S(\bar{x},\bar{\Delta}x,\gamma)}{\partial\gamma_{k}}\Big{|}_{(\bar{x},\bar{\Delta}x,\gamma)={\hat{\chi}}}$
$\displaystyle=\Xi_{k}(0)\qquad\qquad\qquad\qquad\forall k\in[1\\!:\\!K].$
(53)

Cone $N({\hat{\bar{x}}};\widetilde{\mathcal{C}})+\mathfrak{P}_{\lim}$ and
subgradients of $\operatorname*{lsc}_{\widetilde{\mathcal{C}}}{W}_{\sup}$

First, define the sets $\widetilde{\mathcal{C}}$ and $\bar{\mathcal{X}}$ as
follows:

$\displaystyle\widetilde{\mathcal{C}}\equiv\bar{\mathcal{X}}\stackrel{{\scriptstyle\triangle}}{{=}}\mathbb{R}\times\mathcal{C}_{\operatorname*{\textrm{home}}}$
$\displaystyle\textrm{\ in the proof of Theorem\leavevmode\nobreak\ \ref{7}},$
$\displaystyle\widetilde{\mathcal{C}}\stackrel{{\scriptstyle\triangle}}{{=}}\mathbb{R}\times(G\cap\mathcal{C}_{\operatorname*{\textrm{home}}}),\
\bar{\mathcal{X}}\stackrel{{\scriptstyle\triangle}}{{=}}\mathbb{R}\times G$
$\displaystyle\textrm{\ in the proof of Theorem\leavevmode\nobreak\ \ref{8}}.$

In both cases, we also have
${\hat{\bar{x}}}\in\widetilde{\mathcal{C}}\subset\mathbb{R}\times\mathcal{C}_{\operatorname*{\textrm{home}}}$
and

$\displaystyle N({\hat{\bar{x}}};\bar{\mathcal{X}})\subset
N({\hat{\bar{x}}};\widetilde{\mathcal{C}})=\\{0\\}\times
N({\hat{y}}(0);\mathcal{C}_{\operatorname*{\textrm{home}}}).$ (54)

Note that a pair $(\lambda,-\psi(0))\in\bar{\mathbb{X}}^{*}$ satisfies (7) if
this pair lies in $\\{0\\}\times
N({\hat{y}}(0);\mathcal{C}_{\operatorname*{\textrm{home}}})+\mathfrak{P}_{\lim}$,
where

$\mathfrak{P}_{\lim}\stackrel{{\scriptstyle\triangle}}{{=}}\operatorname*{Limsup}_{n\uparrow\infty}\operatorname*{co}\operatorname*{cl}\Big{\\{}\lambda\big{(}1,\frac{\partial{\hat{J}}}{\partial
x}(x;\theta)\big{)}\,\Big{|}\,\,\lambda>0,\
\bar{x}=(r,x)\in\bar{\mathcal{X}},\qquad\qquad\qquad\qquad\\\
||\bar{x}-{\hat{\bar{x}}}||<\varkappa^{n},\
|{\hat{J}}(x;\theta)-{\hat{J}}({\hat{y}}(0);\theta)|<\varkappa^{n},\
\theta>1/\varkappa^{n}\Big{\\}}.$

Second, define the functions ${W}_{\theta}:\bar{\mathbb{X}}\to\mathbb{R}$
$(\theta>0)$ as follows:

$\displaystyle{W}_{\theta}(r,x)\stackrel{{\scriptstyle\triangle}}{{=}}r+{\hat{J}}(x;\theta)-{\hat{J}}({\hat{y}}(0);\theta)=\bar{e}^{*}_{0}\mathrm{z}(\bar{x},0,{\hat{u}};\theta)-\bar{e}^{*}_{0}\mathrm{z}({\hat{\bar{x}}},0,{\hat{u}};\theta)\quad\forall\bar{x}=(r,x)\in\bar{\mathbb{X}}.$

It is easy to see that ${W}_{\theta}({\hat{\bar{x}}})=0$. Now, these maps are
Fréchet differentiable in $(r,x)$ on
$\mathbb{R}\times{\hat{\mathbb{G}}}_{\exists}$ by $(H3)$. In particular, by
(49), $\frac{\partial{W}_{\theta}(\bar{x})}{\partial\bar{x}}$ is well defined
for all $\bar{x}=(w_{\omega},b_{\omega})$ and $\theta>0$ if
$\omega\in\Omega_{\varkappa}$. Further, for all
$\bar{x}=(r,x)\in\mathbb{R}\times{\hat{\mathbb{G}}}_{\exists}$, we obtain
$\partial{W}_{\theta}(r,x)=\\{(1,\frac{\partial{\hat{J}}}{\partial
x}(x;\theta))\\}$ and

$\displaystyle{N}(r,{x},{W}_{\theta}(r,{x});\operatorname*{epi}{W}_{\theta})={\hat{N}}_{0}(r,{x},{W}_{\theta}(r,{x});\operatorname*{epi}{W}_{\theta})=\bigcup_{\lambda\geq
0}\lambda\big{(}1,\frac{\partial{\hat{J}}}{\partial x}(x;\theta),-1\big{)}.$
(55)

At last, define functions ${W}_{\sup}$,
$\operatorname*{lsc}_{\widetilde{\mathcal{C}}}{W}_{\sup}$, and
$\operatorname*{lsc}_{\bar{\mathcal{X}}}{W}_{\sup}$ from $\bar{\mathbb{X}}$ to
$\mathbb{R}$ by rules (24) and (25). Now, the equality
${W}_{\sup}({\hat{\bar{x}}})=0$ is trivial.

Under the assumption
$\operatorname*{lsc}_{\widetilde{\mathcal{C}}}{W}_{\sup}({\hat{\bar{x}}})={W}_{\sup}({\hat{\bar{x}}})=0$,
now we claim that

$\partial\operatorname*{lsc}\nolimits_{\widetilde{\mathcal{C}}}{W}_{\sup}({\hat{\bar{x}}})\subset\\{0\\}\times
N({\hat{y}}(0);\mathcal{C}_{\operatorname*{\textrm{home}}})+\bigcup_{(1,q)\in\mathfrak{P}_{\lim}}\\{(1,q)\\},\\\
\partial^{\infty}\operatorname*{lsc}\nolimits_{\widetilde{\mathcal{C}}}{W}_{\sup}({\hat{\bar{x}}})\subset\\{0\\}\times
N({\hat{y}}(0);\mathcal{C}_{\operatorname*{\textrm{home}}})+\bigcup_{(0,q)\in\mathfrak{P}_{\lim}}\\{(0,q)\\}.$
(56)

We will prove the equality
$\operatorname*{lsc}_{\widetilde{\mathcal{C}}}{W}_{\sup}({\hat{\bar{x}}})={W}_{\sup}({\hat{\bar{x}}})=0$
in the next step.

To verify (56) for the proof of Theorem 4.1 under the assumption
$\operatorname*{lsc}_{\widetilde{\mathcal{C}}}{W}_{\sup}({\hat{\bar{x}}})=0$,
note that ${\widetilde{\mathcal{C}}}$ is convex and the maps
${W}_{\sup}({\hat{\bar{x}}})$ are Fréchet differentiable. Then, by Lemma 6.7
for ${{\mathcal{X}}}={\widetilde{\mathcal{C}}}$, we have shown (41). Hence, by
(5c), (5d), and (55), the set
$N({\hat{\bar{x}}},0;\operatorname*{epi}\operatorname*{lsc}\nolimits_{\widetilde{\mathcal{C}}}{W}_{\sup})$
is contained in

$\displaystyle\qquad
N({\hat{\bar{x}}};{\widetilde{\mathcal{C}}})\times\\{0\\}+\bigcap_{n\in\mathbb{N}}\operatorname*{co}\operatorname*{cl}\bigcup_{\begin{subarray}{c}\bar{x}\in{\widetilde{\mathcal{C}}},\
||\bar{x}-{\hat{\bar{x}}}||<\varkappa^{n},\\\
|{W}_{\theta}(\bar{x})|<\varkappa^{n},\
\theta>1/\varkappa^{n}\end{subarray}}{\hat{N}}_{0}(\bar{x},{W}_{\theta}(\bar{x});\operatorname*{epi}\operatorname*{lsc}\nolimits_{\widetilde{\mathcal{C}}}{W}_{\theta})$
$\displaystyle=\\{0\\}\times
N({\hat{y}}(0);\mathcal{C}_{\operatorname*{\textrm{home}}})\times\\{0\\}+\bigcap_{n\in\mathbb{N}}\operatorname*{co}\operatorname*{cl}\bigcup_{\begin{subarray}{c}\bar{x}=(r,x)\in\bar{\mathcal{X}},\
||\bar{x}-{\hat{\bar{x}}}||<\varkappa^{n},\\\
|{W}_{\theta}(\bar{x})|<\varkappa^{n},\
\theta>1/\varkappa^{n}\end{subarray}}\big{\\{}\lambda\big{(}\bar{e}_{0}^{*}A_{x,\theta}(0),-1\big{)}\big{\\}}$
$\displaystyle=\\{0\\}\times
N({\hat{y}}(0);\mathcal{C}_{\operatorname*{\textrm{home}}})\times\\{0\\}+\bigcup_{(\lambda,q)\in\mathfrak{P}_{\lim}}\big{\\{}\big{(}\lambda,q,-\lambda\big{)}\big{\\}}.$

So, by (5a) and (5b), we have proved (56) under the assumption
$\operatorname*{lsc}_{\widetilde{\mathcal{C}}}{W}_{\sup}({\hat{\bar{x}}})={W}_{\sup}({\hat{\bar{x}}})=0$
in the proof of Theorem 4.1.

To verify (56) for the proof of Theorem 4.2, note that, first, decreasing $G$
if necessary, there is no loss of generality in supposing that $G$ and
$\bar{\mathcal{X}}$ are convex. Second, since the lower semicontinuous map
(12) coincides with $W_{\sup}$ on $\bar{\mathcal{X}}$, one has
$\operatorname*{lsc}_{\bar{\mathcal{X}}}W_{\sup}|_{\bar{\mathcal{X}}}=W_{\sup}|_{\bar{\mathcal{X}}}$
and
$\operatorname*{lsc}_{\bar{\mathcal{X}}}W_{\sup}({\hat{\bar{x}}})=W_{\sup}({\hat{\bar{x}}})=0$.
Again, applying Lemma 6.7, we obtain (41), and this inclusion entails the
inclusions

$\partial\operatorname*{lsc}\nolimits_{\bar{\mathcal{X}}}{W}_{\sup}({\hat{\bar{x}}})\subset
N({\hat{\bar{x}}};\bar{\mathcal{X}})+\\!\\!\\!\\!\bigcup_{(1,q)\in\mathfrak{P}_{\lim}}\\!\\!\\!\\!\big{\\{}(1,q)\big{\\}},\
\partial^{\infty}\operatorname*{lsc}\nolimits_{\bar{\mathcal{X}}}{W}_{\sup}({\hat{\bar{x}}})\subset
N({\hat{\bar{x}}};\bar{\mathcal{X}})+\\!\\!\\!\\!\bigcup_{(0,q)\in\mathfrak{P}_{\lim}}\\!\\!\\!\\!\big{\\{}(0,q)\big{\\}}.$
(57)

Hence the assumption (9) yields the following constraint qualification
condition:

$\big{(}-N({\hat{\bar{x}}};\widetilde{\mathcal{C}})\big{)}\cap\bigcup_{(0,q)\in\mathfrak{P}_{\lim}}\\{(0,q)\\}\stackrel{{\scriptstyle(\ref{2485})}}{{\subset}}\\{0\\}\times\Big{(}\big{(}-N({\hat{y}}(0);\mathcal{C}_{\operatorname*{\textrm{home}}})\big{)}\cap\bigcup_{(0,q)\in\mathfrak{P}_{\lim}}\\{q\\}\Big{)}\stackrel{{\scriptstyle(\ref{gad4})}}{{\subset}}\\{(0,0)\\}.$

Then, first, one has
$\operatorname*{lsc}_{\bar{\mathcal{X}}}W_{\sup}=W_{\sup}$ on
$\bar{\mathcal{X}}$ by (25); second, because
$\operatorname*{cl}\widetilde{\mathcal{C}}$ is a closed subset of
$\bar{\mathcal{X}}$, we also have
$\operatorname*{lsc}_{\bar{\mathcal{X}}}W_{\sup}=\operatorname*{lsc}_{\widetilde{\mathcal{C}}}W_{\sup}$
on ${\tilde{\mathcal{C}}}$ and
$\operatorname*{lsc}_{\widetilde{\mathcal{C}}}W_{\sup}=\operatorname*{lsc}_{\bar{\mathcal{X}}}W_{\sup}+\delta_{\operatorname*{cl}\widetilde{\mathcal{C}}}$.
Further,
$N({\hat{\bar{x}}};\widetilde{\mathcal{C}})=N({\hat{\bar{x}}};\widetilde{\mathcal{C}}\cap\bar{\mathcal{X}})\supset
N({\hat{\bar{x}}};\widetilde{\mathcal{C}})+N({\hat{\bar{x}}};\bar{\mathcal{X}})$
holds. Since the corresponding constraint qualification condition holds, by
[31, Theorem 2.19], we have
$\partial\operatorname*{lsc}_{\widetilde{\mathcal{C}}}W_{\sup}({\hat{\bar{x}}})=\partial\operatorname*{lsc}_{\bar{\mathcal{X}}}W_{\sup}({\hat{\bar{x}}})+N({\hat{\bar{x}}};\widetilde{\mathcal{C}})$
and
$\partial^{\infty}\operatorname*{lsc}_{\widetilde{\mathcal{C}}}W_{\sup}({\hat{\bar{x}}})=\partial^{\infty}\operatorname*{lsc}_{\bar{\mathcal{X}}}W_{\sup}({\hat{\bar{x}}})+N({\hat{\bar{x}}};\widetilde{\mathcal{C}})$.
Combining it, (57), and (54) gives (56) under the conditions of Theorem 4.2.

The constrained optimization problem in a finite-dimensional space

First, consider an $\omega=(0,\Delta
x,\gamma)\in\\{0\\}\times(\mathcal{C}_{0}-{\hat{y}}(0))\times[0;\varkappa]^{K}$;
for all $\theta>T$ combining
${\hat{u}}|_{[T;\infty)}={u}_{\omega}|_{[T;\infty)}$ with
$\mathrm{z}((w_{\omega},b_{\omega}),0,{\hat{u}};T)=z_{\omega}(T)$ leads to

$\displaystyle
J(y_{\omega}(0),0,u_{\omega};\theta)-{\hat{J}}({\hat{y}}(0);\theta)$
$\displaystyle=J(y_{\omega}(0),0,u_{\omega};T)+J(y_{\omega}(T),T,{\hat{u}};\theta)-{\hat{J}}({\hat{y}}(0);\theta)$
$\displaystyle=J(y_{\omega}(0),0,u_{\omega};T)\\!-\\!{\hat{J}}(b_{\omega};T)+{\hat{J}}(b_{\omega};\theta)-{\hat{J}}({\hat{y}}(0);\theta)$
$\displaystyle=w_{\omega}+{W}_{\theta}(0,b_{\omega})={W}_{\theta}(w_{\omega},b_{\omega}).$

Passing to the upper pointwise limit, we obtain

$\limsup_{\theta\uparrow\infty}\Big{[}J(y_{\omega}(0),0,u_{\omega};\theta)-{\hat{J}}({\hat{y}}(0);\theta)\Big{]}={W}_{\sup}(w_{\omega},b_{\omega}).$

Due to the definition of the locally weakly overtaking criterion (see (3c)),
combining (49) with (3b) yields

$l(y_{\omega}(0))+{W}_{\sup}(w_{\omega},b_{\omega})\geq l({\hat{y}}(0))$

if the process $(y_{\omega},u_{\omega})$ is admissible, i.e., if
$b_{\omega}\in\mathcal{C}_{\operatorname*{\textrm{home}}}$ and $\Delta
x=y_{\omega}(0)-{\hat{y}}(0)$ lies in $\mathcal{C}_{0}-{\hat{y}}(0)$. So,
${\hat{\chi}}=({\hat{\bar{x}}},(0,0),0)\in\bar{\mathbb{X}}\times\bar{\mathbb{X}}\times\mathbb{R}^{K}$
is optimal in the problem

minimize $\displaystyle{l}({\hat{y}}(0)+\Delta x)+{W}_{\sup}(\bar{x})$ subject
to $\displaystyle\bar{x}=(w_{\omega},b_{\omega}),\
\bar{x}\in\mathbb{R}\times\mathcal{C}_{\operatorname*{\textrm{home}}},\
{\omega}\\!=\\!(r,\Delta
x,\gamma)\in\\{0\\}\times(\mathcal{C}_{0}-{\hat{y}}(0))\times[0;\varkappa]^{K}.$

Note that, by the definition of $S$, one has $\bar{x}=(w_{\omega},b_{\omega})$
iff $S(\bar{x},\omega)=0$ holds. Now, since lower semicontinuous function
$\operatorname*{lsc}_{\widetilde{\mathcal{C}}}{W}_{\sup}$ equals $+\infty$ on
$\bar{\mathbb{X}}\setminus\operatorname*{cl}\widetilde{\mathcal{C}}$, due to
the continuity of functions $l$ and $(w_{\cdot},b_{\cdot})$, the point
${\hat{\chi}}=({\hat{\bar{x}}},(0,0),0)\in\widetilde{\mathcal{C}}\times\\{0\\}\times(\mathcal{C}_{0}-{\hat{y}}(0))\times[0;\varkappa]^{K}$
is also a minimizer of the problem

minimize $\displaystyle{l}({\hat{y}}(0)+\Delta
x)+\operatorname*{lsc}\nolimits_{\widetilde{\mathcal{C}}}{W}_{\sup}(\bar{x})$
subject to $\displaystyle S(\bar{x},\omega)=0,\ \bar{x}\in\bar{\mathbb{X}},\
{\omega}=(r,\Delta
x,\gamma)\in\\{0\\}\times(\mathcal{C}_{0}-{\hat{y}}(0))\times[0;\varkappa]^{K}.$
(58)

Further, we also obtain
$\operatorname*{lsc}_{\widetilde{\mathcal{C}}}{W}_{\sup}({\hat{\chi}})={W}_{\sup}({\hat{\chi}})$
and we have just proved (56).

Define the function
$R:\bar{\mathbb{X}}\times\bar{\mathbb{X}}\times\mathbb{R}^{K}\to\mathbb{R}\cup\\{+\infty\\}$
by the rule

$\displaystyle R(\bar{x},r,{\Delta}x,\gamma)$
$\displaystyle\stackrel{{\scriptstyle\triangle}}{{=}}{l}({\hat{y}}(0)+\Delta
x)+\operatorname*{lsc}\nolimits_{\widetilde{\mathcal{C}}}{W}_{\sup}(\bar{x})$

whenever $(\bar{x},r,{\Delta}x,\gamma)$. Now, since $l$ is Lipschitz
continuous and $\operatorname*{lsc}_{\widetilde{\mathcal{C}}}{W}_{\sup}$ is
independent of $\Delta x$, by the sum rule [31, Theorem 2.19], we obtain

$\partial
R({\hat{\chi}})\subset\partial\operatorname*{lsc}\nolimits_{\widetilde{\mathcal{C}}}{W}_{\sup}({\hat{\bar{x}}})\times\\{0\\}\times\partial{l}({\hat{y}}(0))\times\\{0\\},\
\
\partial^{\infty}R({\hat{\chi}})\subset\partial^{\infty}\operatorname*{lsc}\nolimits_{\widetilde{\mathcal{C}}}{W}_{\sup}({\hat{\bar{x}}})\times\\{(0,0,0)\\}$
(59)

for ${\hat{\chi}}=({\hat{\bar{x}}},(0,0),0)$. In particular, we obtain the
constraint qualification condition

$(-\partial^{\infty}R({\hat{\chi}}))\cap
N({\hat{\chi}};\bar{\mathbb{X}}\times\\{0\\}\times(\mathcal{C}_{0}-{\hat{y}}(0))\times[0;\varkappa]^{K})\subset\\{(0,0,0,0)\\}.$
(60)

Define
$\bar{e}_{1}^{*}\stackrel{{\scriptstyle\triangle}}{{=}}(0,1,0,\dots,0)$,
$\bar{e}_{2}^{*}\stackrel{{\scriptstyle\triangle}}{{=}}(0,0,1,0,\dots,0)$,…,$\bar{e}_{m}^{*}\stackrel{{\scriptstyle\triangle}}{{=}}(0,0,\dots
0,1)\in\bar{\mathbb{X}}^{*}$, where
$m=\dim{\mathbb{X}}=\dim\bar{\mathbb{X}}-1$. Then, the point ${\hat{\chi}}$ is
a local solution of the problem

minimize $\displaystyle\ R(\bar{x},r,{\Delta}x,\gamma)={l}({\hat{y}}(0)+\Delta
x)+\operatorname*{lsc}\nolimits_{\widetilde{\mathcal{C}}}{W}_{\sup}(\bar{x})$
subject to $\displaystyle\ \bar{e}_{0}^{*}S(\bar{x},r,{\Delta}x,\gamma)=0,\
\bar{e}_{1}^{*}S(\bar{x},r,{\Delta}x,\gamma)=0,\dots,\bar{e}_{m}^{*}S(\bar{x},r,{\Delta}x,\gamma)=0,$
$\displaystyle\
(\bar{x},r,{\Delta}x,\gamma)\in\bar{\mathbb{X}}\times\\{0\\}\times(\mathcal{C}_{0}-{\hat{y}}(0))\times[0;\varkappa]^{K}.$

The necessary conditions for this constrained optimization problem

Since all functions $\bar{e}_{i}S$ are Lipschitz continuous and the constraint
qualification condition (60) also holds, thanks to the Lagrange multiplier
rule [20, Theorem 1(b)], [21, Corollary 2.6], there exist a vector
$\mu=(\mu_{0},\dots,\mu_{m})\in\bar{\mathbb{X}}^{*}$ and a nonnegative number
${\hat{\lambda}}$ satisfying $({\hat{\lambda}},\mu)\neq 0$ and

$0\in{\hat{\lambda}}\partial
R({\hat{\chi}})+\partial^{\infty}R({\hat{\chi}})+N({\hat{\chi}};\bar{\mathbb{X}}\times\\{0\\}\times(\mathcal{C}_{0}-{\hat{y}}(0))\times[0;\varkappa]^{K})+\sum_{i=0}^{m}\mu_{i}({\partial}(\bar{e}_{i}^{*}S)({\hat{\chi}})\cup{\partial}(-\bar{e}_{i}^{*}S)({\hat{\chi}})).$

Due to the inclusion (59) and equalities (5a) and (5b), we obtain

$(0,0,0,0)\in\big{(}{\hat{\lambda}}\partial\operatorname*{lsc}\nolimits_{\widetilde{\mathcal{C}}}{W}_{\sup}({\hat{\bar{x}}})+\partial^{\infty}\operatorname*{lsc}\nolimits_{\widetilde{\mathcal{C}}}{W}_{\sup}({\hat{\bar{x}}})\big{)}\times\\{0\\}\times{\hat{\lambda}}\partial{l}({\hat{y}}(0))\times\\{0\\}\\\
+\\{0\\}\times\mathbb{R}\times
N({\hat{y}}(0);\mathcal{C}_{0})\times(-\mathbb{R}_{+})^{K}+\sum_{i=0}^{m}{\partial}(\mu_{i}\sigma_{i}\bar{e}_{i}^{*}S)({\hat{\chi}})$

for some signs $\sigma_{i}\in\\{1,-1\\}$, $i\in[0\\!:\\!m]$. Put
${\hat{\mu}}\stackrel{{\scriptstyle\triangle}}{{=}}(\mu_{0}\sigma_{0},\dots,\mu_{m}\sigma_{m})$.
Due to the upper-estimate (56) and the equality
$\mathfrak{P}_{\lim}+\mathfrak{P}_{\lim}=\mathfrak{P}_{\lim}$, one can find a
subgradient

$({\hat{\lambda}},{\hat{q}})\in{\hat{\lambda}}\partial\operatorname*{lsc}\nolimits_{\widetilde{\mathcal{C}}}{W}_{\sup}({\hat{\chi}})+\partial^{\infty}\operatorname*{lsc}\nolimits_{\widetilde{\mathcal{C}}}{W}_{\sup}({\hat{\chi}})\subset\\{0\\}\times
N({\hat{y}}(0);\mathcal{C}_{\operatorname*{\textrm{home}}})+\mathfrak{P}_{\lim}$

satisfying

$\displaystyle(0,0,0,0)\in\\{({\hat{\lambda}},{\hat{q}})\\}\times\mathbb{R}\times\partial({\hat{\lambda}}{l}+\delta_{\mathcal{C}_{0}})({\hat{y}}(0))\times(-\mathbb{R}_{+})^{K}+\sum_{i=0}^{m}{\partial}({\hat{\mu}}_{i}\bar{e}_{i}^{*}S)({\hat{\chi}}).$

According to the equalities $\frac{\partial
S}{\partial{\bar{\Delta}}x}({\hat{\chi}})=E=-\frac{\partial
S}{\partial{\bar{x}}}({\hat{\chi}}),$ $\frac{\partial
S}{\partial\gamma_{k}}({\hat{\chi}})=A^{-1}_{b_{0},T}(0)\bar{\Delta}\bar{f}_{k}$,
and $\sum_{i=0}^{m}{\hat{\mu}}_{i}\bar{e}^{*}_{i}={\hat{\mu}}$, we may rewrite
these relations as follows:

$\displaystyle{\hat{\mu}}\stackrel{{\scriptstyle(\ref{FFF0})}}{{=}}-\sum_{i=1}^{m}{\hat{\mu}}_{i}\bar{e}_{i}^{*}\frac{\partial
S({\hat{\chi}})}{\partial{\bar{x}}}$
$\displaystyle=({\hat{\lambda}},{\hat{q}}),$
$\displaystyle-{\hat{\mu}}\stackrel{{\scriptstyle(\ref{FFF0})}}{{=}}-\sum_{i=1}^{m}{\hat{\mu}}_{i}\bar{e}_{i}^{*}\frac{\partial
S({\hat{\chi}})}{\partial\bar{\Delta}x}$
$\displaystyle\in\mathbb{R}\times\big{(}{\hat{\lambda}}{\partial}{l}({\hat{y}}(0))+N({\hat{y}}(0);\mathcal{C}_{0})\big{)},$
$\displaystyle-{\hat{\mu}}A^{-1}_{b_{0},T}(0)\bar{\Delta}\bar{f}_{k}\stackrel{{\scriptstyle(\ref{FFF0})}}{{=}}-\sum_{i=1}^{m}{\hat{\mu}}_{i}\bar{e}^{*}_{i}\frac{\partial
S({\hat{\chi}})}{\partial\gamma_{k}}$
$\displaystyle\in-\mathbb{R}_{+}\qquad\qquad\qquad\qquad\forall
k\in[1\\!:\\!K].$

Now, the first relation implies ${\hat{\lambda}}={\hat{\mu}}_{0}$. Define the
map ${\hat{\psi}}:\mathbb{R}_{+}\to\mathbb{X}^{*}$ by the rule

${\hat{\mu}}{A}^{-1}_{{\hat{y}}(0),t}(0)=({\hat{\lambda}},-{\hat{\psi}}(t))\qquad\forall
t\geq 0.$

Then, the pair $({\hat{\lambda}},-{\hat{\psi}})$ lies in
$\mathfrak{Q}_{\lim}$; further,
$({\hat{\lambda}},{\hat{\lambda}},-{\hat{\psi}}(0))=({\hat{\lambda}},{\hat{\mu}})\neq
0$ holds. On the one hand, the second relation yields
${\hat{\psi}}(0)\in{\hat{\lambda}}{\partial}{l}({\hat{y}}(0))+N({\hat{y}}(0);\mathcal{C}_{0})$;
therefore, the condition (6) has been proved. On the other hand, for all $k\in
I_{K}$, the third relation implies
$({\hat{\lambda}},-{\hat{\psi}}(0)){A}^{-1}_{{\hat{y}}(0),t_{k}}(0)\big{(}\bar{f}(t_{k},z_{0}(t_{k}),u_{k})-\bar{f}(t_{k},z_{0}(t_{k}),{\hat{u}}(t_{k}))\big{)}\geq
0$, i.e. (45) for all
$(t,\upsilon)\stackrel{{\scriptstyle\triangle}}{{=}}(t_{k},u_{k})$; so, (1c)
has also been proved. At last, the first relation yields

$-{\hat{\psi}}(0)={\hat{q}}\in
N({\hat{y}}(0);\mathcal{C}_{\operatorname*{\textrm{home}}})+\bigcup_{({\hat{\lambda}},q)\in\mathfrak{P}_{\lim}}\\{q\\};$

now, ${\hat{\psi}}(0)$ satisfies (7) by the definition of
$\mathfrak{P}_{\lim}$.

The proofs are completed. ∎

######  Remark 8.

Note that in the proof of Theorem 4.1 the convexity of
$\mathcal{C}_{\operatorname*{\textrm{home}}}$ was applied only to the proving
(56). Hence, without this assumption on
$\mathcal{C}_{\operatorname*{\textrm{home}}}$, we obtain
$({\hat{\lambda}},{\hat{q}},-{\hat{\lambda}})=({\hat{\lambda}},-{\hat{\psi}}(0),-{\hat{\lambda}})\in
N(0,{\hat{y}}(0),0;\operatorname*{epi}\operatorname*{lsc}\nolimits_{\tilde{\mathcal{C}}_{\operatorname*{\textrm{home}}}}{W}_{\sup})$.
Then, under conditions $(H0)$–$(H5)$, for each locally weakly overtaking
optimal process $({\hat{y}},{\hat{u}})$, one finds a nonzero solution
$({\hat{\psi}},{\hat{\lambda}})$ of the corresponding to
$({\hat{y}},{\hat{u}})$ system (1b)–(1c), (6) and (61):

$\displaystyle-({\hat{\psi}}(0),\lambda)\in\operatorname*{co}N\big{(}{\hat{y}}(0),0;\operatorname*{epi}\operatorname*{lsc}\limsup_{\theta\uparrow\infty}\big{[}{\hat{J}}(\cdot;\theta)-{\hat{J}}({\hat{y}}(0);\theta)+\delta_{\mathcal{C}_{\operatorname*{\textrm{home}}}}(\cdot)\big{]}\big{)}.$
(61)

This condition is very weak and very awkward for applications. But, formally,
the necessity of this condition does not demand any asymptotic assumptions
even where asymptotic terminal constraint is active. The necessary condition
alike (61) for problems without asymptotic constraints can be seen in [42,
Theorem 19(i)].

######  Remark 9.

Note that hypothesis $(H2)$ may be omitted from the assumptions of Theorems
4.1 and 4.2. Indeed, since the process $({\hat{y}},{\hat{u}})$ is locally
optimal, by (49), the hypotheses $(H3)$-$(H5)$ guarantee the existence of all
needed motions and payoffs.

##  8 Proofs of Theorems 4.5 and 4.6

We will again prove both theorems simultaneously. The idea of the proof is
briefly described as follows. At the beginning, we replace the control set $U$
by some dense countable set $U^{\textrm{dense}}\subset U$ for betterment of
the verification of [16, Hypothesis 22.25]. Next, following Halkin’s method
[2], we will reduce the verification of all desired relations on $\mathbb{R}$
to its verification on finite intervals $[0;n]$. Then, for fixed $n$,
extending $f$ and $f_{0}$ from $[0;n]$ to $[0;2n]$ by back-track, we will
consider a new dynamics function $f_{n}$ and integrand $f_{0,n}$; this trick
ensures the transfer of the right endpoint to ${\hat{y}}(0)$. Next, we will
rewrite the corresponding criterions (weakly overtaking criterion in Theorem
4.1 and overtaking criterion in Theorem 4.6) in terms of a constrained
optimization problem in a finite-dimensional space. Then, we will reduce this
problem to some finite-horizon control problem. Proceed to Halkin’s method, we
will consider the corresponding Pontryagin Maximum Principle for this control
problem and refine the happened transversality condition by Lemma 6.3 and
Lemma 6.2 in the proof of Theorem 4.5 and the proof of Theorem 4.6,
respectively.

Step 0. Let
$U^{\textrm{dense}}\stackrel{{\scriptstyle\triangle}}{{=}}\\{u^{1},u^{2},\dots\\}$
be a dense countable subset of $U$; fix its numeration. For all
$n\in\mathbb{N}$ define the sets
${U}_{n}(t)\stackrel{{\scriptstyle\triangle}}{{=}}\\{{\hat{u}}(t),u^{1},\dots,u^{n}\\}$
for all positive $t\geq 0$; then, we have

$\displaystyle\operatorname*{Limsup}_{n\to\infty}U_{n}(t)=U\qquad\forall t\geq
0.$ (62)

Then, for all natural $n$, by $\mathcal{U}_{n}$ denote the set of all
admissible control $u$ satisfying $u(t)\in U_{n}(t)$ for all $t\in[0;n]$ and
$u(t)={\hat{u}}(t)$ for all $t>n$. Thus, ${\hat{u}}$ lies in each
$\mathcal{U}_{n}$.

Step 1. We begin by showing that both theorems are implies from the following
assertion:

  * $(A)$ there exist a co-state arc ${\psi}_{n}\in C([0;n],\mathbb{X}^{*})$ and a number ${\lambda}_{n}\in\\{0,1\\}$ such that a nonzero pair $(\psi_{n}(0),\lambda_{n})$ satisfies with $({\hat{y}},{\hat{u}})$ adjoint equation (1b), the all desired transversality conditions (either (6) and (7) or (6) and (15)), and

$\\!\\!\sup_{\upsilon\in
U_{n}(t)}\\!{H}\big{(}{\hat{y}}(t),{\psi}_{n}(t),\upsilon,\lambda_{n}(t),t\big{)}\\!=\\!{H}\big{(}{\hat{y}}(t),{\psi}_{n}(t),{\hat{u}}(t),\lambda_{n},t\big{)}\textrm{\
a.a.\ }t\in[0;n].$ (63)

Assume that assertion $(A)$ is established for all natural $n$. Scaling each
$\psi_{n}$, if necessary, one can also suppose that ${\psi}_{n}(0)=n$ if
$\lambda_{n}=0$. Then, passing from the sequence of $n$ to its certain
subsequence, if necessary, one can assume that either all $\lambda_{n_{k}}$
equal to $1$ and the sequence of $\psi_{n_{k}}(0)$ converges to a vector
$\phi_{0}\in\mathbb{X}^{*}$, or the sequence of $||\psi_{n_{k}}(0)||$ is
unboundedly increasing and the sequence of
$\psi_{n_{k}}(0)/||\psi_{n_{k}}(0)||$ converges to a nonzero vector
$\phi_{0}\in\mathbb{X}^{*}$.

If $\lambda_{n_{k}}\equiv 1$ and $\psi_{n_{k}}(0)\to\phi_{0}$, put
${\hat{\lambda}}\stackrel{{\scriptstyle\triangle}}{{=}}1$ and
$r_{k}\stackrel{{\scriptstyle\triangle}}{{=}}1$. In the case of unbounded
sequence of $\psi_{n_{k}}(0)$ and converging to $\phi_{0}$ sequence of
$\psi_{n_{k}}(0)/||\psi_{n_{k}}(0)||$, put
${\hat{\lambda}}\stackrel{{\scriptstyle\triangle}}{{=}}0$ and
$r_{k}\stackrel{{\scriptstyle\triangle}}{{=}}1/||\psi_{n_{k}}(0)||$.

In both cases, the sequence of $(r_{k}\psi_{n_{k}}(0),r_{k}\lambda_{n_{k}})$
converges to $(\phi_{0},{\hat{\lambda}})$. Now, each pair
$(r_{k}\psi_{n_{k}}(0),r_{k}\lambda_{n_{k}})$ satisfies (1b) and must also
inherit all transversality conditions that are verified for
$({\psi}_{n_{k}},{\lambda}_{n_{k}})$. Further, by the theorem on continuous
dependence of differential equations’ solutions on the initial state, the
sequence of $r_{k}\psi_{n_{k}}$ converges in $\mathbb{R}_{+}$ to a certain
solution ${\hat{\psi}}$ of (1b) with ${\hat{\psi}}(0)=\phi_{0}$ and
$\lambda={\hat{\lambda}}$. Note that this convergence is uniform on an
arbitrary compact interval. Hence the pair $({\hat{\psi}},{\hat{\lambda}})$
also satisfies (1b) on the whole $\mathbb{R}_{+}$. Moreover, taking into
account (62), it follows from (63) for each $n$ that (1c) holds on the whole
$\mathbb{R}_{+}$.

Thus, we need to prove only assertion $(A)$ for all natural $n$.

Step 2. Fix a natural $n$.

Define the dynamics $\bar{f}_{n}:[0;2n]\times\mathbb{X}\times U\to\mathbb{X}$
and the integrand $\bar{f}_{0,n}:[0;2n]\times\mathbb{X}\times U\to\mathbb{R}$
as follows: for all $(t,x,u)\in[0;2n]\times{\mathbb{X}}\times U$,

$\displaystyle(\bar{f}_{n},\bar{f}_{0,n})(t,x,u)=\left\\{\begin{array}[]{cc}(f,f_{0})(t,x,u)&\textrm{if}\
t<n;\\\ -(f,f_{0})(2n-t,x,{\hat{u}}(2n-t))&\textrm{if}\ t\geq
n.\end{array}\right.$

Define also the function
$\bar{H}_{n}:{\mathbb{X}}\times{\mathbb{X}}^{*}\times{U}\times{\mathbb{R}}\times[0,2n]\to{\mathbb{R}}$
by the rule

$\bar{H}_{n}(x,\psi,u,\mu,t)\stackrel{{\scriptstyle\triangle}}{{=}}\psi\bar{f}_{n}\big{(}t,x,u\big{)}+\mu\bar{f}_{0,n}\big{(}t,x,u\big{)}\
$

whenever
$(x,\psi,u,\mu,t)\\!\in\\!{\mathbb{X}}\times{\mathbb{X}}^{*}\times{U}\times{\mathbb{R}}\times[0;2n].$

Now, for every $u\in\mathcal{U}$, $t\in[0;2n]$, and initial state
$x\in\mathbb{X}$, denote by
$(\bar{\mathrm{y}}_{n},\bar{\mathrm{w}}_{n})(x,u;\cdot)$ the unique solution
(on $[0;2n]$) of the Cauchy problem

$\frac{d\bar{y}(\tau)}{d\tau}=\bar{f}_{n}(\tau,\bar{y}(\tau),u(\tau)),\
\frac{d\bar{w}(\tau)}{d\tau}=\bar{f}_{0,n}(\tau,\bar{y}(\tau),u(\tau)),\quad(\bar{y},\bar{w})(0)=(x,0).$

Consider an admissible control process $(y,u)$ such that
${u}|_{[n;\infty)}={\hat{u}}|_{[n;\infty)}$. Let $\check{y}$ be the generated
by optimal control ${\hat{u}}$ motion of (1a) satisfying $y(n)=\check{y}(n)$,
i.e., $\check{y}(\cdot)=\mathrm{y}(y(n),n,{\hat{u}};\cdot)$. It is easy to see
that

$\displaystyle(\bar{\mathrm{y}}_{n},\bar{\mathrm{w}}_{n})\big{(}{y}(0),u;2n\big{)}=\big{(}\check{y}(0),J({y}(0),0,{u};n)-J\big{(}\check{y}(0),0,{\hat{u}};n\big{)}\big{)}.$
(64)

In particular,
$(\bar{\mathrm{y}}_{n},\bar{\mathrm{w}}_{n})\big{(}x,{\hat{u}};2n\big{)}=(x,0)$
holds whenever $x$. Define

${\hat{\bar{y}}}_{n}(t)\stackrel{{\scriptstyle\triangle}}{{=}}\bar{\mathrm{y}}\big{(}{\hat{y}}(0),{\hat{u}};t\big{)},\
{\hat{\bar{w}}}_{n}(t)\stackrel{{\scriptstyle\triangle}}{{=}}\bar{\mathrm{w}}\big{(}{\hat{y}}(0),{\hat{u}};t\big{)}\qquad\forall
t\in[0;2n].$

Then, ${\hat{\bar{y}}}_{n}(2n)={\hat{\bar{y}}}_{n}(0)={\hat{y}}(0)$ and
${\hat{\bar{w}}}_{n}(2n)={\hat{\bar{w}}}_{n}(0)=0$ hold. Set also
${\hat{\bar{s}}}=0\in\mathbb{R}.$

Step 3. Consider maps $\mathbb{X}\ni x\mapsto
W_{\theta}(x)\stackrel{{\scriptstyle\triangle}}{{=}}{\hat{J}}(x;\theta)-{\hat{J}}({\hat{y}}(0);\theta)$
for all $\theta>0$. Define their limits $W_{\sup}$, $W_{\inf}$ from
$\mathbb{X}$ to $\mathbb{R}\cup\\{-\infty,+\infty\\}$ by rule (24). Note that

$W_{\theta}({\hat{y}}(0))=W_{\sup}({\hat{y}}(0))=W_{\inf}({\hat{y}}(0))=0\qquad\forall\theta>0.$
(65)

Define also the functions
$\operatorname*{lsc}\nolimits_{\mathcal{C}_{\operatorname*{\textrm{home}}}}W_{\sup}$
and
$\operatorname*{lsc}\nolimits_{\mathcal{C}_{\operatorname*{\textrm{home}}}}W_{\inf}$
by rules (25). Set

$\displaystyle W_{extr}\stackrel{{\scriptstyle\triangle}}{{=}}W_{\sup}$
$\displaystyle\textrm{\ and\
}\operatorname*{lsc}\nolimits_{\mathcal{C}_{\operatorname*{\textrm{home}}}}W_{extr}\stackrel{{\scriptstyle\triangle}}{{=}}\operatorname*{lsc}\nolimits_{\mathcal{C}_{\operatorname*{\textrm{home}}}}W_{\sup}\textrm{\
in the proof of Theorem\leavevmode\nobreak\ \ref{9}},$ $\displaystyle
W_{extr}\stackrel{{\scriptstyle\triangle}}{{=}}W_{\inf}$
$\displaystyle\textrm{\ and\
}\operatorname*{lsc}\nolimits_{\mathcal{C}_{\operatorname*{\textrm{home}}}}W_{extr}\stackrel{{\scriptstyle\triangle}}{{=}}\operatorname*{lsc}\nolimits_{\mathcal{C}_{\operatorname*{\textrm{home}}}}W_{\inf}\textrm{\
in the proof of Theorem\leavevmode\nobreak\ \ref{8}}.$

Consider some $x\in\mathcal{C}_{0}$ and control $u\in\mathcal{U}_{n}$. By
$y(\cdot)$ denote the generated by control ${u}$ motion of (1a) with $x=y(0)$.
Let $\check{y}$ be the generated by optimal control ${\hat{u}}$ motion of (1a)
satisfying $y(n)=\check{y}(n)$. Now, by (64), we have
$\check{y}(0)=\bar{\mathrm{y}}_{n}((y(0),0),0,{u};2n)$. Accordingly, by
definition of $J$, for the admissible control process $(y,u)$, we obtain

$\displaystyle J(y(0),0,u;\theta)\\!-\\!{\hat{J}}({\hat{y}}(0);\theta)$
$\displaystyle\;=\;J(y(0),0,u;n)+J(\check{y}(n),n,{\hat{u}};\theta)\\!-\\!{\hat{J}}({\hat{y}}(0);\theta)$
$\displaystyle\;=\;J(y(0),0,u;n)\\!-\\!{\hat{J}}(\check{y}(0);n)+J(\check{y}(0),0,{\hat{u}};\theta)\\!-\\!{\hat{J}}({\hat{y}}(0);\theta)$
$\displaystyle\stackrel{{\scriptstyle(\ref{11301})}}{{=}}\bar{\mathrm{w}}_{n}(x,{u};2n)+J\big{(}\bar{\mathrm{y}}_{n}(x,{u};2n),0,{\hat{u}};\theta\big{)}\\!-\\!{\hat{J}}\big{(}{\hat{y}}(0);\theta\big{)}$

for all $\theta>n$. Passing to the corresponding limit as
$\theta\uparrow\infty$, we have

$\displaystyle l({\hat{y}}(0))\\!\\!\\!\\!\\!\\!$ $\displaystyle\leq$
$\displaystyle\\!\\!\\!\\!\\!l(y(0))\\!+\\!\bar{\mathrm{w}}_{n}(x,{u};2n)\\!+\\!W_{\sup}(\bar{\mathrm{y}}_{n}(x,{u};2n))\textrm{\,if\,}({\hat{y}},{\hat{u}})\textrm{\,is\,weakly\,overtaking\,optimal,}$
$\displaystyle l({\hat{y}}(0))\\!\\!\\!\\!\\!\\!$ $\displaystyle\leq$
$\displaystyle\\!\\!\\!\\!\\!l(y(0))\\!+\\!\bar{\mathrm{w}}_{n}(x,{u};2n)\\!+\\!W_{\inf}(\bar{\mathrm{y}}_{n}(x,{u};2n))\textrm{\;if\;}({\hat{y}},{\hat{u}})\textrm{\,is\;overtaking\;optimal.}$

In the proof of Theorem 4.5 as well as the proof of Theorem 4.6, we obtain

$\displaystyle l({\hat{y}}(0))\leq
l(y(0))+\bar{\mathrm{w}}_{n}(y(0),{u};2n)+W_{extr}(\bar{\mathrm{y}}_{n}(y(0),{u};2n))$

for all admissible control processes $(y,u)$. Further, the infimum of the
right side of this inequality is attained at
$(y,u)=\big{(}{\hat{y}},{\hat{u}}\big{)}$ and is equalled to
$l({\hat{y}}(0))$. So, the point
$\big{(}{\hat{y}}(0),{\hat{w}}(2n),{\hat{\bar{y}}}(2n))=({\hat{y}}(0),0,{\hat{y}}(0)\big{)}$
has to

$\textrm{ minimize\ }l(x_{0})+w+W_{extr}(x_{1})\ \textrm{subject to}\
(x_{0},w,x_{1})\in\mathcal{Q}_{n},x_{1}\in\mathcal{C}_{\operatorname*{\textrm{home}}}.$

Here,
$\mathcal{Q}_{n}\stackrel{{\scriptstyle\triangle}}{{=}}\big{\\{}\big{(}x_{0},\bar{\mathrm{w}}_{n}(x_{0},u;2n),\bar{\mathrm{y}}_{n}(x_{0},u;2n)\big{)}\in{\mathbb{X}\times\mathbb{R}\times\mathbb{X}\,\big{|}\,x_{0}\in\mathcal{C}_{0}},u\in\mathcal{U}_{n}\big{\\}}.$

Hence, since the lower semicontinuous function
$\operatorname*{lsc}\nolimits_{\mathcal{C}_{\operatorname*{\textrm{home}}}}W_{extr}$
equals $+\infty$ for all $x\notin\mathcal{C}_{\operatorname*{\textrm{home}}}$,
the point
$\big{(}{\hat{y}}(0),{\hat{w}}(2n),{\hat{\bar{y}}}(2n))=({\hat{y}}(0),0,{\hat{y}}(0)\big{)}\in\mathcal{Q}_{n}$
has to

$\textrm{minimize\
}l(x_{0})+w+\operatorname*{lsc}\nolimits_{\mathcal{C}_{\operatorname*{\textrm{home}}}}W_{extr}(x_{1})\
\textrm{\ subject to\ }(x_{0},w,x_{1})\in\operatorname*{cl}\mathcal{Q}_{n}.$

For each admissible control process $(y,{u})$ with $u\in\mathcal{U}_{n}$, the
process
$\big{(}({\bar{\mathrm{y}}}_{n},{\bar{\mathrm{w}}}_{n})(y(0),{u};\cdot);{u}(\cdot)\big{)}$
satisfies $(y(0),{w}(2n),y(2n))\in\mathcal{Q}_{n}$ by the definition of the
set $\mathcal{Q}_{n}$. Since a pair $(x,s)$ in
$\operatorname*{epi}\operatorname*{lsc}_{\mathcal{C}_{\operatorname*{\textrm{home}}}}{W}_{extr}$
satisfies the inequality
$s\geq\operatorname*{lsc}_{\mathcal{C}_{\operatorname*{\textrm{home}}}}{W}_{extr}(x)$,
the process
$\big{(}{\hat{\bar{y}}}_{n},{\hat{\bar{w}}}_{n},{\hat{{s}}}=0;{\hat{u}}\big{)}$
is an optimal solution for the problem

minimize $\displaystyle l(\bar{y}(0))+\bar{w}(2n)+s(2n)$ subject to
$\displaystyle\frac{d\bar{y}(t)}{dt}=\bar{f}_{n}(t,\bar{y}(t),u(t)),\
\frac{d\bar{w}(t)}{dt}=\bar{f}_{0,n}(t,\bar{y}(t),u(t)),\ \frac{ds(t)}{dt}=0,\
t\in[0;2n],$
$\displaystyle\bar{w}(0)=0,\quad\bar{y}(0)\in\operatorname*{cl}\mathcal{C}_{0},\quad(\bar{y}(2n),s(2n))\in\operatorname*{epi}\operatorname*{lsc}\nolimits_{\mathcal{C}_{\operatorname*{\textrm{home}}}}{W}_{extr},\quad
u\in\mathcal{U}_{n}.$

Step 4. Now we may apply the Pontryagin Maximum Principle [16, Theorem 22.26],
because its assumptions [16, Hypothesis 22.25] are implied from $(H4)$ and
$(H5^{\prime})$. Due to [16, Theorem 22.26], there exist a co-state arc
$({\mu}_{w},\mu_{y},\mu_{s})\in
C([0;2n],(\mathbb{R}\times\mathbb{X}\times\mathbb{R})^{*})$ and a number
$\lambda_{n}\in\\{0,1\\}$ with

$({\mu}_{w}(t),\mu_{y}(t),\mu_{s}(t),\lambda_{n})\neq 0\qquad\forall
t\in[0;2n]$

such that the following transversality conditions hold

$\displaystyle({\mu}_{w},\mu_{y},\mu_{s})(0)$
$\displaystyle\in\\{0\\}\times\lambda_{n}\partial
l({\hat{\bar{y}}}(0))\times\\{0\\}+\mathbb{R}\times
N({\hat{\bar{y}}}(0);\operatorname*{cl}\mathcal{C}_{0})\times\\{0\\}$
$\displaystyle-({\mu}_{w},\mu_{y},\mu_{s})(2n)$
$\displaystyle\in\lambda_{n}(1,0,1)+\\{0\\}\times
N\big{(}({\hat{\bar{y}}},{\hat{s}})(2n);\operatorname*{epi}\operatorname*{lsc}\nolimits_{\mathcal{C}_{\operatorname*{\textrm{home}}}}{W}_{extr}),$

and such that the co-state arc $({\mu}_{w},\mu_{y},\mu_{s})$ satisfies the
adjoint system:

$\displaystyle\frac{d\mu_{y}(t)}{dt}$
$\displaystyle=-\frac{\partial\bar{H}_{n}}{\partial
x}\big{(}{\hat{\bar{y}}}_{n}(t),\psi_{n}(t),{\hat{u}}(t),\mu_{y}(t),t\big{)},\
\frac{d\mu_{w}(t)}{dt}=\frac{d\mu_{s}(t)}{dt}=0\textrm{\ a.a.\ }t\in[0;2n],$

as well as the Pontryagin maximum condition on $[0;2n]$; in particular, one
has

$\sup_{\upsilon\in
U_{n}(t)}\\!\bar{H}_{n}\big{(}{\hat{\bar{y}}}_{n}(t),\mu_{y}(t),\upsilon,\mu_{w}(t),t\big{)}\\!=\\!\bar{H}_{n}\big{(}{\hat{\bar{y}}}_{n}(t),\mu_{y}(t),{\hat{u}}(t),\mu_{w}(t),t\big{)}\;\textrm{a.a.}\;t\in[0;n].$
(66)

The adjoint system entails that $\mu_{w}$ and $\mu_{s}$ are constant; further,
$\mu_{y}(0)=\mu_{y}(2n)$ holds. The first transversality condition yields
$\mu_{s}\equiv 0$. Then, according to
${\hat{\bar{y}}}(0)={\hat{\bar{y}}}(2n)={\hat{y}}(0)$ and ${\hat{s}}=0$, we
may rewrite the transversality conditions as follows:

$\displaystyle\mu_{y}(0)$ $\displaystyle\in\lambda_{n}\partial
l({\hat{y}}(0))+N({\hat{y}}(0);\operatorname*{cl}\mathcal{C}_{0}),$
$\displaystyle-({\mu}_{w},\mu_{y},0)(0)$
$\displaystyle\in\lambda_{n}(1,0,1)+\\{0\\}\times
N\big{(}{\hat{y}}(0),0;\operatorname*{epi}\operatorname*{lsc}\nolimits_{\mathcal{C}_{\operatorname*{\textrm{home}}}}{W}_{extr}).$

Set $\psi_{n}(\cdot)\stackrel{{\scriptstyle\triangle}}{{=}}\mu_{y}(\cdot)$,
then the first transversality condition implies (6), the second transversality
condition leads to $\mu_{w}=-\lambda_{n}$ and

$-(\psi_{n}(0),\lambda_{n})\in
N\big{(}{\hat{y}}(0),0;\operatorname*{epi}\operatorname*{lsc}\nolimits_{\mathcal{C}_{\operatorname*{\textrm{home}}}}{W}_{extr});\qquad$
(67)

further, for every $t\geq 0$, since
$({\mu}_{w}(t),\mu_{y}(t),\mu_{s}(t),\lambda_{n})=(-\lambda_{n},\psi_{n}(t),0,\lambda_{n})$
is nonzero, the pair $(\psi_{n}(t),\lambda_{n})$ is also nonzero.

Note that
${\hat{\bar{y}}}_{n}|_{[0;n]}={\hat{\bar{z}}}_{n}|_{[0;n]}={\hat{y}}|_{[0;n]}$
holds. Then, on the one hand,

$\displaystyle\bar{H}_{n}\big{(}{\hat{\bar{y}}}_{n}(t),\mu_{y}(t),\upsilon,\mu_{w}(t),t\big{)}={H}\big{(}{\hat{y}}(t),\psi_{n}(t),\upsilon,\lambda_{n},t\big{)}$

holds for all $v\in U$ and almost all $t\in[0;n]$. Therefore, the Pontryagin
maximum condition (66) on $[0;n]$ implies (63) for $\psi_{n}$. On the other
hand, the co-state arc $\psi_{n}|_{[0;n]}$ is the solution of (1b) with
$\lambda=\lambda_{n}$. Thus, the conditions (6), (63) have also been proved
and we need to consider only transversality conditions (7) and (15).

To prove Theorem 4.5, note that $W_{extr}=W_{\sup}$ coincides with map (12) on
$G\cap\mathcal{C}_{\operatorname*{\textrm{home}}}$. Since
$W_{\sup}({\hat{y}}(0))=0$ by (65), the desired inclusion (7) for (67) holds
true by Lemma 6.3. Theorem 4.1 has been proved.

To prove Theorem 4.6, note that $W_{extr}=W_{\inf}$ coincides with map (14) on
$G\cap\mathcal{C}_{\operatorname*{\textrm{home}}}$. Since
$W_{\inf}({\hat{y}}(0))=0$ by (65), Lemma 6.2 entails the desired inclusion
(15) for (67). Theorem 4.6 has been proved.

## Some questions instead of a conclusion

The key feature of this paper is that we obtain the boundary condition (7),
necessary for the weakly overtaking criterion, without any a priori
assumptions on asymptotics if the terminal constraints are absent. The
obtained condition cuts out the unique co-state arc under assumption (17);
however, one would like to have asymptotic assumptions that guarantee such
uniqueness but are easier to test than (17). Moreover, although formula (7),
as an analog of the Clarke subdifferential, matches in its form the necessary
conditions of optimality that are customary in the variational analysis, it is
only convenient to apply it to the simplest models. Example 5.1 is a case in
point. Then, this condition needs to be improved. In particular, any simple
assumptions making these transversality conditions into sufficient conditions
would be useful.

Next, under the conditions of Theorem 4.5, the overtaking optimality gets the
necessary condition (15): a boundary condition that is significantly stronger
than (7). Is it due to a greater regularity in the problem and the assumption
(14), or is it in the essence of the overtaking optimality? This is another
question the author does not know the answer to.

Furthermore, Theorem 4.1, following a similar assumption in Lemma 6.7,
requires the convexity of $\mathcal{C}_{\operatorname*{\textrm{home}}}$. It is
possible that in future research this assumption will be removed as Theorem
4.5 and Lemma 6.2. Since, formally, the convexity of
$\mathcal{C}_{\operatorname*{\textrm{home}}}$ becomes an asymptotic
assumption; therefore, I consider that it must be excluded.

## Acknowledgements

I would like to express my gratitude to Boris Mordukhovich for a valuable
discussion in during the writing this article.

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