+\biggl{(}\int_{0}^{1}\biggl{(}\int_{\mathbb{R}\times[0,\infty)}\bigl{\langle}\mathcal{V}^{\epsilon}_{t},B^{\epsilon,\eta}_{t}(x)\bigr{\rangle}\partial_{x}\varphi(x,y)\,\mathrm{d}\mu_{t}(x,y)\biggr{)}\mathrm{d}\eta\biggr{)}\,\mathrm{d}t\end{split}$
(4.7)

for $\varphi\in H^{1}(\mathbb{R}^{2})$. Our goal is to show that the family
$(\mathcal{V}^{\epsilon})_{\epsilon}$ is Cauchy in
$C_{\mathbb{F}}^{2}([0,T];L^{2}(\mathbb{R}^{2}))\cap
L_{\mathbb{F}}^{2}([0,T];H^{1,0}(\mathbb{R}^{2}))$.

We begin with a uniform bound on $\sup_{0\leq t\leq
T}\lVert\mathcal{V}^{\epsilon}_{t}\rVert_{L^{2}}^{2}$. Clearly, the SPDE (4.7)
has the same form as SPDE (4.1) with $\beta$ and $h$ replaced by
$\mathfrak{b}^{\epsilon}$ and

$\mathfrak{h}^{\epsilon}_{t}(\varphi)=\bigl{\langle}\mu_{t},\partial_{g}b_{1}(t,\cdot)h_{t}\partial_{x}\varphi\bigr{\rangle}\,\mathrm{d}\eta+\int_{0}^{1}\biggl{(}\int_{\mathbb{R}\times[0,\infty)}\bigl{\langle}\mathcal{V}^{\epsilon}_{t},B^{\epsilon,\eta}_{t}(x)\bigr{\rangle}\partial_{x}\varphi(x,y)\,\mathrm{d}\mu_{t}(x,y)\biggr{)}\mathrm{d}\eta$

for $\varphi\in H^{1,0}(\mathbb{R}^{2})$, respectively. Hence, we can apply
the estimate (4.2) to see that

$\sup_{0\leq s\leq
t}\lVert\mathcal{V}^{\epsilon}_{s}\rVert_{L^{2}}^{2}+\int_{0}^{t}\lVert\mathcal{V}^{\epsilon}_{s}\rVert_{H^{1,0}}^{2}\,\mathrm{d}s\leq
C\int_{0}^{t}\bigl{(}\lvert\langle\mathcal{V}^{\epsilon}_{s},\mathfrak{b}^{\epsilon}_{s}\partial_{x}\mathcal{V}^{\epsilon}_{s}\rangle\rvert+\lvert\mathfrak{h}^{\epsilon}_{s}(\mathcal{V}^{\epsilon}_{s})\rvert\bigr{)}\,\mathrm{d}s.$
(4.8)

Let us estimate the two terms on the right-hand side. For any $\delta>0$ we
find that
$\lvert\langle\mathcal{V}^{\epsilon}_{s},\mathfrak{b}^{\epsilon}_{s}\partial_{x}\mathcal{V}^{\epsilon}_{s}\rangle\rvert\leq\frac{C_{b}^{2}}{2\delta}\lVert\mathcal{V}^{\epsilon}_{s}\rVert_{L^{2}}^{2}+\frac{\delta}{2}\lVert\mathcal{V}^{\epsilon}_{s}\rVert_{H^{1,0}}^{2}$.
Next, it holds that

$\displaystyle\lvert\mathfrak{h}^{\epsilon}_{s}(\mathcal{V}^{\epsilon}_{s})\rvert$
$\displaystyle\leq\frac{C_{b}^{2}\lVert
h\rVert_{L^{\infty}}^{2}}{2\delta}\lVert\mu_{s}\rVert_{L^{2}}^{2}+\frac{\delta}{2}\lVert\mathcal{V}^{\epsilon}_{s}\rVert_{H^{1,0}}^{2}+\lVert\mathcal{V}^{\epsilon}_{s}\rVert_{L^{2}}\lVert\mu_{s}\rVert_{L^{2}}\lVert\partial_{x}\mathcal{V}^{\epsilon}_{s}\rVert_{L^{2}}\lVert
h_{b}\rVert_{L^{2}}$ $\displaystyle\leq\frac{C_{b}^{2}\lVert
h\rVert_{L^{\infty}}^{2}+\lVert
h_{b}\rVert_{L^{2}}^{2}\lVert\mathcal{V}^{\epsilon}_{s}\rVert_{L^{2}}^{2}}{2\delta}\lVert\mu_{s}\rVert_{L^{2}}^{2}+\delta\lVert\mathcal{V}^{\epsilon}_{s}\rVert_{H^{1,0}}^{2},$

where we used that
$\bigl{\langle}\mathcal{V}^{\epsilon}_{s},B^{\epsilon,\eta}_{s}(x)\bigr{\rangle}\leq\lVert\mathcal{V}^{\epsilon}_{s}\rVert_{L^{2}}\lVert
h_{b}\rVert_{L^{2}}$ by Assumption 2.10 (iii). Note that here $h_{b}$ does not
refer to the random function
$h\mathpunct{:}[0,T]\times\Omega\times\mathbb{R}^{2}\to\mathbb{R}^{d_{G}}$,
but is the dominating function from Assumption 2.10 (iii). Substituting this
and the previous estimate into Equation (4.8), choosing $\delta=\frac{1}{3C}$,
rearranging, and finally applying Grönwall’s inequality implies that

$\sup_{0\leq t\leq
T}\lVert\mathcal{V}^{\epsilon}_{t}\rVert_{L^{2}}^{2}+\frac{1}{2}\int_{0}^{T}\lVert\mathcal{V}^{\epsilon}_{t}\rVert_{H^{1,0}}^{2}\,\mathrm{d}t\leq
C$

for a possibly enlarged constant $C$ independent of $\epsilon$.

Using the uniform bound on the family $(\mathcal{V}^{\epsilon})_{\epsilon>0}$
we can show that it is Cauchy in
$C_{\mathbb{F}}^{2}([0,T];L^{2}(\mathbb{R}^{2}))\cap
L_{\mathbb{F}}^{2}([0,T];H^{1,0}(\mathbb{R}^{2}))$. Indeed, let us define
$\Delta^{\epsilon,\delta}=\mathcal{V}^{\epsilon}-\mathcal{V}^{\delta}$. Note
that again, the SPDE satisfied by $\Delta^{\epsilon,\delta}$ is of the same
form as SPDE (4.1). This time with $\beta$ and $h$ replaced by
$\mathfrak{b}^{\epsilon}$ and

$\displaystyle\mathfrak{h}^{\epsilon,\delta}_{t}(\varphi)$
$\displaystyle=\mathfrak{h}^{\epsilon,\delta,1}_{t}(\varphi)+\mathfrak{h}^{\epsilon,\delta,2}_{t}(\varphi)+\mathfrak{h}^{\epsilon,\delta,3}_{t}(\varphi)$
$\displaystyle=\bigl{\langle}\mathcal{V}^{\delta}_{t},(\mathfrak{b}^{\epsilon}_{t}-\mathfrak{b}^{\delta}_{t})\partial_{x}\varphi\bigr{\rangle}$
$\displaystyle\ \ \
+\int_{0}^{1}\biggl{(}\int_{\mathbb{R}\times[0,\infty)}\bigl{\langle}\Delta^{\epsilon,\delta}_{t},B^{\epsilon,\eta}_{t}(x)\bigr{\rangle}\partial_{x}\varphi(x,y)\,\mathrm{d}\mu_{t}(x,y)\biggr{)}\mathrm{d}\eta$
$\displaystyle\ \ \
+\int_{0}^{1}\biggl{(}\int_{\mathbb{R}\times[0,\infty)}\bigl{\langle}\mathcal{V}^{\delta}_{t},B^{\epsilon,\eta}_{t}(x)-B^{\epsilon,\eta}_{t}(x)\bigr{\rangle}\partial_{x}\varphi(x,y)\,\mathrm{d}\mu_{t}(x,y)\biggr{)}\mathrm{d}\eta.$

for $\varphi\in H^{1,0}(\mathbb{R}^{2})$, respectively. As before, we bound
the terms
$\lvert\langle\Delta^{\epsilon,\delta}_{s},\mathfrak{b}^{\epsilon}_{s}\partial_{x}\Delta^{\epsilon,\delta}_{s}\rangle\rvert$
and
$\lvert\mathfrak{h}^{\epsilon,\delta}_{s}(\Delta^{\epsilon,\delta}_{s})\rvert$
appearing in the estimate for $\Delta^{\epsilon,\delta}$ provided by (4.2).
Analogous to the above, for any $\delta>0$, we have
$\lvert\langle\Delta^{\epsilon,\delta}_{s},\mathfrak{b}^{\epsilon}_{s}\partial_{x}\Delta^{\epsilon,\delta}_{s}\rangle\rvert\leq\frac{C_{b}^{2}}{2\delta}\lVert\Delta^{\epsilon,\delta}_{s}\rVert_{L^{2}}^{2}+\frac{\delta}{2}\lVert\Delta^{\epsilon,\delta}_{s}\rVert_{H^{1,0}}^{2}$.
Next, by the triangle inequality
$\lvert\mathfrak{h}^{\epsilon,\delta}_{s}(\Delta^{\epsilon,\delta}_{s})\rvert\leq\lvert\mathfrak{h}^{\epsilon,\delta,1}_{s}(\Delta^{\epsilon,\delta}_{s})\rvert+\lvert\mathfrak{h}^{\epsilon,\delta,2}_{s}(\Delta^{\epsilon,\delta}_{s})\rvert+\lvert\mathfrak{h}^{\epsilon,\delta,3}_{s}(\Delta^{\epsilon,\delta}_{s})\rvert$
and we estimate the three terms on the right-hand side separately. For the
first, we get

$\displaystyle\lvert\mathfrak{h}^{\epsilon,\delta,1}_{s}(\Delta^{\epsilon,\delta}_{s})\rvert$
$\displaystyle\leq\frac{1}{2\delta}\lVert\mathcal{V}^{\delta}_{s}(\mathfrak{b}^{\epsilon}_{s}-\mathfrak{b}^{\delta}_{s})\rVert_{L^{2}}^{2}+\frac{\delta}{2}\lVert\Delta^{\epsilon,\delta}_{s}\rVert_{H^{1,0}}^{2}$
$\displaystyle\leq\frac{C_{b}^{2}\lVert\mathcal{V}^{\delta}_{s}\rVert_{L^{2}}^{2}}{2\delta}\bigl{(}\lVert\mu^{\epsilon}_{s}-\mu^{\delta}_{s}\rVert_{L^{2}}^{2}+\lVert
h_{s}\rVert_{L^{\infty}}^{2}(\epsilon-\delta)^{2}\bigr{)}+\frac{\delta}{2}\lVert\Delta^{\epsilon,\delta}_{s}\rVert_{H^{1,0}}^{2},$
(4.9)

where we used the Lipschitz continuity of $b$ in $v$ (cf. Assumption 2.4 (i),
which holds by Assumption 2.7), the linearity of $b$ in the control argument,
and that
$\lVert\nu^{\epsilon}_{s}-\nu^{\delta}_{s}\rVert_{L^{2}}\leq\lVert\mu^{\epsilon}_{s}-\mu^{\delta}_{s}\rVert_{L^{2}}$.
For the second term, we estimate

$\displaystyle\lvert\mathfrak{h}^{\epsilon,\delta,2}_{s}(\Delta^{\epsilon,\delta}_{s})\rvert$
$\displaystyle\leq\lVert\Delta^{\epsilon,\delta}_{s}\rVert_{L^{2}}\lVert\mu_{s}\rVert_{L^{2}}\lVert\partial_{x}\Delta^{\epsilon,\delta}_{s}\rVert_{L^{2}}\lVert
h_{b}\rVert_{L^{2}}^{2}$ $\displaystyle\leq\frac{\lVert
h_{b}\rVert_{L^{2}}^{2}\lVert\mu_{s}\rVert_{L^{2}}^{2}}{2\delta}\lVert\Delta^{\epsilon,\delta}_{s}\rVert_{L^{2}}^{2}+\frac{\delta}{2}\lVert\Delta^{\epsilon,\delta}_{s}\rVert_{H^{1,0}}^{2},$
(4.10)

where as above we used that
$\bigl{\langle}\mathcal{V}^{\epsilon}_{s},B^{\epsilon,\eta}_{s}(x)\bigr{\rangle}\leq\lVert\mathcal{V}^{\epsilon}_{s}\rVert_{L^{2}}\lVert
h_{b}\rVert_{L^{2}}$ by Assumption 2.10 (iii). Lastly, it holds that

$\displaystyle\lvert\mathfrak{h}^{\epsilon,\delta,3}_{s}(\Delta^{\epsilon,\delta}_{s})\rvert$
$\displaystyle\leq\frac{\lVert\mathcal{V}^{\delta}_{s}\rVert_{L^{2}}^{2}}{2\delta}\int_{0}^{1}\biggl{(}\int_{\mathbb{R}\times[0,\infty)}\bigl{\lVert}B^{\epsilon,\eta}_{s}(x)-B^{\delta,\eta}_{s}(x)\bigr{\rVert}_{L^{2}}^{2}\lvert\mu_{s}(x,y)\rvert^{2}\,\mathrm{d}x\mathrm{d}y\biggr{)}\,\mathrm{d}\eta$
$\displaystyle\ \ \
+\frac{\delta}{2}\lVert\Delta^{\epsilon,\delta}_{s}\rVert_{H^{1,0}}^{2}$
$\displaystyle=\frac{\lVert\mathcal{V}^{\delta}_{s}\rVert_{L^{2}}^{2}}{2\delta}I^{\epsilon,\delta}_{s}+\frac{\delta}{2}\lVert\Delta^{\epsilon,\delta}_{s}\rVert_{H^{1,0}}^{2}.$
(4.11)

We fix $\delta=\frac{1}{4C}$ and insert (4.1), (4.1), (4.1), and the bound on
$\lvert\langle\Delta^{\epsilon,\delta}_{s},\mathfrak{b}^{\epsilon}_{s}\partial_{x}\Delta^{\epsilon,\delta}_{s}\rangle\rvert$
into the estimate for $\Delta^{\epsilon,\delta}$ provided by (4.2). Then we
rearrange and apply Grönwall’s inequality to obtain

$\displaystyle\begin{split}\sup_{0\leq t\leq
T}\lVert\Delta^{\epsilon,\delta}_{s}\rVert_{L^{2}}^{2}+\frac{1}{2}\int_{0}^{T}\lVert\Delta^{\epsilon,\delta}_{s}\rVert_{H^{1,0}}^{2}\,\mathrm{d}t\leq
C\int_{0}^{T}\Bigl{(}\lVert\mu^{\epsilon}_{t}-\mu^{\delta}_{t}\rVert_{L^{2}}^{2}+(\epsilon-\delta)^{2}+I^{\delta,\epsilon}_{t}\Bigr{)}\,\mathrm{d}t\end{split}$
(4.12)

for a possibly enlarged $C$. Let us show that the right-hand side tends to
zero as $\epsilon$, $\delta\to 0$. First, we have that
$\lVert\mu^{\epsilon}_{t}-\mu^{\delta}_{t}\rVert_{L^{2}}\leq\epsilon\lVert\mathcal{V}^{\epsilon}_{t}\rVert_{L^{2}}+\delta\lVert\mathcal{V}^{\delta}_{t}\rVert_{L^{2}}$,
which is bounded uniformly in $\epsilon$ and $\delta$ and tends to zero as
$\epsilon$, $\delta\to 0$. Hence, it remains to take care of the quantity
$I^{\delta,\epsilon}_{t}$. It is bounded by $4\lVert
h_{b}\rVert_{L^{2}}^{2}\lVert\mu_{t}\rVert_{L^{2}}^{2}$, so if we prove that
$I^{\delta,\epsilon}_{t}$ vanishes as $\epsilon$, $\delta\to 0$, then the
dominated convergence theorem allows us to conclude the same for the right-
hand side of (4.12). By Assumption 2.12 (i) the map $\mathcal{M}^{2}_{\leq
1}(\mathbb{R})\cap L^{2}(\mathbb{R})\ni v\mapsto Db_{0}(t,x,v)(x^{\prime})$ is
continuous with respect to $\lVert\cdot\rVert_{L^{2}}$, so by definition of
$B^{\epsilon,\eta}_{t}$ we have
$\bigl{\lvert}B^{\epsilon,\eta}_{t}(x)(x^{\prime},y^{\prime})-B^{\delta,\eta}_{t}(x)(x^{\prime},y^{\prime})\bigr{\rvert}\to
0$ as $\epsilon$, $\delta\to 0$. Since by Assumption 2.10 (iii) it holds that
$\lvert B^{\epsilon,\eta}_{t}(x)(x^{\prime},y^{\prime})\rvert\leq
e^{-y^{\prime}}h_{b}(x^{\prime})$ and $e^{-\cdot}h_{b}\in
L^{2}(\mathbb{R}\times[0,\infty])$, we can conclude that
$\bigl{\lVert}B^{\epsilon,\eta}_{s}(x)-B^{\delta,\eta}_{s}(x)\bigr{\rVert}_{L^{2}}^{2}\to
0$. This readily implies $I^{\epsilon,\delta}_{t}\to 0$ as $\epsilon$ and
$\delta$ are taken to zero. Thus, it follows that
$(\mathcal{V}^{\epsilon})_{\epsilon>0}$ is Cauchy in
$C_{\mathbb{F}}^{2}([0,T];L^{2}(\mathbb{R}^{2}))\cap
L_{\mathbb{F}}^{2}([0,T];H^{1,0}(\mathbb{R}^{2}))$.

Let $\mathcal{V}$ denote the limit of $(\mathcal{V}^{\epsilon})_{\epsilon>0}$
in $C_{\mathbb{F}}^{2}([0,T];L^{2}(\mathbb{R}^{2}))\cap
L_{\mathbb{F}}^{2}([0,T];H^{1,0}(\mathbb{R}^{2}))$. Using the convergence of
$\mathcal{V}^{\epsilon}$ to $\mathcal{V}$, we can pass to the limit in SPDE
(4.7) to see that $\mathcal{V}$ solves SPDE (4.4). Since $\mu^{\epsilon}_{t}$
and $\mu_{t}$ are supported in $\mathbb{R}\times[0,\infty)$ the same holds for
$\mathcal{V}^{\epsilon}_{t}$ and its limit $\mathcal{V}_{t}$. This concludes
the proof. ∎

###  4.2 Necessary Stochastic Maximum Principle

The goal of this subsection is to establish the necessary SMP. For that, we
first show the existence of an adjoint process satisfying BSPDE (2.16), i.e.
we prove Proposition 2.11. Next, we use the adjoint process and the variation
$\mathcal{V}$ from the previous subsection to derive an expression for the
derivative of the cost $J_{\text{cl}}$ defined in (2.14). Inserting the
convexity of $h\mapsto b(t,x,v,h)p+e^{-y}f(t,x,v,h)$ implied by Assumption 2.6
into this expression yields the necessary SMP.

###### Proof of Proposition 2.11.

We intend to apply Theorem 2.3, so we simply have to show that the
coefficients $\tilde{a}$, $\tilde{H}$, $\tilde{F}$, $\lambda$, $\sigma_{0}$,
and $\tilde{\beta}$ as well as the terminal condition $\tilde{\psi}$ defined
by $\tilde{a}_{t}(x,y)=a_{t}(x)$,
$\tilde{H}_{t}(x,y,r,p)=\tilde{K}^{\mu}_{t}\bigl{(}x,y,p,g_{t}(x,y)\bigr{)}$,
$\tilde{F}_{t}(x,y,u,v)=\tilde{F}^{\mu}_{t}(x,y,v)$,
$\tilde{\beta}_{t}(x,y)=0$, and $\psi(x,y)=e^{-y}D\psi(\nu_{T})(x)$ for
$(t,\omega,x,y,r,p,u,v)\in[0,T]\times\Omega\times\mathbb{R}\times(\ell,\infty)\times\mathbb{R}^{2}\times
L^{2}(\mathbb{R}\times(\ell,\infty))^{2}$ satisfy the properties stated for
$a$, $H$, $F$, $\lambda$, $\alpha$, $\beta$, and $\psi$ in Assumption 2.2.
This is obvious for $\tilde{a}$, $\lambda$, $\sigma_{0}$, $\beta$, and $\psi$
so we will concentrate on $\tilde{H}$ and $\tilde{F}$. Let us begin with the
growth conditions Assumption 2.2 (iii). We have for
$(t,\omega,x,y,r,p,u,v)\in[0,T]\times\Omega\times\mathbb{R}\times(\ell,\infty)\times\mathbb{R}^{2}\times
L^{2}(\mathbb{R}\times(\ell,\infty))^{2}$ that

$\displaystyle\lvert\tilde{H}_{t}(x,y,r,p)\rvert$
$\displaystyle\leq\bigl{\lvert}b\bigl{(}t,x,\nu_{t},g_{t}(x,y)\bigr{)}p\bigr{\rvert}+\bigl{\lvert}e^{-y}f\bigl{(}t,x,\nu_{t},g_{t}(x,y)\bigr{)}\bigr{\rvert}$
$\displaystyle\leq C_{b}\lvert p\rvert+e^{-y}h_{f}(x)$

for $h_{f}\in L^{2}(\mathbb{R})$ by Assumption 2.10 (i) and

$\displaystyle\lvert\tilde{F}_{t}(x,y,u,v)\rvert$
$\displaystyle\leq\bigl{\lvert}\bigl{\langle}\mu_{t},e^{-y}Db_{0}(t,\cdot,\nu_{t})(x)v\bigr{\rangle}\bigr{\rvert}+\bigl{\lvert}\bigl{\langle}\mu_{t},e^{-y}D\tilde{f}_{0}(t,\cdot,\cdot,\nu_{t})(x)\bigr{\rangle}\bigr{\rvert}$
$\displaystyle\leq e^{-y}h_{b}(x)\lVert\mu_{t}\rVert_{L^{2}}\lVert
v\rVert_{L^{2}}+e^{-y}h_{f}(x)$

for $h_{b}\in L^{2}(\mathbb{R})$ by Assumption 2.10 (iii). Since
$\lVert\mu_{t}\rVert_{L^{2}}\leq C\lVert\mu_{0}\rVert_{L^{2}}<\infty$ by
Proposition 2.8, the right-hand side is of the desired form. Next, we verify
the Lipschitz conditions from Assumption 2.2 (i). For $(t,\omega,x,y,r,p,u,v)$
as above and
$(r^{\prime},p^{\prime},u^{\prime},v^{\prime})\in\mathbb{R}^{2}\times
L^{2}(\mathbb{R}\times(\ell,\infty))^{2}$, it holds that

$\displaystyle\lvert\tilde{H}_{t}(x,y,r,p)-\tilde{H}_{t}(x,y,r^{\prime},p^{\prime})\rvert=\bigl{\lvert}b\bigl{(}t,x,\nu_{t},g_{t}(x,y)\bigr{)}p-b\bigl{(}t,x,\nu_{t},g_{t}(x,y)\bigr{)}p^{\prime}\bigr{\rvert}\leq
C_{b}\lvert p-p^{\prime}\rvert$

by the boundedness of $b$ and

$\displaystyle\lvert\tilde{F}_{t}(x,y,u,v)-\tilde{F}_{t}(x,y,u^{\prime},v^{\prime})\rvert$
$\displaystyle=\bigl{\lvert}\bigl{\langle}\mu_{t},e^{-y}Db_{0}(t,\cdot,\nu_{t})(x)v\bigr{\rangle}-\bigl{\langle}\mu_{t},e^{-y}Db_{0}(t,\cdot,\nu_{t})(x)v^{\prime}\bigr{\rangle}\bigr{\rvert}$
$\displaystyle\leq e^{-y}h_{b}(x)\lVert\mu_{t}\rVert_{L^{2}}\lVert
v-v^{\prime}\rVert_{L^{2}}$

by Assumption 2.10 (iii). These bounds are as desired, so we can apply Theorem
2.3. ∎

We will use the adjoint process $(\tilde{u},\tilde{q},\tilde{m})$ determined
by BSPDE (2.16) to derive an expression for the Gâteaux derivative of the cost
function $J_{\text{cl}}$.

######  Proposition 4.3.

Let Assumptions 2.7, 2.10, and 2.12 be satisfied. Then for any control
$g\in\mathcal{G}_{\mathbb{F}}^{2}$ and any $\mathbb{F}$-progressively
measurable random function
$h\mathpunct{:}[0,T]\times\Omega\times\mathbb{R}^{2}\to\mathbb{R}^{d_{G}}$,
such that $g+\epsilon h$ maps into $G$, it holds that

$\frac{\mathrm{d}}{\mathrm{d}\epsilon}J_{\textup{cl}}(g+\epsilon
h)\Big{|}_{\epsilon=0}=\mathbb{E}\int_{0}^{T}\bigl{\langle}\mu_{t},\partial_{g}\tilde{K}^{\mu}_{t}(\cdot,\cdot,\partial_{x}\tilde{u}_{t},g_{t})h_{t}\bigr{\rangle}\,\mathrm{d}t.$
(4.13)

###### Proof.

Set $g^{\epsilon}=g+\epsilon h$ and denote the solutions to SPDE (2.10)
corresponding to the inputs $g^{\epsilon}$ and $g$ by $\mu^{\epsilon}$ and
$\mu$, respectively. Further, let
$\mathcal{V}^{\epsilon}=\frac{1}{\epsilon}(\mu^{\epsilon}-\mu)$ and let
$\mathcal{V}$ be the limit of $(\mathcal{V}^{\epsilon})_{\epsilon>0}$ in
$C_{\mathbb{F}}^{2}([0,T];L^{2}(\mathbb{R}^{2}))\cap
L_{\mathbb{F}}^{2}([0,T];H^{1,0}(\mathbb{R}^{2}))$, which exists by
Proposition 4.2. Then, we decompose

$\displaystyle\begin{split}J_{\textup{cl}}(g^{\epsilon}&)-J_{\textup{cl}}(g)\\\
&=\mathbb{E}\int_{0}^{T}\bigl{\langle}\mu^{\epsilon}_{t}-\mu_{t},\tilde{f}(t,\cdot,\cdot,\nu^{\epsilon}_{t},g^{\epsilon}_{t})\bigr{\rangle}\,\mathrm{d}t+\mathbb{E}\int_{0}^{T}\bigl{\langle}\mu_{t},\tilde{f}_{1}(t,\cdot,\cdot,g^{\epsilon}_{t})-\tilde{f}_{1}(t,\cdot,\cdot,g_{t})\bigr{\rangle}\,\mathrm{d}t\\\
&\ \ \
+\mathbb{E}\int_{0}^{T}\bigl{\langle}\mu_{t},\tilde{f}_{0}(t,\cdot,\cdot,\nu^{\epsilon}_{t})-\tilde{f}_{0}(t,\cdot,\cdot,\nu_{t})\bigr{\rangle}\,\mathrm{d}t+\mathbb{E}[\psi(\nu^{\epsilon}_{T})-\psi(\nu_{T})].\end{split}$
(4.14)

We divide the equation by $\epsilon$ and then take the limit as $\epsilon$
tends to zero. Owing to the continuity of the map $(x,v,g^{\prime})\mapsto
f(t,x,v,g^{\prime})$ guaranteed by Assumption 2.4 (ii) (which is in place by
Assumption 2.7) as well as the growth condition on $f$ from Assumption 2.10
(i), we can pass to the limit in the first expression on the right-hand side
above. For the second term, we use Assumption 2.12 (ii) and apply the
fundamental theorem of calculus, whereby

$\displaystyle\tilde{f}_{1}\bigl{(}t,x,y,g^{\epsilon}_{t}(x,y)\bigr{)}-\tilde{f}_{1}\bigl{(}t,x,y,g_{t}(x,y)\bigr{)}=\epsilon\int_{0}^{1}\partial_{g}f_{1}\bigl{(}(t,x,g_{t}(x,y)+\epsilon\eta
h_{t}(x,y)\bigr{)}h_{t}(x,y)\,\mathrm{d}\eta.$

Since $g^{\prime}\mapsto\partial_{g}f_{1}(t,x,g^{\prime})$ is continuous and
bounded, we conclude by the dominated convergence theorem that the second
expression tends to
$\mathbb{E}\int_{0}^{T}\bigl{\langle}\mu_{t},\partial_{g}\tilde{f}_{1}(t,\cdot,\cdot,g_{t})h_{t}\bigr{\rangle}\,\mathrm{d}t$.
Next, by Assumption 2.10 (ii) and Lemma 4.1, for
$(t,x,y)\in[0,T]\times\mathbb{R}\times[0,\infty)$ the map
$\mathcal{P}^{2}(\mathbb{R}\times[0,\infty))\ni
m\mapsto\tilde{f}_{0}(t,x,y,\mathcal{S}(m))$ has the linear functional
derivative $(x^{\prime},y^{\prime})\mapsto
e^{-y^{\prime}}D\tilde{f}_{0}(t,x,y,\mathcal{S}(m))(x^{\prime})$, where
$x^{\prime}\mapsto D\tilde{f}_{0}(t,x,y,v)(x^{\prime})$ denotes the linear
functional derivative of the map $\mathcal{M}^{2}_{\leq 1}(\mathbb{R})\ni
v\mapsto D\tilde{f}_{0}(t,x,y,v)$. Hence, we can write

$\displaystyle\tilde{f}_{0}(t,x,y,\nu^{\epsilon}_{t})-\tilde{f}_{0}(t,x,y,\nu_{t})$
$\displaystyle=\int_{0}^{1}\Bigl{\langle}\mu^{\epsilon}_{t}-\mu_{t},e^{-\cdot}D\tilde{f}_{0}\bigl{(}t,x,y,\eta\nu^{\epsilon}_{t}+(1-\eta)\nu_{t}\bigr{)}\Bigr{\rangle}\,\mathrm{d}\eta.$

Since $x^{\prime}\mapsto D\tilde{f}_{0}(t,x,y,v)(x^{\prime})$ is bounded by a
function $h_{f}\in L^{2}(\mathbb{R})$ by Assumption 2.10 (iii) and
$\mathcal{M}^{2}_{\leq 1}(\mathbb{R})\cap L^{2}(\mathbb{R})\ni v\mapsto
D\tilde{f}_{0}(t,x,y,v)(x^{\prime})$ is continuous with respect to
$\lVert\cdot\rVert_{L^{2}}$ by Assumption 2.12 (i), divided by $\epsilon$ the
quantity on the right-hand side above approaches
$\bigl{\langle}\mathcal{V}_{t},e^{-\cdot}D\tilde{f}_{0}(t,x,y,\nu_{t})\bigr{\rangle}$.
Consequently, the third summand in (4.14) converges to
$\mathbb{E}\int_{0}^{T}\int_{\mathbb{R}\times[0,\infty)}\bigl{\langle}\mathcal{V}_{t},e^{-\cdot}D\tilde{f}_{0}(t,x,y,\nu_{t})\bigr{\rangle}\,\mathrm{d}\mu_{t}(x,y)\,\mathrm{d}t$.
The same reason allows us to pass to the limit as $\epsilon\to 0$ in the last
term on the right-hand side of Equation (4.14), so putting everything
together, we obtain

$\displaystyle\begin{split}\frac{\mathrm{d}}{\mathrm{d}\epsilon}&J_{\textup{cl}}(g+\epsilon
h)\Big{|}_{\epsilon=0}\\\
&=\mathbb{E}\int_{0}^{T}\bigl{\langle}\mathcal{V}_{t},\tilde{f}(t,\cdot,\cdot,\nu_{t},g_{t})\bigr{\rangle}\,\mathrm{d}t+\mathbb{E}\int_{0}^{T}\bigl{\langle}\mu_{t},\partial_{g}\tilde{f}_{1}(t,\cdot,\cdot,g_{t})h_{t}\bigr{\rangle}\,\mathrm{d}t\\\
&\ \ \
+\mathbb{E}\int_{0}^{T}\int_{\mathbb{R}\times[0,\infty)}\bigl{\langle}\mathcal{V}_{t},e^{-\cdot}D\tilde{f}_{0}(t,x,y,\nu_{t})\bigr{\rangle}\,\mathrm{d}\mu_{t}(x,y)\,\mathrm{d}t+\mathbb{E}\bigl{\langle}\mathcal{V}_{T},e^{-\cdot}D\psi(\nu_{T})\bigr{\rangle}.\end{split}$
(4.15)

Next, we apply the generalisation of Itô’s formula from Theorem 3.3 with
$\ell=-1$ to $\langle\mathcal{V}_{t},\tilde{u}_{t}\rangle$. This implies that

$\displaystyle\bigl{\langle}\mathcal{V}_{T},e^{-\cdot}$ $\displaystyle
D\psi(\nu_{T})\bigr{\rangle}$
$\displaystyle=\langle\mathcal{V}_{0},\tilde{u}_{0}\rangle+\int_{0}^{T}\Bigl{\langle}b(t,\cdot,\nu_{t},g_{t})\mathcal{V}_{t}-\partial_{x}(a_{t}\mathcal{V}_{t})+\partial_{g}b_{1}(t,\cdot)h_{t}\mu_{t},\partial_{x}\tilde{u}_{t}\Bigr{\rangle}\,\mathrm{d}t$
$\displaystyle\ \ \
+\int_{0}^{T}\biggl{(}\int_{\mathbb{R}\times[0,\infty)}\bigl{\langle}\mathcal{V}_{t},e^{-\cdot}Db_{0}(t,x,\nu_{t})\bigr{\rangle}\partial_{x}\tilde{u}_{t}(x,y)\,\mathrm{d}\mu_{t}(x,y)\biggr{)}\,\mathrm{d}t$
$\displaystyle\ \ \
-\int_{0}^{T}\Bigl{(}\bigl{\langle}b_{t}(t,\cdot,\nu_{t},g_{t})\partial_{x}\tilde{u}_{t}+\tilde{f}(t,\cdot,\cdot,\nu_{t},g_{t})+\tilde{F}^{\mu}_{t}(\cdot,\cdot,\partial_{x}\tilde{u}_{t}),\mathcal{V}_{t}\bigr{\rangle}\Bigr{)}$
$\displaystyle\ \ \
+\int_{0}^{T}\Bigl{(}\langle\partial_{x}\tilde{u}_{t},\partial_{x}(a_{t}\mathcal{V}_{t})\rangle-\langle\tilde{q}_{t},\sigma_{0}(t)\partial_{x}\mathcal{V}_{t}\rangle\Bigr{)}\,\mathrm{d}t+\int_{0}^{T}\langle\partial_{x}\mathcal{V}_{t},\sigma_{0}(t)\tilde{q}_{t}\rangle\,\mathrm{d}t$
$\displaystyle\ \ \
+\int_{0}^{T}\bigl{(}\langle\mathcal{V}_{t},\sigma_{0}(t)\partial_{x}\tilde{u}_{t}\rangle+\langle\tilde{q}_{s},\mathcal{V}_{t}\rangle\bigr{)}\,\mathrm{d}W_{t}+\langle\tilde{m}_{T},\mathcal{V}_{T}\rangle$
$\displaystyle=\int_{0}^{T}\Bigl{(}\bigl{\langle}\mu_{t},\partial_{g}b_{1}(t,\cdot)h_{t}\partial_{x}\tilde{u}_{t}\bigr{\rangle}-\langle\mathcal{V}_{t},\tilde{f}(t,\cdot,\cdot,\nu_{t},g_{t})\rangle\Bigr{)}\,\mathrm{d}t$
(4.16) $\displaystyle\ \ \
-\int_{0}^{T}\biggl{(}\int_{\mathbb{R}\times[0,\infty)}\bigl{\langle}\mathcal{V}_{t},e^{-\cdot}D\tilde{f}_{0}\bigl{(}t,x,\nu_{t},g_{t}(x,y)\bigr{)}\bigr{\rangle}\,\mathrm{d}\mu_{t}(x,y)\biggr{)}\,\mathrm{d}t$
$\displaystyle\ \ \
+\int_{0}^{T}\bigl{(}\langle\mathcal{V}_{t},\sigma_{0}(t)\partial_{x}\tilde{u}_{t}\rangle+\langle\tilde{q}_{s},\mathcal{V}_{t}\rangle\bigr{)}\,\mathrm{d}W_{t}+\langle\tilde{m}_{T},\mathcal{V}_{T}\rangle,$

where we used that $\mathcal{V}_{0}=0$ and that

$\displaystyle\tilde{F}^{\mu}_{t}(x^{\prime},y^{\prime},\partial_{x}\tilde{u}_{t})-\int_{\mathbb{R}\times[0,\infty)}e^{-y^{\prime}}$
$\displaystyle
Db_{0}(t,x,\nu_{t})(x^{\prime})\partial_{x}\tilde{u}_{t}(x,y)\,\mathrm{d}\mu_{t}(x,y)$
$\displaystyle=\int_{\mathbb{R}\times[0,\infty)}e^{-y^{\prime}}D\tilde{f}_{0}(t,x,y,\nu_{t})(x^{\prime})\partial_{x}\tilde{u}_{t}(x,y)\,\mathrm{d}\mu_{t}(x,y)$

in the second equality. Now, we take expectations on both sides of (4.2), so
that the martingale terms disappear, and then plug the expression for
$\mathbb{E}\bigl{\langle}\mathcal{V}_{T},e^{-\cdot}D\psi(\nu_{T})\bigr{\rangle}$
provided by the right-hand side of (4.2) into Equation (4.15). This implies
that

$\frac{\mathrm{d}}{\mathrm{d}\epsilon}J_{\textup{cl}}(g+\epsilon
h)\Big{|}_{\epsilon=0}=\mathbb{E}\int_{0}^{T}\Bigl{\langle}\mu_{t},\partial_{g}b_{1}(t,\cdot,\nu_{t})h_{t}\partial_{x}\tilde{u}_{t}+\partial_{g}\tilde{f}_{1}(t,\cdot,\cdot,\nu_{t})h_{t}\Bigr{\rangle}\,\mathrm{d}t,$

which yields the desired expression (4.13). ∎

From (4.13) and the convexity of the Hamiltonian in the control argument,
guaranteed by Assumption 2.6 (which is in place by Assumption 2.12), we can
deduce the necessary SMP.

###### Proof of Theorem 2.13.

Let $g$, $h\in\mathcal{G}_{\mathbb{F}}^{2}$. Since $G$ is convex, for any
$\epsilon>0$ the random function $g^{\epsilon}=(1-\epsilon)g+\epsilon
h=g+\epsilon(h-g)$ maps into $G$. Hence, we can apply Proposition 4.3 with $h$
replaced by $h-g$, which implies that

$\displaystyle\frac{\mathrm{d}}{\mathrm{d}\epsilon}J_{\textup{cl}}(g^{\epsilon})\Big{|}_{\epsilon=0}$
$\displaystyle=\mathbb{E}\int_{0}^{T}\bigl{\langle}\mu_{t},\partial_{g}\tilde{K}^{\mu}_{t}(\cdot,\cdot,\partial_{x}\tilde{u}_{t},g_{t})(h_{t}-g_{t})\bigr{\rangle}\,\mathrm{d}t$
$\displaystyle\leq\mathbb{E}\int_{0}^{T}\bigl{\langle}\mu_{t},\tilde{K}^{\mu}_{t}(\cdot,\cdot,\partial_{x}\tilde{u}_{t},h_{t})-\tilde{K}^{\mu}_{t}(\cdot,\cdot,\partial_{x}\tilde{u}_{t},g_{t})\bigr{\rangle}\,\mathrm{d}t,$

where we used that the function
$h^{\prime}\mapsto\tilde{K}^{\mu}_{t}\bigl{(}x,y,\partial_{x}\tilde{u}_{t}(x,y),h^{\prime}\bigr{)}$
is convex by Assumption 2.6. On the other hand, we know that $g$ is a
minimiser of the cost functional $J_{\text{cl}}$, so that

$\frac{\mathrm{d}}{\mathrm{d}\epsilon}J_{\textup{cl}}(g^{\epsilon})\Big{|}_{\epsilon=0}=\lim_{\epsilon\to
0}\frac{1}{\epsilon}\bigl{(}J_{\textup{cl}}(g+\epsilon(h-g))-J_{\textup{cl}}(g)\bigr{)}\geq
0.$

Combining the previous two inequalities yields

$\mathbb{E}\int_{0}^{T}\bigl{\langle}\mu_{t},\tilde{K}^{\mu}_{t}(\cdot,\cdot,\partial_{x}\tilde{u}_{t},h_{t})-\tilde{K}^{\mu}_{t}(\cdot,\cdot,\partial_{x}\tilde{u}_{t},g_{t})\bigr{\rangle}\,\mathrm{d}t\geq
0.$ (4.17)

Now, for any $\mathbb{F}$-progressively measurable set
$A\subset[0,T]\times\Omega\times\mathbb{R}^{2}$ and any $h_{\ast}\in G$ we can
define an $h\in\mathcal{G}_{\mathbb{F}}^{2}$ by $h_{t}(x,y)=h_{\ast}$ if
$(t,\omega,x,y)\in A$ and $h_{t}(x,y)=g_{t}(x,y)$ otherwise. Applying (4.17)
with this choice of $h$ implies that

$\mathbb{E}\int_{0}^{T}\int_{\mathbb{R}^{2}}\mathbf{1}_{A}(t,\cdot,x,y)\Bigl{(}\tilde{K}^{\mu}_{t}\bigl{(}x,y,\partial_{x}\tilde{u}_{t}(x,y),h_{\ast}\bigr{)}-\tilde{K}^{\mu}_{t}\bigl{(}x,y,\partial_{x}\tilde{u}_{t}(x,y),g_{t}(x,y)\bigr{)}\Bigr{)}\,\mathrm{d}\mu_{t}(x,y)\,\mathrm{d}t$

is nonnegative. Since $A$ and $h_{\ast}$ were arbitrary, we conclude that for
all $h_{\ast}\in G$ it holds that

$\tilde{K}^{\mu}_{t}\bigl{(}x,y,\partial_{x}\tilde{u}_{t}(x,y),h_{\ast}\bigr{)}\geq\tilde{K}^{\mu}_{t}\bigl{(}x,y,\partial_{x}\tilde{u}_{t}(x,y),g_{t}(x,y)\bigr{)}$

for $\mu_{t}$-a.e. $(x,y)\in\mathbb{R}\times[0,\infty)$ and
$\operatorname{Leb}\otimes\mathbb{P}$-a.e. $(t,\omega)\in[0,T]\times\Omega$. ∎

###  4.3 Existence and Uniqueness of the Semilinear BSPDEs (2.18) and (2.22)

In this section we establish the existence of a unique strong solution
$(u,q,m)$ of BSPDE (2.18). From this process $(u,q,m)$ we construct the unique
strong solution $(\tilde{u},\tilde{q},\tilde{m})$ of BSPDE (2.22) by setting
$\tilde{u}_{t}(x,y)=e^{-y}u_{t}(x)$, $\tilde{q}_{t}(x,y)=e^{-y}q_{t}(x)$, and
$\tilde{m}_{t}(x,y)=e^{-y}m_{t}(x)$ for
$(t,\omega,x,y)\in[0,T]\times\Omega\times\mathbb{R}\times(0,\infty)$.

###### Proof of Proposition 2.14.

We apply Theorem 2.3, so we simply have to verify Assumption 2.2 for the
coefficients $a$, $H$, $F$, $\alpha$, and $\beta$ defined by
$H_{t}(x,r,p)=H^{\nu}_{t}(x,r,p)$, $F_{t}(x,u,v)=F^{\nu}_{t}(x,u,v)$,
$\alpha_{t}=\sigma_{0}(t)$, and $\beta_{t}(x)=0$ for
$(t,\omega,x,r,p,u,v)\in[0,T]\times\Omega\times\mathbb{R}^{3}\times
L^{2}(\mathbb{R})^{2}$. We only check the growth and Lipschitz conditions,
Assumptions 2.2 (iii) and (i), for $H$. The coefficient $F$ can be dealt with
in the same manner as with $\tilde{F}$ in the proof of Proposition 2.11 and
the properties of the remaining coefficients are obvious. For
$(t,\omega,x,r,p)\in[0,T]\times\Omega\times\mathbb{R}^{3}$ it holds that

$\displaystyle\lvert H_{t}(x,r,p)\rvert$
$\displaystyle\leq\biggl{\lvert}\inf_{g\in
G}\bigl{(}b(t,x,\nu_{t},g)p+f(t,x,\nu_{t},g)\bigr{)}\biggr{\rvert}+\lvert\lambda_{t}(x)r\rvert$
$\displaystyle\leq C_{b}\lvert p\rvert+h_{f}(x)+C_{\lambda}\lvert r|,$

where the boundedness of $b$ and $\lambda$ is guaranteed by Assumption 2.7
(ii) and $h_{f}\in L^{2}(\mathbb{R})$ by Assumption 2.10 (i), so that $H$
satisfies Assumption 2.2 (iii). Next, if in addition
$(r^{\prime},p^{\prime})\in\mathbb{R}^{2}$, then

$\displaystyle\lvert H_{t}$
$\displaystyle(x,r,p)-H_{t}(x,r^{\prime},p^{\prime})\rvert$
$\displaystyle\leq\biggl{\lvert}\inf_{g\in
G}\bigl{(}b(t,x,\nu_{t},g)p+f(t,x,\nu_{t},g)\bigr{)}-\inf_{g\in
G}\bigl{(}b(t,x,\nu_{t},g)p^{\prime}+f(t,x,\nu_{t},g)\bigr{)}\biggr{\rvert}$
$\displaystyle\ \ \ +\lvert\lambda_{t}(x)(r-r^{\prime})\rvert$
$\displaystyle\leq\sup_{g\in
G}\Bigl{\lvert}b(t,x,\nu_{t},g)p+f(t,x,\nu_{t},g)-\bigl{(}b(t,x,\nu_{t},g)p^{\prime}+f(t,x,\nu_{t},g)\bigr{)}\Bigr{\rvert}+C_{\lambda}\lvert
r-r^{\prime}\rvert$ $\displaystyle\leq C_{b}\lvert
p-p^{\prime}\rvert+C_{\lambda}\lvert r-r^{\prime}\rvert,$

so that $H$ also satisfies the Lipschitz condition from Assumption 2.2 (i).
Thus, we can apply Theorem 2.3 to obtain a unique strong solution to BSPDE
(2.16). ∎

###### Proof of Proposition 2.15.

We first show that the process
$(\tilde{u},\tilde{q},\tilde{m})\in\mathbb{H}(\mathbb{R}\times(0,\infty))$
given by $\tilde{u}_{t}(x,y)=e^{-y}u_{t}(x)$,
$\tilde{q}_{t}(x,y)=e^{-y}q_{t}(x)$, $\tilde{m}_{t}(x,y)=e^{-y}m_{t}(x)$ for
$(t,\omega,x,y)\in[0,T]\times\Omega\times\mathbb{R}\times(0,\infty)$, where
$(u,q,m)$ is the unique strong solution of BSPDE (2.18), solves BSPDE (2.22).
Let $\varphi\in H^{1}_{0}(\mathbb{R}\times(0,\infty))$ and define
$\varphi_{0}\in H^{1}(\mathbb{R})$ by
$\varphi_{0}(x)=\int_{(0,\infty)}e^{-y}\varphi(x,y)\,\mathrm{d}y$. Then, using
BSPDE (2.18), we find

$\displaystyle\langle\tilde{u}_{t},\varphi\rangle$ $\displaystyle=\langle
u_{t},\varphi_{0}\rangle$ $\displaystyle=\langle
D\psi,\varphi_{0}\rangle+\int_{t}^{T}\Bigl{\langle}H^{\nu}_{s}\bigl{(}\cdot,u_{s}(\cdot),\partial_{x}u_{s}(\cdot)\bigr{)}+F^{\nu}_{s}(\cdot,\partial_{x}u_{s}),\varphi_{0}\Bigr{\rangle}\,\mathrm{d}s$
$\displaystyle\ \ \
-\int_{t}^{T}\bigl{\langle}\partial_{x}u_{s},\partial_{x}(a_{s}\varphi_{0})\bigr{\rangle}+\bigl{\langle}q_{s},\sigma_{0}(s)\partial_{x}\varphi_{0}\bigr{\rangle}\,\mathrm{d}s-\int_{t}^{T}\langle
q_{s},\varphi_{0}\rangle\,\mathrm{d}W_{s}-\langle
m_{T}-m_{t},\varphi_{0}\rangle$ $\displaystyle=\langle
e^{-\cdot}D\psi,\varphi\rangle+\int_{t}^{T}\Bigl{\langle}H^{\nu}_{s}\bigl{(}\cdot,u_{s}(\cdot),\partial_{x}u_{s}(\cdot)\bigr{)}+F^{\nu}_{s}(\cdot,\partial_{x}u_{s}),\varphi_{0}\Bigr{\rangle}\,\mathrm{d}s$
(4.18) $\displaystyle\ \ \
-\int_{t}^{T}\bigl{\langle}\partial_{x}\tilde{u}_{s},\partial_{x}(a_{s}\varphi)\bigr{\rangle}+\langle\tilde{q}_{s},\sigma_{0}(s)\partial_{x}\varphi\rangle\,\mathrm{d}s-\int_{t}^{T}\langle\tilde{q}_{s},\varphi\rangle\,\mathrm{d}W_{s}-\langle\tilde{m}_{T}-\tilde{m}_{t},\varphi\rangle.$

The terminal condition and the second line are already of the desired form, so
it remains to compute the two terms involving $H^{\nu}$ and $F^{\nu}$. For the
first expression, we write

$\displaystyle\begin{split}&\bigl{\langle}H^{\nu}_{s}(\cdot,u_{s}(\cdot),\partial_{x}u_{s}(\cdot)\bigr{)},\varphi_{0}\bigr{\rangle}\\\
&=\int_{\mathbb{R}\times(0,\infty)}e^{-y}\Bigl{(}\inf_{h\in
G}\bigl{(}b(s,x,\nu_{s},h)\partial_{x}u_{s}(x)+f(s,x,\nu_{s},h)\bigr{)}-\lambda_{s}(x)u_{s}(x)\Bigr{)}\varphi(x,y)\,\mathrm{d}x\mathrm{d}y.\end{split}$
(4.19)

Now, if $\mu_{t}(x,y)>0$ then

$\displaystyle e^{-y}\inf_{h\in G}\bigl{(}b(s,x,\nu_{s},h)$
$\displaystyle\partial_{x}u_{t}(x)+f(s,x,\nu_{s},h)\bigr{)}$
$\displaystyle=\inf_{h\in
G}\bigl{(}b(s,x,\nu_{s},h)\partial_{x}\tilde{u}_{t}(x,y)+\tilde{f}(s,x,y,\nu_{s},h)\bigr{)}$
$\displaystyle=\inf_{h\in
G}\tilde{K}^{\mu}_{s}\bigl{(}x,y,\partial_{x}\tilde{u}_{s}(x,y),h\bigr{)}.$

If, on the other hand, $\mu_{t}(x,y)=0$ we compute

$\displaystyle e^{-y}$ $\displaystyle\inf_{h\in
G}\bigl{(}b(s,x,\nu_{s},h)\partial_{x}u_{t}(x)+f(s,x,\nu_{s},h)\bigr{)}$
$\displaystyle=e^{-y}\Bigl{(}b\bigl{(}s,x,\nu_{s},g_{\ast}\bigl{(}s,x,\partial_{x}u_{s}(x)\bigr{)}\bigr{)}\partial_{x}u_{s}(x,y)+f\bigl{(}s,x,\nu_{s},g_{\ast}\bigl{(}s,x,\partial_{x}u_{s}(x)\bigr{)}\bigr{)}\Bigr{)}$
$\displaystyle=\Bigl{(}b\bigl{(}s,x,g_{\ast}\bigl{(}s,x,\nu_{s},\partial_{x}u_{s}(x)\bigr{)}\bigr{)}\partial_{x}\tilde{u}_{s}(x,y)+\tilde{f}\bigl{(}s,x,y,g_{\ast}\bigl{(}s,x,\nu_{s},\partial_{x}u_{s}(x)\bigr{)}\bigr{)}\Bigr{)}$
$\displaystyle=\tilde{K}^{\mu}_{s}\bigl{(}x,y,\partial_{x}\tilde{u}_{s}(x,y),g_{\ast}\bigl{(}s,x,\partial_{x}u_{s}(x)\bigr{)}\bigr{)},$

so together we get

$\displaystyle e^{-y}\Bigl{(}\inf_{h\in G}\bigl{(}b(s,x,\nu_{s},h)$
$\displaystyle\partial_{x}u_{t}(x)+f(s,x,\nu_{s},h)\bigr{)}$
$\displaystyle=\mathbf{1}_{\mu_{s}(x,y)>0}\inf_{h\in
G}\tilde{K}^{\mu}_{s}\bigl{(}x,y,\partial_{x}\tilde{u}_{s}(x,y),h\bigr{)}$
$\displaystyle\ \ \
+\mathbf{1}_{\mu_{s}(x,y)=0}\tilde{K}^{\mu}_{s}\bigl{(}x,y,\partial_{x}\tilde{u}_{s}(x,y),g_{\ast}\bigl{(}s,x,\partial_{x}u_{s}(x)\bigr{)}\bigr{)}$
$\displaystyle=\tilde{H}^{\mu}_{s}\bigl{(}x,y,\partial_{x}\tilde{u}_{s}(x,y)\bigr{)},$

where in the last step we simply apply the definition of $\tilde{H}^{\mu}$
from (2.21). Inserting this equality into (4.19) yields

$\displaystyle\bigl{\langle}H^{\nu}_{s}(\cdot,u_{s}(\cdot),$
$\displaystyle\partial_{x}u_{s}(\cdot)\bigr{)},\varphi_{0}\bigr{\rangle}$
$\displaystyle=\int_{\mathbb{R}\times(0,\infty)}\Bigl{(}\tilde{H}^{\mu}_{s}\bigl{(}x,y,\partial_{x}\tilde{u}_{s}(x,y)\bigr{)}+\lambda_{s}(x)\partial_{y}\tilde{u}_{s}(x,y)\Bigr{)}\varphi(x,y)\,\mathrm{d}x\mathrm{d}y$
$\displaystyle=\bigl{\langle}\tilde{H}^{\mu}_{s}\bigl{(}\cdot,\cdot,\partial_{x}\tilde{u}_{s}(\cdot,\cdot)\bigr{)},\varphi\bigr{\rangle}-\langle\tilde{u}_{s},\lambda_{s}\partial_{y}\varphi\rangle,$

where we used in the first equality that
$-e^{-y}u_{s}(x)=\partial_{y}\tilde{u}_{s}(x,y)$ and applied integration by
parts in the second equality, using that the boundary terms vanish since
$\varphi(0)=0$. For the second term, we get

$\displaystyle\bigl{\langle}F^{\nu}_{s}($ $\displaystyle
x,\partial_{x}u_{s}),\varphi_{0}\bigr{\rangle}$
$\displaystyle=\int_{\mathbb{R}}\Bigl{(}\bigl{\langle}\nu_{s},Db_{0}(s,\cdot,\nu_{s})(x)\partial_{x}u_{s}\bigr{\rangle}+\bigl{\langle}\nu_{s},Df_{0}(s,\cdot,\nu_{s})(x)\bigr{\rangle}\Bigr{)}\varphi_{0}(x)\,\mathrm{d}x$
$\displaystyle=\int_{\mathbb{R}\times(0,\infty)}\Bigl{(}\bigl{\langle}\mu_{s},e^{-y}Db_{0}(s,\cdot,\nu_{s})(x)\partial_{x}\tilde{u}_{s}\bigr{\rangle}+\bigl{\langle}\mu_{s},e^{-y}D\tilde{f}_{0}(s,\cdot,\cdot,\nu_{s})(x)\bigr{\rangle}\Bigr{)}\varphi(x,y)\,\mathrm{d}x\mathrm{d}y$
$\displaystyle=\bigl{\langle}\tilde{F}^{\mu}_{s}(\cdot,\cdot,\partial_{x}\tilde{u}_{s}),\varphi\bigr{\rangle},$

where we used in the second equality that $\nu_{s}=\mathcal{S}(\mu_{s})$.
Inserting the previous two equalities into (4.3) yields

$\displaystyle\langle\tilde{u}_{t},\varphi\rangle$ $\displaystyle=\langle
e^{-\cdot}D\psi,\varphi\rangle+\int_{t}^{T}\Bigl{\langle}\tilde{H}^{\mu}_{s}\bigl{(}\cdot,\cdot,\partial_{x}\tilde{u}_{s}(\cdot,\cdot)\bigr{)}+\tilde{F}^{\mu}_{s}(\cdot,\cdot,\partial_{x}\tilde{u}_{s}),\varphi\Bigr{\rangle}\,\mathrm{d}s$
$\displaystyle\ \ \
-\int_{t}^{T}\bigl{\langle}\partial_{x}\tilde{u}_{s},\partial_{x}(a_{s}\varphi)\bigr{\rangle}+\langle\tilde{u}_{s},\lambda_{s}\partial_{y}\varphi\rangle+\langle\tilde{q}_{s},\sigma_{0}(s)\partial_{x}\varphi\rangle\,\mathrm{d}s$
$\displaystyle\ \ \
-\int_{t}^{T}\langle\tilde{q}_{s},\varphi\rangle\,\mathrm{d}W_{s}-\langle\tilde{m}_{T}-\tilde{m}_{t},\varphi\rangle.$

Since $\varphi\in H^{1}_{0}(\mathbb{R}\times(0,\infty))$ was arbitrary,
$(\tilde{u},\tilde{q},\tilde{m})\in\mathbb{H}(\mathbb{R}\times(0,\infty))$
satisfies BSPDE (2.22).

To show that $(\tilde{u},\tilde{q},\tilde{m})$ is the unique solution to BSPDE
(2.22), we apply Theorem 2.3. Hence, we need to verify that the coefficients
of BSPDE (2.22) satisfy Assumption 2.2. We already did that for all
coefficients except $\tilde{H}^{\mu}$ in the proof of Proposition 2.11. The
assumptions on $\tilde{H}^{\mu}$ can be verified in the same manner as the
assumptions on $H$ in the proof of Proposition 2.14. This concludes the proof.
∎

From Proposition 2.15 and the necessary SMP, Theorem 2.13, we can then deduce
Corollary 2.16 in the manner outlined above the statement of Corollary 2.16.
Hence, there exists an optimal control
$g^{\ast}\in\mathcal{G}_{\mathbb{F}}^{1}$ which takes the form
$g^{\ast}_{t}(x)=g_{\ast}\bigl{(}t,x,\partial_{x}u_{t}(x)\bigr{)}$ for
$x\in\mathbb{R}$ and $\mathbb{P}\otimes\operatorname{Leb}$-a.e.
$(t,\omega)\in[0,T]\times\Omega$. Our next goal is to extend the existence of
such an optimal semiclosed-loop control independent of the intensity to more
general coefficients and cost functions.

###  4.4 Extending Equivalence to General Coefficients and Cost Functions

Throughout this subsection Assumptions 2.4 and 2.6 are in place and we assume
that $\mathcal{L}(\xi,\zeta)\in L^{2}(\mathbb{R})$. We first establish a weak
stability result for SPDE (2.10) with varying coefficients and inputs. Let
$(b^{n})_{n}$, $(\lambda^{n})_{n}$, $(f^{n})_{n}$, and $(\psi^{n})_{n}$ be
sequences of functions such that

  1. (i)

$b^{n}$, $\lambda^{n}$, $f^{n}$, and $\psi^{n}$ satisfy the same properties as
$b$, $\lambda$, $f$, and $\psi$ in Assumptions 2.7, 2.10, and 2.12;

  2. (ii)

there exists a $C>0$ such that for
$(t,x,v,g)\in[0,T]\times\mathbb{R}\times\mathcal{M}^{2}_{\leq
1}(\mathbb{R})\times G$ we have

$\displaystyle\lvert b^{n}(t,x,v,g)\rvert+\lvert\lambda^{n}(t,x)\rvert$
$\displaystyle\leq C(1+\lvert x\rvert+M_{2}(v)),$ $\displaystyle\lvert
f^{n}(t,x,v,g)\rvert+\lvert\psi^{n}(v)\rvert$ $\displaystyle\leq C(1+\lvert
x\rvert^{2}+M^{2}_{2}(v));$

  3. (iii)

$b^{n}(t,x,v,g)$, $\lambda^{n}(t,x)$, $f^{n}(t,x,v,g)$, and $\psi^{n}(v)$ tend
to $b(t,x,v,g)$, $\lambda(t,x)$, $f(t,x,v,g)$, and $\psi(v)$ uniformly for
$(t,x,v,g)$ in compact subsets of
$[0,T]\times\mathbb{R}\times\mathcal{M}^{2}_{\leq 1}(\mathbb{R})\times G$;

  4. (iv)

the map $G\ni g\mapsto f^{n}_{1}(t,x,g)$ is strictly convex for all
$(t,x)\in[0,T]\times\mathbb{R}$.

Since $b^{n}$, $\lambda^{n}$, $f^{n}$, and $\psi^{n}$ satisfy the same
properties as $b$, $\lambda$, $f$, and $\psi$ in Assumptions 2.7, 2.10, and
2.12, we can apply Corollary 2.16, where we replace the coefficients and costs
$b$, $\lambda$, $f$, and $\psi$ of the control problem by $b^{n}$,
$\lambda^{n}$, $f^{n}$, and $\psi^{n}$ and the initial condition
$\mathcal{L}(\xi,\zeta)$ by $\mathcal{L}(\xi,\zeta/n)\in
L^{2}(\mathbb{R}^{2})$. The latter ensures that the cumulative intensity will
be started from zero in the limit. We will use the assumptions formulated
above, in particular the convergence condition (iii), to show that the
existence of an optimal semiclosed-loop control independent of the intensity
transfers from the control problem with coefficients and costs $b^{n}$,
$\lambda^{n}$, $f^{n}$, and $\psi^{n}$ to the limiting control problem with
coefficients and costs $b$, $\lambda$, $f$, and $\psi$.

For $n\geq 1$ we define the cost functionals

$J_{\text{cl}}^{n}(g)=\mathbb{E}\biggl{[}\int_{0}^{T}\bigl{\langle}\mu^{n}_{t},\tilde{f}^{n}(t,\cdot,\cdot,\nu^{n}_{t},g_{t})\bigr{\rangle}\,\mathrm{d}t+\psi^{n}(\nu^{n}_{T})\biggr{]},$
(4.20)

for $g\in\mathcal{G}_{\mathbb{F}}^{2}$, where $\mu^{n}=(\mu^{n}_{t})_{0\leq
t\leq T}$ satisfies SPDE (2.10) with $b$ and $\lambda$ replaced by $b^{n}$ and
$\lambda^{n}$, respectively, initial condition
$\mu^{n}_{0}=\mathcal{L}(\xi,\zeta/n)$, and control $g$. For a control
$g\in\mathcal{G}_{\mathbb{F}}^{1}$, which is independent of the intensity, we
can rewrite $J_{\text{cl}}^{n}(g)$ as
$\mathbb{E}\bigl{[}\int_{0}^{T}\bigl{\langle}\nu^{n}_{t},f^{n}(t,\cdot,\nu^{n}_{t},g_{t})\bigr{\rangle}\,\mathrm{d}t+\psi^{n}(\nu^{n}_{T})\bigr{]}$,
where $\nu^{n}=(\nu^{n}_{t})_{0\leq t\leq T}$ is a solution to SPDE (2.23)
with $b$ and $\lambda$ replaced by $b^{n}$ and $\lambda^{n}$, respectively,
initial condition $\nu^{n}_{0}=\mathbb{E}[e^{-\zeta/n}]\mathcal{L}(\xi)$, and
control $g$. The definition of the initial condition is to be understood in
the sense that for all $\varphi\in C_{b}(\mathbb{R})$ it holds that
$\langle\nu^{n}_{0},\varphi\rangle=\mathbb{E}[e^{-\zeta/n}]\mathbb{E}[\varphi(\xi)]$.

Let $(E,d)$ be a complete separable metric space. We let $\mathbb{M}_{\leq
T}^{2}(E)$ denote the space of square-integrable measures $m$ on $[0,T]\times
E$ such that $m([s,t]\times E)\leq t-s$ for all $0\leq s\leq t\leq T$. Here
square-integrable means that $\int_{E}d^{2}(x,x_{0})\,\mathrm{d}m(x)$ for a
fixed $x_{0}\in E$.

We say that a random variable $\mathfrak{m}$ with values in $\mathbb{M}_{\leq
T}^{2}(E)$ is $\mathbb{F}$-progressively measurable if for any $t\in[0,T]$,
the random variable $\mathfrak{m}([0,s]\times A)$ is
$\mathcal{B}([0,t])\otimes\mathcal{F}_{t}$-measurable for all $s\in[0,t]$ and
$A\in\mathcal{B}(E)$.

######  Proposition 4.4.

Let the sequences $(b^{n})$, $(\lambda^{n})$, $(f^{n})$, and $(\psi^{n})$
satisfy Properties (i) to (iv) above. Fix a sequence $(g^{n})_{n\geq 1}$ in
$\mathcal{G}_{\mathbb{F}}^{1}$ and denote the solution to SPDE (2.23) with
coefficients $b$ and $\lambda$ replaced by $b^{n}$ and $\lambda^{n}$,
respectively, control $g^{n}$, and initial condition
$\nu^{n}_{0}=\mathbb{E}[e^{-\zeta/n}]\mathcal{L}(\xi)$ by
$\nu^{n}=(\nu^{n}_{t})_{0\leq t\leq T}$. Define the $\mathbb{M}_{\leq
T}^{2}(\mathbb{R}\times G)$-valued random variable $\mathfrak{m}^{n}$ by

$\int_{[0,T]\times\mathbb{R}\times
G}\varphi(t,x,g)\,\mathrm{d}\mathfrak{m}^{n}(t,x,g)=\int_{[0,T]}\langle\nu^{n}_{t},\varphi(t,\cdot,g_{n})\rangle\,\mathrm{d}t$

for $\varphi\in C_{b}([0,T]\times\mathbb{R}\times G)$. Then the sequence
$(\mathfrak{m}^{n},\nu^{n},W)$ is tight on $\mathbb{M}_{\leq
T}^{2}(\mathbb{R}\times G)\times C([0,T];\mathcal{M}^{1}_{\leq
1}(\mathbb{R}))\times C([0,T])$ and any subsequential limit
$(\mathfrak{m},\nu,W)$, which we assume is defined on the same probability
space with the same Brownian motion $W$, solves the SPDE

$\mathrm{d}\langle\nu_{t},\varphi\rangle=\int_{\mathbb{R}\times
G}\mathcal{L}\varphi(t,x,\nu_{t},g)\,\mathrm{d}\mathfrak{m}(t,x,g)+\langle\nu_{t},\sigma_{0}(t)\partial_{x}\varphi\rangle\,\mathrm{d}W_{t}$
(4.21)

for $\varphi\in C^{2}_{c}(\mathbb{R})$ with initial condition
$\nu_{0}=\mathcal{L}(\xi)$. Along the chosen subsequence the cost
$J_{\textup{cl}}^{n}(g^{n})$ converges to

$\mathbb{E}\biggl{[}\int_{[0,T]\times\mathbb{R}\times
G}f(t,x,\nu_{t},g)\,\mathrm{d}\mathfrak{m}(t,x,g)+\psi(\nu_{T})\biggr{]}.$

Note that assuming that $(\mathfrak{m},\nu,W)$ is defined on
$(\Omega,\mathcal{F},\mathbb{P})$ means that we might have to enlarge the
probability space. We do this in such a way that $(\mathfrak{m},\nu,W)$ are
independent of $(\xi,B)$.

###### Proof of Proposition 4.4.

It is not difficult to see that $\nu^{n}$ is given by
$\nu^{n}_{t}=\mathcal{S}(\mu^{n}_{t})$, where $\mu^{n}=(\mu^{n}_{t})_{0\leq
t\leq T}$ solves SPDE (2.10) with input $g^{n}$, coefficients $b$ and
$\lambda$ replaced by $b^{n}$ and $\lambda^{n}$, respectively, and initial
condition $\mu^{n}=\mathcal{L}(\xi,\zeta/n)$. By Proposition 2.8, extending
the weak setup $\mathbb{S}^{\ast}$ if necessary, we can find a solution
$(X^{n},\Lambda^{n})$ of McKean–Vlasov SDE (2.8) with initial condition
$(X^{n}_{0},\Lambda^{n}_{0})=(\xi,\zeta/n)$ such that
$\mu^{n}_{t}=\mathcal{L}(X^{n}_{t},\Lambda^{n}_{t}|\mathcal{F}_{T})$.

Step 1: We first show that the sequence $(\mathfrak{m}^{n})_{n\geq 1}$ is
tight on $\mathbb{M}_{\leq T}^{2}(\mathbb{R}\times G)$. By [Lac15, Proposition
B.1] it is enough to show that the sequence of measures $(m^{n})_{n}$ on
$[0,T]\times\mathbb{R}\times G$ defined by
$m^{n}(A)=\mathbb{E}[\mathfrak{m}^{n}(A)]$ for
$A\in\mathcal{B}([0,T]\times\mathbb{R}\times G)$ is compact in
$\mathbb{M}_{\leq T}^{2}(\mathbb{R}\times G)$ and that the family
$(M_{2}^{2}(\mathfrak{m}^{n}))_{n\geq 1}$ is uniformly integrable. To
establish the former, by Lemma A.2, it is enough to show that the sequence
$(m^{n})_{n}$ is uniformly square-integrable. Now, we have that

$\displaystyle\sup_{n\geq 1}\int_{[0,T]\times\mathbb{R}\times
G}\mathbf{1}_{\lvert x\rvert\geq c}\lvert
x\rvert\,\mathrm{d}m^{n}(t,x,g)\leq\int_{0}^{T}\sup_{n\geq
1}\mathbb{E}[\mathbf{1}_{\lvert X^{n}_{t}\rvert\geq c}\lvert
X^{n}_{t}\rvert^{2}]\,\mathrm{d}t.$

Hence, if we can prove that the family $(X^{n}_{t})_{n\geq 1}$ is uniformly
square-integrable for all $t\in[0,T]$, the same holds for $(m^{n})_{n}$.
Before we establish the former, let us consider the second criterion above,
namely that $(M_{2}^{2}(\mathfrak{m}^{n}))_{n\geq 1}$ is uniformly integrable.
Since $[0,T]$ and $G$ are compact, uniform integrability of
$(M_{2}^{2}(\mathfrak{m}^{n}))_{n\geq 1}$ is equivalent to uniform
integrability of
$\bigl{(}\int_{0}^{T}\langle\nu^{n}_{t},\lvert\cdot\rvert^{2}\rangle\,\mathrm{d}t\bigr{)}_{n\geq
1}$. We can estimate

$\int_{0}^{T}\langle\nu^{n}_{t},\lvert\cdot\rvert^{2}\rangle\,\mathrm{d}t\leq\int_{0}^{T}\mathbb{E}[\lvert
X^{n}_{t}\rvert^{2}|\mathcal{F}_{T}]\,\mathrm{d}t\leq
T\mathbb{E}\bigl{[}(\lvert
X^{n}\rvert^{\ast}_{T})^{2}\big{|}\mathcal{F}_{T}\bigr{]},$

which shows that uniform integrability of
$\bigl{(}\int_{0}^{T}\langle\nu^{n}_{t},\lvert\cdot\rvert^{2}\rangle\,\mathrm{d}t\bigr{)}_{n\geq
1}$ is implied by uniform integrability of $\mathbb{E}\bigl{[}(\lvert
X^{n}\rvert^{\ast}_{T})^{2}\big{|}\mathcal{F}_{T}\bigr{]}$. Now, both the
uniform square-integrability of $(X^{n}_{t})_{n\geq 1}$ for all $t\in[0,T]$
and the uniform integrability of $\bigl{(}\mathbb{E}\bigl{[}(\lvert
X^{n}\rvert^{\ast}_{T})^{2}\big{|}\mathcal{F}_{T}\bigr{]}\bigr{)}_{n\geq 1}$
are implied by the uniform square-integrability of $(\lvert
X^{n}\rvert^{\ast}_{T})_{n\geq 1}$, so that is what we will show. Using
standard estimates for the McKean–Vlasov SDE satisfied by
$(X^{n},\Lambda^{n})$ and, subsequently, applying Grönwall’s inequality, we
find

$(\lvert X^{n}\rvert^{\ast}_{t})^{2}\leq
C\lvert\xi\rvert^{2}+C\int_{0}^{t}(1+M_{2}^{2}(\nu^{n}_{s}))\,\mathrm{d}s+CZ^{n}_{t},$
(4.22)

where $Z^{n}_{t}=\sup_{0x\leq s\leq
t}\bigl{\lvert}\int_{0}^{s}\sigma(r,X^{n}_{r})\,\mathrm{d}B_{r}+\int_{0}^{s}\sigma_{0}(r)\,\mathrm{d}W_{r}\bigr{\rvert}^{2}$.
Further estimating $M^{2}_{2}(\nu^{n}_{s})\leq\mathbb{E}[\lvert
X^{n}_{s}\rvert^{2}|\mathcal{F}_{T}]\leq\mathbb{E}[(\lvert
X^{n}\rvert^{\ast}_{t})^{2}|\mathcal{F}_{T}]$ for $s\in[0,t]$, taking
conditional expectation with respect to $\mathcal{F}_{T}$ on both sides of
(4.22), and then applying Grönwall’s inequality again yields that
$\mathbb{E}[(\lvert X^{n}\rvert^{\ast}_{T})^{2}|\mathcal{F}_{T}]\leq
C+C\mathbb{E}[Z^{n}_{T}|\mathcal{F}_{T}]$ for a possibly enlarged constant
$C>0$. Since the diffusion coefficients $\sigma$ and $\sigma_{0}$ are bounded,
the sequence $(\mathbb{E}[Z^{n}_{T}|\mathcal{F}_{T}])_{n\geq 1}$ is uniformly
integrable, so that the same is true for $(\mathbb{E}[(\lvert
X^{n}\rvert^{\ast}_{T})^{2}|\mathcal{F}_{T}])_{n\geq 1}$ and, therefore,
$(\sup_{0\leq t\leq T}M_{2}^{2}(\nu^{n}_{t}))_{n\geq 1}$. Thus, in view of
(4.22), the family $(\lvert X^{n}\rvert^{\ast}_{T})_{n\geq 1}$ is uniformly
integrable.

Next, we establish the tightness of $(\nu^{n})_{n\geq 1}$ on
$C([0,T];\mathcal{M}^{1}_{\leq 1}(\mathbb{R}))$. By Theorem 8.6 of Chapter 3
in [EK86] to prove tightness it is enough to show that (i) $(\nu^{n}_{t})_{n}$
is tight on $\mathcal{M}^{1}_{\leq 1}(\mathbb{R})$ for all $t\in[0,T]$ and
(ii) there exists a sequence of random variables $(\zeta_{n})_{n}$ with
$\sup_{n}\mathbb{E}\zeta_{n}<\infty$ such that for all $0\leq s\leq t\leq T$,
we have
$\mathbb{E}\bigl{[}d_{1}(\nu^{n}_{t},\nu^{n}_{s})\big{|}\mathcal{F}_{s}\bigr{]}\leq\lvert
t-s\rvert^{1/2}\mathbb{E}[\zeta_{n}|\mathcal{F}_{s}]$, where we recall that
$d_{1}$ is introduced in Subsection 2.2. We already showed (i) and the second
property is easily established, using the expression
$d_{1}(v_{1},v_{2})=\sup_{\lVert\varphi\rVert_{\text{Lip}}}\langle
v_{1}-v_{2},\varphi\rangle$. We do not provide the details here, but note that
the required estimates can be deduced along the same lines as the proof of
Proposition 3.4 in [HJ23].

Step 2: Let $(\mathfrak{m},\nu,W)$ be a weak subsequential limit of
$(\mathfrak{m}^{n},\nu^{n},W)_{n\geq 1}$. We show that $(\mathfrak{m},\nu,W)$
solves SPDE (4.21). To achieve this we reformulate the SPDEs satisfied by
$(\mathfrak{m}^{n},\nu^{n},W)$ and $(\mathfrak{m},\nu,W)$ as martingale
problems and deduce the stability of those martingale problems. Let us define
the generator $\mathcal{L}^{n}$ by

$\mathcal{L}^{n}\varphi(t,x,v,g)=b^{n}(t,x,v,g)\partial_{x}\varphi(x)-\lambda^{n}(t,x)\varphi(x)+a_{t}(x)\partial_{x}^{2}\varphi(x)$

for $\varphi\in C_{b}^{2}(\mathbb{R})$ and
$(t,x,v,g)\in[0,T]\times\mathbb{R}\times\mathcal{M}^{2}_{\leq
1}(\mathbb{R})\times G$. By assumption $\nu^{n}$ satisfies SPDE (2.23) with
$b$ and $\lambda$ replaced by $b^{n}$ and $\lambda^{n}$, respectively, initial
condition $\nu^{n}_{0}=\mathbb{E}[e^{-\zeta/n}]\mathcal{L}(\xi)$, and control
$g^{n}$. We can rewrite this as

$\mathrm{d}\langle\nu^{n}_{t},\varphi\rangle=\int_{\mathbb{R}\times
G}\mathcal{L}^{n}\varphi(t,x,\nu^{n}_{t},g)\,\mathrm{d}\mathfrak{m}^{n}(t,x,g)+\langle\nu^{n}_{t},\sigma_{0}(t)\partial_{x}\varphi\rangle\,\mathrm{d}W_{t}$

for $\varphi\in C^{2}_{c}(\mathbb{R})$. We formulate this SPDE as a martingale
problem. For $k\geq 1$, let $\Phi\in C^{2}_{c}(\mathbb{R}^{k+1})$ and
$\varphi\in C^{2}_{c}(\mathbb{R};\mathbb{R}^{k})$, and let $\langle
v,\varphi\rangle$ denote the vector $(\langle
v,\varphi_{1}\rangle,\dots,\langle v,\varphi_{k}\rangle)$ for
$v\in\mathcal{M}^{2}_{\leq 1}(\mathbb{R})$. We define the differential
operator $\mathbb{L}^{n}$ acting on $\Phi$ and $\varphi$ by

$\displaystyle\mathbb{L}^{n}(\Phi,\varphi)(t,x,w,v,g)$
$\displaystyle=\sum_{i=1}^{k}\partial_{y_{i}}\Phi(\langle
v,\varphi\rangle,w)\mathcal{L}^{n}\varphi_{i}(t,x,v,g)+\sum_{i,j=1}^{k}\frac{1}{2}\partial^{2}_{y_{i}y_{j}}\Phi(\langle
v,\varphi\rangle,w)\langle
v,\sigma_{0}(t)\partial_{x}\varphi_{j}\rangle\sigma_{0}(t)\partial_{x}\varphi_{i}(x)$
$\displaystyle\ \ \
+\sum_{i=1}^{k}\frac{1}{2}\partial^{2}_{y_{i}y_{k+1}}\Phi(\langle
v,\varphi\rangle,w)\sigma_{0}(t)\partial_{x}\varphi_{i}(x)$

for $(t,x,w,v,g)\in[0,T]\times\mathbb{R}^{2}\times\mathcal{M}^{2}_{\leq
1}(\mathbb{R})\times G$. We define the operator $\mathbb{L}$ analogously, only
replacing the appearance of $\mathcal{L}^{n}$ in the definition of
$\mathbb{L}^{n}$ by $\mathcal{L}$. Now, applying Itô’s formula to
$\Phi(\langle\nu^{n}_{t},\varphi\rangle,W_{t})$ implies that

$\displaystyle\mathrm{d}\Phi(\langle\nu^{n}_{t},\varphi\rangle,W_{t})$
$\displaystyle=\int_{\mathbb{R}\times
G}\mathbb{L}^{n}(\Phi,\varphi)(t,x,W_{t},\nu^{n}_{t},g)\,\mathrm{d}\mathfrak{m}^{n}(t,x,g)+\frac{1}{2}\partial^{2}_{y_{k+1}}\Phi(\langle\nu^{n}_{t},\varphi\rangle,W_{t})\,\mathrm{d}t$
$\displaystyle\ \ \
+\sum_{i=1}^{k}\partial_{y_{i}}\Phi(\langle\nu^{n}_{t},\varphi\rangle,W_{t})\langle\nu^{n}_{t},\sigma_{0}(t)\partial_{x}\varphi_{i}\rangle\,\mathrm{d}W_{t}+\partial_{y_{k+1}}\Phi(\langle\nu^{n}_{t},\varphi\rangle,W_{t})\,\mathrm{d}W_{t}.$

In other words, the process $\mathcal{M}^{n}(\Phi,\varphi)$ defined by

$\displaystyle\mathcal{M}^{n}_{t}(\Phi,\varphi)$
$\displaystyle=\Phi(\langle\nu^{n}_{t},\varphi\rangle,W_{t})-\int_{[0,t]\times\mathbb{R}\times
G}\mathbb{L}^{n}(\Phi,\varphi)(s,x,W_{s},\nu^{n}_{s},g)\,\mathrm{d}\mathfrak{m}^{n}(s,x,g)$
$\displaystyle\ \ \
-\int_{0}^{t}\frac{1}{2}\partial^{2}_{y_{k+1}}\Phi(\langle\nu^{n}_{s},\varphi\rangle,W_{s})\,\mathrm{d}s$

is a martingale for any $\Phi\in C^{2}_{c}(\mathbb{R}^{k})$ and $\varphi\in
C^{2}_{c}(\mathbb{R};\mathbb{R}^{k})$. On the one hand, as in [HJ23, Lemma
A.7], we can show that for any convergent sequence $(m^{n},v^{n},w^{n})_{n\geq
1}$ in $\mathbb{M}^{2}_{\leq T}(\mathbb{R}\times G)\times
C([0,T];\mathcal{M}^{2}_{\leq 1}(\mathbb{R}))\times C([0,T])$ with limit
$(m,v,w)$, the expression

$\displaystyle\int_{[0,t]\times\mathbb{R}\times
G}\mathbb{L}^{n}(\Phi,\varphi)(s,x,w^{n}_{s},v^{n}_{s},g)\,\mathrm{d}m^{n}(s,x,g)+\int_{0}^{t}\frac{1}{2}\partial^{2}_{y_{k+1}}\Phi(\langle
v^{n}_{s},\varphi\rangle,w^{n}_{s})\,\mathrm{d}s$

converges to

$\displaystyle\int_{[0,t]\times\mathbb{R}\times
G}\mathbb{L}(\Phi,\varphi)(s,x,w_{s},v_{s},g)\,\mathrm{d}m(s,x,g)+\int_{0}^{t}\frac{1}{2}\partial^{2}_{y_{k+1}}\Phi(\langle
v_{s},\varphi\rangle,w_{s})\,\mathrm{d}s,$

and $\Phi(\langle v^{n}_{t},\varphi\rangle,w^{n}_{t})$ tends to $\Phi(\langle
v_{t},\varphi\rangle,w_{t})$. On the other hand, it is not difficult to see
that $(\mathcal{M}^{n}(\Phi,\varphi))_{n\geq 1}$ is tight on $C([0,T])$.
Consequently, we see that along a further subsequence
$\mathcal{M}^{n}(\Phi,\varphi)$ converges weakly to the process
$\mathcal{M}(\Phi,\varphi)$ given by

$\displaystyle\mathcal{M}_{t}(\Phi,\varphi)$
$\displaystyle=\Phi(\langle\nu_{t},\varphi\rangle,W_{t})-\int_{[0,t]\times\mathbb{R}\times
G}\mathbb{L}(\Phi,\varphi)(s,x,W_{s},\nu_{s},g)\,\mathrm{d}\mathfrak{m}(s,x,g)$
$\displaystyle\ \ \
-\int_{0}^{t}\frac{1}{2}\partial^{2}_{y_{k+1}}\Phi(\langle\nu_{s},\varphi\rangle,W_{s})\,\mathrm{d}s.$

Moreover, since the local martingale property is stable under weak convergence
and the expectation $\mathbb{E}\lvert\mathcal{M}_{T}(\Phi,\varphi)\rvert^{2}$
is finite, it follows that $\mathcal{M}(\Phi,\varphi)$ is a martingale for all
$\Phi\in C^{2}_{c}(\mathbb{R}^{k})$ and $\varphi\in
C^{2}_{c}(\mathbb{R};\mathbb{R}^{k})$. By Theorem II.7.1 from [IW89] we can
conclude that

$\langle\nu_{t},\varphi_{i}\rangle=\langle
v_{0},\varphi_{i}\rangle+\int_{[0,t]\times\mathbb{R}\times
G}\mathcal{L}\varphi_{i}(s,x,\nu_{s},g)\,\mathrm{d}\mathfrak{m}(s,x,g)+\int_{0}^{t}\langle\nu_{s},\sigma_{0}(s)\partial_{x}\varphi_{i}\rangle\,\mathrm{d}W_{s}$

holds a.s. for any finite collection $\varphi_{1}$, …, $\varphi_{k}$ of
function in $C^{2}_{c}(\mathbb{R})$. Since $C^{2}_{c}(\mathbb{R})$ is
separable when equipped with the norm
$\varphi\mapsto\lVert\varphi\rVert_{\infty}+\lVert\partial_{x}\varphi\rVert_{\infty}+\lVert\partial_{x}^{2}\varphi\rVert_{\infty}$
and both sides of the above equation vary continuously as a function of
$\varphi_{i}$, we conclude that $(\mathfrak{m},\nu,W)$ satisfies SPDE (4.21).

Step 3: It remains to establish the convergence of the costs
$J^{n}_{\text{cl}}(g^{n})$. Note that again proceeding as in [HJ23, Lemma
A.7], we can show that along the subsequence along which
$(\mathfrak{m}^{n},\nu^{n},W)$ converges weakly to $(\mathfrak{m},\nu,W)$, the
integral

$\int_{0}^{T}\bigl{\langle}\nu^{n}_{t},f^{n}(t,\cdot,\nu^{n}_{t},g^{n}_{t})\bigr{\rangle}\,\mathrm{d}t+\psi(\nu^{n}_{T})=\int_{0,T]\times\mathbb{R}\times
G}f^{n}(t,x,\nu^{n}_{t},g)\,\mathrm{d}\mathfrak{m}^{n}(t,x,g)+\psi(\nu^{n}_{T})$

converges weakly to $\int_{0,T]\times\mathbb{R}\times
G}f(t,x,\nu_{t},g)\,\mathrm{d}\mathfrak{m}(t,x,g)+\psi(\nu_{T})$. Here, we
crucially use the uniform convergence of $f^{n}$ and $\psi^{n}$ on compact
subsets of $[0,T]\times\mathbb{R}\times\mathcal{M}^{2}_{\leq
1}(\mathbb{R})\times G$, Property (iii), as well as the fact that convergent
sequences in $\mathbb{M}_{\leq T}^{2}(\mathbb{R}\times G)$ are uniformly
square-integrable together with the uniform quadratic growth of $f^{n}$ and
$\psi^{n}$, Property (ii). Next, from Step 1 we know that the sequences
$(M_{2}^{2}(\mathfrak{m}^{n}))_{n\geq 1}$ and $(M_{2}^{2}(\nu^{n}_{T}))_{n\geq
1}$ are uniformly integrable. Hence, using Property (ii) again, allows us to
upgrade the weak subsequential convergence of
$\int_{0}^{T}\bigl{\langle}\nu^{n}_{t},f^{n}(t,\cdot,\nu^{n}_{t},g^{n}_{t})\bigr{\rangle}\,\mathrm{d}t+\psi(\nu^{n}_{T})$
to subsequential convergence in mean. Thus, $J^{n}_{\text{cl}}(g^{n})$
converges subsequentially to
$\mathbb{E}\bigl{[}\int_{0,T]\times\mathbb{R}\times
G}f(t,x,\nu_{t},g)\,\mathrm{d}\mathfrak{m}(t,x,g)+\psi(\nu_{T})\bigr{]}$. This
concludes the proof. ∎

Let $(\mathfrak{m},\nu,W)$ be as in the statement of Proposition 4.4. Firstly,
let us note that the random variables $\mathfrak{m}$ and $\nu$ are not
necessarily $\mathbb{F}$-progressively measurable, so we have to introduce a
different filtration. We let $\mathbb{F}^{\mathfrak{m},W}$ denote the
completion of the filtration

$\Bigl{(}\sigma\bigl{(}\mathfrak{m}([0,s]\times
A),\,W_{s}\mathpunct{:}A\in\mathcal{B}(\mathbb{R}\times G),\,0\leq s\leq
t\bigr{)}\Bigr{)}_{0\leq t\leq T}.$

The random measure $\mathfrak{m}$ is by definition
$\mathbb{F}^{\mathfrak{m},W}$-progressively measurable. Moreover, owing to
Lemma A.3 there is a family of random probability measures
$(\Gamma_{t,x})_{(t,x)\in[0,T]\times\mathbb{R}}$ on $G$ such that
$\mathrm{d}\mathfrak{m}(t,x,g)=\mathrm{d}\Gamma_{t,x}(g)\mathrm{d}\nu_{t}(x)\mathrm{d}t$.
In particular, the process $\nu$ is $\mathbb{F}^{\mathfrak{m},W}$-adapted as
well. In addition, it follows from [Lac15, Lemma 3.2] that the map
$\Gamma\mathpunct{:}[0,T]\times\Omega\times\mathbb{R}\to\mathcal{P}(G)$ can be
chosen in an $\mathbb{F}^{\mathfrak{m},W}$-predictable manner. We now define
$\mathbb{S}^{\ast}=(\Omega,\mathcal{F},\mathbb{P},\mathbb{F}^{\mathfrak{m},\xi,B,W},\mathbb{F}^{\mathfrak{m},W},\xi,0,B,W)$,
where $\mathbb{F}^{\mathfrak{m},\xi,W,B}$ is the completion of
$\mathbb{F}^{\mathfrak{m},W}\lor\mathbb{F}^{\xi,B}$ and $\mathbb{F}^{\xi,B}$
denotes the completion of the filtration generated by $\xi$ and $B$. One can
check that $\mathbb{S}^{\ast}$ is a weak setup.

######  Remark 4.5.

Let us note that Equation (4.21) can be viewed as a relaxed formulation for
the control of the stochastic Fokker–Planck equation (2.23). Here a relaxed
control is a progressively measurable $\mathbb{M}_{\leq
T}^{2}(\mathbb{R}\times G)$-valued process $\mathfrak{m}$ such that the flow
of measures $\nu=(\nu_{t})_{0\leq t\leq T}$ obtained through the
disintegration
$\mathrm{d}\mathfrak{m}(t,x,g)=\mathrm{d}\Gamma_{t,x}(g)\,\mathrm{d}\nu_{t}(x)\mathrm{d}t$
satisfies SPDE (4.21). This idea can be extended to arbitrary (stochastic)
Fokker–Planck equations or, more generally, measure-valued (S)PDEs, such as
those arising in filtering theory, and may be useful to establish the
existence of optimal controls.

Next, since Assumption 2.6 is in place, as in the proof of [Lac15, Theorem
3.7], we can find $g^{\ast}\in\mathcal{G}_{\mathbb{F}^{\mathfrak{m},W}}^{1}$
such that

$\int_{G}b(t,x,\nu_{t},g)\,\mathrm{d}\Gamma_{t,x}(g)=b\bigl{(}t,x,\nu_{t},g^{\ast}_{t}(x)\bigr{)},\quad\int_{G}f(t,x,\nu_{t},g)\,\mathrm{d}\Gamma_{t,x}(g)\geq
f(t,x,\nu_{t},g^{\ast}_{t}(x)).$ (4.23)

This allows us to rewrite SPDE (4.21) as

$\mathrm{d}\langle\nu_{t},\varphi\rangle=\bigl{\langle}\nu_{t},\mathcal{L}\varphi(t,\cdot,\nu_{t},g^{\ast}_{t})\bigr{\rangle}\,\mathrm{d}t+\langle\nu_{t},\sigma_{0}(t)\partial_{x}\varphi\rangle\,\mathrm{d}W_{t}$
(4.24)

for $\varphi\in C^{2}_{c}(\mathbb{R})$. For the associated cost we obtain the
upper bound

$\mathbb{E}\biggl{[}\int_{0}^{T}\bigl{\langle}\nu_{t},f(t,\cdot,\nu_{t},g^{\ast}_{t})\bigr{\rangle}\,\mathrm{d}t+\psi(\nu_{T})\biggr{]}\leq\mathbb{E}\biggl{[}\int_{[0,T]\times\mathbb{R}\times
G}f(t,x,\nu_{t},g)\,\mathrm{d}\mathfrak{m}(t,x,g)+\psi(\nu_{T})\biggr{]}.$
(4.25)

######  Proposition 4.6.

Let $\nu$ and $g^{\ast}$ be as above. Then, extending the weak setup
$\mathbb{S}^{\ast}$ if necessary, there exists a solution $(X,\Lambda)$ to
McKean–Vlasov SDE (2.8) with initial conditions $X_{0}=\xi$, $\Lambda_{0}=0$,
subprobability distribution
$\langle\nu_{t},\varphi\rangle=\mathbb{E}[e^{\Lambda_{t}}\varphi(X_{t})|\mathcal{F}_{T}]$,
and control $\gamma^{\ast}=(\gamma^{\ast}_{t})_{0\leq t\leq T}$ given by
$\gamma^{\ast}_{t}=g^{\ast}_{t}(X_{t})$. That is, $\gamma^{\ast}$ is a
semiclosed-loop control with feedback function $g^{\ast}$ and
$J_{\mathbb{S}^{\ast}}(\gamma^{\ast})=\mathbb{E}\bigl{[}\int_{0}^{T}\bigl{\langle}\nu_{t},f(t,\cdot,\nu_{t},g^{\ast}_{t})\bigr{\rangle}\,\mathrm{d}t+\psi(\nu_{T})\bigr{]}$.

###### Proof.

Let us consider the random linear SPDE

$\mathrm{d}\langle\tilde{\mu}_{t},\varphi\rangle=\bigl{\langle}\tilde{\mu}_{t},\mathcal{L}_{2}\varphi(t,\cdot,\nu_{t},g^{\ast}_{t})\bigr{\rangle}\,\mathrm{d}t+\langle\tilde{\mu}_{t},\sigma_{0}(t)\partial_{x}\varphi\rangle\,\mathrm{d}W_{t}$

for $\varphi\in C^{2}_{c}(\mathbb{R}^{2})$ with initial condition
$\tilde{\mu}_{0}=\mathcal{L}(\xi,\zeta)$. Note that unlike in SPDE (2.10), the
flow of subprobabilities $\nu$ is fixed and not determined by the SPDE. Hence,
the SPDE is not nonlinear but linear with random coefficients. Since
$\mathcal{L}(\xi,\zeta)\in L^{2}(\mathbb{R}^{2})$, using similar arguments as
in the proof of Proposition 2.8, we can show that this SPDE has a unique
strong solution. Moreover, if we define $\tilde{\nu}=(\tilde{\nu}_{t})_{0\leq
t\leq T}$ by
$\tilde{\nu}_{t},=\mathbb{E}[e^{-\zeta}]^{-1}\mathcal{S}(\tilde{\mu}_{t})$,
then $\tilde{\nu}$ solves the random linear SPDE

$\mathrm{d}\langle\tilde{\nu}_{t},\varphi\rangle=\bigl{\langle}\tilde{\nu}_{t},\mathcal{L}\varphi(t,\cdot,\nu_{t},g^{\ast}_{t})\bigr{\rangle}\,\mathrm{d}t+\langle\tilde{\nu}_{t},\sigma_{0}(t)\partial_{x}\varphi\rangle\,\mathrm{d}W_{t}$

for $\varphi\in C^{2}_{c}(\mathbb{R})$ with initial condition
$\tilde{\nu}_{0}=\mathcal{L}(\xi)$. This SPDE again has a unique solution, so
since $\nu$ solves the same linear SPDE we obtain that $\tilde{\nu}=\nu$. In
other words, it holds that
$\nu_{t}=\mathbb{E}[e^{-\zeta}]^{-1}\mathcal{S}(\tilde{\mu}_{t})$.

Now, by [LSZ20, Theorem 3.1], extending the weak setup if necessary, we can
find a solution $(X,\tilde{\Lambda})$ to the random SDE

$\mathrm{d}X_{t}=b\bigl{(}t,X_{t},\nu_{t},g^{\ast}_{t}(X_{t})\bigr{)}\,\mathrm{d}t+\sigma(t,X_{t})\,\mathrm{d}B_{t}+\sigma_{0}(t)\,\mathrm{d}W_{t},\quad\mathrm{d}\tilde{\Lambda}_{t}=\lambda(t,X_{t})\,\mathrm{d}t$

with initial conditions $X_{0}=\xi$, $\tilde{\Lambda}_{0}=\zeta$, such that
$\mathcal{L}(X_{t},\tilde{\Lambda}_{t}|\mathcal{F}_{T})=\tilde{\mu}_{t}$. Note
that $\nu$ is a fixed input and not determined by the SDE. We define
$\Lambda_{t}=\int_{0}^{t}\lambda(s,X_{s})\,\mathrm{d}s$ and set
$\mu_{t}=\mathcal{L}(X_{t},\Lambda_{s}|\mathcal{F}_{T})$. Then it follows that

$\langle\mathcal{S}(\mu_{t}),\varphi\rangle=\mathbb{E}[e^{-\Lambda_{t}}\varphi(X_{t})|\mathcal{F}_{T}]=\mathbb{E}[e^{-\zeta}]^{-1}\mathbb{E}\bigl{[}e^{-\tilde{\Lambda}_{t}}\varphi(X_{t})\big{|}\mathcal{F}_{T}\bigr{]}=\mathbb{E}[e^{-\zeta}]^{-1}\mathcal{S}(\tilde{\mu}_{t})=\nu_{t},$

where we used that $\zeta$ is independent of $X$, $\Lambda$, and
$\mathcal{F}_{T}$. Hence, $(X,\Lambda)$ is a solution to McKean–Vlasov SDE
(2.8). This, in particular, implies that

$J_{\mathbb{S}}(\gamma^{\ast})=\mathbb{E}\biggl{[}\int_{0}^{T}\bigl{\langle}\nu_{t},f(t,\cdot,\cdot,\nu_{t},g^{\ast}_{t})\bigr{\rangle}\,\mathrm{d}t+\psi(\nu_{T})\biggr{]},$

where $\gamma^{\ast}=(\gamma^{\ast}_{t})_{0\leq t\leq T}$ is defined by
$\gamma^{\ast}_{t}=g^{\ast}_{t}(X_{t})$. ∎

We can use the subsequential convergence established in Proposition 4.4 and
the representation from Proposition 4.6 to show that if the controls
$g^{n}\in\mathcal{G}_{\mathbb{F}}^{1}$ in the statement of Proposition 4.4 are
chosen optimally, then a random function
$g^{\ast}\in\mathcal{G}_{\mathbb{F}}^{1}$ chosen according to the criterion
(4.23) yields an optimal semiclosed-loop control which is independent of the
intensity.

For convenience, in the following we let $V^{n}$ denote the infimum of
$J^{n}_{\textup{cl}}(g)$ over control $g\in\mathcal{G}_{\mathbb{F}}^{2}$. From
our discussion in Section 2.2 we know that $V^{n}$ coincides with the value of
the strong formulation introduced at the beginning of Section 2.2 with $b$,
$\lambda$, $f$, and $\psi$ replaced by $b^{n}$, $\lambda^{n}$, $f^{n}$, and
$\psi^{n}$, respectively, and initial condition $(\xi,\zeta/n)$.

######  Proposition 4.7.

Choose the controls $g^{n}\in\mathcal{G}_{\mathbb{F}}^{1}$ in the statement of
Proposition 4.4 optimally. Then the control $\gamma^{\ast}$ from Proposition
4.6 is an optimal semiclosed-loop control and the feedback function
$g^{\ast}\in\mathcal{G}_{\mathbb{F}^{\mathfrak{m},W}}^{1}$ does not depend on
the intensity.

Note that optimal controls $g^{n}\in\mathcal{G}_{\mathbb{F}}^{1}$, i.e.
controls for which $J^{n}_{\textup{cl}}(g^{n})=V^{n}_{\textup{cl}}$, exist by
Corollary 2.16.

###### Proof of Proposition 4.7.

Note that regardless of which weakly convergent subsequence
$(\mathfrak{m}^{n_{k}},\nu^{n_{k}},W)_{k\geq 1}$ we choose in Proposition 4.4,
in view of Equation (4.25) and Proposition 4.6, we always obtain that

$\lim_{k\to\infty}J_{\text{cl}}^{n_{k}}(g^{n_{k}})=\mathbb{E}\biggl{[}\int_{[0,T]\times\mathbb{R}\times
G}f(t,x,\nu_{t},g)\,\mathrm{d}\mathfrak{m}(t,x,g)+\psi(\nu_{T})\biggr{]}\geq
J_{\mathbb{S}^{\ast}}(\gamma^{\ast})\geq V_{\mathbb{S}^{\ast}}=V.$

Hence, we find that $\liminf_{n\to\infty}V^{n}\geq V$.

Next, let us deduce the converse inequality. To achieve this we exploit the
equivalence between the strong formulation introduced at the beginning of
Section 2.2 and the SPDE formulation that we study here. Let us denote the
cost functional for the strong formulation with $b$, $\lambda$, $f$, and
$\psi$ replaced by $b^{n}$, $\lambda^{n}$, $f^{n}$, and $\psi^{n}$,
respectively, and initial condition $(\xi,\zeta/n)$ by $J^{n}$. Then
$J_{\textup{cl}}(g^{n})=V^{n}=\inf_{\gamma}J^{n}(\gamma)$, where the infimum
is over $\mathbb{F}^{\xi,\zeta,B,W}$-progressively measurable $G$-valued
processes $\gamma=(\gamma_{t})_{0\leq t\leq T}$, and
$V=\inf_{\gamma}J(\gamma)$, where the infimum is over
$\mathbb{F}^{\xi,B,W}$-progressively measurable $G$-valued processes
$\gamma=(\gamma_{t})_{0\leq t\leq T}$. Now, let us choose an
$\mathbb{F}^{\xi,B,W}$-progressively measurable $G$-valued processes
$\gamma=(\gamma_{t})_{0\leq t\leq T}$ with $J(\gamma)\leq V+\epsilon$. It
holds that $J_{\textup{cl}}^{n}(g^{n})=V^{n}\leq J^{n}(\gamma)$, so if we can
show that $\lim_{n\to\infty}J^{n}(\gamma)=J(\gamma)$, then we have proved that

$V\leq\limsup_{n\to\infty}V^{n}\leq\limsup_{n\to\infty}J^{n}(\gamma)=J(\gamma)\leq
V+\epsilon.$

Letting $\epsilon\to 0$, then yields the equality $V\leq
J_{\mathbb{S}^{\ast}}(\gamma^{\ast})\leq\lim_{n\to\infty}J_{\textup{cl}}^{n}(g^{n})=\lim_{n\to\infty}V^{n}=V$.
Hence, $\gamma^{\ast}$ is an optimal semiclosed-loop control whose feedback
function $g^{\ast}$ is independent of the intensity.

It remains to prove that $J^{n}(\gamma)\to J(\gamma)$ for the
$\mathbb{F}^{\xi,B,W}$-progressively measurable $G$-valued process $\gamma$
from above. Let $(X^{n},\Lambda^{n})$ be the solution to McKean–Vlasov SDE
(2.5) with coefficients $b$ and $\lambda$ replaced by $b^{n}$ and
$\lambda^{n}$, respectively, initial condition $(\xi,\zeta/n)$, and control
$\gamma$. By $(X,\Lambda)$ we denote the solution to McKean–Vlasov SDE (2.5)
with initial condition $(\xi,0)$ and control $\gamma$. Then owing to the
uniform in $n$ linear growth of $b^{n}$ and $\lambda^{n}$ and the locally
uniform convergence of $b^{n}$ to $b$ and $\lambda^{n}$ to $\lambda$ (cf.
Properties (iii) and (ii)), we can show that
$\lim_{n\to\infty}\mathbb{E}(\lvert X^{n}-X\rvert^{\ast}_{T})^{2}=0$. The only
difficulty in obtaining this convergence lies in the fact that the mapping
$\mathcal{S}$ defined in (2.12) is only locally Lipschitz continuous as a map
$\mathcal{P}^{2}(\mathbb{R}\times[0,\infty))\to\mathcal{M}^{2}_{\leq
1}(\mathbb{R})$, where $\mathcal{P}^{2}(\mathbb{R}\times[0,\infty))$ is
equipped with the $2$-Wasserstein distance and $\mathcal{M}^{2}_{\leq
1}(\mathbb{R})$ with the metric $d_{1}$. However, this hurdle can be overcome
through stopping time arguments. We do not provide the details here and
instead refer to [HJ23, Propositions 3.12 and A.2]. Once we know that
$(X^{n})_{n\geq 1}$ converges to $X$ in $L^{2}$-$\sup$, the convergence of the
costs $J^{n}(\gamma)$ to $J(\gamma)$ follows straightforwardly from the
uniform quadratic growth assumptions placed on $f^{n}$ and $\psi^{n}$ as well
as the locally uniform convergence of $f^{n}$ to $f$ and $\psi^{n}$ to $\psi$
(cf. Properties (iii) and (ii)). This concludes the proof. ∎

Finally, to prove Theorem 2.17 we wish to apply Proposition 4.7. Starting from
the coefficients $b$ and $\lambda$ as well as cost functions $f$ and $\psi$
satisfying Assumptions 2.4 and 2.6, we want to mollify and cut off these
functions appropriately to obtain maps $b^{n}$, $\lambda^{n}$, $f^{n}$, and
$\psi^{n}$ that satisfy Properties (i) to (iv) outlined at the beginning of
this subsection. If we find such maps, then Proposition 4.7 implies the
existence of an optimal semiclosed-loop control independent of the intensity.
To construct the desired maps, we use Proposition A.4 and Proposition A.5.

###### Proof of Theorem 2.17.

Owing to Assumption 2.4, the functions $b$ and $f$ decompose as
$b(t,x,v,g)=b_{0}(t,x,v)+b_{1}(t,x,g)$ and
$f(t,x,v,g)=f_{0}(t,x,v)+f_{1}(t,x,g)$ for
$(t,x,v,g)\in[0,T]\times\mathbb{R}\times\mathcal{M}^{2}_{\leq
1}(\mathbb{R})\times G$. Now, we can apply Proposition A.4 to obtain a
sequence $(f_{1}^{n})_{n\geq 1}$ that converges locally uniformly to $f_{1}$.
Further, we define the sequence $(b_{1}^{n})_{n\geq 1}$ by
$b_{1}^{n}(t,x,g)=b(t,-n\lor x\land n,g)$. Next, by Proposition A.5 we can
find sequences $(\tilde{b}_{0}^{n})_{n\geq 1}$, $(\tilde{f}_{0}^{n})_{n\geq
1}$, and $(\psi^{n})_{n\geq 1}$ that converge locally uniformly to $b_{0}$,
$f_{0}$, and $\psi$, respectively. We then set
$b_{0}^{n}(t,x,v)=(\eta_{1/n}\ast\tilde{b}_{0}^{n}(t,\cdot,g))(x)$ and
$f_{0}^{n}(t,x,v)=(\eta_{1/n}\ast\tilde{f}_{0}^{n}(t,\cdot,g))(x)$, where
$\eta\mathpunct{:}\mathbb{R}\to\mathbb{R}$ is the standard mollifier and
$\eta_{\epsilon}(x)=\frac{1}{\epsilon}\eta(\frac{x}{\epsilon})$. This ensures
the Lipschitz continuity of $(x,v)\mapsto b_{0}^{n}(t,x,v)$ uniformly in
$t\in[0,T]$ and the uniform continuity of $(x,v)\mapsto f_{0}^{n}(t,x,v)$ for
$(t,x,v)$ in compact subsets of
$[0,T]\times\mathbb{R}\times\mathcal{M}^{2}_{\leq 1}(\mathbb{R})$ without
destroying Properties (i) to (iv) from Proposition A.5. Then, we define the
functions $b^{n}$, $f^{n}$, and $\lambda^{n}$ by
$b^{n}(t,x,v,g)=b_{0}^{n}(t,x,v)+b_{1}^{n}(t,x,g)$,
$f^{n}(t,x,v,g)=f_{0}^{n}(t,x,v)+f_{1}^{n}(t,x,g)$, and
$\lambda^{n}(t,x)=\lambda(t,x)\land n$ for
$(t,x,v,g)\in[0,T]\times\mathbb{R}\times\mathcal{M}^{2}_{\leq
1}(\mathbb{R})\times G$. It is straightforward to check that the functions
$b^{n}$, $\lambda^{n}$, $f^{n}$, and $\psi^{n}$ satisfy Properties (i) to
(iv). Hence, as stated above, we can conclude with Proposition 4.7. ∎

##  Appendix A Appendix

###  A.1 Auxiliary Results for Section 3

######  Lemma A.1.

Let $\mathcal{H}$ be a separable Hilbert space. Suppose that $(u^{n})_{n\geq
1}$ is a sequence in $D_{\mathbb{F}}^{2}([0,T];\mathcal{H})$ that converges to
some process $u\in D_{\mathbb{F}}^{2}([0,T];\mathcal{H})$, and that
$(m^{n})_{n\geq 1}$, $(\tilde{m}^{n})_{n\geq 1}$ are sequences in
$\mathcal{M}_{\mathbb{F}}^{2}([0,T];\mathcal{H})$ that converge to processes
$m$, $\tilde{m}\in\mathcal{M}_{\mathbb{F}}^{2}([0,T];\mathcal{H})$,
respectively. Next, let $F_{1}\mathpunct{:}\mathcal{H}\to\mathcal{H}$ and
$F_{2}\mathpunct{:}\mathcal{H}\to B_{2}(\mathcal{H})$ be continuous and
bounded on bounded sets of $\mathcal{H}$. Then

$\displaystyle\int_{0}^{t}\langle
F_{1}(u^{n}_{s-}),\mathrm{d}m^{n}_{s}\rangle_{\mathcal{H}}$
$\displaystyle\to\int_{0}^{t}\langle
F_{1}(u_{s-}),\mathrm{d}m_{s}\rangle_{\mathcal{H}},$ (A.1)
$\displaystyle\int_{0}^{t}\operatorname{tr}\bigl{(}F_{2}(u^{n}_{s-})\,\mathrm{d}[[m^{n},\tilde{m}^{n}]]^{c}_{s}\bigr{)}$
$\displaystyle\to\int_{0}^{t}\operatorname{tr}\bigl{(}F_{2}(u_{s-})\,\mathrm{d}[[m,\tilde{m}]]^{c}_{s}\bigr{)}$
(A.2)

uniformly in $t\in[0,T]$ in probability.

###### Proof.

We prove the convergences in (A.1) and (A.2) separately. For the first one we
write

$\displaystyle\begin{split}\int_{0}^{t}\langle
F_{1}(u^{n}_{s-})&,\mathrm{d}m^{n}_{s}\rangle_{\mathcal{H}}-\int_{0}^{t}\langle
F_{1}(u_{s-}),\mathrm{d}m_{s}\rangle_{\mathcal{H}}\\\
&=\int_{0}^{t}\bigl{\langle}F_{1}(u^{n}_{s-})-F_{1}(u_{s-}),\mathrm{d}m^{n}_{s}\bigr{\rangle}_{\mathcal{H}}+\int_{0}^{t}\bigl{\langle}F_{1}(u_{s-}),\mathrm{d}(m^{n}_{s}-m_{s})\bigr{\rangle}_{\mathcal{H}}.\end{split}$
(A.3)

We will deal with both terms on the right-hand side in sequence. Our goal is
to prove that both converge to zero in probability. For $k\geq 1$ let
$\pi_{k}\mathpunct{:}\mathcal{H}\to\mathcal{H}$ denote the projection onto the
centred ball $B_{k}$ of radius $k$ in $\mathcal{H}$. Now, since
$\mathbb{E}\sup_{0\leq s\leq t}\lVert
u^{n}_{s}-u_{s}\rVert_{\mathcal{H}}^{2}\to 0$, for any $\delta>0$ we can find
$K\geq 1$ such that for all $k\geq K$ it holds that

$\displaystyle\mathbb{P}\Bigl{(}\pi_{k}(u^{n}_{s-})=u^{n}_{s-},\,\pi_{k}(u_{s-})=u_{s-}\
\text{for }s\in(0,t]\Bigr{)}$ $\displaystyle\leq\mathbb{P}\biggl{(}\sup_{0\leq
s\leq t}\lVert u^{n}_{s}\rVert_{\mathcal{H}}^{2}\lor\sup_{0\leq s\leq t}\lVert
u_{s}\rVert_{\mathcal{H}}^{2}\leq k\biggr{)}$ $\displaystyle\geq 1-\delta.$

Hence, for any $\epsilon>0$ we obtain

$\displaystyle\begin{split}\mathbb{P}\biggl{(}\sup_{0\leq t\leq
T}\biggl{\lvert}&\int_{0}^{t}\bigl{\langle}F_{1}(u^{n}_{s-})-F_{1}(u_{s-}),\mathrm{d}m^{n}_{s}\bigr{\rangle}_{\mathcal{H}}\biggr{\rvert}>\epsilon\biggr{)}\\\
&\leq\mathbb{P}\biggl{(}\sup_{0\leq t\leq
T}\biggl{\lvert}\int_{0}^{t}\bigl{\langle}F_{1}(\pi_{k}(u^{n}_{s-}))-F_{1}(\pi_{k}(u_{s-})),\mathrm{d}m^{n}_{s}\bigr{\rangle}_{\mathcal{H}}\biggr{\rvert}>\epsilon\biggr{)}+\delta.\end{split}$
(A.4)

Next, using Proposition 3.1 we find that

$\displaystyle\mathbb{P}\biggl{(}\sup_{0\leq t\leq
T}\biggl{\lvert}\int_{0}^{t}\bigl{\langle}F_{1}$
$\displaystyle(\pi_{k}(u^{n}_{s-}))-F_{1}(\pi_{k}(u_{s-})),\mathrm{d}m^{n}_{s}\bigr{\rangle}_{\mathcal{H}}\biggr{\rvert}>\epsilon\biggr{)}$
$\displaystyle\leq\frac{1}{\epsilon}\mathbb{E}\sup_{0\leq t\leq
T}\biggl{\lvert}\int_{0}^{t}\bigl{\langle}F_{1}(\pi_{k}(u^{n}_{t-}))-F_{1}(\pi_{k}(u_{t-})),\mathrm{d}m^{n}_{s}\bigr{\rangle}_{\mathcal{H}}\biggr{\rvert}$
$\displaystyle\leq\frac{C_{1}}{\epsilon}\mathbb{E}\biggl{(}\int_{0}^{T}\bigl{\lVert}F_{1}(\pi_{k}(u^{n}_{t-}))-F_{1}(\pi_{k}(u_{t-}))\bigr{\rVert}_{\mathcal{H}}^{2}\,\mathrm{d}[m^{n}]_{t}\biggr{)}^{1/2}$
$\displaystyle\leq\frac{C_{1}}{\epsilon}\mathbb{E}\biggl{[}\sup_{0\leq t\leq
T}\bigl{\lVert}F_{1}(\pi_{k}(u^{n}_{t}))-F_{1}(\pi_{k}(u_{t}))\bigr{\rVert}_{\mathcal{H}}\bigl{(}[m^{n}]_{T}-[m^{n}]_{0}\bigr{)}^{1/2}\biggr{]}$
$\displaystyle\leq\frac{C_{1}^{2}}{\delta\epsilon^{2}}\mathbb{E}\sup_{0\leq
t\leq
T}\bigl{\lVert}F_{1}(\pi_{k}(u^{n}_{t}))-F_{1}(\pi_{k}(u_{t}))\bigr{\rVert}_{\mathcal{H}}^{2}+\delta\mathbb{E}\bigl{(}[m^{n}]_{T}-[m^{n}]_{0}\bigr{)}.$
(A.5)

Now, since $m^{n}$ converges to $m$ in
$\mathcal{M}_{\mathbb{F}}^{2}([0,T];\mathcal{H})$, the quantity
$\mathbb{E}\bigl{(}[m^{n}]_{T}-[m^{n}]_{0}\bigr{)}$ is bounded uniformly in
$n\geq 1$ by some constant $C>0$. Moreover, the map $F_{1}\circ\pi_{k}$ is
continuous and bounded, so since $\mathbb{E}\sup_{0\leq t\leq T}\lVert
u^{n}_{t}-u_{t}\rVert_{\mathcal{H}}^{2}\to 0$ by the dominated convergence
theorem, we may choose $n\geq 1$ large enough so that $\mathbb{E}\sup_{0\leq
t\leq
T}\bigl{\lVert}F_{1}(\pi_{k}(u^{n}_{t}))-F_{1}(\pi_{k}(u_{t}))\bigr{\rVert}_{\mathcal{H}}^{2}\leq\frac{\delta^{2}\epsilon^{2}}{C_{1}^{2}}$.
Plugging this into (A.1) and combining it with Equation (A.4) implies that

$\mathbb{P}\biggl{(}\sup_{0\leq t\leq
T}\biggl{\lvert}\int_{0}^{t}\bigl{\langle}F_{1}(u^{n}_{s-})-F_{1}(u_{s-}),\mathrm{d}m^{n}_{s}\bigr{\rangle}_{\mathcal{H}}\biggr{\rvert}>\epsilon\biggr{)}\leq(2+C)\delta.$

Since $\delta$ and $\epsilon$ were arbitrary this implies that
$\int_{0}^{t}\bigl{\langle}F_{1}(u^{n}_{s-})-F_{1}(u_{s-}),\mathrm{d}m^{n}_{s}\bigr{\rangle}_{\mathcal{H}}$
converges uniformly in $t\in[0,T]$ to zero in probability.

Next, let us consider the second term on the right-hand side of (A.3). Arguing
as above for any $\delta>0$ we can choose $K\geq 1$ and $k\geq K$ such that
for any $\epsilon>0$,

$\displaystyle\mathbb{P}\biggl{(}\biggl{\lvert}\int_{0}^{t}\bigl{\langle}F_{1}$
$\displaystyle(u_{s-}),\mathrm{d}(m^{n}_{s}-m_{s})\bigr{\rangle}_{\mathcal{H}}\biggr{\rvert}>\epsilon\biggr{)}$
$\displaystyle\leq\frac{1}{\epsilon}\mathbb{E}\biggl{\lvert}\int_{0}^{t}\bigl{\langle}F_{1}(\pi_{k}(u_{s-})),\mathrm{d}(m^{n}_{s}-m_{s})\bigr{\rangle}_{\mathcal{H}}\biggr{\rvert}+\delta$
$\displaystyle\leq\frac{\delta}{c_{k}^{2}}\mathbb{E}\sup_{0\leq t\leq T}\lVert
F_{1}(\pi_{k}(u_{t}))\rVert_{\mathcal{H}}^{2}+\frac{C_{1}^{2}c_{k}^{2}}{\delta\epsilon^{2}}\mathbb{E}\bigl{[}[m^{n}]_{t}-[m^{\epsilon}]_{0}-([m]_{t}-[m]_{0})\bigr{]}+\delta$
$\displaystyle\leq
2\delta+\frac{C_{1}^{2}c_{k}^{2}}{\delta\epsilon^{2}}\mathbb{E}\bigl{[}[m^{n}]_{t}-[m^{n}]_{0}-([m]_{t}-[m]_{0})\bigr{]},$

where $c_{k}=\sup_{h\in B_{k}}\lVert F_{1}(h)\rVert_{\mathcal{H}}$. Then we
choose $\epsilon$ large enough such that
$\mathbb{E}\bigl{[}[m^{n}]_{t}-[m^{n}]_{0}-([m]_{t}-[m]_{0})\bigr{]}\leq\frac{\delta^{2}\epsilon^{2}}{C_{1}^{2}c_{n}^{2}}$
to obtain $\mathbb{P}\bigl{(}\sup_{0\leq t\leq
T}\bigl{\lvert}\int_{0}^{t}\bigl{\langle}F_{1}(u_{s-}),\mathrm{d}(m^{n}_{s}-m_{s})\bigr{\rangle}_{\mathcal{H}}\bigr{\rvert}>\epsilon\bigr{)}\leq
3\delta$, which means that
$\int_{0}^{t}\bigl{\langle}F_{1}(u_{s-}),\mathrm{d}(m^{n}_{s}-m_{s})\bigr{\rangle}_{\mathcal{H}}$
converges uniformly in $t\in[0,T]$ to zero in probability as well. This
concludes our treatment of the convergence in (A.1).

We continue with (A.2). By appealing to the polarisation identity
$[[m,\tilde{m}]]=\frac{1}{4}([[m+\tilde{m}]]-[[m-\tilde{m}]])$, we can reduce
to the case $\tilde{m}^{n}=m^{n}$ and $\tilde{m}=m$. Then similarly to
Equation (A.3), we decompose

$\displaystyle\int_{0}^{t}\operatorname{tr}\bigl{(}$ $\displaystyle
F_{2}(u^{n}_{s-})\,\mathrm{d}[[m^{n}]]^{c}_{s}\bigr{)}-\int_{0}^{t}\operatorname{tr}\bigl{(}F_{2}(u_{s-})\,\mathrm{d}[[m]]^{c}_{s}\bigr{)}$
$\displaystyle=\int_{0}^{t}\operatorname{tr}\bigl{(}(F_{2}(u^{n}_{s-})-F_{2}(u_{s-}))\,\mathrm{d}[[m^{n}]]^{c}_{s}\bigr{)}+\int_{0}^{t}\operatorname{tr}\bigl{(}F_{2}(u_{s-})\,\mathrm{d}([[m^{n}]]^{c}_{s}-[[m]]^{c}_{s})\bigr{)}.$

For the first term we apply the inequality

$\sup_{0\leq t\leq
T}\biggl{\lvert}\int_{0}^{t}\operatorname{tr}\bigl{(}(F_{2}(u^{n}_{s-})-F_{2}(u_{s-}))\,\mathrm{d}[[m^{n}]]^{c}_{s}\bigr{)}\biggr{\rvert}\leq\int_{0}^{T}\bigl{\lVert}F_{2}(u^{n}_{t-})-F_{2}(u_{t-})\bigr{\rVert}\,\mathrm{d}[m^{n}]^{c}_{t},$
(A.6)

where $\bigl{\lVert}F_{2}(u^{n}_{t-})-F_{2}(u_{t-})\bigr{\rVert}$ denotes the
operator norm of
$F_{2}(u^{n}_{t-})-F_{2}(u_{t-})\mathpunct{:}\mathcal{H}\to\mathcal{H}$ and
then proceed as above to show that the right-hand side converges to zero in
probability as $n\to\infty$. To deal with the second term, we write
$[[m^{n}]]^{c}_{s}-[[m]]^{c}_{s}=[[m^{n}-m,m^{n}]]^{c}_{s}+[[m,m^{n}-m]]^{c}_{s}$
and then apply the Kunita–Watanabe inequality to obtain

$\displaystyle\sup_{0\leq t\leq
T}\biggl{\lvert}\int_{0}^{t}\operatorname{tr}\bigl{(}F_{2}$
$\displaystyle(u_{s-})\,\mathrm{d}([[m^{n}]]^{c}_{s}-[[m]]^{c}_{s})\bigr{)}\biggr{\rvert}$
$\displaystyle\leq\sqrt{2}\biggl{(}\int_{0}^{T}\lVert
F_{2}(u_{t-})\rVert\,\mathrm{d}([m^{n}]_{t}+[m]_{t})\biggr{)}^{1/2}\biggl{(}\int_{0}^{T}\lVert
F_{2}(u_{t-})\rVert\,\mathrm{d}[m^{n}-m]_{t}\biggr{)}^{1/2}$
$\displaystyle\leq\delta\int_{0}^{T}\lVert
F_{2}(u_{t-})\rVert\,\mathrm{d}([m^{n}]_{t}+[m]_{t})+\frac{1}{2\delta}\int_{0}^{T}\lVert
F_{2}(u_{t-})\rVert\,\mathrm{d}[m^{n}-m]_{t}.$

The first expression on the right-hand can be made arbitrarily small by
choosing $\delta>0$ appropriately. The second term tends to zero in
probability as $n\to\infty$. Hence, $\sup_{0\leq t\leq
T}\bigl{\lvert}\int_{0}^{t}\operatorname{tr}\bigl{(}F_{2}(u_{s-})\,\mathrm{d}([[m^{n}]]^{c}_{s}-[[m]]^{c}_{s})\bigr{)}\bigr{\rvert}\to
0$ in probability, which concludes the proof. ∎

###  A.2 Auxiliary Results for Section 4

Let $(E,d)$ be a complete separable metric space. We let $\mathbb{M}_{\leq
T}^{2}(E)$ denote the space of square-integrable measures $m$ on $[0,T]\times
E$, i.e. those with $\int_{[0,T]\times
E}d^{2}(x,x_{0})\,\mathrm{d}m(x)<\infty$ for some fixed $x_{0}\in E$, such
that $m([s,t]\times E)\leq t-s$ for all $0\leq s\leq t\leq T$. We equip
$\mathbb{M}_{\leq T}^{2}(E)$ with the metric

$d_{2}(m^{1},m^{2})=TW_{2}\bigl{(}\tfrac{1}{T}\tilde{m}^{1},\tfrac{1}{T}\tilde{m}^{2}\bigr{)}+\bigl{\lvert}m^{1}([0,T]\times
E)-m^{2}([0,T]\times E)\bigr{\rvert},$

where $\tilde{m}^{i}$ is defined by
$\mathrm{d}\tilde{m}^{i}(t,x)=\mathrm{d}m^{i}(t,x)+(m^{i}_{t}(E)-1)\mathrm{d}\delta_{x_{0}}\,\mathrm{d}t$,
where $(m^{i}_{t})_{0\leq t\leq T}\in L^{1}([0,T])$ is the weak derivative of
the map $t\mapsto m^{i}([0,t]\times E)$. Here $W_{2}$ denotes the
$2$-Wasserstein distance on the space of square-integrable probability
measures on $[0,T]\times E$.

######  Lemma A.2.

A sequence in $\mathbb{M}_{\leq T}^{2}(E)$ is compact if and only if it is
tight and uniformly square-integrable.

###### Proof.

This follows essentially from [Vil09, Theorem 6.9]. ∎

######  Lemma A.3.

Let $(E^{\prime},d^{\prime})$ be another complete separable metric space and
let $(m^{n})_{n}$ be a convergent sequence in $\mathbb{M}_{\leq T}^{2}(E\times
E^{\prime})$ with limit $m$. Assume that there exists a family of probability
measures $(p^{n}_{t,x})_{(t,x)\in[0,T]\times E}$ such that
$\mathrm{d}m^{n}(t,x,y)=\mathrm{d}p^{n}_{t,x}(y)\mathrm{d}\pi^{\\#}m^{n}(t,x)$,
where $\pi\mathpunct{:}[0,T]\times E\times E^{\prime}\to[0,T]\times E$ is the
projection onto $[0,T]\times E$. Then the same holds for $m$, i.e. we find a
family of probability measures $(p_{t,x})_{(t,x)\in[0,T]\times E}$ such that
$\mathrm{d}m(t,x,y)=\mathrm{d}p_{t,x}(y)\mathrm{d}\pi^{\\#}m(t,x)$.

###### Proof.

Let us define the space $E_{\ast}=E\times\\{0\\}\cup\\{(x_{0},1)\\}$, which we
endow with the product metric
$d_{\ast}((x,y),(x^{\prime},y^{\prime}))=d(x,x^{\prime})+\lvert
y-y^{\prime}\rvert$ for $x$, $x^{\prime}\in E$ and $y$,
$y^{\prime}\in\\{0,1\\}$. We will identify the elements $x$ of $E$ with
$(x,0)\in E_{\ast}$ and view $\ast=(x_{0},1)\in E_{\ast}$ as an isolated
point. Now, it holds that $\pi^{\\#}m^{n}([s,t]\times E)\leq t-s$ for any
$0\leq s\leq t\leq T$, so there exists a flow of subprobability measures
$(v^{n}_{t})_{0\leq t\leq T}$, such that
$\mathrm{d}\pi^{\\#}(t,x)=\mathrm{d}v^{n}_{t}(x)\mathrm{d}t$. Using this
disintegration, to any $m^{n}$, we can associate a measure $\bar{m}^{n}$ on
$[0,T]\times E_{\ast}\times E^{\prime}$ by
$\mathrm{d}\bar{m}^{n}(t,x,y)=\mathrm{d}m^{n}(t,x,y)+\mathrm{d}\delta_{y_{0}}(y)(1-v^{n}_{t}(E))\mathrm{d}\delta_{\ast}(x)\mathrm{d}t$
for some fixed $y_{0}\in E^{\prime}$. The family $(\bar{m}^{n})_{n}$ is tight
and we let $\bar{m}$ denote its limit along some convergent subsequence
$(n_{k})_{k}$. Since $\frac{1}{T}\bar{m}$ is a probability measure and
$\bar{m}([s,t]\times E_{\ast}\times E^{\prime})=t-s$ for $0\leq s\leq t\leq
T$, we can disintegrate $\bar{m}$ as
$\mathrm{d}\bar{m}(t,x,y)=\mathrm{d}p_{t,x}(y)\mathrm{d}\bar{v}_{t}(x)\mathrm{d}t$
for families of probability measures $(p_{t,x})_{(t,x)\in[0,T]\times
E_{\ast}}$ and $(\bar{v}_{t})_{0\in[0,T]}$. We define the flow of
subprobability measures $(v_{t})_{0\in[0,T]}$ by $v_{t}(A)=\bar{v}_{t}(A)$ for
$A\in\mathcal{B}(E)$ and claim that
$\mathrm{d}m(t,x,y)=\mathrm{d}p_{t,x}(y)\mathrm{d}v_{t}(x)\mathrm{d}t$, which
yields the desired disintegration of $m$. Let $\varphi\mathpunct{:}[0,T]\times
E\times E^{\prime}\to\mathbb{R}$ be continuous and bounded and define
$\bar{\varphi}\in C_{b}([0,T]\times E_{\ast}\times E^{\prime})$ by
$\bar{\varphi}(t,x,y)=\varphi(t,x,y)$ if $x\in E$ and $\bar{\varphi}(t,x,y)=0$
if $x=\ast$. Then, it holds that

$\displaystyle\langle m,\varphi\rangle$
$\displaystyle=\lim_{k\to\infty}\langle
m^{n_{k}},\varphi\rangle=\lim_{k\to\infty}\langle\bar{m}^{n_{k}},\bar{\varphi}\rangle=\langle\bar{m},\bar{\varphi}\rangle$
$\displaystyle=\int_{0}^{T}\int_{E}\int_{E^{\prime}}\bar{\varphi}(t,x,y)\,\mathrm{d}p_{t,x}(y)\,\mathrm{d}\bar{v}_{t}(x)\,\mathrm{d}t+\int_{0}^{T}\biggl{(}\int_{E^{\prime}}\bar{\varphi}(t,\ast,y)\,\mathrm{d}p_{t,x}(y)\biggr{)}\bar{v}_{t}(\\{\ast\\})\,\mathrm{d}t$
$\displaystyle=\int_{0}^{T}\int_{E}\int_{E^{\prime}}\varphi(t,x,y)\,\mathrm{d}p_{t,x}(y)\,\mathrm{d}v_{t}(x)\,\mathrm{d}t.$

Since $\varphi$ was arbitrary, this proves the claim. ∎

###  A.3 Regularisation of Functions

######  Proposition A.4.

Let $\varphi\mathpunct{:}[0,T]\times\mathbb{R}\times G\to\mathbb{R}$ satisfy
the same conditions as $f_{1}$ in Assumptions 2.4 and 2.6. Then we can find a
sequence $(\varphi_{n})_{n}$ of functions
$\varphi_{n}\mathpunct{:}[0,T]\times\mathbb{R}\times\mathbb{R}^{d}\to\mathbb{R}$
such that

  1. (i)

$\varphi_{n}(t,\cdot,\cdot)$ converges to $\varphi(t,\cdot,\cdot)$ uniformly
on compact subsets of $\mathbb{R}\times G$ for all $t\in[0,T]$;

  2. (ii)

$\lvert\varphi_{n}(t,x,g)\rvert\leq C^{\prime}(1+\lvert x\rvert^{2})$ for a
constant $C^{\prime}>0$ independent of $n$;

  3. (iii)

there exists a function $h_{n}\in L^{2}(\mathbb{R})$ with
$\lvert\varphi_{n}(t,x,g)\rvert\leq h_{n}(x)$ for all
$(t,x,g)\in[0,T]\times\mathbb{R}\times G$;

  4. (iv)

the map $g\mapsto\varphi_{n}(t,x,g)$ is strictly convex and continuous
differentiable, and $\partial_{g}\varphi_{n}$ is bounded.

###### Proof.

We construct an approximation of $\varphi$ in two steps. Let $n\geq 1$ and
define $\tilde{\varphi}_{n}\mathpunct{:}\mathbb{R}^{d}\to\mathbb{R}$ through
$\inf$-convolution, i.e. we set $\tilde{\varphi}_{n}(t,x,g)=\inf_{h\in
G}\bigl{(}\varphi(t,x,h)+n\lvert g-h\rvert^{2}\bigr{)}$. Since
$g\mapsto\varphi(t,x,g)$ is convex, the same holds for
$g\mapsto\tilde{\varphi}_{n}(t,x,g)$. The second step is to mollify
$\tilde{\varphi}_{n}$. Let $\eta\mathpunct{:}\mathbb{R}^{d}\to\mathbb{R}$ be
the standard mollifier and set
$\eta_{\epsilon}(x)=\frac{1}{\epsilon^{d}}\eta(\frac{x}{\epsilon})$. We define
$\tilde{\varphi}_{n}^{\epsilon}(t,x,g)=(\eta_{\epsilon}\ast\tilde{\varphi}_{n}(t,x,\cdot)(g)$
and finally set
$\varphi_{n}^{\epsilon}(t,x,g)=e^{-x^{2}/n}\bigl{(}\tilde{\varphi}_{n}^{\epsilon}(t,x,g)+\frac{1}{n}\lvert
g\rvert^{2}\bigr{)}$. We will now choose $\epsilon=\epsilon_{n}$ to ensure the
locally uniform convergence of $\varphi_{n}^{\epsilon_{n}}$. We estimate

$\displaystyle\begin{split}\lvert\varphi_{n}^{\epsilon}(t,x,g)-\varphi(t,x,g)\rvert&\leq\lvert\varphi_{n}^{\epsilon}(t,x,g)-\tilde{\varphi}_{n}^{\epsilon}(t,x,g)\rvert+\lvert\tilde{\varphi}_{n}^{\epsilon}(t,x,g)-\tilde{\varphi}_{n}(t,x,g)\rvert\\\
&\ \ \ \lvert\tilde{\varphi}_{n}(t,x,g)-\varphi(t,x,g)\rvert.\end{split}$
(A.7)

We will estimate the three terms on the right-hand side above. Let us fix a
compact subset $K$ of $\mathbb{R}$ and let $\omega_{K}$ be a modulus of
continuity of $g\mapsto\varphi(t,x,g)$ which is uniform in $[0,T]\times K$,
i.e. $\lvert\varphi(t,x,g)-\varphi(t,x,g^{\prime})\rvert\leq\omega_{K}(\lvert
g-g^{\prime}\rvert)$ for all $g$, $g^{\prime}\in G$ and $(t,x)\in[0,T]\times
K$. Let us start with the third term on the right-hand side of (A.7). For
$g\in G$, we estimate

$\displaystyle\lvert\tilde{\varphi}_{n}(t,x,g)-\varphi(t,x,g)\rvert$
$\displaystyle=\biggl{\lvert}\inf_{h\in G}\varphi(t,x,h)+n\lvert
g-h\rvert^{2}-\varphi(t,x,g)\bigg{\rvert}$
$\displaystyle\leq\inf_{\delta\in[0,\text{diam}(G)]}-\omega_{K}(\delta)+n\delta^{2}.$

Define
$\delta_{n}=\sup\\{\delta\in[0,\text{diam}(G)]\mathpunct{:}-\omega_{K}(\delta)+n\delta^{2}\leq
0\\}$. Clearly, it holds that $\delta_{n}\to 0$, so that

$\lvert\tilde{\varphi}_{n}(t,x,g)-\varphi(t,x,g)\rvert\leq\inf_{\delta\in[0,\text{diam}(G)]}-\omega_{K}(\delta)+n\delta^{2}\leq\omega_{K}(\delta_{n})\to
0$

as $n\to\infty$. Moreover, from the definition of $\tilde{\varphi}_{n}$, it
immediately follows that
$\lvert\tilde{\varphi}_{n}(t,x,g)\rvert\leq\lvert\varphi(t,x,g)\rvert\leq
C(1+\lvert x\rvert^{2})$ for $(t,x,g)\in[0,T]\times\mathbb{R}\times G$.

Next, we consider the second expression on the right-hand side of (A.7). We
have

$\displaystyle\lvert\tilde{\varphi}_{n}^{\epsilon}(t,x,g)-\tilde{\varphi}_{n}(t,x,g)\rvert$
$\displaystyle\leq\int_{\mathbb{R}^{d}}\eta_{\epsilon}(g-h)\bigl{\lvert}\tilde{\varphi}_{n}(t,x,h)-\tilde{\varphi}_{n}(t,x,g)\bigr{\rvert}\,\mathrm{d}h$
$\displaystyle\leq\sup_{h\in
B_{\epsilon}(g)}\bigl{\lvert}\tilde{\varphi}_{n}(t,x,h)-\tilde{\varphi}_{n}(t,x,g)\bigr{\rvert}$
$\displaystyle\leq\sup_{h\in B_{\epsilon}(g)}\sup_{y\in G}n\bigl{\lvert}\lvert
g-y\rvert^{2}-\lvert h-y\rvert^{2}\bigr{\rvert},$

where $B_{\epsilon}(h)$ denotes the ball of radius $\epsilon$ centered at
$h\in\mathbb{R}^{d}$. Since $G$ is compact, the expression in the third line
above tends to zero as $\epsilon\to 0$ uniformly over $g\in G$. Consequently,
we may choose $\epsilon_{n}>0$ such that $\epsilon_{n}\to 0$ and
$\lvert\tilde{\varphi}_{n}^{\epsilon_{n}}(t,x,g)-\tilde{\varphi}_{n}(t,x,g)\rvert\leq\epsilon_{n}$
for all $(t,x,g)\in[0,T]\times\mathbb{R}\times G$.

Lastly, we bound the first term on the right-hand side of (A.7). We have

$\lvert\varphi_{n}^{\epsilon}(t,x,g)-\tilde{\varphi}_{n}^{\epsilon}(t,x,g)\rvert\leq(1-e^{-x^{2}/n})\lvert\tilde{\varphi}_{n}^{\epsilon}(t,x,g)\rvert+\frac{1}{n}\sup_{g\in
G}\lvert g\rvert^{2}.$ (A.8)

Now, it holds that

$\lvert\tilde{\varphi}_{n}^{\epsilon}(t,x,g)\rvert\leq\int_{\mathbb{R}}\eta_{\epsilon}(h-g)\lvert\tilde{\varphi}_{n}(t,x,h)\rvert\,\mathrm{d}h\leq\sup_{h\in
B_{\epsilon}(g)}\lvert\tilde{\varphi}_{n}(t,x,h)\rvert\leq C(1+\lvert
x\rvert^{2}).$ (A.9)

Inserting this into (A.8) yields

$\lvert\varphi_{n}^{\epsilon}(t,x,g)-\tilde{\varphi}_{n}^{\epsilon}(t,x,g)\rvert\leq
C(1-e^{-x^{2}/n})(1+\lvert x\rvert^{2})+\frac{1}{n}\sup_{h\in G}\lvert
h\rvert^{2}$

and the bound on the right-hand side converges to zero uniformly in
$\epsilon>0$ and $x\in K$. Putting everything together implies that
$\varphi_{n}\mathpunct{:}[0,T]\times\mathbb{R}\times G\to G$ defined by
$\varphi_{n}^{\epsilon_{n}}(t,x,g)=\varphi_{n}(t,x,g)$ converges to $\varphi$
uniformly on compact subsets of $[0,T]\times\mathbb{R}\times G$, so Property
(i) holds. Moreover, by (A.9) we obtain that

$\displaystyle\lvert\varphi_{n}(t,x,g)\rvert$ $\displaystyle\leq
e^{-x^{2}/n}\lvert\tilde{\varphi}_{n}^{\epsilon_{n}}(t,x,g)\rvert+e^{-x^{2}/n}\frac{1}{n}\lvert
g\rvert^{2}\leq Ce^{-x^{2}/n}(1+\lvert
x\rvert^{2})+e^{-x^{2}/n}\frac{1}{n}\sup_{h\in G}\lvert h\rvert^{2}$

for all $(t,x,g)\in[0,T]\times\mathbb{R}\times G$, which establishes Property
(ii). Next, note that the left-hand side is in $L^{2}(\mathbb{R})$ as a
function of $x$, so Property (iii) is satisfied as well. Lastly, the strict
convexity and continuous differentiability of $g\mapsto\varphi_{n}(t,x,g)$
follow by construction and the boundedness of $\partial_{g}\varphi_{n}(t,x,g)$
is a consequence of the prefactor $e^{-x^{2}/n}$. This establishes Property
(iv) and concludes the proof. ∎

######  Proposition A.5.

Let $\varphi\mathpunct{:}[0,T]\times\mathbb{R}\times\mathcal{M}^{2}_{\leq
1}(\mathbb{R})\to\mathbb{R}$ be a measurable map such that
$(x,v)\mapsto\varphi(t,x,v)$ is uniformly continuous for $(t,x,v)$ in compact
subsets of $[0,T]\times\mathbb{R}\times\mathcal{M}^{2}_{\leq 1}(\mathbb{R})$
and $\lvert\varphi(t,x,v)\rvert\leq C(1+\lvert x\rvert+M_{2}^{p}(v))$ for some
$p\in[1,2]$ and $C>0$. Then there exists a sequence $(\varphi_{n})_{n}$ of
measurable function
$\varphi_{n}\mathpunct{:}[0,T]\times\mathbb{R}\times\mathcal{M}^{2}_{\leq
1}(\mathbb{R})\to\mathbb{R}$ such that

  1. (i)

$\varphi_{n}(t,\cdot,\cdot)$ converges to $\varphi(t,\cdot,\cdot)$ uniformly
on compact subsets of $\mathbb{R}\times\mathcal{M}^{2}_{\leq 1}(\mathbb{R})$
for all $t\in[0,T]$;

  2. (ii)

$\lvert\varphi_{n}(t,x,v)\rvert\leq C^{\prime}(1+\lvert
x\rvert^{p}+M_{2}^{p}(v))$ for a constant $C^{\prime}>0$ independent of $n$;

  3. (iii)

there exists a function $h_{n}\in C_{c}(\mathbb{R})$ with
$\lvert\varphi_{n}(t,x,v)\rvert\leq h_{n}(x)$ for all
$(t,x,v)\in[0,T]\times\mathbb{R}\times\mathcal{M}^{2}_{\leq 1}(\mathbb{R})$;

  4. (iv)

$v\mapsto\varphi_{n}(t,x,v)$ admits a linear functional derivative
$D\varphi_{n}\mathpunct{:}[0,T]\times\mathbb{R}\times\mathcal{M}^{2}_{\leq
1}(\mathbb{R})\times\mathbb{R}\to\mathbb{R}$ such that $\lvert
D\varphi_{n}(t,x,v)(y)\rvert\leq h_{n}(y)$ for all
$(t,x,v)\in[0,T]\times\mathbb{R}\times\mathcal{M}^{2}_{\leq 1}(\mathbb{R})$
and $y\in\mathbb{R}$ and $\mathcal{M}^{2}_{\leq 1}(\mathbb{R})\cap
L^{2}(\mathbb{R})\ni v\mapsto D\varphi_{n}(t,x,v)(y)$ is continuous with
respect to $\lVert\cdot\rVert_{L^{2}}$.

###### Proof.

We construct an approximation of $\varphi$ in three steps. First we pick
$n\geq 1$ and $k\geq 1$ and set $x_{i}=-n+i2^{-k}$ for $i=0$, …, $2n2^{k}=L$.
Then we let $(\psi_{i})_{i=1,\dots,L-1}$ be a partition of unity subordinated
to the open cover $((x_{i}-2^{-k},x_{i}+2^{-k}))_{i=1,\dots,L-1}$ and set
$\kappa_{n}(x)=\sum_{i=1}^{L-1}\psi_{i}(x)$ for $x\in\mathbb{R}$. Now, for
$v\in\mathcal{M}^{2}_{\leq 1}(\mathbb{R})$ we define the subprobability
measures $v_{n}\in\mathcal{M}^{2}_{\leq 1}(\mathbb{R})$ by
$\mathrm{d}v_{n}(x)=\kappa_{n}(x)\,\mathrm{d}v(x)$ and
$v_{n,k}\in\mathcal{M}^{2}_{\leq 1}(\mathbb{R})$ by
$\mathrm{d}v_{n,k}(x)=\sum_{i=1}^{L-1}\varpi_{i}(v)\bar{\psi}_{i}(x)\,\mathrm{d}x$,
where $\varpi_{i}(v)=\int_{\mathbb{R}}\psi_{i}(x)\,\mathrm{d}v(x)$ and
$\bar{\psi}_{i}(x)=c_{i}\psi_{i}(x)$ with
$c_{i}^{-1}=\int_{\mathbb{R}}\psi_{i}(x)\,\mathrm{d}x$. Next, for
$w\in[0,\infty)^{L-1}$ we let $m_{w}\in\mathcal{M}(\mathbb{R})$ be defined by
$\mathrm{d}m_{w}(x)=\sum_{i=1}w_{i}\bar{\psi}_{i}(x)\,\mathrm{d}x$. Then, we
introduce the function
$\Phi_{n,k}\mathpunct{:}[0,T]\times\mathbb{R}\times\mathbb{R}^{L-1}\to\mathbb{R}$
given by $\Phi_{n,k}(t,x,w)=\varphi(t,x,m_{w^{+}})$, where $w^{+}=(w_{i}\lor
0)_{i=1,\dots,L-1}$ for $w\in\mathbb{R}^{L-1}$. Finally, for $\epsilon>0$ we
set
$\Phi_{n,k}^{\epsilon}(t,x,w)=(\eta_{\epsilon}\ast\Phi_{n,k}(t,x,\cdot))(w)$
and $\varphi_{n,k}^{\epsilon}(t,x,v)=\Phi_{n,k}^{\epsilon}(t,x,\varpi(v))$,
where $\eta\mathpunct{:}\mathbb{R}^{L-1}\to\mathbb{R}$ is the standard
mollifier and
$\eta_{\epsilon}(x)=\frac{1}{\epsilon^{L-1}}\eta(\frac{x}{\epsilon})$ for
$x\in\mathbb{R}^{L-1}$. Our goal is to show that choosing first
$\epsilon=\epsilon_{n,k}$ small enough and then $k=k_{n}$ sufficiently large,
the resulting sequence $(\varphi_{n})_{n\geq 1}$, defined by
$\varphi_{n}(t,x,v)=\kappa_{n}(x)\varphi_{n,k_{n}}^{\epsilon_{n,k_{n}}}(t,x,v)$,
satisfies Properties (i), (ii), and (iv).

Let us start with Property (i). Note that for $n\geq 1$, on the set
$\mathcal{K}_{n}$ of measures $v\in\mathcal{M}^{2}_{\leq 1}(\mathbb{R})$ such
that $\operatorname{supp}v\subset[-n,n]$ the metrics $d_{0}$ and $d_{2}$ are
strongly equivalent, so there exists a constant $c_{n}>0$ such that $d_{2}\leq
c_{n}d_{0}$. Now, for $k\geq 1$, and $\epsilon>0$, we can write

$\displaystyle\begin{split}\lvert\varphi_{n,k}^{\epsilon}(t,x,v)-\varphi(t,x,v)\rvert&\leq\lvert\varphi_{n,k}^{\epsilon}(t,x,v)-\kappa_{n}(x)\varphi(t,x,v_{n,k})\rvert+\lvert\varphi(t,x,v_{n,k})-\varphi(t,x,v_{n})\rvert\\\
&\ \ \
+\lvert\kappa_{n}(x)\varphi(t,x,v_{n})-\varphi(t,x,v)\rvert.\end{split}$
(A.10)

We inspect the three terms in turn. For the first, we obtain

$\displaystyle\lvert\varphi_{n,k}^{\epsilon}(t,x,v$
$\displaystyle)-\kappa_{n}(x)\varphi(t,x,v_{n,k})\rvert$
$\displaystyle\leq\int_{\mathbb{R}^{L-1}}\eta_{\epsilon}(\varpi(v)-w)\bigl{\lvert}\Phi_{n,k}(t,x,w)-\Phi_{n,k}(t,x,\varpi(v))\bigr{\rvert}\,\mathrm{d}w$
$\displaystyle\leq\sup_{w\in
B_{\epsilon}(\varpi(v))}\lvert\varphi_{n,k}(t,x,m_{w^{+}})-\varphi(t,x,v_{n,k})|,$
(A.11)

where $B_{\epsilon}(w)$ is the ball of radius $\epsilon$ centered at
$w\in\mathbb{R}^{L-1}$. For $w\in B_{\epsilon}(\varpi(v))$, we estimate

$\displaystyle d_{2}(m_{w^{+}},v_{n,k})$ $\displaystyle\leq
c_{n}d_{0}(m_{w^{+}},v_{n,k})$
$\displaystyle=c_{n}\sup_{\lVert\Phi\rVert_{\text{bLip}\leq 1}}\langle
m_{w^{+}}-v_{n,k},\Phi\rangle$
$\displaystyle=c_{n}\sup_{\lVert\Phi\rVert_{\text{bLip}\leq
1}}\sum_{i=1}^{L-1}\int_{\mathbb{R}}((w_{i}\lor
0)-\varpi_{i}(v))\Phi(x)\bar{\psi}_{i}(x)\,\mathrm{d}x$ $\displaystyle\leq
c_{n}\sum_{i=1}^{L-1}\lvert w_{i}-\varpi_{i}(v)\rvert$ $\displaystyle\leq
c_{n}(L-1)\epsilon.$ (A.12)

For $t\in[0,T]$, let us define the modulus of continuity $\omega_{n}$ by

$\omega_{n}(\delta)=\sup\biggl{\\{}\sup_{(t,x)\in[0,T]\times[-n,n]}\lvert\varphi(t,x,v)-\varphi(t,x,v^{\prime})\rvert\mathpunct{:}v,v^{\prime}\in\mathcal{K}_{n},\,d_{2}(v,v^{\prime})\leq\delta\biggr{\\}}.$

By assumption, this modulus is continuous at zero, i.e. $\omega_{n}(\delta)\to
0$ as $\delta\to 0$. Inserting (A.3) into (A.3) and applying $\omega_{n}$
yields

$\bigl{\lvert}\varphi_{n,k}^{\epsilon}(t,x,v)-\kappa_{n}(x)\varphi(t,x,v_{n,k})\bigr{\rvert}\leq\omega_{n}\bigl{(}c_{n}(2n2^{k}-1)\epsilon\bigr{)}.$

Now, we chose $\epsilon_{n,k}>0$ such that
$\omega_{n}\bigl{(}c_{n}(2n2^{k}-1)\epsilon_{n,k}\bigr{)}\leq\frac{1}{n}\bigr{\\}}$,
whence
$\omega_{n}\bigl{(}c_{n}(2n2^{k}-1)\epsilon_{n,k}\bigr{)}=\frac{1}{n}\to 0$.
This finishes our analysis of the first summand of (A.10).

We estimate the second term in (A.10) by

$\displaystyle\lvert\varphi(t,x,v_{n,k})-\varphi(t,x,v_{n})\rvert\leq\omega^{n}_{t}(d_{2}(v_{n,k},v_{n}))\leq\omega_{n}(c_{n}d_{0}(v_{n,k},v_{n})).$

We bound the $d_{0}$-metric appearing on the right-hand side by

$\displaystyle d_{0}(v_{n,k},v_{k})$
$\displaystyle=\sup_{\lVert\Phi\rVert_{\text{bLip}\leq 1}}\langle
v_{n,k}-v_{k},\Phi\rangle$
$\displaystyle=\sup_{\lVert\Phi\rVert_{\text{bLip}\leq
1}}\sum_{i=1}^{L-1}\biggl{(}\int_{\mathbb{R}}\varpi_{i}(v)\Phi(x)\bar{\psi}_{i}(x)\,\mathrm{d}x-\int_{\mathbb{R}}\Phi(x)\psi_{i}(x)\,\mathrm{d}v(x)\biggr{)}$
$\displaystyle\leq\sup_{\lVert\Phi\rVert_{\text{bLip}\leq
1}}\sum_{i=1}^{L-1}\biggl{(}\int_{\mathbb{R}}\varpi_{i}(v)\Phi(x_{i})\bar{\psi}_{i}(x)\,\mathrm{d}x-\int_{\mathbb{R}}\Phi(x_{i})\psi_{i}(x)\,\mathrm{d}v(x)\biggr{)}$
$\displaystyle\ \ \
+\sum_{i=1}^{L-1}\biggl{(}\int_{\mathbb{R}}\varpi_{i}(v)\lvert
x_{i}-x\rvert\bar{\psi}_{i}(x)\,\mathrm{d}x+\int_{\mathbb{R}}\lvert
x_{i}-x\rvert\psi_{i}(x)\,\mathrm{d}v(x)\biggr{)}$ $\displaystyle\leq
2^{-k}v(\mathbb{R}),$

where we note that the expression in the third line vanishes. In view of the
previous display, we obtain
$\lvert\varphi(t,x,v_{n,k})-\varphi(t,x,v_{n})\rvert\leq\omega_{n}(c_{n}2^{-k})$.
We set $k_{n}=\inf\\{k\geq
1\mathpunct{:}\omega_{n}(c_{n}2^{-k})\leq\frac{1}{n}\\}$.

Let us now study the last term in (A.10). Fix compact subsets $K$ and
$\mathcal{K}$ of $\mathbb{R}$ and $\mathcal{M}_{\leq 1}^{2}(\mathbb{R})$,
respectively and let $\omega_{t}$ be a modulus of continuity of
$K\times\mathcal{K}\ni(x,v)\mapsto\varphi(t,x,v)$. We estimate

$d_{2}^{2}(v_{n},v)\leq\int_{\mathbb{R}}(1+\lvert
x\rvert^{2})(1-\kappa_{n}(x))\,\mathrm{d}v(x),$ (A.13)

where we use the coupling $\pi\in\mathcal{P}^{2}(\mathbb{R}^{2})$ of
$\tilde{v}=v+(1-v(\mathbb{R}))\delta_{0}$ and
$\tilde{v}_{n}=v_{n}+(1-v_{n}(\mathbb{R}))\delta_{0}$ defined by

$\displaystyle\int_{\mathbb{R}^{2}}h(x,y)\,\mathrm{d}\pi(x,y)$
$\displaystyle=\int_{\mathbb{R}}h(x,x)\,\mathrm{d}v_{n}(x)+(1-v_{n}(\mathbb{R}))\int_{\mathbb{R}}h(x,0)(1-\kappa_{n}(x))\,\mathrm{d}v(x)$
$\displaystyle\ \ \ +(1-v(\mathbb{R}))(1-v_{n}(\mathbb{R}))h(0,0)$

for $h\in C_{b}(\mathbb{R}^{2})$ to estimate the $2$-Wasserstein distance
between $\tilde{v}$ and $\tilde{v}_{n}$. The right-hand side of Equation
(A.13) vanishes uniformly in $v\in\mathcal{K}$ as $n\to\infty$. Thus, we
obtain that

$\displaystyle\lvert\kappa_{n}(x)\varphi(t,x,$ $\displaystyle
v_{n})-\kappa_{n}(x)\varphi(t,x,v)\rvert+\lvert\kappa_{n}(x)\varphi(t,x,v)-\varphi(t,x,v)\rvert$
$\displaystyle\leq\omega_{t}(d_{2}(v_{n},v))+C\bigl{(}1+\lvert
x\rvert^{p}+M_{2}^{p}(v)\bigr{)}\mathbf{1}_{x\notin[-(n-1),n-1]}\to 0$

uniformly over $(x,v)\in K\times\mathcal{K}$. Setting
$\varphi_{n}(t,x,v)=\varphi_{n,k_{n}}^{\epsilon_{n}}(t,x,v)$ with
$\epsilon_{n}=\epsilon_{n,k_{n}}$ and putting everything together yields
$\bigl{\lvert}\varphi_{n}(t,x,v)-\varphi(t,x,v)\bigr{\rvert}\to 0$ uniformly
over $(x,v)$ in compact subsets of $\mathbb{R}\times\mathcal{M}^{2}_{\leq
1}(\mathbb{R})$ as $n\to\infty$.

For Property (ii), similarly to (A.10), we write

$\displaystyle\lvert\varphi_{n}(t,x,v)\rvert$
$\displaystyle\leq\lvert\varphi_{n}(t,x,v)-\kappa_{n}(x)\varphi(t,x,v_{n})\rvert+\lvert\kappa_{n}(x)\varphi(t,x,v_{n})\rvert$
$\displaystyle\leq\bigl{\lvert}\varphi_{n}(t,x,v)-\kappa_{n}(x)\varphi(t,x,v_{n,k_{n}})\bigr{\rvert}+\bigl{\lvert}\varphi(t,x,v_{n,k_{n}})-\varphi(t,x,v_{n})\bigr{\rvert}$
$\displaystyle\ \ \ +C\bigl{(}1+\lvert x\rvert^{p}+M_{2}^{p}(v_{n})\bigr{)}$
$\displaystyle\leq\frac{2}{n}+C\bigl{(}1+\lvert
x\rvert^{2}+M_{2}^{p}(v_{n})\bigr{)}$ $\displaystyle\leq(C+2)(1+\lvert
x\rvert^{p}+M_{2}^{p}(v_{n})\bigr{)},$

so choosing $C^{\prime}=C+2$ gives the uniform growth bound. Moreover, we can
further estimate the right-hand side above by $(C+2)(1+\lvert
x\rvert^{p}+n^{p}\bigr{)}$. Then noting that $\lvert\varphi_{n}(t,x,v)\rvert$
vanishes outside of $[-n,n]$, immediately gives Property (iii).

It remains to establish Property (iv). Let us first prove that
$v\mapsto\varphi_{n}(t,x,v)$ has a linear functional derivative. For $v$,
$v^{\prime}\in\mathcal{M}^{2}_{\leq 1}(\mathbb{R})$, setting
$\Phi_{n}(t,x,w)=\Phi_{n,k_{n}}^{\epsilon_{n}}(t,x,w)$, we compute

$\displaystyle\varphi_{n}(t,x,v)-\varphi_{n}(t,x,v^{\prime})$
$\displaystyle=\sum_{i=1}^{L-1}\int_{0}^{1}\partial_{w_{i}}\Phi_{n}\bigl{(}t,x,s\varpi(v)+(1-s)\varpi(v^{\prime})\bigr{)}(\varpi(v)-\varpi(v^{\prime}))\,\mathrm{d}s$
$\displaystyle=\sum_{i=1}^{L-1}\int_{0}^{1}\partial_{w_{i}}\Phi_{n}\bigl{(}t,x,w(sv+(1-s)v^{\prime})\bigr{)}\int_{\mathbb{R}}\psi_{i}(y)\,\mathrm{d}(v-v^{\prime})(y)\,\mathrm{d}s$
$\displaystyle=\int_{0}^{1}\bigl{\langle}v-v^{\prime},D\varphi_{n}\bigl{(}t,x,sv+(1-s)v^{\prime}\bigr{)}\bigr{\rangle}\,\mathrm{d}s,$

where we use in the second equality that the function $v\mapsto\varpi(v)$ is
linear and where we defined
$D\varphi_{n}(t,x,v)(y)=\sum_{i=1}^{K-1}\partial_{w_{i}}\Phi_{n}(t,x,\varpi(v))\psi_{i}(y)$
for $y\in\mathbb{R}$ and $v\in\mathcal{M}^{2}_{\leq 1}(\mathbb{R})$. For
$y\in\mathbb{R}$, we have that $\lvert
D\varphi_{n}(t,x,v)(y)\rvert\leq\sum_{i=1}^{L-1}\lvert\partial_{w_{i}}\Phi_{n}(t,x,\varpi(v))\rvert\psi_{i}(y)$
and for $i=1$, …, $K$ it holds with $\epsilon_{n}=\epsilon_{n,k_{n}}$ that

$\displaystyle\lvert\partial_{w_{i}}$
$\displaystyle\Phi_{n}(t,x,\varpi(v))\rvert$
$\displaystyle\leq\int_{\mathbb{R}^{L-1}}\bigl{\lvert}\partial_{w_{i}}\eta_{\epsilon_{n}}(\varpi(v)-w)\bigr{\rvert}\kappa_{n}(x)\lvert\Phi_{n,k_{n}}(t,x,w)\rvert\,\mathrm{d}w$
$\displaystyle\leq\lVert\partial_{w_{i}}\eta_{\epsilon_{n}}\rVert_{\infty}\kappa_{n}(x)\sup_{w\in
B_{\epsilon_{n}}(\varpi(v))}\lvert\Phi_{n,k_{n}}(t,x,w)\rvert$
$\displaystyle\leq\lVert\partial_{w_{i}}\eta_{\epsilon_{n}}\rVert_{\infty}\kappa_{n}(x)\biggl{(}\sup_{w\in
B_{\epsilon_{n}}(\varpi(v))}\omega_{n}(d_{2}(m_{w^{+}},v_{n,k}))+\omega_{n}(d_{2}(v_{n,k},v_{n}))+\lvert\varphi(t,x,v_{n})\rvert\biggr{)}$
$\displaystyle\leq\lVert\partial_{w_{i}}\eta_{\epsilon_{n}}\rVert_{\infty}\bigl{(}\tfrac{2}{n}+\kappa_{n}(x)C(1+\lvert
x\rvert^{p}+n^{p})\bigr{)}.$

Hence, $\lvert
D\varphi_{n}(t,x,v)(y)\rvert\leq\lVert\partial_{w_{i}}\eta_{\epsilon_{n}}\rVert_{\infty}\bigl{(}\tfrac{2}{n}+C(1+2n^{p})\bigr{)}\sum_{i=1}^{L-1}\psi_{i}(y)=h_{n}(y)$.
Clearly, the function $h_{n}$ has compact support and $\lVert
h_{n}\rVert_{\infty}\leq C_{n}(1+C)$ for an appropriately chosen $C_{n}>0$.
Lastly, the continuity of $\mathcal{M}^{2}_{\leq 1}(\mathbb{R})\cap
L^{2}(\mathbb{R})\ni v\mapsto D\varphi(t,x,v)(y)$ with respect to the norm
$\lVert\cdot\rVert_{L^{2}}$ is a simple consequence of the truncation and
mollification. ∎

###  A.4 Control of Conditional Processes

We briefly describe the setup of the control of conditional processes. On a
filtered probability space $(\Omega,\mathcal{F},\mathbb{F},\mathbb{P})$
equipped with a $d$-dimensional $\mathbb{F}$-Brownian motion $W=(W_{t})_{t\geq
0}$ and a $d$-dimensional $\mathcal{F}_{0}$-measurable random variable $\xi$,
we consider the SDE

$\mathrm{d}X_{t}=\alpha_{t}\,\mathrm{d}t+\mathrm{d}W_{t},\quad X_{0}=\xi$
(A.14)

for a $d$-dimensional Brownian motion $W=(W_{t})_{t\geq 0}$ and an
$\mathbb{F}$-progressively measurable $\mathbb{R}^{d}$-valued process
$\alpha=(\alpha_{t})_{t\geq 0}$ such that
$\mathbb{E}\int_{0}^{T}\lvert\alpha_{t}\rvert^{2}\,\mathrm{d}t<\infty$. We
call $\mathfrak{a}=(\Omega,\mathcal{F},\mathbb{F},\mathbb{P},W,\alpha)$ a weak
control and denote the space of all such objects by $\mathbb{A}$. The cost
functional is given by

$J(\mathfrak{a})=\int_{0}^{T}\mathbb{E}[f(t,X_{t},\alpha_{t})|\tau>t]\,\mathrm{d}t+\mathbb{E}[g(X_{T})|\tau>T],$
(A.15)

where $\tau=\inf\\{t>0\mathpunct{:}X_{t}\notin D\\}$ for an open subset
$D\subset\mathbb{R}^{d}$ with $C^{1}$-boundary and $\mathbb{E}$ denotes the
expectation with respect to $\mathbb{P}$. We let the value $V$ be the infimum
of $J(\mathfrak{a})$ over weak controls $\mathfrak{a}\in\mathbb{A}$. We could
consider an extended setup with a drift and diffusion coefficient that depend
on time and the location of the diffusion or even its conditional law
$\mu_{t}=\mathbb{P}(X_{t}\in\cdot|\tau>t)$. The general case introduces
additional difficulties, so for the purpose of illustration we stick to this
simple setting.

######  Assumption A.6.

Let $f\mathpunct{:}[0,T]\times\mathbb{R}^{d}\times\mathbb{R}^{d}\to\mathbb{R}$
and $g\mathpunct{:}\mathbb{R}^{d}\to\mathbb{R}$ be measurable and assume there
exists $C>0$ such that

  1. (i)

for all $(t,x,a)$ we have

$\lvert f(t,x,a)\rvert+\lvert g(x)\rvert\leq C(1+\lvert x\rvert^{2}+\lvert
a\rvert^{2});$

  2. (ii)

for all $t$ the maps $(x,a)\mapsto f(t,x,a)$ and $x\mapsto g(x)$ are
continuous;

  3. (iii)

for all $(t,x)$ the map $a\mapsto f(t,x,a)$ is convex.

Our first goal is to prove the equivalence between the weak formulation just
introduced and the closed-loop formulation. A closed-loop control is a
measurable function
$a\mathpunct{:}[0,\infty)\times\mathbb{R}^{d}\to\mathbb{R}^{d}$ such that the
SDE

$\mathrm{d}X_{t}=a(t,X_{t})\,\mathrm{d}t+\mathrm{d}W_{t},\quad X_{0}=\xi$
(A.16)

has a weak solution on some filtered probability space
$(\Omega,\mathcal{F},\mathbb{F},\mathbb{P})$ equipped with a $d$-dimensional
$\mathbb{F}$-Brownian motion $W$ and an $\mathcal{F}_{0}$-measurable random
variable $\xi$, for which $\mathbb{E}\int_{0}^{T}\lvert
a(t,X_{t})\rvert^{2}\,\mathrm{d}t<\infty$. Clearly, SDE (A.16) exhibits
uniqueness in law for any measurable function
$a\mathpunct{:}[0,T]\times\mathbb{R}^{d}\to\mathbb{R}^{d}$, so we can
introduce the cost functional

$J_{\text{cl}}(a)=\int_{0}^{T}\mathbb{E}\bigl{[}f\bigl{(}t,X_{t},a(t,X_{t})\bigr{)}\big{|}\tau>t\bigr{]}\,\mathrm{d}t+\mathbb{E}[g(X_{T})|\tau>T]$
(A.17)

as well as the associated value $V_{\text{cl}}$, which is simply the infimum
of $J_{\text{cl}}(a)$ over all closed-loop controls. It is clear that any
closed-loop control $a$ induces a weak control $\mathfrak{a}$ through the
assignment $\alpha_{t}=a(t,X_{t})$, so that $V\leq V_{\text{cl}}$. We obtain
the reverse inequality through a mimicking argument.

######  Proposition A.7.

Let Assumption A.6 be satisfied. For any $K>0$ and any weak control
$\mathfrak{a}=(\Omega,\mathcal{F},\mathbb{F},\mathbb{P},W,\alpha)$ with
$\lvert\alpha_{t}\rvert\leq K$ for $\operatorname{Leb}\otimes\mathbb{P}$-a.e.
$(t,\omega)\in[0,T]\times\Omega$, there exists a closed-loop control $a$ with
corresponding weak solution
$(\tilde{\Omega},\tilde{\mathcal{F}},\tilde{\mathbb{F}},\tilde{\mathbb{P}},\tilde{W},\tilde{\xi},\tilde{X})$
to SDE (A.16) such that
$\mathbb{P}(X_{t}\in\cdot|\tau>t)=\tilde{\mathbb{P}}(\tilde{X}_{t}\in\cdot|\tilde{\tau}>t)$
for all $t\in[0,T]$ and $J_{\textup{cl}}(a)\leq J(\mathfrak{a})$, where
$\tilde{\tau}=\inf\\{t>0\mathpunct{:}\tilde{X}_{t}\notin D\\}$. In particular,
it holds that $V_{\textup{cl}}=V$.

###### Proof.

Let $\mathfrak{a}$ be as in the statement of the theorem and define $X^{\tau}$
by $X^{\tau}_{t}=X_{\tau\land t}$. The process $X^{\tau}$ satisfies the SDE

$\mathrm{d}X^{\tau}_{t}=\chi_{D}(X^{\tau}_{t})\alpha_{t}\,\mathrm{d}t+\chi_{D}(X^{\tau}_{t})\,\mathrm{d}W_{t},\quad
X^{\tau}_{0}=\xi,$

where $\chi_{D}\mathpunct{:}\mathbb{R}^{d}\to\\{0,1\\}$ is given by
$\chi_{D}(x)=\mathbf{1}_{x\in D}$. Now, proceeding as in the proof of Theorem
3.7 in [Lac15], using the convexity of $f$ in the control argument guaranteed
by Assumption A.6 (iii), we can find a measurable function
$a\mathpunct{:}[0,\infty)\times\mathbb{R}^{d}\to\mathbb{R}^{d}$ bounded by $K$
such that

$\displaystyle\begin{split}\mathbb{E}\bigl{[}\chi_{D}(X^{\tau}_{t})\alpha_{t}\big{|}X^{\tau}_{t}\bigr{]}&=\chi_{D}(X^{\tau}_{t})a(t,X^{\tau}_{t}),\\\
\mathbb{E}\bigl{[}\chi_{D}(X^{\tau}_{t})f(t,X^{\tau}_{t},\alpha_{t})\big{|}X^{\tau}_{t}\bigr{]}&\geq\chi_{D}(X^{\tau}_{t})f\bigl{(}t,X^{\tau}_{t},a(t,X^{\tau}_{t})\bigr{)}\end{split}$
(A.18)

for $t\in[0,T]$. We may assume that $a(t,x)=0$ for all $x\in\mathbb{R}^{d}$ if
$t>T$. Then the mimicking theorem by Brunick and Shreve [BS13] yields a
filtered probability space
$(\tilde{\Omega},\tilde{\mathcal{F}},\tilde{\mathbb{F}},\tilde{\mathbb{Q}})$
equipped with a $d$-dimensional $\tilde{\mathbb{F}}$-Brownian motion
$\tilde{B}$ and a $d$-dimensional $\tilde{\mathbb{F}}$-adapted process
$\tilde{Y}$ such that

$\mathrm{d}\tilde{Y}_{t}=\chi_{D}(\tilde{Y}_{t})a(t,\tilde{Y}_{t})\,\mathrm{d}t+\chi_{D}(\tilde{Y}_{t})\,\mathrm{d}\tilde{B}_{t},\quad\tilde{Y}_{0}\sim\mathcal{L}(\xi)$

and
$\mathcal{L}^{\tilde{\mathbb{P}}}(\tilde{Y}_{t})=\mathcal{L}(X^{\tau}_{t})$.
The latter implies together with (A.18) that

$\displaystyle J(\mathfrak{a})$
$\displaystyle=\int_{0}^{T}\frac{\mathbb{E}\bigl{[}\chi_{D}(X^{\tau}_{t})f(t,X^{\tau}_{t},\alpha_{t})\bigr{]}}{\mathbb{E}\chi_{D}(X^{\tau}_{t})}\,\mathrm{d}t+\frac{\mathbb{E}[\chi_{D}(X^{\tau}_{T})g(X^{\tau}_{T})]}{\mathbb{E}\chi_{D}(X^{\tau}_{T})}$
$\displaystyle\geq\int_{0}^{T}\frac{\mathbb{E}\bigl{[}\chi_{D}(X^{\tau}_{t})f\bigl{(}t,X^{\tau}_{t},a(t,X^{\tau}_{t})\bigr{)}\bigr{]}}{\mathbb{E}\chi_{D}(X^{\tau}_{t})}\,\mathrm{d}t+\frac{\mathbb{E}[\chi_{D}(X^{\tau}_{T})g(X^{\tau}_{T})]}{\mathbb{E}\chi_{D}(X^{\tau}_{T})}$
$\displaystyle=\int_{0}^{T}\frac{\mathbb{E}^{\tilde{\mathbb{Q}}}\bigl{[}\chi_{D}(\tilde{Y}_{t})f\bigl{(}t,\tilde{Y}_{t},a(t,\tilde{Y}_{t})\bigr{)}\bigr{]}}{\mathbb{E}^{\tilde{\mathbb{Q}}}\chi_{D}(\tilde{Y}_{t})}\,\mathrm{d}t+\frac{\mathbb{E}^{\tilde{\mathbb{Q}}}[\chi_{D}(\tilde{Y}_{T})g(\tilde{Y}_{T})]}{\mathbb{E}^{\tilde{\mathbb{Q}}}\chi_{D}(\tilde{Y}_{T})}.$
(A.19)

We wish to extend $\tilde{Y}$ to a solution of SDE (A.16). For that, we
perform the change of measure

$\frac{\mathrm{d}\tilde{\mathbb{P}}}{\mathrm{d}\tilde{\mathbb{Q}}}\bigg{|}_{\tilde{\mathcal{F}}}=\mathcal{E}\biggl{(}\int_{0}^{\cdot}(1-\chi_{D}(\tilde{Y}_{t}))a(t,\tilde{X}_{t})\cdot\mathrm{d}\tilde{B}_{t}\biggr{)}_{\infty},$

where the process $\tilde{X}=(\tilde{X}_{t})_{t\geq 0}$ is given by
$\tilde{X}_{t}=\tilde{Y}_{t}$ if $t<\tilde{\tau}$ and
$\tilde{X}_{t}=\tilde{Y}_{\tilde{\tau}}+\tilde{B}_{t}-\tilde{B}_{\tilde{\tau}}$
otherwise. By Girsanov’s theorem, the process
$\tilde{W}=(\tilde{W}_{t})_{t\geq 0}$ defined by
$\tilde{W}_{t}=\tilde{B}_{t}-\int_{0}^{t}(1-\chi_{D}(\tilde{Y}_{s}))a(s,\tilde{X}_{s})\,\mathrm{d}s$
is a $d$-dimensional $\tilde{\mathbb{F}}$-Brownian motion under
$\tilde{\mathbb{P}}$. Moreover, we claim that $\tilde{X}$ is a weak solution
to SDE (A.16) on
$(\tilde{\Omega},\tilde{\mathcal{F}},\tilde{\mathbb{F}},\tilde{\mathbb{P}})$
with Brownian motion $\tilde{W}$ and initial condition $\tilde{\xi}$. Indeed,
$\tilde{X}$ clearly satisfies SDE (A.16) on $[0,\tilde{\tau})$, since
$\tilde{W}=\tilde{B}$ on $[0,\tilde{\tau})$ and $\tilde{Y}$ solves SDE (A.16)
on $[0,\tilde{\tau})$ with Brownian motion $\tilde{B}$. Next, assume that
$t>\tilde{\tau}$. Then, it holds that

$\displaystyle\tilde{X}_{t}$
$\displaystyle=\tilde{Y}_{\tilde{\tau}}+\tilde{B}_{t}-\tilde{B}_{\tilde{\tau}}$
$\displaystyle=\tilde{Y}_{0}+\int_{0}^{\tilde{\tau}}\chi_{D}(\tilde{Y}_{s})a(s,\tilde{Y}_{s})\,\mathrm{d}s+\tilde{B}_{\tilde{\tau}}+\tilde{B}_{t}-\tilde{B}_{\tilde{\tau}}$
$\displaystyle=\tilde{Y}_{0}+\int_{0}^{\tilde{\tau}}a(s,\tilde{X}_{s})\,\mathrm{d}s+\int_{0}^{t}(1-\chi_{D}(\tilde{Y}_{s}))a(s,\tilde{X}_{s})\,\mathrm{d}s+\tilde{W}_{t}$
$\displaystyle=\tilde{Y}_{0}+\int_{0}^{t}a(s,\tilde{X}_{s})\,\mathrm{d}s+\tilde{W}_{t},$

so $\tilde{X}$ also solves SDE (A.16) for $t>\tilde{\tau}$. Moreover, in view
of Equation (A.4) we have that

$\displaystyle J(\mathfrak{a})$
$\displaystyle\geq\int_{0}^{T}\frac{\mathbb{E}^{\tilde{\mathbb{P}}}\bigl{[}\chi_{D}(\tilde{X}_{t})f\bigl{(}t,\tilde{X}_{t},a(t,\tilde{X}_{t})\bigr{)}\bigr{]}}{\mathbb{E}^{\tilde{\mathbb{P}}}\chi_{D}(\tilde{X}_{t})}\,\mathrm{d}t+\frac{\mathbb{E}^{\tilde{\mathbb{P}}}[\chi_{D}(\tilde{X}_{T})g(\tilde{X}_{T})]}{\mathbb{E}^{\tilde{\mathbb{P}}}\chi_{D}(\tilde{X}_{T})}=J_{\text{cl}}(a),$

where we used that $\tilde{\mathbb{Q}}$ and $\tilde{\mathbb{P}}$ coincide on
$\tilde{\mathcal{F}}_{\tilde{\tau}\land T}$. It remains to show that the
conditional laws of $X_{t}$ and $\tilde{X}_{t}$ coincide. This follows from
$\mathcal{L}^{\tilde{\mathbb{P}}}(\tilde{Y}_{t})=\mathcal{L}(X^{\tau}_{t})$,
since for $t\in[0,T]$ we have

$\displaystyle\mathbb{P}(X_{t}\in A|\tau>t)$
$\displaystyle=\frac{\mathbb{E}[\mathbf{1}_{\tau>t}\mathbf{1}_{X_{t}\in
A}]}{\mathbb{E}\mathbf{1}_{\tau>t}}=\frac{\mathbb{E}[\chi_{D}(X^{\tau}_{t})\mathbf{1}_{X^{\tau}_{t}\in
A}]}{\mathbb{E}\chi_{D}(X^{\tau}_{t})}=\frac{\mathbb{E}^{\tilde{\mathbb{Q}}}[\chi_{D}(\tilde{Y}_{t})\mathbf{1}_{\tilde{Y}_{t}\in
A}]}{\mathbb{E}^{\tilde{\mathbb{Q}}}\chi_{D}(\tilde{Y}_{t})}$
$\displaystyle=\frac{\mathbb{E}^{\tilde{\mathbb{P}}}[\chi_{D}(\tilde{X}^{\tilde{\tau}}_{t})\mathbf{1}_{\tilde{X}^{\tilde{\tau}}_{t}\in
A}]}{\mathbb{E}^{\tilde{\mathbb{P}}}\chi_{D}(\tilde{X}^{\tilde{\tau}}_{t})}=\tilde{\mathbb{P}}(\tilde{X}_{t}\in
A|\tilde{\tau}>t),$

where $\tilde{\tau}=\inf\\{t>0\mathpunct{:}\tilde{Y}\notin
D\\}=\inf\\{t>0\mathpunct{:}\tilde{X}\notin D\\}$.

The last statement follows from a simple limit procedure. Let us choose a
sequence of weak controls
$\mathfrak{a}^{k}=(\Omega^{k},\mathcal{F}^{k},\mathbb{F}^{k},W^{k},\alpha^{k})$
such that $\lvert\alpha_{t}^{k}\rvert\leq k$ and
$\lim_{k\to\infty}J(\mathfrak{a}^{k})=V$. Such a sequence exists owing to the
continuity condition from Assumption A.6 (ii). According to what we proved
above, for each $k\geq 1$, there exists a closed-loop control $a^{k}$ such
that $J_{\text{cl}}(a^{k})=J(\mathfrak{a}^{k})$. Consequently, we get

$V_{\text{cl}}\leq\limsup_{k\to\infty}J_{\text{cl}}(a^{k})=\limsup_{k\to\infty}J(\mathfrak{a}^{k})=V,$

which concludes the proof. ∎

## Acknowledgement

This research has been supported by the EPSRC Centre for Doctoral Training in
Mathematics of Random Systems: Analysis, Modelling and Simulation
(EP/S023925/1).

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