) The change in total mechanical energy (KE + PE) of a system, due to net external work done on a system.
Although you have referred to the work energy theorem as the the net work done on a "system", you will find that the most most common statement of the theorem is the "net work done on a particle or object equals its change in kinetic energy". In my view, the use of the term "system" in connection with the work energy theorem, as is sometimes done, is problematic because a "system" can undergo a change potential energy (PE) as well as kinetic energy (KE) as a result of net work on the system, whereas an particle or object alone cannot undergo a change in PE. And that's because PE is a system property and not a property of an object alone.

$\Delta K_{sys} < 0$ for inelastic collisions

For inelastic collisions conservation of mechanical KE does not apply. You can use the WET to analyze collisions with respect to the average impact force during a collision. An example is given here: http://hyperphysics.phy-astr.gsu.edu/hbase/carcr.html#cc1

$\Delta K_{sys} > 0$ for superelastic collisions

Here the increase in KE has to do with conservation of mechanical energy of an isolated system (concept 2 above), which can be stated as, in the absence of external work,
$$\Delta KE+\Delta PE=0$$
So for the "superelastic" collision, the increase in KE simply equals the decrease in PE stored in the system (elastic potential energy in the case of the example of sprung mousetraps, or stored chemical potential energy in the case of explosives, though chemical potential energy is not "mechanical" energy)

How can the work of these internal forces change the system's kinetic
energy if such a change can only be attributed to the work of external
forces?

Because the change in the KE of a system is not only attributable to external forces. The change in KE plus the change in PE of a system is.  Internal forces can convert KE to PE or PE to KE such that for the system $\Delta KE+\Delta PE=0$.

As far as I know, the conservativity of a force only relates to the
work it does being path-independent or equivalently, that it is the
gradient of some potential function, which is irrelevant to kinetic
energies.

Conservative forces are not irrelevant to kinetic energies. A change in PE equals the negative of the work done by a conservative force. For an isolated system, not subjected to dissipative forces, an increase/decrease in PE must be accompanied by a decrease/increase in KE of an equal amount.
Hope this helps.