Proof. For ?? , it is sufficient to show that for a given \(\pi\in\Pi^\nu\), there exists \(\underline{\pi}\in\underline{\Pi}^\nu\) such that \[\begin{align} v_{\underline{\pi}}(x, y) \geq v_\pi(x, y), \quad (x, y )\in\mathbb{R}\times E. \label{Lem30595001a} \end{align}\tag{9}\] For fixed \(\pi\in\Pi^\nu\), we define \(\pi^\prime\in\Pi^\nu\) as \[\begin{align} L^{\pi^\prime}_t=L^{\pi}_t,\quad R^{\pi^\prime}_t=-\!\left( \inf_{s \in[0, t]}(X_s-L^{\pi}_s)\land 0 \right),\qquad t\geq 0. \label{ubs} \end{align}\tag{10}\] Then, the strategy \(\pi^\prime\) satisfies the second equation of 8 . From condition (iii) in the definition of strategies in \(\Pi^\nu\) and 10 , we have, for \(t\geq 0\), \[\begin{align} R^{\pi}_t-R^{\pi^\prime}_t=R^{\pi}_t+\inf_{s \in[0, t]}(X_s-L^{\pi}_s) \geq\inf_{s \in[0, t]}(X_s-L^{\pi}_s+R^{\pi}_s)\geq 0. \end{align}\] Thus, we have, for \((x, y)\in\mathbb{R}\times E\), \[\begin{align} v_{\pi^\prime}(x, y)- v_{\pi}(x, y) =&\beta{\mathbf{E}}_{(x, y)}\!\left[ \int_{[0,\infty)}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}R^{\pi}_t-\int_{[0,\infty)}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}R^{\pi^\prime}_t \right]\\ =&\beta{\mathbf{E}}_{(x, y)}\!\left[ \int_0^\infty {\mathbf{q}}(Y_s) e^{-{\mathfrak{q}}(s)} ( R^{\pi}_s-R^{\pi^\prime}_s){\mathrm{d}}s \right]\geq 0. \label{ineq95002} \end{align}\tag{11}\] We define \(\underline{\pi}\in\Pi^\nu\) as \[\begin{align} L^{\underline{\pi}}_t=\sum_{s \in[0, t]}\Delta L^{\pi^\prime}_s\land \!\left\{ (U^{\pi^\prime}_{s-}+\Delta X_s)\lor 0 \right\},\quad R^{\underline{\pi}}_t=R^{\pi^\prime}_t -\widetilde{L}^{\pi^\prime,\underline{\pi}}_t,\qquad t\geq 0, \end{align}\] where \(\widetilde{L}^{\pi^\prime,\underline{\pi}}_t:=\sum_{s\in[0,t]}(\Delta L^{\pi^\prime}_s-\Delta L^{\underline{\pi}}_s)\) for \(t\geq 0\). Then, the resulting controlled processes corresponding to \(\pi^\prime\) and \(\underline{\pi}\) coincide, and it is easy to verify that \(\underline{\pi}\in \underline{\Pi}^\nu\). By the definition of \(\underline{\pi}\), we have, for \((x, y)\in\mathbb{R}\times E\), \[\begin{align} &v_{\underline{\pi}}(x, y)-v_{\pi^\prime}(x, y)\\ &\;\;={\mathbf{E}}_{(x, y)}\bigg[\int_{[0,\infty)}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}L^{\underline{\pi}}_t-\int_{[0,\infty)}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}L^{\pi^\prime}_t -\beta\!\left( \int_{[0,\infty)}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}R^{\underline{\pi}}_t-\int_{[0,\infty)}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}R^{\pi^\prime}_t \right)\bigg]\\ &\;\;={\mathbf{E}}_{(x, y)}\!\left[ (\beta-1)\int_{[0,\infty)}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}\widetilde{L}^{\pi^\prime,\underline{\pi}}_t \right]\geq 0, \label{ineq95001} \end{align}\tag{12}\] where the last inequality follows from the fact that the process \(\!\left\{ \widetilde{L}^{\pi^\prime,\underline{\pi}}_t:t\geq 0 \right\}\) is non-decreasing. By 11 and 12 , we have 9 .

If \(\pi\in\Pi^\nu\) does not belong to \(\underline{\Pi}^\nu\), there exists \((x, y) \in \mathbb{R}\times E\) such that the event that 8 fails has positive \({\mathbf{P}}_{(x, y)}\)-probability. Let us fix such a \((x, y)\). First, we assume that the second equation in 8 fails with positive probability under the probability measure \({\mathbf{P}}_{(x, y)}\). Then, with positive \({\mathbf{P}}_{(x,y)}\)-probability, the set of times at which \(R^{\pi}_t-R^{\pi^\prime}_t=R^{\pi}_t+\inf_{s \in[0, t]}(X_s-L^{\pi^\prime}_s)>0\) has strictly positive Lebesgue measure, and hence the last term in 11 is strictly positive. Thus, \(\pi^\prime\) strictly improves upon \(\pi\), and \(\pi\) is not an optimal strategy. Next, we assume that the second equation in 8 always holds, whereas the first inequality in 8 fails with positive probability under the probability measure \({\mathbf{P}}_{(x, y)}\). In this case, \(\pi\) and \(\pi^\prime\) coincide. However, since the last term in 12 is strictly positive, \(\underline{\pi}\) yields a strictly larger value than \(\pi\), and \(\pi\) is not an optimal strategy. The proof is complete. ◻

Proof. By 8 , we have, for \(\pi\in\underline{\Pi}^\nu\), \[\begin{align} \beta\int_{[0,T^\nu]}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}R^{\pi}_t =\beta\int_{[0,T^\nu]}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}((-\underline{X}_t)\lor 0) , \label{3039502} \end{align}\tag{13}\] for all \((x,y)\in\mathbb{R}\times E\), \({\mathbf{P}}_{(x, y)}\)-a.s., where \(\underline{X}_t:= \inf_{s \in[0, t]} X_s\) for \(t\geq 0\). Moreover, \[\begin{align} \begin{aligned} &\text{if no dividend is paid, the value of the }\\ &\qquad\qquad\qquad\qquad\qquad\text{resulting controlled process at time }T^\nu\text{ is }X^0_{T^\nu}, \end{aligned} \label{setsumei} \end{align}\tag{14}\] where \(X^0_t:= X_t - \!\left( \underline{X}_t \land 0 \right)\) for \(t\geq 0\). Let us define a measurable function \(g^z_f\) on \([0,z]\times E\) for \(z \in[0, \infty)\) by \(g^z_f(x, y)=x+ f(z-x, y)\). Then \(x \mapsto g^z_f(x, y)\) has a left derivative, denoted by \(x \mapsto g^{z\prime}_{f-}(x, y)\), which satisfies \(g^{z\prime}_{f-}(x, y) = 1-f^\prime_+(z-x, y)\) for \((x, y)\in[0,z]\times E\). Thus, we have, for \(y\in E\), \[\begin{align} g^{z\prime}_{f-}(x, y) \begin{cases} > 0, \qquad & x \in [0, z-\overline{{\mathbf{b}}}_f (y)), \\ = 0,\qquad & x \in [z-\overline{{\mathbf{b}}}_f (y), z-\underline{{\mathbf{b}}}_f (y)], \\ < 0, \qquad & x\in (z-\underline{{\mathbf{b}}}_f (y), z]. \end{cases} \label{Lem30595001} \end{align}\tag{15}\] Therefore, for \(y \in E\), \(x\mapsto g^{z}_{f}(x, y)\) is strictly increasing on \([0, z-\overline{{\mathbf{b}}}_f (y)]\) and strictly decreasing on \([z-\underline{{\mathbf{b}}}_f (y), z]\), and thus for every \({y^\prime}\in E\) and for \(m_{y^\prime}\)-a.e. \(y\in E\), \[\begin{align} g^z_f((z-{\mathbf{b}}_f (y))\lor 0, y)\geq g^z_f(x, y),\qquad x \in [0,z]. \label{ineqg} \end{align}\tag{16}\] By 14 , the definition of \(\pi^\nu_{{\mathbf{b}}_f}\) and 16 , we have, for \(\pi\in\underline{\Pi}^\nu\), \[\begin{align} L^{\pi^\nu_{{\mathbf{b}}_f}}_{T^\nu}+ f(U^{\pi^\nu_{{\mathbf{b}}_f}}_{T^\nu}, Y_{T^\nu}) =& g^{X^0_{T^\nu}}_f((X^0_{T^\nu}-{\mathbf{b}}_f (Y_{T^\nu}))\lor 0, Y_{T^\nu}) \\ \geq & g^{X^0_{T^\nu}}_f(L^\pi_{T^\nu}, Y_{T^\nu}) =L^{\pi}_{T^\nu}+ f(U^{\pi}_{T^\nu}, Y_{T^\nu}), \label{3039501} \end{align}\tag{17}\] for all \((x,y)\in\mathbb{R}\times E\), \({\mathbf{P}}_{(x, y)}\)-a.s. By 13 and 17 , we have for \(\pi\in\underline{\Pi}^\nu\) and \((x, y)\in\mathbb{R}\times E\), \[\begin{align} \mathcal{V}^{\nu}_{\pi^\nu_{{\mathbf{b}}_f}}f(x, y)\geq \mathcal{V}^{\nu}_{\pi}f(x, y). \end{align}\] The proof is complete. ◻

Proof. We first define, for \(g\in \Gamma^\nu\) and \(y \in E\), \[\begin{align} \widetilde{g} (x, y):= & \begin{cases} g(x, y), \qquad &x\leq \underline{{\mathbf{b}}}_g(y), \\ g(\underline{{\mathbf{b}}}_g(y), y)+ (x-\underline{{\mathbf{b}}}_g(y)) , \qquad &x> \underline{{\mathbf{b}}}_g(y), \end{cases} \\ =&\begin{cases} g(x, y), \qquad &x\leq \overline{{\mathbf{b}}}_g(y), \\ g(\overline{{\mathbf{b}}}_g(y), y)+ (x-\overline{{\mathbf{b}}}_g(y)) , \qquad &x> \overline{{\mathbf{b}}}_g(y). \end{cases} \label{tilde95f} \end{align}\tag{18}\] We fix \(f \in \Gamma^\nu\). Note that for \(y \in E\), the function \(x\mapsto\widetilde{f}(x, y)\) is concave and the right derivative \(x\mapsto\widetilde{f}^\prime_+(x, y)\) is equal to \(1\) for \(x\geq \underline{{\mathbf{b}}}_f(y)\) by the definition of \(\underline{{\mathbf{b}}}_f\). By Lemma 9, we have, for \(x \in\mathbb{R}\), \[\begin{align} \mathcal{V}^{\nu}&f(x,y )= \mathcal{V}^{\nu}_{{\pi^\nu_{\overline{{\mathbf{b}}}_f}}}f (x, y) \\ &= {\mathbf{E}}_{(x, y)}\!\left[ e^{-{\mathfrak{q}}(T^\nu)} L^{{\pi^\nu_{\overline{{\mathbf{b}}}_f}}}_{T^\nu} -\beta\int_{[0,T^\nu]}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}R^{{\pi^\nu_{\overline{{\mathbf{b}}}_f}}}_t+ e^{-{\mathfrak{q}}(T^\nu)} f(U^{{\pi^\nu_{\overline{{\mathbf{b}}}_f}}}_{T^\nu}, Y_{T^\nu}) \right]\\ &= {\mathbf{E}}_{(x, y)}\!\left[ -\beta\int_{[0,T^\nu]}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}((-\underline{X}_t) \lor 0)+ e^{-{\mathfrak{q}}(T^\nu)} \widetilde{f}(X^0_{T^\nu}, Y_{T^\nu}) \right]. \label{optimal95b} \end{align}\tag{19}\]

We confirm that \(\mathcal{V}^{\nu} f\) satisfies condition (iii) for membership in \(\Gamma^\nu\). We define, for \((x,y)\in\mathbb{R}\times E\), \[\begin{align} v^L_0 (x, y):=& {\mathbf{E}}_{(x, y)}\!\left[ \int_{[0,\infty) }e^{-{\mathfrak{q}}(t)} {\mathrm{d}}L^{0}_t \right] =x\lor0+v^L_{\pi^\nu_0}(0, y), \\ v^R_0 (x, y):= &{\mathbf{E}}_{(x, y)}\!\left[ \int_{[0,\infty) }e^{-{\mathfrak{q}}(t)} {\mathrm{d}}R^{0}_t \right], \end{align}\] where \(\pi^\nu_0\) is the MMPCB strategy with barrier \(0\). The derivative of the map \(x\mapsto v^L_0 (x, y)\) on \((0,\infty)\) is equal to \(1\). Hence, by the definition of \(\widetilde{f}\) and condition (iii) for membership in \(\Gamma^\nu\), we have \[\begin{align} v^L_0 (x, y) \geq \widetilde{f}(x,y)\geq -\beta v^R_0 (x, y),\qquad (x, y) \in \mathbb{R}\times E. \end{align}\] Therefore, by 19 , 6 at \(T^\nu\) and ?? , we have, for \((x, y)\in [0,\infty)\times E\), \[\begin{align} \mathcal{V}^{\nu} f (x, y) &\leq {\mathbf{E}}_{(x, y)}\!\left[ e^{-{\mathfrak{q}}(T^\nu)} \widetilde{f}(X^0_{T^\nu}, Y_{T^\nu}) \right] \\ &\leq{\mathbf{E}}_{(x, y)}\!\left[ e^{-{\mathfrak{q}}(T^\nu)} v^L_0 (X^0_{T^\nu}, Y_{T^\nu}) \right] =v^L_{\pi^\nu_0}(x, y)\leq v^L_0(x, y). \end{align}\] On the other hand, by 19 , 6 at \(T^\nu\) and ?? , \[\begin{align} \mathcal{V}^{\nu} f (0, y)& = {\mathbf{E}}_{(0, y)}\!\left[ -\beta\int_{[0,T^\nu]}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}((-\underline{X}_t) \lor 0)+ e^{-{\mathfrak{q}}(T^\nu)} \widetilde{f}(X^0_{T^\nu}, Y_{T^\nu}) \right] \\ &\geq {\mathbf{E}}_{(0, y)}\!\left[ -\beta\int_{[0,T^\nu]}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}R^{\pi^\nu_0}_t-\beta e^{-{\mathfrak{q}}(T^\nu)} v^R_0(0, Y_{T^\nu}) \right] =-\beta v^R_{\pi^\nu_0}(0,y) \geq-\beta v^R_0(0, y). \end{align}\]

We confirm condition (i) for membership in \(\Gamma^\nu\). We compute the right derivative \(x\mapsto {(\mathcal{V}^{\nu}f)}^\prime_+(x, y)\) for fixed \(y\in E\). For \(\varepsilon\in\mathbb{R}\), we define \(X^{(\varepsilon)}:=\{X^{(\varepsilon)}_t : t\geq 0\}\) as \[\begin{align} X^{(\varepsilon)}_t := X_t+\varepsilon, \qquad t\geq 0, \end{align}\] and define, for measurable function \({\mathbf{b}}\), \[\begin{align} &\;\;\;\tau^{(\varepsilon), -}_0:=\inf\{t>0:X^{(\varepsilon)}_t<0\},\qquad \tau^{(\varepsilon), -}_{0+}:=\inf\{t\geq 0:X^{(\varepsilon)}_t\leq0\}, \\ T^{(\varepsilon),+}_{\mathbf{b}}:&=\min\{T^\nu_k>0: X^{(\varepsilon)}_{T^\nu_k}>{\mathbf{b}}{(Y_{T^\nu_k})} \}, \qquad T^{(\varepsilon),+}_{{\mathbf{b}}-}:=\min\{T^\nu_k>0: X^{(\varepsilon)}_{T^\nu_k}\geq {\mathbf{b}}{(Y_{T^\nu_k})} \}, \\ &\qquad \qquad \underline{X}^{(\varepsilon)}_t := \inf_{s \in [0, t]}X^{(\varepsilon)}_s, \qquad X^{0,(\varepsilon)}_t:=X^{(\varepsilon)}_t-(\underline{X}^{(\varepsilon)}_t \land 0) \qquad t\geq 0. \end{align}\] We also write \(L^{\pi^{(\varepsilon)}_{{\mathbf{b}}}}:=\{L^{\pi^{(\varepsilon)}_{{\mathbf{b}}}}_t : t\geq 0\}\), \(R^{\pi^{(\varepsilon)}_{{\mathbf{b}}}}:=\{R^{\pi^{(\varepsilon)}_{{\mathbf{b}}}}_t : t\geq 0\}\) and \(U^{\pi^{(\varepsilon)}_{{\mathbf{b}}}}:=\{U^{\pi^{(\varepsilon)}_{{\mathbf{b}}}}_t : t\geq 0\}\) for processes representing the dividend, capital injection and controlled processes when the MMPCB strategy with barrier \({\mathbf{b}}\) is applied to \(X^{(\varepsilon)}\). Then, by 1 , the following assertion (\(\sharp\)) can be proved by adapting the proof of [11], which treats the case of Lévy processes.

• We fix a measurable function \({\mathbf{b}}\). For \(\varepsilon>0\) and \((x, y)\in\mathbb{R}\times E\), we have, \({\mathbf{P}}_{(x, y)}\)-a.s., \[\begin{align} L^{{\pi}^{(\varepsilon)}_{{\mathbf{b}}}}_t- L^{{\pi}^{(0)}_{{\mathbf{b}}}}_t=0, \; R^{{\pi}^{(\varepsilon)}_{{\mathbf{b}}}}_t- R^{{\pi}^{(0)}_{{\mathbf{b}}}}_t=0, \; U^{{\pi}^{(\varepsilon)}_{{\mathbf{b}}}}_t-U^{{\pi}^{(0)}_{{\mathbf{b}}}}_t= \varepsilon, \quad t \in [0, {T}^{ (\varepsilon), +}_{\mathbf{b}}\land {\tau}^{ (0), -}_0), \end{align}\] and the process \(t\mapsto L^{{\pi}^{(\varepsilon)}_{{\mathbf{b}}}}_t-L^{{\pi}^{(0)}_{{\mathbf{b}}}}_t\in[0, \varepsilon]\) is non-decreasing and the processes \(t\mapsto R^{{\pi}^{(\varepsilon)}_{{\mathbf{b}}}}_t-R^{{\pi}^{(0)}_{{\mathbf{b}}}}_t\in[-\varepsilon, 0]\) and \(t\mapsto U^{{\pi}^{(\varepsilon)}_{{\mathbf{b}}}}_t-U^{{\pi}^{(0)}_{{\mathbf{b}}}}_t\in[0,\varepsilon]\) are non-increasing on \((0, \infty)\). In particular, if \({T}^{ (0), +}_{{\mathbf{b}}-} \leq {\tau}^{ (0), -}_0\), \[\begin{align} L^{{\pi}^{(\varepsilon)}_{{\mathbf{b}}}}_t- L^{{\pi}^{(0)}_{{\mathbf{b}}}}_t=\varepsilon, \quad R^{{\pi}^{(\varepsilon)}_{{\mathbf{b}}}}_t- R^{{\pi}^{(0)}_{{\mathbf{b}}}}_t=0,\quad U^{{\pi}^{(\varepsilon)}_{{\mathbf{b}}}}_t-U^{{\pi}^{(0)}_{{\mathbf{b}}}}_t= 0, \quad t \in [{T}^{ (0), +}_{{\mathbf{b}}-}, \infty) , \end{align}\] and if \({\tau}^{ (\varepsilon), -}_{0+}<{T}^{ (\varepsilon), +}_{\mathbf{b}}\), \[\begin{align} L^{{\pi}^{(\varepsilon)}_{{\mathbf{b}}}}_t- L^{{\pi}^{(0)}_{{\mathbf{b}}}}_t=0, \quad R^{{\pi}^{(\varepsilon)}_{{\mathbf{b}}}}_t- R^{{\pi}^{(0)}_{{\mathbf{b}}}}_t=-\varepsilon,\quad U^{{\pi}^{(\varepsilon)}_{{\mathbf{b}}}}_t-U^{{\pi}^{(0)}_{{\mathbf{b}}}}_t= 0, \quad t \in [ {\tau}^{ (\varepsilon), -}_{0+}, \infty), \end{align}\]

Then, for \(x \in \mathbb{R}\) and \(\varepsilon > 0\), we have \[\begin{align} &\mathcal{V}^{\nu}f(x+\varepsilon ,y )- \mathcal{V}^{\nu}f(x,y )\\ &\;\;={\mathbf{E}}_{(x, y)}\Bigg[ e^{-{\mathfrak{q}}(T^\nu)} \!\left( \widetilde{f}(X^{0,(\varepsilon)}_{T^\nu}, Y_{T^\nu})-\widetilde{f}(X^{0,(0)}_{T^\nu}, Y_{T^\nu}) \right) +{\beta}\int_{[0,T^\nu]}e^{-{\mathfrak{q}}(t)} {\mathrm{d}} (((-X_t)\lor 0)\land \varepsilon) \Bigg],\label{Lem30695001} \end{align}\tag{20}\] where we used 19 . By (\(\sharp\)), we have \[\begin{align} {\mathbf{E}}_{(x, y)}&\!\left[ e^{-{\mathfrak{q}}(T^\nu)} \!\left( \widetilde{f}(X^{(\varepsilon)}_{T^\nu}, Y_{T^\nu})-\widetilde{f}(X^{(0)}_{T^\nu}, Y_{T^\nu}) \right); \underline{X}^{(0)}_{T^\nu}\geq 0 \right]\\ &\leq {\mathbf{E}}_{(x, y)}\!\left[ e^{-{\mathfrak{q}}(T^\nu)} \!\left( \widetilde{f}(X^{0,(\varepsilon)}_{T^\nu}, Y_{T^\nu})-\widetilde{f}(X^{0,(0)}_{T^\nu}, Y_{T^\nu}) \right) \right]\\ &\leq{\mathbf{E}}_{(x, y)}\!\left[ e^{-{\mathfrak{q}}(T^\nu)} \!\left( \widetilde{f}(X^{(\varepsilon)}_{T^\nu}, Y_{T^\nu})-\widetilde{f}(X^{(0)}_{T^\nu} , Y_{T^\nu}) \right); \underline{X}^{(\varepsilon)}_{T^\nu}> 0 \right], \label{Lem30695002} \end{align}\tag{21}\] and \[\begin{align} \beta\varepsilon {\mathbf{E}}_{(x, y)}\!\left[ e^{-{\mathfrak{q}}(\tau^{(\varepsilon),-}_{0+})};\underline{X}^{(\varepsilon)}_{T^\nu}\leq 0 \right] &\leq {\mathbf{E}}_{(x, y)}\!\left[ {\beta}\int_{[0,T^\nu]}e^{-{\mathfrak{q}}(t)} {\mathrm{d}} (((-X_t)\lor 0)\land \varepsilon) \right] \\ &\leq \beta\varepsilon {\mathbf{E}}_{(x, y)}\!\left[ e^{-{\mathfrak{q}}(\tau^{(0),-}_0)};\underline{X}^{(0)}_{T^\nu}< 0 \right]. \label{Lem30695003} \end{align}\tag{22}\] Combining 20 , 21 and 22 , and using the fact that \(\lim_{\varepsilon\downarrow0}\tau^{(\varepsilon),-}_{0+}=\tau^{(0),-}_0\), \({\mathbf{P}}_{(x, y)}\)-a.s., we have, for \(x\in\mathbb{R}\), \[\begin{align} {(\mathcal{V}^{\nu}f)}^\prime_+(x ,y )&:= \lim_{\varepsilon\downarrow0} \frac{\mathcal{V}^{\nu}f(x+\varepsilon ,y )- \mathcal{V}^{\nu}f(x,y )}{\varepsilon}\\ &={\mathbf{E}}_{(x, y)}\!\left[ e^{-{\mathfrak{q}}(T^\nu)} \widetilde{f}^\prime_+(X_{T^\nu}, Y_{T^\nu}); \underline{X}^{(0)}_{T^\nu}\geq 0 \right] +\beta {\mathbf{E}}_{(x, y)}\!\left[ e^{-{\mathfrak{q}}(\tau^{-}_0)};\underline{X}^{(0)}_{T^\nu}< 0 \right]\\ &={\mathbf{E}}_{(0, y)}\!\left[ e^{-{\mathfrak{q}}(T^\nu\land\tau^{(x),-}_0)} \!\left( \widetilde{f}^\prime_+(X^{(x)}_{T^\nu}, Y_{T^\nu})1_{\{\underline{X}^{(x)}_{T^\nu}\geq 0 \}} +\beta 1_{\{\underline{X}^{(x)}_{T^\nu}< 0 \}} \right) \right]. \label{Lem30695004} \end{align}\tag{23}\] Since \(x\mapsto e^{-{\mathfrak{q}}(T^\nu\land\tau^{(x),-}_0)}\) and \(x\mapsto \widetilde{f}^\prime_+(X^{(x)}_{T^\nu}, Y_{T^\nu})1_{\{\underline{X}^{(x)}_{T^\nu}\geq 0 \}} +\beta1_{\{\underline{X}^{(x)}_{T^\nu}< 0 \}}\) are non-negative and non-increasing, the right derivative \(x\mapsto {(\mathcal{V}^{\nu}f)}^\prime_+(x ,y )\) exists and is also non-increasing. It follows from this fact and [23] that the function \(x\mapsto \mathcal{V}^{\nu} f(x, y)\) is concave.

By 23 and since \(\widetilde{f}^\prime_+(x, y)\in[1, \beta]\) for \((x, y)\in[0, \infty)\times E\), \(\mathcal{V}^\nu f\) satisfies condition (ii) for membership in \(\Gamma^\nu\). The proof is complete. ◻

Proof. We prove the second inequality of ?? . By the definition of \(\mathcal{V}^{\nu, (n)}\), Lemma 9 and 6 at each \(T^\nu_k\) with \(k\in\{1, 2, \dots, n\}\), we have \[\begin{align} \mathcal{V}^{\nu, (n)} {v_0}(x, y) = v_{\pi^\ast_n}(x, y),\qquad (x, y)\in\mathbb{R}\times E, \label{Lem30795003} \end{align}\tag{25}\] where \(\pi^\ast_n\in\underline{\Pi}^\nu\) is a strategy defined analogously to (\(\star\)) in Section [sec:SubSec203], where at each time \(T^\nu_k\), a dividend is paid corresponding to the excess over \(\underline{{\mathbf{b}}}_{\mathcal{V}^{\nu,(n-k)}v_0}(Y_{T^\nu_k})\), for \(k = 1, \dots, n\). Thus, by the definition of \(V^\nu\), the second inequality of ?? holds.

We prove the first inequality of ?? . We prove by induction that, for \(k\in \{ 0, 1, \dots, n\}\), the following holds \[\begin{align} v_{\pi_n}(x, y)\leq {\mathbf{E}}_{(x, y)}\Bigg[\int_{[0,T^\nu_{n-k}]}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}(L^{\pi_n}_t &-\beta R^{\pi_n}_t) +e^{-{\mathfrak{q}}(T^\nu_{n-k})}\mathcal{V}^{\nu, (k)}{v_0}(U^{\pi_n}_{T^\nu_{n-k}}, Y_{T^\nu_{n-k}})\Bigg], \label{induction95inequality} \end{align}\tag{26}\] for \((x, y)\in\mathbb{R}\times E\). We have 26 with equality in place of \(\leq\) when \(k=0\) by 6 at \(T^\nu_n\). We assume that \(n\geq 2\) and 26 holds for \(k=l \in\{0, 1, \dots, n-1\}\) and prove it for \(k=l+1\). By 26 with \(k=l\), we have \[\begin{align} v_{\pi_n}(x, y)\leq{\mathbf{E}}_{(x, y)}\Bigg[\int_{[0,T^\nu_{n-l-1}]}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}L^{\pi_n}_t&+ e^{-{\mathfrak{q}}(T^\nu_{n-l})} \Delta L^{\pi_n}_{T^\nu_{n-l}} \\ -\beta\int_{[0,T^\nu_{n-l}]}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}R^{\pi_n}_t &+e^{-{\mathfrak{q}}(T^\nu_{n-l})}\mathcal{V}^{\nu, (l)}{v_0}(U^{\pi_n}_{T^\nu_{n-l}}, Y_{T^\nu_{n-l}})\Bigg] , \label{Lem30795001} \end{align}\tag{27}\] for \((x,y)\in\mathbb{R}\times E\). As in the proof of 17 , we have \[\begin{align} \mathcal{V}^{\nu, (l)}{v_0}(U^{\pi_n}_{(n-l-)}\land \underline{{\mathbf{b}}}_{\mathcal{V}^{\nu, (l)}v_0}(Y_{T^\nu_{n-l}}), Y_{T^\nu_{n-l}}) +&\left(U^{\pi_n}_{(n-l-)}-\underline{{\mathbf{b}}}_{\mathcal{V}^{\nu, (l)}v_0}(Y_{T^\nu_{n-l}})\right)\lor 0\\ &\geq \Delta L^{\pi_n}_{T^\nu_{n-l}}+\mathcal{V}^{\nu, (l)} {v_0}(U^{\pi_n}_{T^\nu_{n-l}}, Y_{T^\nu_{n-l}}). \label{Lem30795002} \end{align}\tag{28}\] Thus, we have \[\begin{align} &v_{\pi_n}(x, y)\leq{\mathbf{E}}_{(x, y)}\Bigg[\int_{[0,T^\nu_{n-l-1}]}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}L^{\pi_n}_t -\beta\int_{[0,T^\nu_{n-l}]}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}R^{\pi_n}_t+e^{-{\mathfrak{q}}(T^\nu_{n-l})}\\ &\quad\times\left(\left(U^{\pi_n}_{(n-l-)}-\underline{{\mathbf{b}}}_{\mathcal{V}^{\nu, (l)}v_0}(Y_{T^\nu_{n-l}})\right)\lor 0\right) +e^{-{\mathfrak{q}}(T^\nu_{n-l})} \mathcal{V}^{\nu, (l)}{v_0}(U^{\pi_n}_{(n-l-)}\land \underline{{\mathbf{b}}}_{\mathcal{V}^{\nu, (l)}v_0}(Y_{T^\nu_{n-l}}), Y_{T^\nu_{n-l}})\Bigg]\\ &={\mathbf{E}}_{(x, y)}\!\left[ \int_{[0,T^\nu_{n-l-1}]}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}(L^{\pi_n}_t -\beta R^{\pi_n}_t) +e^{-{\mathfrak{q}}(T^\nu_{n-l-1})}\mathcal{V}^{\nu, (l+1)}{v_0}(U^{\pi_n}_{T^\nu_{n-l-1}}, Y_{T^\nu_{n-l-1}}) \right], \end{align}\] where the inequality follows from 27 and 28 , and the equality follows from 6 at \(T^\nu_{n-l-1}\). Therefore, we obtain 26 with \(k=l+1\). Since 26 with \(k=n\) is equivalent to the first inequality of ?? , the proof is complete. ◻

Proof. By ?? , we have \[\begin{align} \limsup_{n\to\infty}\mathcal{V}^{\nu, (n)} {v_0}(x, y)\leq V^{\nu}(x, y),\qquad (x, y)\in\mathbb{R}\times E. \label{Lem30995004} \end{align}\tag{29}\] For \((x, y)\in\mathbb{R}\times E\) and \(\varepsilon>0\), we may choose the strategy \(\pi^{(x, y),\varepsilon}\in\underline{\Pi}^\nu\) such that \[\begin{align} 0\leq V^{\nu}(x, y)-v_{\pi^{(x, y),\varepsilon}}(x, y)\leq\varepsilon. \label{Lem30995001} \end{align}\tag{30}\] For \(n\in\mathbb{N}\), \((x, y)\in\mathbb{R}\times E\) and \(\varepsilon>0\), we have \[\begin{align} V^{\nu}(x, y)-\mathcal{V}^{\nu, (n)} {v_0}(x, y)\leq &V^{\nu}(x, y)-v_{\pi^{(x, y),\varepsilon}_n}(x, y)\\ \leq&V^{\nu}(x, y)-v_{\pi^{(x, y),\varepsilon}}(x, y) +|v_{\pi^{(x, y),\varepsilon}}(x, y)-v_{\pi^{(x, y),\varepsilon}_n}(x, y)| \\ \leq&\varepsilon +|v_{\pi^{(x, y),\varepsilon}}(x, y)-v_{\pi^{(x, y),\varepsilon}_n}(x, y)|, \label{Lem30995002} \end{align}\tag{31}\] where the first inequality follows from Lemma 11 and the last inequality follows from 30 . By taking the limit of 31 as \(n\to \infty\) and by Lemma 12, we have \[\begin{align} V^{\nu}(x, y)-\liminf_{n\to\infty}\mathcal{V}^{\nu, (n)} {v_0}(x, y) \leq\varepsilon. \label{Lem30995003} \end{align}\tag{32}\] Since 32 holds for any \(\varepsilon>0\) and \((x,y)\in\mathbb{R}\times E\), we have \[\begin{align} V^{\nu}(x, y)-\liminf_{n\to\infty}\mathcal{V}^{\nu, (n)} {v_0}(x, y) \leq0, \qquad (x, y)\in\mathbb{R}\times E. \label{Lem30995005} \end{align}\tag{33}\] By 29 and 33 , we obtain ?? . Since \(\mathcal{V}^{\nu, (n)} {v_0} \in \Gamma^\nu\) for \(n\in\mathbb{N}\), it follows from ?? and [23] that \(V^\nu\) satisfies conditions (i) and (iii) for membership in \(\Gamma^\nu\). Using also [24], we see that \(V^\nu\) satisfies condition (ii) for membership in \(\Gamma^\nu\). The proof is complete. ◻

Proof of (\(\Leftarrow\)) in Lemma 14. (1) For \({\mathbf{b}}_{V^\nu}\in\widehat{\Xi}^\nu_{V^\nu}\), we prove \[\begin{align} \mathcal{V}^{\nu}_{\pi^\nu_{{\mathbf{b}}_{V^{\nu}}}}V^{\nu}(x, y)= V^{\nu}(x, y),\qquad (x, y)\in\mathbb{R}\times E. \label{Lem31095003} \end{align}\tag{34}\] By the definition of \(\mathcal{V}^\nu\) and ?? , we have \[\begin{align} \mathcal{V}^\nu_{\pi^\nu_{{\mathbf{b}}_{V^\nu}}} \mathcal{V}^{\nu, (n)} {v_0}(x, y) \leq \mathcal{V}^{\nu, (n+1)} {v_0}(x, y) \leq V^{\nu}(x, y),\qquad (x, y)\in\mathbb{R}\times E. \label{Lem31095005} \end{align}\tag{35}\] By the dominated convergence theorem, ?? and 35 , we have \[\begin{align} \mathcal{V}^\nu_{\pi^\nu_{{\mathbf{b}}_{V^{\nu}}}} V^{\nu}(x, y) =\lim_{n\to\infty} \mathcal{V}^\nu_{\pi^\nu_{{\mathbf{b}}_{V^\nu}}} \mathcal{V}^{\nu, (n)} {v_0}(x, y) \leq V^{\nu}(x, y),\qquad (x, y)\in\mathbb{R}\times E. \label{Lem31095001} \end{align}\tag{36}\] Let \(\widetilde{\Pi}^\nu\) be the set of strategies \(\pi\) belonging to \(\underline{\Pi}^\nu\) such that \[\begin{align} L^{\pi}_t=\sum_{k\in\mathbb{N}} \left(( U^\pi_{(k-)} - {\mathbf{b}}^{\pi}_k(Y_{T^\nu_k}) )\lor 0\right) 1_{[0,t]}(T^\nu_k) ,\qquad t\geq 0, \end{align}\] for a set of non-negative measurable functions \(\{{\mathbf{b}}^{\pi}_n:n\in\mathbb{N}\}\) on \(E\). Since \(\pi^\ast_n\) defined in the proof of Lemma 11 belongs to \(\widetilde{\Pi}^\nu\), 25 and ?? imply that \[\begin{align} V^{\nu}(x, y ) = \sup_{\pi\in\widetilde{\Pi}^\nu}v_\pi(x, y), \qquad (x, y)\in\mathbb{R}\times E. \end{align}\] For \(n \in\mathbb{N}\) and a strategy \(\pi \in \widetilde{\Pi}^\nu\), we define a strategy \(\pi_{(n)} \in \widetilde{\Pi}^\nu\) to satisfy \({\mathbf{b}}^{\pi_{(n)}}_k\equiv {\mathbf{b}}^{\pi}_{k+n}\) for \(k\in\mathbb{N}\). Then, we have, for \(\pi \in \widetilde{\Pi}^\nu\), \[\begin{align} \mathcal{V}^{\nu}_{\pi^\nu_{{\mathbf{b}}_{V^{\nu}}}}V^{\nu}(x, y)=\mathcal{V}^{\nu}V^{\nu}(x, y)\geq \mathcal{V}^{\nu} v_{\pi_{(1)}}(x, y) \geq v_\pi(x, y),\qquad (x, y)\in\mathbb{R}\times E. \end{align}\] By taking the supremum over \(\pi\), we have \[\begin{align} \mathcal{V}^{\nu}_{\pi^\nu_{{\mathbf{b}}_{V^{\nu}}}}V^{\nu}(x, y)\geq V^{\nu}(x, y),\qquad (x, y)\in\mathbb{R}\times E. \label{Lem31095002} \end{align}\tag{37}\] By 36 and 37 , we obtain 34 .

(2) We prove the optimality of the strategies included in \(\Pi^{\nu,\ast}_{V^\nu}\). We fix \(\pi\in\Pi^{\nu,\ast}_{V^\nu}\). Using the definitions of \(\underline{{\mathbf{b}}}_{V^\nu}\), \(\overline{{\mathbf{b}}}_{V^\nu}\), and \(\widehat{\Xi}^\nu_{V^\nu}\) together with 34 and 19 , we have, for \((x, y)\in \mathbb{R}\times E\), \[\begin{align} V^\nu(x, y) =& {\mathbf{E}}_{(x, y)}\!\left[ -\beta\int_{[0,T^\nu]}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}((-\underline{X}_t) \lor 0)+ e^{-{\mathfrak{q}}(T^\nu)} \widetilde{V}^\nu(X^0_{T^\nu}, Y_{T^\nu}) \right]\\ =& {\mathbf{E}}_{(x, y)}\!\left[ e^{-{\mathfrak{q}}(T^\nu)} \Delta L^\pi_{T^\nu} -\beta\int_{[0,T^\nu]}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}R^{\pi}_t+ e^{-{\mathfrak{q}}(T^\nu)} V^\nu(U^\pi_{T^\nu}, Y_{T^\nu}) \right]. \end{align}\] Iterating this identity \(n\in\mathbb{N}\) times and using 6 , we have \[\begin{align} V^\nu(x, y) =& {\mathbf{E}}_{(x, y)}\!\left[ \int_{[0, T^\nu_n]}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}L^\pi_t -\beta\int_{[0,T^\nu_n]}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}R^{\pi}_t+ e^{-{\mathfrak{q}}(T^\nu_n)} V^\nu(U^\pi_{T^\nu_n}, Y_{T^\nu_n}) \right]. \end{align}\] Thus, we have, for \((x, y)\in\mathbb{R}\times E\) and \(n\in\mathbb{N}\), \[\begin{align} &0\leq V^{\nu}(x, y)-v_{\pi}(x, y)\\ &={\mathbf{E}}_{(x, y)}\!\left[ e^{-{\mathfrak{q}}(T^\nu_n)} v^L_0(U^\pi_{T^\nu_n}, Y_{T^\nu_n}) -\int_{[T^\nu_{n+1}, \infty)}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}L^\pi_t +\beta\int_{(T^\nu_n, \infty) }e^{-{\mathfrak{q}}(t)} {\mathrm{d}}R^{\pi}_t \right]\\ &\leq{\mathbf{E}}_{(x, y)}\!\left[ e^{-{\mathfrak{q}}(T^\nu_n)} \!\left( V^\nu(U^\pi_{T^\nu_n}, Y_{T^\nu_n})+\beta v^R_0(U^\pi_{T^\nu_n}, Y_{T^\nu_n}) \right) \right] \leq{\mathbf{E}}_{(x, y)}\!\left[ e^{-{\mathfrak{q}}(T^\nu_n)} \!\left( U^{\pi}_{T^\nu_n}+(1+\beta)B^\nu \right) \right], \label{divide} \end{align}\tag{38}\] where the second inequality follows from 6 and ?? , and the last inequality follows from ?? . For sufficiently large \(\ell\in\mathbb{N}\) such that \(e^{-q(k-1)}k\leq e^{-\frac{q\ell}{2}}\) for \(k>\ell\), we have \[\begin{align} {\mathbf{E}}_{(x, y)}\!\left[ e^{-{\mathfrak{q}}(T^\nu_n)} U^{\pi}_{T^\nu_n} \right]&\leq {\mathbf{E}}_{(0, y)}\!\left[ e^{-q T^\nu_n}\!\left( (x\lor 0)+X_{T^\nu_n}-\underline{X}_{T^\nu_n} \right) \right] \\ &\leq 2{\mathbf{E}}_{(0, y)}\!\left[ e^{-q T^\nu_n}\!\left( (x\lor 0)+\sup_{s\in[0,T^\nu_n]}|X_{s}| \right) \right] \\ &\leq 2\sum_{k\in\mathbb{N}} e^{-q(k-1)} {\mathbf{E}}_{(0, y)}\!\left[ (x\lor 0)+\sup_{s\in[0,k]}|X_{s}|; T^\nu_n\in (k-1, k] \right] \\ &\leq 2\sum_{k\in\mathbb{N}} e^{-q(k-1)} ((x\lor0)+kM) {\mathbf{P}}(T^\nu_n\in (k-1, k])\\ &\leq 2\sum_{k=1}^\ell e^{-q(k-1)} ((x\lor0)+kM) {\mathbf{P}}(T^\nu_n\in (k-1, k])\\ &\;\;+ 2e^{-q\ell} (x\lor0){\mathbf{P}}(T^\nu_n\in (\ell, \infty))+2e^{-\frac{q\ell}{2}} M{\mathbf{P}}(T^\nu_n\in (\ell, \infty)), \end{align}\] where the first inequality follows from the fact that \(\pi\in\underline{\Pi}^\nu\). By taking the limit as \(n\to\infty\), we have \[\begin{align} \limsup_{n\to\infty}{\mathbf{E}}_{(x, y)}\!\left[ e^{-{\mathfrak{q}}(T^\nu_n)} U^{\pi}_{T^\nu_n} \right]\leq 2e^{-q\ell} (x\lor0)+2e^{-\frac{q\ell}{2}} M. \label{limsup95bound} \end{align}\tag{39}\] Since 39 is true for any large enough \(\ell\in\mathbb{N}\), we have \[\begin{align} \limsup_{n\to\infty}{\mathbf{E}}_{(x, y)}\!\left[ e^{-{\mathfrak{q}}(T^\nu_n)} U^{\pi}_{T^\nu_n} \right]=0. \label{limsup95result} \end{align}\tag{40}\] By 38 with \(n\to \infty\) and 40 , we have, for \((x, y)\in\mathbb{R}\times E\), \(V^{\nu}(x, y)-v_{\pi}(x, y)=0\). The proof is complete. ◻

Proof. Since \({\mathbf{b}}(y)\in[\underline{{\mathbf{b}}}_{v_{\pi^\nu_{\mathbf{b}}}}(y),\overline{{\mathbf{b}}}_{v_{\pi^\nu_{\mathbf{b}}}}(y)]\) for \(y\in E\) and by 6 at \(T^\nu\), we have, for \((x, y) \in \mathbb{R}\times E\), \[\begin{align} \mathcal{V}^{\nu}v_{\pi^\nu_{\mathbf{b}}}(x, y)= \mathcal{V}^{\nu}_{\pi^\nu_{{\mathbf{b}}}}v_{\pi^\nu_{\mathbf{b}}}(x, y) =v_{\pi^\nu_{\mathbf{b}}}(x, y). \label{tokucho} \end{align}\tag{41}\] For \(\pi\in\underline{\Pi}^\nu\), \(n\in\mathbb{N}\) and \(f\in\Gamma^\nu\), we write \[\begin{align} \mathcal{V}^{\nu,(n)}_\pi f (x, y):=\mathcal{V}^{\nu}_\pi(\mathcal{V}^{\nu,(n-1)}_\pi f) (x, y),\qquad (x, y)\in\mathbb{R}\times E, \end{align}\] inductively, if it is well-defined, where \(\mathcal{V}^{\nu, (0)}_\pi f(x, y):=f(x, y)\) for \((x, y)\in\mathbb{R}\times E\). It is easy to verify that for \(n\in\mathbb{N}\), \(\mathcal{V}^{\nu, (n)}_{\pi^\nu_{\underline{{\mathbf{b}}}_{V^\nu}}}v_{\pi^\nu_{\mathbf{b}}}\) is well-defined and \[\begin{align} \mathcal{V}^{\nu, (n)}_{\pi^\nu_{\underline{{\mathbf{b}}}_{V^\nu}}}v_{\pi^\nu_{\mathbf{b}}}(x, y) \leq v_{\pi^\nu_{\mathbf{b}}}(x, y),\qquad (x, y)\in\mathbb{R}\times E, \label{pib95ineq} \end{align}\tag{42}\] by induction. In fact, by the definition of \(\mathcal{V}^\nu\) and 41 , we have \[\begin{align} \mathcal{V}^{\nu}_{\pi^\nu_{\underline{{\mathbf{b}}}_{V^\nu}}}v_{\pi^\nu_{\mathbf{b}}}(x, y) \leq \mathcal{V}^\nu v_{\pi^\nu_{\mathbf{b}}}(x, y)=v_{\pi^\nu_{\mathbf{b}}}(x, y), \end{align}\] so \(\mathcal{V}^{\nu}_{\pi^\nu_{\underline{{\mathbf{b}}}_{V^\nu}}}v_{\pi^\nu_{\mathbf{b}}}\) is well-defined and 42 with \(n=1\) holds. If \(\mathcal{V}^{\nu, (k)}_{\pi^\nu_{\underline{{\mathbf{b}}}_{V^\nu}}}\) is well-defined and 42 with \(n=k\) is true, then by the assumption, the definition of \(\mathcal{V}^\nu\) and 41 , \[\begin{align} \mathcal{V}^{\nu, (k+1)}_{\pi^\nu_{\underline{{\mathbf{b}}}_{V^\nu}}}v_{\pi^\nu_{\mathbf{b}}}(x, y) \leq \mathcal{V}^{\nu}_{\pi^\nu_{\underline{{\mathbf{b}}}_{V^\nu}}}v_{\pi^\nu_{\mathbf{b}}}(x, y) \leq \mathcal{V}^\nu v_{\pi^\nu_{\mathbf{b}}}(x, y)=v_{\pi^\nu_{\mathbf{b}}}(x, y), \end{align}\] so \(\mathcal{V}^{\nu, (k+1)}_{\pi^\nu_{\underline{{\mathbf{b}}}_{V^\nu}}}v_{\pi^\nu_{\mathbf{b}}}\) is well-defined and 42 with \(n=k+1\) holds. Since \(v_{\pi^\nu_{\mathbf{b}}}\) is not greater than \(V^\nu\), we have, for \(n\in\mathbb{N}\) and \((x, y)\in\mathbb{R}\times E\), \[\begin{align} &0\leq V^{\nu}(x, y) -v_{\pi^\nu_{\mathbf{b}}}(x, y) \leq \mathcal{V}^{\nu, (n)}_{\pi^\nu_{\underline{{\mathbf{b}}}_{V^\nu}}}V^\nu(x, y) -\mathcal{V}^{\nu, (n)}_{\pi^\nu_{\underline{{\mathbf{b}}}_{V^\nu}}}v_{\pi^\nu_{\mathbf{b}}}(x, y) \\ &={\mathbf{E}}_{(x, y)}\!\left[ e^{-{\mathfrak{q}}(T^\nu_n)} \!\left( V^\nu(U^{\pi^\nu_{\underline{{\mathbf{b}}}_{V^{\nu}}}}_{T^\nu_n}, Y_{T^\nu_n})-v_{\pi^\nu_{\mathbf{b}}}(U^{\pi^\nu_{\underline{{\mathbf{b}}}_{V^{\nu}}}}_{T^\nu_n}, Y_{T^\nu_n}) \right) \right] \leq{\mathbf{E}}_{(x, y)}\!\left[ e^{-{\mathfrak{q}}(T^\nu_n)} \!\left( U^{\pi^\nu_{\underline{{\mathbf{b}}}_{V^{\nu}}}}_{T^\nu_n}+2B^\nu \right) \right] . \label{ThmB0195002} \end{align}\tag{43}\] Here, the second inequality follows from 34 and 42 , the equality from 6 , and the last inequality from ?? . By 43 with \(n\to\infty\) and 40 , we have, for \((x, y)\in \mathbb{R}\times E\), \(V^{\nu}(x, y) -v_{\pi^\nu_{\mathbf{b}}}(x, y)=0\). The proof is complete. ◻

Proof. We prove (1) “if \({\mathbf{b}}\in\Xi^\nu_{V^\nu}\), then \({\mathbf{b}}\in\Xi^\nu\)” and (2) “if \({\mathbf{b}}\in\Xi^\nu\), then \({\mathbf{b}}\in\Xi^\nu_{V^\nu}\)” in turn.

(1) We assume \({\mathbf{b}}\in\Xi^\nu_{V^\nu}\) and prove \({\mathbf{b}}\in\Xi^\nu\). By the definition of \(\Xi^\nu_{V^\nu}\), we have, for \(y\in E\), \[\begin{align} \lim_{\varepsilon\downarrow 0}\frac{v_{{\pi}^{\nu}_{{\mathbf{b}}}}({\mathbf{b}}(y)+\varepsilon, y)-v_{{\pi}^{\nu}_{{\mathbf{b}}}}({\mathbf{b}}(y), y)}{\varepsilon} \leq 1\leq \lim_{\varepsilon\downarrow 0}\frac{v_{{\pi}^{\nu}_{{\mathbf{b}}}}({\mathbf{b}}(y), y)-v_{{\pi}^{\nu}_{{\mathbf{b}}}}({\mathbf{b}}(y)-\varepsilon, y)}{\varepsilon}. \label{maxMMPCB} \end{align}\tag{44}\] By ?? , we have, for \(y\in E\), \[\begin{align} \lim_{\varepsilon\downarrow0}& \frac{v_{{\pi}^{\nu}_{{\mathbf{b}}}}({\mathbf{b}}(y)+\varepsilon, y)-v_{{\pi}^{\nu}_{{\mathbf{b}}}}({\mathbf{b}}(y), y)}{\varepsilon}\\ &={\mathbf{E}}_{({\mathbf{b}}(y),y )}\!\left[ e^{-{\mathfrak{q}}({T}^{ \nu, +}_{{\mathbf{b}}-})} ;{T}^{ \nu, +}_{{\mathbf{b}}-} \leq {\tau}^{ -}_0 \right]+ \beta{\mathbf{E}}_{({\mathbf{b}}(y),y )}\!\left[ e^{-{\mathfrak{q}}({\tau}^{ -}_0)} ;{\tau}^{ -}_0<{T}^{ \nu, +}_{{\mathbf{b}}-} \right]. \label{right95derivative9502} \end{align}\tag{45}\] By employing an argument similar to the proof of Lemma 15, we have, for \(y \in E\), \[\begin{align} \lim_{\varepsilon\downarrow0}& \frac{v_{{\pi}^{\nu}_{{\mathbf{b}}}}({\mathbf{b}}(y), y)-v_{{\pi}^{\nu}_{{\mathbf{b}}}}({\mathbf{b}}(y)-\varepsilon, y)}{\varepsilon} =\lim_{\varepsilon\downarrow0} \frac{v_{{\pi}^{(0)}_{{\mathbf{b}}}}({\mathbf{b}}(y), y)-v_{{\pi}^{(-\varepsilon)}_{{\mathbf{b}}}}({\mathbf{b}}(y), y) }{\varepsilon} \\ &={\mathbf{E}}_{({\mathbf{b}}(y),y )}\!\left[ e^{-{\mathfrak{q}}({T}^{ \nu, +}_{{\mathbf{b}}})} ;{T}^{ \nu, +}_{{\mathbf{b}}} < {\tau}^{ -}_{0+} \right]+ \beta{\mathbf{E}}_{({\mathbf{b}}(y),y )}\!\left[ e^{-{\mathfrak{q}}({\tau}^{ -}_{0+})} ;{\tau}^{ -}_{0+}< {T}^{ \nu, +}_{{\mathbf{b}}} \right]. \label{left95derivative} \end{align}\tag{46}\] By 44 , 45 and 46 , we obtain \({\mathbf{b}}\in\Xi^\nu\).

(2) We assume \({\mathbf{b}}\in\Xi^\nu\) and prove \({\mathbf{b}}\in\Xi^\nu_{V^\nu}\). By the proof of Lemma 15, we have, for \((x, y) \in \mathbb{R}\times E\), \[\begin{align} v^{L\prime}_{\pi^\nu_{\mathbf{b}}+}(x, y) := &\lim_{\varepsilon\downarrow0}\frac{v^L_{\pi^\nu_{\mathbf{b}}}(x+\varepsilon, y)-v^L_{\pi^\nu_{\mathbf{b}}}(x, y)}{\varepsilon} ={\mathbf{E}}_{(x, y)}\!\left[ e^{-{\mathfrak{q}}(T^{\nu,+}_{{\mathbf{b}}-})};T^{\nu,+}_{{\mathbf{b}}-} \leq\tau^-_0 \right], \\ v^{R\prime}_{\pi^\nu_{\mathbf{b}}+}(x, y) := &\lim_{\varepsilon\downarrow0}\frac{v^R_{\pi^\nu_{\mathbf{b}}}(x+\varepsilon, y)-v^R_{\pi^\nu_{\mathbf{b}}}(x, y)}{\varepsilon} =-{\mathbf{E}}_{(x, y)}\!\left[ e^{-{\mathfrak{q}}(\tau^-_0)};\tau^-_0<T^{\nu,+}_{{\mathbf{b}}-} \right]. \end{align}\] Both \(x\mapsto v^{L\prime}_{\pi^\nu_{\mathbf{b}}+}(x, y)\) and \(x\mapsto v^{R\prime}_{\pi^\nu_{\mathbf{b}}+}(x, y)\) are non-decreasing for any \(y\in E\). Hence, by [23], the functions \(x\mapsto v^L_{\pi^\nu_{\mathbf{b}}}(x, y)\) and \(x\mapsto v^R_{\pi^\nu_{\mathbf{b}}}(x, y)\) are convex, and their right derivatives are Radon–Nikodym densities of these functions, respectively, for any \(y \in E\). Thus, for fixed \(y\in E\), the function \(x\mapsto v^\prime_{\pi^\nu_{\mathbf{b}}+}(x, y)\), defined by \(v^\prime_{\pi^\nu_{\mathbf{b}}+}(x, y):=v^{L\prime}_{\pi^\nu_{\mathbf{b}}+}(x, y)-\beta v^{R\prime}_{\pi^\nu_{\mathbf{b}}+}(x, y)\), is a density of \(x\mapsto v_{\pi^\nu_{\mathbf{b}}}(x , y)\). By the identity in ?? and 6 , we have, for \((x, y)\in\mathbb{R}\times E\), \[\begin{align} &v^\prime_{\pi^\nu_{\mathbf{b}}+}(x, y)={\mathbf{E}}_{(x, y)}\!\left[ e^{-{\mathfrak{q}}(T^{\nu,+}_{{\mathbf{b}}-})}\varrho^{\nu, {\mathbf{b}}}_{p_{\mathbf{b}}}( {\mathbf{b}}(Y_{T^{\nu,+}_{{\mathbf{b}}-}}), Y_{T^{\nu,+}_{{\mathbf{b}}-}});T^{\nu,+}_{{\mathbf{b}}-} \leq\tau^-_0 \right] +\beta {\mathbf{E}}_{(x, y)}\!\left[ e^{-{\mathfrak{q}}(\tau^-_0)};\tau^-_0<T^{\nu,+}_{{\mathbf{b}}-} \right]\\ &\;=\sum_{k\in\mathbb{N}}{\mathbf{E}}_{(x, y)}\!\left[ e^{-{\mathfrak{q}}(T^{\nu}_k)}\varrho^{\nu, {\mathbf{b}}}_{p_{\mathbf{b}}}( {\mathbf{b}}(Y_{T^{\nu}_k}), Y_{T^{\nu}_k});T^{\nu,+}_{{\mathbf{b}}-}=T^\nu_k, T^{\nu}_k \leq\tau^-_0 \right] +\beta {\mathbf{E}}_{(x, y)}\!\left[ e^{-{\mathfrak{q}}(\tau^-_0)};\tau^-_0<T^{\nu,+}_{{\mathbf{b}}-} \right]\\ &\;=\sum_{k\in\mathbb{N}}\beta{\mathbf{E}}_{(x, y)}\!\left[ e^{-{\mathfrak{q}}(\kappa^{\nu,{\mathbf{b}}}_{p_{\mathbf{b}},0})};T^{\nu,+}_{{\mathbf{b}}-}=T^\nu_k, T^{\nu}_k \leq\tau^-_0 \right] +\beta {\mathbf{E}}_{(x, y)}\!\left[ e^{-{\mathfrak{q}}(\tau^-_0)};\tau^-_0<T^{\nu,+}_{{\mathbf{b}}-} \right] =\rho^{\nu, {\mathbf{b}}}_{p_{\mathbf{b}}}( x, y). \label{v95right95derivative} \end{align}\tag{47}\] Since \(x\mapsto v^\prime_{\pi^\nu_{\mathbf{b}}+}(x, y)\) is non-increasing by 47 and Lemma 18, \(x\mapsto v_{\pi^\nu_{\mathbf{b}}}(x, y)\) is concave. In addition, \(v_{\pi^\nu_{\mathbf{b}}}\) satisfies condition (ii) for membership in \(\Gamma^\nu\) by 47 and condition (iii) for membership in \(\Gamma^\nu\) by ?? . Hence, \(v_{\pi^\nu_{\mathbf{b}}}\) belongs to \(\Gamma^\nu\). By Lemma 19, we have \({\mathbf{b}}(y)\in[\underline{{\mathbf{b}}}_{v_{\pi^\nu_{\mathbf{b}}}}(y), \overline{{\mathbf{b}}}_{v_{\pi^\nu_{\mathbf{b}}}}(y)]\) for \(y\in E\). By Lemma 17, \(\pi^\nu_{\mathbf{b}}\) is an optimal strategy and thus we have \(\underline{{\mathbf{b}}}_{v_{\pi^\nu_{\mathbf{b}}}}(y)=\underline{{\mathbf{b}}}_{V^\nu}(y)\) and \(\overline{{\mathbf{b}}}_{v_{\pi^\nu_{\mathbf{b}}}}(y)=\overline{{\mathbf{b}}}_{V^\nu}(y)\) for \(y \in E\), which implies \({\mathbf{b}}\in\Xi^\nu_{V^\nu}\). The proof is complete. ◻

Proof of Theorem 6. It suffices to prove that, for each fixed \(a> 0\), \[\begin{align} B_a:=\{y\in E: \underline{{\mathbf{b}}}^\nu(y)>a\} \text{ is finely open with respect to }Y. \label{finely95open} \end{align}\tag{48}\] Fix \(y \in B_a\). Then, we can also choose \(x \in (a, \underline{{\mathbf{b}}}^\nu(y))\) and \(\widetilde{\Omega}\in\mathcal{F}\) such that \({\mathbf{P}}_{(x, y)}(\widetilde{\Omega})=1\) and \(t \mapsto (X_t(\omega), Y_t(\omega))\) and \(t\mapsto \rho^{\nu,\underline{{\mathbf{b}}}^\nu}_{p_{\underline{{\mathbf{b}}}^\nu}} (X_t(\omega), Y_t(\omega))\) are both right-continuous on \([0, \tau^-_{0+})\) for all \(\omega\in\widetilde{\Omega}\) by Lemma 21. By the definitions of \(x\), \(y\) and \(\widetilde{\Omega}\), together with the identity \(\underline{{\mathbf{b}}}^\nu=\underline{{\mathbf{b}}}_{V^\nu}\) and the fact that \(x\mapsto \rho^{\nu, \underline{{\mathbf{b}}}^\nu}_{p_{\underline{{\mathbf{b}}}^\nu}}( x, y)\) is the density of \(x\mapsto V^\nu( x, y)\), we have \(\rho^{\nu, \underline{{\mathbf{b}}}^\nu}_{p_{\underline{{\mathbf{b}}}^\nu}}(x, y)>1\), and for \(\omega\in \widetilde{\Omega}\), we may choose \(\delta>0\) such that \[\begin{align} X_t(\omega)>a, \quad \rho^{\nu,\underline{{\mathbf{b}}}^\nu}_{p_{\underline{{\mathbf{b}}}^{\nu}}} (X_t(\omega), Y_t(\omega))>1 \qquad t \in [0, \delta), \end{align}\] which implies \[\begin{align} \underline{{\mathbf{b}}}^\nu( Y_t(\omega))>a, \text{ and thus }Y_t(\omega)\in B_a, \text{ for }t\in[0, \delta). \end{align}\] Therefore, we have \[\begin{align} {\mathbf{P}}^{Y,(x)}_{y}(T_{ E\backslash B_a}>0) \geq {\mathbf{P}}^{Y,(x)}_{y}(\widetilde{\Omega})=1, \end{align}\] where \(T_A:=\inf\{t>0:Y_t \in A \}\) for \(A\in\mathcal{B}(E)\). This proves 48 . Since 48 holds for any \(a \in \mathbb{R}\), the function \(\underline{{\mathbf{b}}}^\nu\) is finely lower semi-continuous.

The proof of the fine upper semi-continuity of \(\overline{{\mathbf{b}}}^\nu\) is analogous, with the relevant inequalities reversed, and is therefore omitted.

The proof is complete. ◻

Proof of Theorem 23. By the definition of \(\underline{{\mathbf{b}}}_{V^{\frac{1}{2^n}}}\), it suffices to prove that, for \(n\in\mathbb{N}\), \[\begin{align} V^{\frac{1}{2^n}\prime}_{+}(x, y) \leq V^{\frac{1}{2^{n+1}}\prime}_{+}(x, y),\qquad (x,y)\in \mathbb{R}\times E, \label{barrier95result} \end{align}\tag{50}\] where \(V^{\frac{1}{2^n}\prime}_+\) is the right derivative of \(V^{\frac{1}{2^n}}\).

When \(\nu=\delta_{\frac{1}{2^n}}\), we use the notation obtained by replacing \(\nu\) with \(\frac{1}{2^n}\) in the symbols introduced in Section [sec:Optimality95pcb]. In addition, we write \(V^{\frac{1}{2^n}}_{ (k)}\) for \(\mathcal{V}^{\frac{1}{2^n}, (k)} {v_0}\) for \(n\in\mathbb{N}\) and \(k\in\mathbb{N}\cup\{0\}\). Since \(x\mapsto V^{\frac{1}{2^n}}(x, y)\) and \(x\mapsto V^{\frac{1}{2^n}}_{(k)}(x, y)\) are concave for \(n, k\in\mathbb{N}\) and \(y \in E\), Lemma 13, [23] and [24] imply \(\lim_{k\to\infty}{V^{\frac{1}{2^n}\prime}_{(k)+}}(x, y)={V^{\frac{1}{2^n}\prime}_+}(x, y)\) for Lebesgue-a.e. \(x\) for any \(n\in\mathbb{N}\) and \(y \in E\). Therefore, if we show that for \(n \in \mathbb{N}\), \[\begin{align} V^{\frac{1}{2^n}\prime}_{(k)+}(x, y) \leq V^{\frac{1}{2^{n+1}}\prime}_{(2k)+}(x, y),\qquad k\in\mathbb{N}\cup\{0 \},~(x,y)\in \mathbb{R}\times E, \label{barrier95convergence} \end{align}\tag{51}\] then 50 follows. In the following, we prove 51 by induction on \(k\).

For \(k=0\), since both \({V^{\frac{1}{2^n}}_{(0)}}\) and \({V^{\frac{1}{2^{n+1}}}_{(0)}}\) equal \(v_0\), 51 holds. Assuming that 51 holds for \(k=l\in\mathbb{N}\cup\{0\}\), we show that 51 also holds for \(k=l+1\). For \(V^{\frac{1}{2^{n}}}_{(l)}\) and \(V^{\frac{1}{2^{n+1}}}_{(2l)}\), we define \(\widetilde{V}^{\frac{1}{2^{n}}}_{(l)}\) and \(\widetilde{V}^{\frac{1}{2^{n+1}}}_{(2l)}\) in the same way as 18 . Then, it follows that \[\begin{align} \widetilde{V}^{\frac{1}{2^n}\prime}_{(l)+}(x, y) \leq \widetilde{V}^{\frac{1}{2^{n+1}}\prime}_{(2l)+}(x, y),\qquad (x,y)\in \mathbb{R}\times E, \label{ass95inequality} \end{align}\tag{52}\]

For \(f\in\Gamma^{\frac{1}{2^{n+1}}}\) and \((x,y)\in\mathbb{R}\times E\), we define an operator \(\mathcal{W}^{\frac{1}{2^{n+1}}}\) on \(\Gamma^{\frac{1}{2^{n+1}}}\) by \[\begin{align} \mathcal{W}^{\frac{1}{2^{n+1}}} f (x, y)= {\mathbf{E}}_{(x, y)}\!\left[ -\beta\int_{[0,\frac{1}{2^{n+1}}]}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}\!\left( (-\underline{X}_t)\lor 0 \right)+ e^{-{\mathfrak{q}}(\frac{1}{2^{n+1}})} f(X^0_{\frac{1}{2^{n+1}}}, Y_{\frac{1}{2^{n+1}}}) \right]. \end{align}\] Note that the operator \(\mathcal{W}^{\frac{1}{2^{n+1}}}\) coincides with the operator \(\mathcal{V}^{\frac{1}{2^{n+1}}}_\pi\) when \(\pi\) pays no dividends. Focusing on condition (iii) in the definition of the class \(\Gamma^\nu\), a slight extension of ?? shows that \(\Gamma^{\frac{1}{2^n}}\subset \Gamma^{\frac{1}{2^{n+1}}}\) holds, and that \(\mathcal{W}^{\frac{1}{2^{n+1}}}\) can also be applied to the functions in \(\Gamma^{\frac{1}{2^n}}\). By applying almost the same method as in the proof of 23 , we obtain the following expression for the right derivative of \(x \mapsto \mathcal{W}^{\frac{1}{2^{n+1}}}f(x, y)\), denoted by \({(\mathcal{W}^{\frac{1}{2^{n+1}}}f)}^\prime_+\), for \((x,y)\in\mathbb{R}\times E\), \[\begin{align} {(\mathcal{W}^{\frac{1}{2^{n+1}}}f)}^\prime_+&(x ,y ) = {\mathbf{E}}_{(x, y)}\!\left[ e^{-{\mathfrak{q}}(\frac{1}{2^{n+1}}\land\tau^{-}_0)} \!\left( f^\prime_+(X_{\frac{1}{2^{n+1}}}, Y_{\frac{1}{2^{n+1}}})1_{\{\underline{X}_{\frac{1}{2^{n+1}}}\geq 0\}} +\beta 1_{\{\underline{X}_{\frac{1}{2^{n+1}}}<0\}} \right) \right]. \label{147295operator95derivative} \end{align}\tag{53}\] By applying the same method as in the proof of Lemma 10, we can verify that \(\mathcal{W}^{\frac{1}{2^{n+1}}}f\in \Gamma^{\frac{1}{2^{n+1}}}\) for \(f \in \Gamma^{\frac{1}{2^{n+1}}}\). By 23 and 53 , we have, for \(f, g\in \Gamma^{\frac{1}{2^{n+1}}}\) with \(f^\prime_+(x, y)\leq g^\prime_+(x, y)\) for \((x,y)\in\mathbb{R}\times E\), \[\begin{align} {(\mathcal{W}^{\frac{1}{2^{n+1}}}f)}^\prime_+(x ,y )\leq {(\mathcal{V}^{\frac{1}{2^{n+1}}}g)}^\prime_+(x ,y ), \qquad (x,y)\in\mathbb{R}\times E. \label{compare95constant} \end{align}\tag{54}\]

By 52 , 53 and the definition of \(V^{\frac{1}{2^{n+1}}}_{(2l+1)}\), we have \[\begin{align} {(\mathcal{W}^{\frac{1}{2^{n+1}}}\widetilde{V}^{\frac{1}{2^n}}_{(l)})}^\prime_+(x ,y )\leq {(\mathcal{W}^{\frac{1}{2^{n+1}}}\widetilde{V}^{\frac{1}{2^{n+1}}}_{(2l)})}^\prime_+(x ,y ) =V^{\frac{1}{2^{n+1}}\prime}_{(2l+1)+}(x, y), \quad (x, y)\in\mathbb{R}\times E. \label{nagaifutoushiki} \end{align}\tag{55}\] By the Markov property at \(\frac{1}{2^{n+1}}\), we have, for \((x,y)\in\mathbb{R}\times E\), \[\begin{align} V^{\frac{1}{2^n}}_{(l+1)}(x, y)=\mathcal{W}^{\frac{1}{2^{n+1}}}\mathcal{W}^{\frac{1}{2^{n+1}}}\widetilde{V}^{\frac{1}{2^n}}_{(l)}(x, y). \label{3bunkai} \end{align}\tag{56}\] Using 56 , the fact that \(V^{\frac{1}{2^{n+1}}}_{(2l+2)}\) is equal to \(\mathcal{V}^{\frac{1}{2^{n+1}}}V^{\frac{1}{2^{n+1}}}_{(2l+1)}\), 55 and 54 , we have 51 with \(k=l+1\). The proof is complete. ◻

Proof of Lemma 7. We write the resulting controlled process of the strategy \((L^0, R^0)\) as \(U^0:=\{U^0_t : t\geq 0\}\). First, we prove in (1) that for \(\pi\in\underline{\Pi}^\nu\), \[\begin{align} L^\pi_t \leq L^0_t,\qquad R^\pi_t \leq R^0_t,\qquad t\geq 0, \label{0strategybound} \end{align}\tag{59}\] for all \((x,y)\in \mathbb{R}\times E\), \({\mathbf{P}}_{(x, y)}\)-a.s. Then, we obtain ?? since we have \[\begin{align} &\int_{[0,\infty)}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}L^0_t-\int_{[0,\infty)}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}L^\pi_t =\int_0^\infty {\mathbf{q}}(Y_s) e^{-{\mathfrak{q}}(s)} ( L^0_s-L^\pi_s){\mathrm{d}}s \geq 0, \tag{60} \\ &\int_{[0,\infty)}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}R^0_t-\int_{[0,\infty)}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}R^\pi_t =\int_0^\infty {\mathbf{q}}(Y_s) e^{-{\mathfrak{q}}(s)} ( R^0_s-R^\pi_s){\mathrm{d}}s \geq 0, \tag{61} \end{align}\] for all \((x,y)\in\mathbb{R}\times E\), \({\mathbf{P}}_{(x, y)}\)-a.s. Second, we prove ?? in (2).

(1) We prove 59 and \[\begin{align} U^\pi_t \geq U^0_t, \qquad t\geq 0 \label{Ubound} \end{align}\tag{62}\] for all \((x,y)\in\mathbb{R}\times E\), \({\mathbf{P}}_{(x, y)}\)-a.s., for a fixed \(\pi\in\underline{\Pi}^\nu\), by induction. For \(k \in \mathbb{N}\cup\{0\}\), we assume that 59 and 62 hold on \([0, T^\nu_k)\), and prove that 59 and 62 hold on \([T^\nu_k, T^\nu_{k+1})\). By the assumption and 8 , we have \[\begin{align} R^\pi_{T^\nu_k}=R^\pi_{T^\nu_k-} -\!\left( U^\pi_{(k-)}\land 0 \right) \leq R^0_{T^\nu_k-} -\!\left\{ (U^0_{T^\nu_k-}+\Delta X_{T^\nu_k} )\land 0 \right\} =R^0_{T^\nu_k}, \label{Lem3039501} \end{align}\tag{63}\] for all \((x,y)\in\mathbb{R}\times E\), \({\mathbf{P}}_{(x, y)}\)-a.s., where \(U^\pi_{(0-)}\) and \(\Delta X_0\) are set equal to \(X_0\), and \(R^\pi_{0-}\), \(R^0_{0-}\) and \(U^0_{0-}\) are equal to \(0\). By the definition of \(U^0\), we have \[\begin{align} U^\pi_{T^\nu_k}\geq 0 = U^0_{T^\nu_k},\label{Lem3039502} \end{align}\tag{64}\] for all \((x,y)\in\mathbb{R}\times E\), \({\mathbf{P}}_{(x, y)}\)-a.s. By 63 and 64 , we have \[\begin{align} L^\pi_{T^\nu_k} =X_{T^\nu_k}+R^\pi_{T^\nu_k}-U^\pi_{T^\nu_k} \leq X_{T^\nu_k}+R^0_{T^\nu_k}-U^0_{T^\nu_k} =L^0_{T^\nu_k}, \end{align}\] for all \((x,y)\in\mathbb{R}\times E\), \({\mathbf{P}}_{(x, y)}\)-a.s., and thus 59 and 62 hold at \(T^\nu_k\). By the definition of \(L^0\), \(R^0\) and \(\pi\) and since 59 and 62 hold at \(T^\nu_k\), we have, for \(t \in (T^\nu_k, T^\nu_{k+1})\), \[\begin{align} L^\pi_t =L^\pi_{T^\nu_k}\leq L^0_{T^\nu_k} =L^0_t, \end{align}\] \[\begin{align} R^\pi_t =R^\pi_{T^\nu_k}-\!\left\{ \!\left( U^\pi_{T^\nu_k}+\inf_{s \in [T^\nu_k, t]}(X_s-X_{T^\nu_k}) \right)\land 0 \right\} \leq R^0_{T^\nu_k}-\!\left\{ \!\left( \inf_{s \in [T^\nu_k, t]}(X_s-X_{T^\nu_k}) \right)\land 0 \right\} =R^0_{t}, \end{align}\] and \[\begin{align} U^\pi_t=U^\pi_{T^\nu_k} +(X_t-X_{T^\nu_k}) -&\!\left\{ \!\left( U^\pi_{T^\nu_k}+\inf_{s \in [T^\nu_k, t]}(X_s-X_{T^\nu_k}) \right)\land 0 \right\}\\ &\geq (X_t-X_{T^\nu_k})-\!\left\{ \!\left( \inf_{s \in [T^\nu_k, t]}(X_s-X_{T^\nu_k}) \right)\land 0 \right\} =U^0_t, \end{align}\] for all \((x,y)\in\mathbb{R}\times E\), \({\mathbf{P}}_{(x, y)}\)-a.s. Thus, 59 and 62 also hold for \(t \in (T^\nu_k, T^\nu_{k+1})\).

(2) Based on the preceding argument, it suffices to establish the existence of \(B^\nu\) that satisfies ?? for \((x, y)\in\mathbb{R}\times E\) with \(L^\pi_t\) and \(R^\pi_t\) replaced by \(L^0_t\) and \(R^0_t\), respectively. By the definition of \(L^0\), 6 and 1 , we have, for \((x, y)\in\mathbb{R}\times E\), \[\begin{align} &{\mathbf{E}}_{(x, y)}\!\left[ \int_{[0,\infty)}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}L^0_t \right] \\ &=x\lor0+ \sum_{k\in\mathbb{N}}{\mathbf{E}}_{(x, y)}\!\left[ e^{-{\mathfrak{q}}(T^\nu_k)} \!\left\{ \!\left( X_{T^\nu_k}-X_{T^\nu_{k-1}} \right)-\inf_{s \in [T^\nu_{k-1},T^\nu_{k}]} \!\left( X_{s}-X_{T^\nu_{k-1}} \right) \right\} \right] \\ &\leq x\lor0+ \sum_{k\in\mathbb{N}}{\mathbf{E}}_{(x, y)}\!\left[ e^{-{\mathfrak{q}}(T^\nu_{k-1})} {\mathbf{E}}_{(X_{T^\nu_{k-1}}, Y_{T^\nu_{k-1}})}\!\left[ e^{-q T^\nu}\!\left( \!\left( X_{T^\nu}-X_0 \right)-\inf_{s \in [0,T^\nu]} \!\left( X_{s}-X_{0} \right) \right) \right] \right] \\ &=x\lor0+ \sum_{k\in\mathbb{N}}{\mathbf{E}}_{(x, y)}\!\left[ e^{-{\mathfrak{q}}(T^\nu_{k-1})} {\mathbf{E}}_{(0, Y_{T^\nu_{k-1}})}\!\left[ e^{-q T^\nu}\!\left( X_{T^\nu}- \underline{X}_{T^\nu} \right) \right] \right]\\ &\leq x \lor0+\sum_{k\in\mathbb{N}} \!\left( \int_0^\infty e^{-qt} \nu ({\mathrm{d}}t) \right)^{k-1} M_L, \label{bound95L} \end{align}\tag{65}\] where \(\underline{X}_t:= \inf_{s \in[0, t]} X_s\) for \(t\geq 0\) and \(M_L:= \sup_{y \in E} {\mathbf{E}}_{(0, y)}\!\left[ e^{-q T^\nu}\!\left( X_{T^\nu}- \underline{X}_{T^\nu} \right) \right]\). By a similar computation, we have, for \((x, y)\in\mathbb{R}\times E\), \[\begin{align} {\mathbf{E}}_{(x, y)}\!\left[ \int_{[0,\infty)}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}R^0_t \right] &\leq -(x\land 0)+\sum_{k\in\mathbb{N}}{\mathbf{E}}_{(x, y)}\!\left[ e^{-{\mathfrak{q}}(T^\nu_{k-1})} {\mathbf{E}}_{(0, Y_{T^\nu_{k-1}})}\!\left[ \int_{(0, T^\nu]} e^{-qt} {\mathrm{d}}\!\left( -\underline{X}_t \right) \right] \right] \\ &\leq -(x\land 0)+\sum_{k\in\mathbb{N}} \!\left( \int_0^\infty e^{-qt} \nu ({\mathrm{d}}t) \right)^{k-1} M_R, \label{bound95R} \end{align}\tag{66}\] where \(M_R:= \sup_{y \in E} {\mathbf{E}}_{(0, y)}\!\left[ \int_{(0, T^\nu]} e^{-qt} {\mathrm{d}}\!\left( -\underline{X}_{t} \right) \right]\). Based on the above discussion, it suffices to prove \(M_L<\infty\) and \(M_R<\infty\) and define \(B^\nu:=(M_L\lor M_R)\sum_{k\in\mathbb{N}} \!\left( \int_0^\infty e^{-qt} \nu ({\mathrm{d}}t) \right)^{k-1}\). By the Markov property and 2 , we have, for \(y \in E\), \[\begin{align} {\mathbf{E}}_{(0, y)}\!\left[ e^{-q T^\nu}\!\left( X_{T^\nu}-\underline{X}_{T^\nu} \right) \right] &\leq 2 {\mathbf{E}}_{(0, y)}\!\left[ e^{-q T^\nu} \sup_{s \in [0 , T^\nu]}|X_s| \right] \\ &= 2\sum_{k\in\mathbb{N}} {\mathbf{E}}_{(0, y)}\!\left[ e^{-q T^\nu} \sup_{s \in [0 , T^\nu]}|X_s|; T^\nu\in (k-1, k] \right] \\ &\leq 2\sum_{k\in\mathbb{N}} e^{-q(k-1)} {\mathbf{E}}_{(0, y)}\!\left[ \sup_{s \in [0 , k]}|X_s| \right] \nu((k-1, k]) \\ &\leq 2\sum_{k\in\mathbb{N}} e^{-q(k-1)} kM \nu((k-1, k]) \leq2M\frac{e^{q-1}}{q} \nu ((0, \infty)) < \infty, \label{upper95sup} \end{align}\tag{67}\] which implies \(M_L<\infty\). Similarly, by the Markov property and 2 , we have, for \(y \in E\), \[\begin{align} {\mathbf{E}}_{(0, y)}\!\left[ \int_{(0, T^\nu]} e^{-qt} {\mathrm{d}}\!\left( -\underline{X}_{t} \right) \right] &\leq \sum_{k\in\mathbb{N}} {\mathbf{E}}_{(0, y)}\!\left[ \int_{(k-1, k]} e^{-qt} {\mathrm{d}}\!\left( -\inf_{s \in [k-1, t]}(X_s-X_{k-1}) \right) \right] \\ &\leq\sum_{k\in\mathbb{N}} e^{-q(k-1)} \sup_{y\in E} {\mathbf{E}}_{(0, y)}\!\left[ \int_{(0, 1]} e^{-qt} {\mathrm{d}}\!\left( -\underline{X}_t \right) \right] \\ &\leq M \sum_{k\in\mathbb{N}} e^{-q(k-1)} =M\frac{1}{1-e^{-q}} < \infty, \end{align}\] which implies \(M_R<\infty\). The proof is complete. ◻

Proof of Lemma 12. For \(n \in\mathbb{N}\) and \((x, y)\in\mathbb{R}\times E\), we have \[\begin{align} &\left| v_\pi(x, y)-v_{\pi_n}(x, y) \right|\\ &=\left| {\mathbf{E}}_{(x, y)}\!\left[ \int_{[T^\nu_{n+1}, \infty)}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}L^{\pi}_t -\beta\int_{[T^\nu_{n+1}, \infty)}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}R^{\pi}_t+\beta\int_{[T^\nu_{n+1}, \infty)}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}R^{\pi_n}_t \right] \right|\\ & \begin{aligned} \leq&{ {\mathbf{E}}_{(x, y)}\!\left[ \int_{[T^\nu_{n+1}, \infty)}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}L^{\pi}_t \right]} +\beta{ {\mathbf{E}}_{(x, y)}\!\left[ \int_{[T^\nu_{n+1}, \infty)}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}R^{\pi}_t \right]}+\beta{ {\mathbf{E}}_{(x, y)}\!\left[ \int_{[T^\nu_{n+1}, \infty)}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}R^{\pi_n}_t \right]}, \end{aligned} \label{Lem30895001} \end{align}\tag{68}\] where the equality follows from 24 . By ?? , 66 and the dominated convergence theorem, we have \[\begin{align} {\mathbf{E}}_{(x, y)}\!\left[ \int_{[T^\nu_{n+1}, \infty)}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}R^{\pi_n}_t \right] &\leq{\mathbf{E}}_{(x, y)}\!\left[ \int_{[T^\nu_{n+1}, \infty)}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}R^{0}_t \right] \to 0 \qquad\text{ as }n\to\infty . \label{Lem30895002} \end{align}\tag{69}\] By ?? and the dominated convergence theorem, we have \[\begin{align} {\mathbf{E}}_{(x, y)}\!\left[ \int_{[T^\nu_{n+1}, \infty)}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}L^{\pi}_t \right]& \to 0\quad\text{ as }n\to \infty, \tag{70} \\ {\mathbf{E}}_{(x, y)}\!\left[ \int_{[T^\nu_{n+1}, \infty)}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}R^{\pi}_t \right]& \to 0\quad\text{ as }n\to \infty. \tag{71} \end{align}\] By 68 , 69 , 70 and 71 , the proof is complete. ◻

Proof of (\(\Rightarrow\)) in Lemma 14. We show that strategies in \(\underline{\Pi}^\nu\) not included in \(\Pi^{\nu,\ast}_{V^\nu}\) are not optimal. Assume that a strategy \(\pi\in \underline{\Pi}^\nu\) does not belong to \(\Pi^{\nu,\ast}_{V^\nu}\). We define \[\begin{align} n_{\neq} := \min\!\left\{ k\in\mathbb{N}: 1_{\{U^\pi_{T^\nu_k}> \overline{{\mathbf{b}}}_{V^\nu}(Y_{T^\nu_k})\}}+ 1_{\{U^\pi_{T^\nu_k}< \underline{{\mathbf{b}}}_{V^\nu}(Y_{T^\nu_k})\}}1_{\{\Delta L^\pi_{T^\nu_k}>0\}}>0 \right\}. \end{align}\] By the definition of \(\pi\), \(T^\nu_{n_{\neq}}<\infty\) with positive probability for some \({\mathbf{P}}_{(x, y)}\) with \((x, y)\in\mathbb{R}\times E\). We fix such a \((x, y)\in\mathbb{R}\times E\) and prove that \[\begin{align} v_\pi(x, y)<V^\nu(x, y). \label{Sec00B95005} \end{align}\tag{72}\] We consider which continuation strategy is optimal after following \(\pi\) up to time \(T^\nu_{n_{\neq}}\). First, we assume that dividend payments according to \(\pi\) are allowed only at the next \(k\in\mathbb{N}\) opportunities after \(T^\nu_{n_{\neq}}\), that no dividends are paid thereafter, and that capital injections are made by reflection at \(0\). Then, the expected NPV of dividends and capital injections is \[\begin{align} u_\pi(k, 0):= {\mathbf{E}}_{(x, y)}&\Bigg[\int_{[0,T^\nu_{n_{\neq}+k-1}]}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}L^\pi_t-\beta\int_{[0,T^\nu_{n_{\neq}+k}]}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}R^\pi_t\\ &+e^{-{\mathfrak{q}}(T^\nu_{n_{\neq}+k})} \!\left( \Delta L^\pi_{T^\nu_{n_{\neq}+k}}+v_0(U^{\pi}_{T^\nu_{n_{\neq}+k}}, Y_{T^\nu_{n_{\neq}+k}}) \right)\Bigg]. \end{align}\] By 16 , \(u_\pi(k, 0)\) is not greater than \[\begin{align} u_\pi&(k, 1)\\ &:= {\mathbf{E}}_{(x, y)}\!\left[ \int_{[0,T^\nu_{n_{\neq}+k-1}]}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}L^\pi_t-\beta\int_{[0,T^\nu_{n_{\neq}+k}]}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}R^\pi_t +e^{-{\mathfrak{q}}(T^\nu_{n_{\neq}+k})} \widetilde{v}_0(U^{\pi}_{({n_{\neq}+k}-)}, Y_{T^\nu_{n_{\neq}+k}}) \right]\\ &={\mathbf{E}}_{(x, y)}\Bigg[\int_{[0,T^\nu_{n_{\neq}+k-2}]}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}L^\pi_t-\beta\int_{[0,T^\nu_{n_{\neq}+k-1}]}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}R^\pi_t\\ &\qquad+e^{-{\mathfrak{q}}(T^\nu_{n_{\neq}+k-1})} \!\left( \Delta L^{\pi}_{T^\nu_{n_{\neq}+k-1}}+\mathcal{V}^{\nu,(1)}v_0(U^{\pi}_{T^\nu_{n_{\neq}+k-1}}, Y_{T^\nu_{n_{\neq}+k-1}}) \right)\Bigg], \end{align}\] where the last equality follows from 6 and 19 . After repeating the same argument \(k-1\) more times, we find that \(u_\pi(k, 0)\) is not greater than \[\begin{align} u_\pi(k, k):={\mathbf{E}}_{(x, y)}&\Bigg[\int_{[0,T^\nu_{n_{\neq}-1}]}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}L^\pi_t-\beta\int_{[0,T^\nu_{n_{\neq}}]}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}R^\pi_t\\ &+e^{-{\mathfrak{q}}(T^\nu_{n_{\neq}})} \!\left( \Delta L^{\pi}_{T^\nu_{n_{\neq}}}+\mathcal{V}^{\nu,(k)}v_0(U^{\pi}_{T^\nu_{n_{\neq}}}, Y_{T^\nu_{n_{\neq}}}) \right)\Bigg]. \end{align}\] By taking the limit as \(k\to\infty\), Lemmas 12, Fatou’s lemma and Lemma 13, we have \[\begin{align} v_\pi(x,y)= \lim_{k\to\infty}u_\pi(k, 0)\leq \limsup_{k\to\infty}u_\pi(k, k) \leq u_\pi(\infty), \label{Sec00B95001} \end{align}\tag{73}\] where \[\begin{align} u_\pi(\infty):={\mathbf{E}}_{(x, y)}&\Bigg[\int_{[0,T^\nu_{n_{\neq}-1}]}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}L^\pi_t-\beta\int_{[0,T^\nu_{n_{\neq}}]}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}R^\pi_t\\ &+e^{-{\mathfrak{q}}(T^\nu_{n_{\neq}})} \!\left( \Delta L^{\pi}_{T^\nu_{n_{\neq}}}+V^\nu(U^{\pi}_{T^\nu_{n_{\neq}}}, Y_{T^\nu_{n_{\neq}}}) \right)\Bigg]. \end{align}\] By the definition of \(T^\nu_{n_{\neq}}\) and recalling 15 , we have on \(\{T^\nu_{n_{\neq}} <\infty\}\), \[\begin{align} \widetilde{V}^\nu(U^{\pi}_{({n_{\neq}}-)}, Y_{T^\nu_{n_{\neq}}})& = g^{U^{\pi}_{(n_{\neq}-)} }_{V^\nu}((U^{\pi}_{(n_{\neq}-)} -\underline{{\mathbf{b}}}_{V^\nu} (Y_{T^\nu_{n_{\neq}}}))\lor 0, Y_{T^\nu_{n_{\neq}}}) \\ &> g^{U^{\pi}_{(n_{\neq}-)} }_{V^\nu}(\Delta L^{\pi}_{T^\nu_{n_{\neq}}}, Y_{T^\nu_{n_{\neq}}}) =\Delta L^{\pi}_{T^\nu_{n_{\neq}}}+V^\nu(U^{\pi}_{T^\nu_{n_{\neq}}}, Y_{T^\nu_{n_{\neq}}}). \label{Sec00B95002} \end{align}\tag{74}\] Since \({\mathbf{P}}_{(x,y)}(T^\nu_{n_{\neq}}<\infty)>0\) and by 73 and 74 , we have \[\begin{align} v_\pi(x,y)<{\mathbf{E}}_{(x, y)}&\Bigg[\int_{[0,T^\nu_{n_{\neq}-1}]}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}L^\pi_t-\beta\int_{[0,T^\nu_{n_{\neq}}]}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}R^\pi_t +e^{-{\mathfrak{q}}(T^\nu_{n_{\neq}})} \widetilde{V}^\nu(U^{\pi}_{({n_{\neq}}-)}, Y_{T^\nu_{n_{\neq}}})\Bigg]. \;\;\;\; \label{Sec00B95003} \end{align}\tag{75}\] We define a strategy \(\pi^\prime\) as the strategy that pays according to \(\pi\) up to just before time \(T^\nu_{n_{\neq}}\), and according to the MMPCB strategy with barrier \(\overline{{\mathbf{b}}}_{V^\nu}\) from time \(T^\nu_{n_{\neq}}\). Then, using the definitions of \(\overline{{\mathbf{b}}}_{V^\nu}\) and \(\widetilde{V}^\nu\), 6 and the fact that \(\pi^\prime\) belongs to \(\Pi^{\nu,\ast}_{V^\nu}\), we have \[\begin{align} {\mathbf{E}}_{(x, y)}&\!\left[ \int_{[0,T^\nu_{n_{\neq}-1}]}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}L^\pi_t-\beta\int_{[0,T^\nu_{n_{\neq}}]}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}R^\pi_t +e^{-{\mathfrak{q}}(T^\nu_{n_{\neq}})} \widetilde{V}^\nu(U^{\pi}_{(n_{\neq}-)}, Y_{T^\nu_{n_{\neq}}}) \right] \\ &={\mathbf{E}}_{(x, y)}\!\left[ \int_{[0,T^\nu_{n_{\neq}}]}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}L^{\pi^\prime}_t-\beta\int_{[0,T^\nu_{n_{\neq}}]}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}R^{\pi^\prime}_t +e^{-{\mathfrak{q}}(T^\nu_{n_{\neq}})} V^\nu(U^{\pi^\prime}_{T^\nu_{n_{\neq}}}, Y_{T^\nu_{n_{\neq}}}) \right]\\ &=\sum_{k\in\mathbb{N}}{\mathbf{E}}_{(x, y)}\!\left[ \int_{[0,T^\nu_{k}]}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}L^{\pi^\prime}_t-\beta\int_{[0,T^\nu_{k}]}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}R^{\pi^\prime}_t+e^{-{\mathfrak{q}}(T^\nu_{k})} V^\nu(U^{\pi^\prime}_{T^\nu_{k}}, Y_{T^\nu_{k}});n_{\neq}=k \right]\\ &\;\;+{\mathbf{E}}_{(x, y)}\!\left[ \int_{[0,\infty)}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}L^{\pi^\prime}_t-\beta\int_{[0,\infty)}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}R^{\pi^\prime}_t;n_{\neq}=\infty \right]\\ &=\sum_{k\in\mathbb{N}}{\mathbf{E}}_{(x, y)}\!\left[ \int_{[0,\infty)}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}L^{\pi^\prime}_t-\beta\int_{[0,\infty)}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}R^{\pi^\prime}_t;n_{\neq}=k \right]\\ &\;\;+{\mathbf{E}}_{(x, y)}\!\left[ \int_{[0,\infty)}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}L^{\pi^\prime}_t-\beta\int_{[0,\infty)}e^{-{\mathfrak{q}}(t)} {\mathrm{d}}R^{\pi^\prime}_t;n_{\neq}=\infty \right]\\ &=v_{\pi^\prime}(x, y)= V^\nu (x, y). \label{Sec00B95004} \end{align}\tag{76}\] By 75 and 76 , we obtain 72 . Hence \(\pi\) is not optimal. The proof is complete. ◻

Proof of Lemma 15. Fix a non-negative measurable function \({\mathbf{b}}\) on \(E\) and \((x, y)\in\mathbb{R}\times E\). Using \((\sharp)\), we have \[\begin{align} \varepsilon{\mathbf{E}}_{(x,y )}\!\left[ e^{-{\mathfrak{q}}({T}^{ (0), +}_{{\mathbf{b}}-})} ;{T}^{ (0), +}_{{\mathbf{b}}-} \leq {\tau}^{ (0), -}_0 \right] \leq&{\mathbf{E}}_{(x,y )}\!\left[ \int_{[0,\infty)} e^{-{\mathfrak{q}}(t)} {\mathrm{d}}(L^{{\pi}^{(\varepsilon)}_{{\mathbf{b}}}}_t-L^{{\pi}^{(0)}_{{\mathbf{b}}}}_t) \right]\\ \leq& \varepsilon{\mathbf{E}}_{(x,y )}\!\left[ e^{-{\mathfrak{q}}({T}^{ (\varepsilon), +}_{\mathbf{b}})} ;{T}^{ (\varepsilon), +}_{\mathbf{b}}< {\tau}^{ (\varepsilon), -}_{0+} \right], \end{align}\] and \[\begin{align} -\varepsilon{\mathbf{E}}_{(x,y )}\!\left[ e^{-{\mathfrak{q}}({\tau}^{ (0), -}_0)} ;{\tau}^{ (0), -}_0<{T}^{ (0), +}_{{\mathbf{b}}-} \right] &\leq {\mathbf{E}}_{(x,y )}\!\left[ \int_{(0,\infty)} e^{-{\mathfrak{q}}(t)} {\mathrm{d}}( R^{{\pi}^{(\varepsilon)}_{{\mathbf{b}}}}_t-R^{{\pi}^{(0)}_{{\mathbf{b}}}}_t) \right]\\ &\leq -\varepsilon{\mathbf{E}}_{(x,y )}\!\left[ e^{-{\mathfrak{q}}({\tau}^{ (\varepsilon), -}_{0+})} ;{\tau}^{ (\varepsilon), -}_{0+}< {T}^{ (\varepsilon), +}_{\mathbf{b}} \right] . \end{align}\] Therefore, we have, for \((x, y) \in \mathbb{R}\times E\), \[\begin{align} v^\prime_{{\pi}^{\nu}_{{\mathbf{b}}}+}(x, y)&=\lim_{\varepsilon\downarrow0} \frac{v_{{\pi}^{\nu}_{{\mathbf{b}}}}(x+\varepsilon, y)-v_{{\pi}^{\nu}_{{\mathbf{b}}}}(x, y)}{\varepsilon} =\lim_{\varepsilon\downarrow0} \frac{v_{{\pi}^{(\varepsilon)}_{{\mathbf{b}}}}(x, y)-v_{{\pi}^{(0)}_{{\mathbf{b}}}}(x, y)}{\varepsilon}\\ &=\lim_{\varepsilon\downarrow0} \frac{1}{\varepsilon} {\mathbf{E}}_{(x,y )}\!\left[ \int_{[0,\infty)} e^{-{\mathfrak{q}}(t)} {\mathrm{d}}( L^{{\pi}^{(\varepsilon)}_{{\mathbf{b}}}}_t-L^{{\pi}^{(0)}_{{\mathbf{b}}}}_t) -\beta \int_{[0,\infty)} e^{-{\mathfrak{q}}(t)} {\mathrm{d}}( R^{{\pi}^{(\varepsilon)}_{{\mathbf{b}}}}_t- R^{{\pi}^{(0)}_{{\mathbf{b}}}}_t) \right] \\ &={\mathbf{E}}_{(x,y )}\!\left[ e^{-{\mathfrak{q}}({T}^{ \nu, +}_{{\mathbf{b}}-})} ;{T}^{ \nu, +}_{{\mathbf{b}}-} \leq {\tau}^{ -}_0 \right]+ \beta{\mathbf{E}}_{(x,y )}\!\left[ e^{-{\mathfrak{q}}({\tau}^{ -}_0)} ;{\tau}^{ -}_0<{T}^{ \nu, +}_{{\mathbf{b}}-} \right]. \end{align}\] The proof is complete. ◻

Proof of Lemma 16. By the version of Corollary 5 in which \(\widehat{\Xi}^\nu\) is replaced by \(\widehat{\Xi}^\nu_{V^\nu}\), the MMPCB strategy with barrier \(\underline{{\mathbf{b}}}_{V^\nu}\) is optimal. By this fact and ?? , we have, for \((x, y) \in [0, \infty) \times E\), \[\begin{align} V^{\nu \prime}_+(x, y) =&\lim_{\varepsilon\downarrow0}\frac{V^{\nu}(x+\varepsilon, y)-V^{\nu}(x, y)}{\varepsilon} =\lim_{\varepsilon\downarrow0}\frac{v_{\pi^\nu_{\underline{{\mathbf{b}}}_{V^\nu}} }(x+\varepsilon, y)-v_{\pi^\nu_{\underline{{\mathbf{b}}}_{V^\nu}} }(x, y)}{\varepsilon}\\ =&{\mathbf{E}}_{(x, y)}\!\left[ e^{-{\mathfrak{q}}(T^{\nu,+}_{\underline{{\mathbf{b}}}_{V^\nu}-})};T^{\nu,+}_{\underline{{\mathbf{b}}}_{V^\nu}-}\leq\tau^-_0 \right] +\beta {\mathbf{E}}_{(x, y)}\!\left[ e^{-{\mathfrak{q}}(\tau^-_0)};\tau^-_0<T^{\nu,+}_{\underline{{\mathbf{b}}}_{V^\nu}-} \right]\\ \leq &{\mathbf{E}}\!\left[ e^{-q T^\nu} \right] +\beta {\mathbf{E}}_{(x, y)}\!\left[ e^{-q\tau^-_0} \right]. \label{Lem31295001} \end{align}\tag{77}\] By 77 , 1 and the dominated convergence theorem, we have, for \(y \in E\), \[\begin{align} \limsup_{x\to\infty}V^{\nu \prime}_+(x, y)\leq {\mathbf{E}}\!\left[ e^{-q T^\nu} \right]<1, \end{align}\] and thus \(\overline{{\mathbf{b}}}_{V^\nu}(y)<\infty\). The proof is complete. ◻

Proof of Lemma 21. We define the process \(M:=\{M_t:t\geq 0\}\) as \[\begin{align} M_t:= e^{-{\mathfrak{q}}(t)-rt }1_{\{t<\tau^-_{0+} \}},\qquad t\geq 0. \end{align}\] Then, \(M\) is a decreasing exact multiplicative functional (for the definition of a decreasing exact multiplicative functional, see, e.g., [21]). Let us define, for any non-negative measurable function \(f\), \(t,\alpha\geq 0\), and \((x,y)\in(0, \infty)\times E\), \[\begin{align} Q_t f(x, y):={\mathbf{E}}_{(x, y)}\!\left[ f(X_t, Y_t)M_t \right], \qquad \end{align}\] and \[\begin{align} U^{(\alpha)}f(x, y):={\mathbf{E}}_{(x, y)}\!\left[ \int_0^\infty e^{-\alpha t}f(X_t, Y_t)M_t {\mathrm{d}}t \right]. \end{align}\] Then, \(\{Q_t: t\geq 0\}\) is a sub-Markov semigroup with resolvent \(\{U^{(\alpha)}: \alpha\geq 0\}\) by [21].

The function \(\rho^{\nu, {\mathbf{b}}}_{p}\) is excessive with respect to \(\{Q_t: t\geq 0\}\) since for \((x,y)\in(0, \infty)\times E\) and \(t\geq 0\), we have \[\begin{align} Q_t \rho^{\nu, {\mathbf{b}}}_{p}(x, y)= & {\mathbf{E}}_{(x, y)}\!\left[ e^{-{\mathfrak{q}}(t)-rt}1_{\{t< \tau^-_{0+}\}}\rho^{\nu, {\mathbf{b}}}_{p}(X_t, Y_t) \right] = {\mathbf{E}}_{(x, y)}\!\left[ e^{-{\mathfrak{q}}(t)}1_{\{t< T^\nu\land\tau^-_{0+}\}}\rho^{\nu, {\mathbf{b}}}_{p}(X_t, Y_t) \right] \\ =&\beta{\mathbf{E}}_{(x, y)}\!\left[ e^{-{\mathfrak{q}}(\kappa^{\nu,{\mathbf{b}}}_{p,0} )}; t<T^\nu \land \tau^-_{0+} \right], \end{align}\] by the Markov property at \(t\) and the memoryless property of \(T^\nu\), and thus \[\begin{align} Q_t \rho^{\nu, {\mathbf{b}}}_{p}(x, y)\leq \rho^{\nu, {\mathbf{b}}}_{p}(x, y),\qquad t\geq 0, \end{align}\] and \[\begin{align} \lim_{t\downarrow0}Q_t \rho^{\nu, {\mathbf{b}}}_{p}(x, y)=\lim_{t\downarrow0} \beta{\mathbf{E}}_{(x, y)}\!\left[ e^{-{\mathfrak{q}}(\kappa^{\nu,{\mathbf{b}}}_{p,0} )}; t<T^\nu \land \tau^-_{0+} \right]= \rho^{\nu, {\mathbf{b}}}_{p}(x, y). \end{align}\] Therefore, the function \(\rho^{\nu, {\mathbf{b}}}_{p}\) is excessive with respect to \(\{U^{(\alpha)}: \alpha\geq 0\}\), so by (iv) of (56.13) in [21], the map \(t\mapsto \rho^{\nu, {\mathbf{b}}}_{p} (X_t, Y_t)\) is right-continuous on \([0, \tau^-_{0+})\), \({\mathbf{P}}_{(x, y)}\)-a.s. for any \((x, y)\in(0, \infty)\times E\). The proof is complete. ◻

Proof of Lemma 18. We fix \(x_1, x_2 \in\mathbb{R}\) with \(x_1< x_2\). We denote by \(\kappa^{(x),{\mathbf{b}}}_{p, 0}\) the counterpart of \(\kappa^{\nu,{\mathbf{b}}}_{p, 0}\) for \(X^{(x)}\) with \(x \in \mathbb{R}\). Here, we compare \(\rho_{p}^{\nu,{\mathbf{b}}}( x_1, y)=\beta{\mathbf{E}}_{(0, y)}\!\left[ e^{-{\mathfrak{q}}(\kappa^{(x_1),{\mathbf{b}}}_{p,0})} \right]\) and \(\rho_{p}^{\nu,{\mathbf{b}}}( x_2, y)=\beta{\mathbf{E}}_{(0, y)}\!\left[ e^{-{\mathfrak{q}}(\kappa^{(x_2),{\mathbf{b}}}_{p,0})} \right]\). By (\(\sharp\)), we have \(U^{{\pi}^{(x_1)}_{{\mathbf{b}}}}_t\leq U^{{\pi}^{(x_2)}_{{\mathbf{b}}}}_t\) and \(R^{{\pi}^{(x_1)}_{{\mathbf{b}}}}_t\geq R^{{\pi}^{(x_2)}_{{\mathbf{b}}}}_t\) for \(t \geq 0\) and \(U^{{\pi}^{(x_1)}_{{\mathbf{b}}}}_t- U^{{\pi}^{(x_2)}_{{\mathbf{b}}}}_t\) and \(R^{{\pi}^{(x_1)}_{{\mathbf{b}}}}_t- R^{{\pi}^{(x_2)}_{{\mathbf{b}}}}_t\) are non-decreasing, \({\mathbf{P}}_{(0, y)}\)-a.s.

On the event \(\{\kappa^{(x_1),{\mathbf{b}}}_{p,0} \geq \kappa^{(x_2),{\mathbf{b}}}_{p,0}\}\), the following three cases exhaust all possibilities \({\mathbf{P}}_{(0,y)}\)-a.s.:

1. If \(U^{{\pi}^{(x_1)}_{{\mathbf{b}}}}_{\kappa^{(x_2),{\mathbf{b}}}_{p,0}-} < U^{{\pi}^{(x_2)}_{{\mathbf{b}}}}_{\kappa^{(x_2),{\mathbf{b}}}_{p,0}-}\), then \(R^{{\pi}^{(x_1)}_{{\mathbf{b}}}}\) increases on \([\kappa^{(x_2),{\mathbf{b}}}_{p,0},\kappa^{(x_2),{\mathbf{b}}}_{p,0}+\varepsilon)\) for any \(\varepsilon>0\), and hence we obtain \(\kappa^{(x_1),{\mathbf{b}}}_{p,0} \leq \kappa^{(x_2),{\mathbf{b}}}_{p,0}\).

2. If \(U^{{\pi}^{(x_1)}_{{\mathbf{b}}}}_{\kappa^{(x_2),{\mathbf{b}}}_{p,0}-} = U^{{\pi}^{(x_2)}_{{\mathbf{b}}}}_{\kappa^{(x_2),{\mathbf{b}}}_{p,0}-}\) and \(R^{{\pi}^{(x_2)}_{{\mathbf{b}}}}\) increases on \([\kappa^{(x_2),{\mathbf{b}}}_{p,0},\kappa^{(x_2),{\mathbf{b}}}_{p,0}+\varepsilon)\) for any \(\varepsilon>0\), then \(R^{{\pi}^{(x_1)}_{{\mathbf{b}}}}\) also increases on such intervals in the same manner as \(R^{{\pi}^{(x_2)}_{{\mathbf{b}}}}\). Hence \(\kappa^{(x_1),{\mathbf{b}}}_{p,0} \leq \kappa^{(x_2),{\mathbf{b}}}_{p,0}\).

3. If \(U^{{\pi}^{(x_1)}_{{\mathbf{b}}}}_{\kappa^{(x_2),{\mathbf{b}}}_{p,0}-} = U^{{\pi}^{(x_2)}_{{\mathbf{b}}}}_{\kappa^{(x_2),{\mathbf{b}}}_{p,0}-}\) and \(R^{{\pi}^{(x_2)}_{{\mathbf{b}}}}\) does not increase on \([\kappa^{(x_2),{\mathbf{b}}}_{p,0},\kappa^{(x_2),{\mathbf{b}}}_{p,0}+\varepsilon)\) for some \(\varepsilon>0\), then \(L^{{\pi}^{(x_2)}_{{\mathbf{b}}}}\) must have increased at some time \(\mu\in (0, \kappa^{(x_2),{\mathbf{b}}}_{p,0})\) for \(U^{{\pi}^{(x_1)}_{{\mathbf{b}}}}\) and \(U^{{\pi}^{(x_2)}_{{\mathbf{b}}}}\) to coincide. Thereafter, the same value of \(\Lambda^p\) is used for both processes when deciding whether the hitting time of \(0\) is included in the corresponding \(\kappa^{(\cdot),{\mathbf{b}}}_{p,0}\). Hence \(\kappa^{(x_1),{\mathbf{b}}}_{p,0} \leq \kappa^{(x_2),{\mathbf{b}}}_{p,0}\).

It follows from these cases that \(\kappa^{(x_1),{\mathbf{b}}}_{p,0} \leq \kappa^{(x_2),{\mathbf{b}}}_{p,0}\) holds \({\mathbf{P}}_{(0,y)}\)-a.s., and therefore \(\rho_{p}^{\nu,{\mathbf{b}}}( x_1, y)\geq \rho_{p}^{\nu,{\mathbf{b}}}( x_2, y)\) holds for \(y \in E\). The proof is complete. ◻

Proof of Lemma 19. We want to prove that there exists a measurable function \(p_{\mathbf{b}}\) from \(E\) to \([0, 1]\) such that the identity in ?? holds. If the identity in ?? holds, then by using 6 , we have, for \((x, y) \in \mathbb{R}\times E\), \[\begin{align} \rho^{\nu, {\mathbf{b}}}_{p_{\mathbf{b}}}&( x, y)\\ =& \beta{\mathbf{E}}_{(x,y)}\!\left[ e^{-{\mathfrak{q}}(\tau^-_0)};\tau^-_0<T^{\nu,+}_{{\mathbf{b}}-} \right] +\sum_{k\in\mathbb{N}}\beta{\mathbf{E}}_{(x,y)}\!\left[ e^{-{\mathfrak{q}}(\kappa^{\nu,{\mathbf{b}}}_{p_{\mathbf{b}},0})};T^{\nu,+}_{{\mathbf{b}}-}=T^\nu_k, T^\nu_k\leq\tau^-_0 \right] \\ =& \beta{\mathbf{E}}_{(x,y)}\!\left[ e^{-{\mathfrak{q}}(\tau^-_0)};\tau^-_0<T^{\nu,+}_{{\mathbf{b}}-} \right] +\sum_{k\in\mathbb{N}}{\mathbf{E}}_{(x,y)}\!\left[ e^{-{\mathfrak{q}}(T^\nu_k)}\varrho^{\nu, {\mathbf{b}}}_{p_{\mathbf{b}}}( {\mathbf{b}}(Y_{T^{\nu}_k}), Y_{T^{\nu}_k});T^{\nu,+}_{{\mathbf{b}}-}=T^\nu_k, T^\nu_k\leq\tau^-_0 \right] \\ =& \beta{\mathbf{E}}_{(x,y)}\!\left[ e^{-{\mathfrak{q}}(\tau^-_0)};\tau^-_0<T^{\nu,+}_{{\mathbf{b}}-} \right] +{\mathbf{E}}_{(x,y)}\!\left[ e^{-{\mathfrak{q}}(T^{\nu,+}_{{\mathbf{b}}-})}\varrho^{\nu, {\mathbf{b}}}_{p_{\mathbf{b}}}( {\mathbf{b}}(Y_{T^{\nu,+}_{{\mathbf{b}}-}}), Y_{T^{\nu,+}_{{\mathbf{b}}-}});T^{\nu,+}_{{\mathbf{b}}-}\leq\tau^-_0 \right]\\ =& \beta{\mathbf{E}}_{(x,y)}\!\left[ e^{-{\mathfrak{q}}(\tau^-_0)};\tau^-_0<T^{\nu,+}_{{\mathbf{b}}-} \right] +{\mathbf{E}}_{(x,y)}\!\left[ e^{-{\mathfrak{q}}(T^{\nu,+}_{{\mathbf{b}}-})};T^{\nu,+}_{{\mathbf{b}}-}\leq\tau^-_0 \right]. \label{LemB02a95001} \end{align}\tag{78}\] By Lemma 18 and 78 , and since \({\mathbf{b}}\) belongs to \(\Xi^\nu\), we obtain the inequality in ?? .

(1) We now define candidates for \(p_{\mathbf{b}}\) satisfying the required condition. For a measurable function \(p\) and \(y \in E\), we have \[\begin{align} {\mathbf{E}}_{({\mathbf{b}}(y),y)} \!\left[ e^{-{\mathfrak{q}}(\mathcal{T}^{1}_{p,{\mathbf{b}}})}; \mathcal{T}^{1}_{p,{\mathbf{b}}}\leq\varkappa^{\nu,{\mathbf{b}}}_{p,0} \right]+\beta& {\mathbf{E}}_{({\mathbf{b}}(y),y)} \!\left[ e^{-{\mathfrak{q}}(\varkappa^{\nu,{\mathbf{b}}}_{p,0})}; \varkappa^{\nu,{\mathbf{b}}}_{p,0}<\mathcal{T}^{1}_{p,{\mathbf{b}}} \right] \\ &=(1-p(y))\rho^{\nu,1}_{{\mathbf{b}}}(y) + p(y)\rho^{\nu,2}_{{\mathbf{b}}}(y). \label{LemB0295006} \end{align}\tag{79}\] Since \(\rho^{\nu,1}_{{\mathbf{b}}}( y)\leq1\) and \(\rho^{\nu,2}_{{\mathbf{b}}}( y)\geq 1\), the value \[\begin{align} p_{\mathbf{b}}(y) :=\inf\{a \in [0, 1]: (1-a)\rho^{\nu,1}_{{\mathbf{b}}}(y) + a\rho^{\nu,2}_{{\mathbf{b}}}(y)=1\}\in[0,1], \end{align}\] is well-defined for \(y\in E\), and the function \(p_{\mathbf{b}}\) is measurable and satisfies \[\begin{align} (1-p_{\mathbf{b}}(y))\rho^{\nu,1}_{{\mathbf{b}}}(y) + p_{\mathbf{b}}(y)\rho^{\nu,2}_{{\mathbf{b}}}(y)=1,\qquad y\in E. \label{LemB0295007} \end{align}\tag{80}\]

(2) We define, for \(n\in\mathbb{N}\) and \(y\in E\), \[\begin{align} \varrho^{\nu, {\mathbf{b}}}_{p_{\mathbf{b}},(n)}( y):=& {\mathbf{E}}_{({\mathbf{b}}(y),y)} \!\left[ e^{-{\mathfrak{q}}(\mathcal{T}^{1}_{p_{\mathbf{b}},{\mathbf{b}}})}\varrho^{\nu, {\mathbf{b}}}_{p_{\mathbf{b}},(n-1)}( Y_{\mathcal{T}^{1}_{p_{\mathbf{b}},{\mathbf{b}}}}); \mathcal{T}^{1}_{p_{\mathbf{b}},{\mathbf{b}}}\leq\varkappa^{\nu,{\mathbf{b}}}_{p_{\mathbf{b}},0} \right]\\ &\qquad +\beta {\mathbf{E}}_{({\mathbf{b}}(y),y)} \!\left[ e^{-{\mathfrak{q}}(\varkappa^{\nu,{\mathbf{b}}}_{p_{\mathbf{b}},0})}; \varkappa^{\nu,{\mathbf{b}}}_{p_{\mathbf{b}},0}<\mathcal{T}^{1}_{p_{\mathbf{b}},{\mathbf{b}}} \right], \label{LemB0295001} \end{align}\tag{81}\] where \(\varrho^{\nu, {\mathbf{b}}}_{p_{\mathbf{b}},(0)}( y):=1\) for \(y\in E\). We prove \[\begin{align} \lim_{n\to\infty}\varrho^{\nu, {\mathbf{b}}}_{p_{\mathbf{b}},(n)}( y)=\varrho^{\nu, {\mathbf{b}}}_{p_{\mathbf{b}}}( {\mathbf{b}}(y), y),\qquad y\in E. \label{LemB0295004} \end{align}\tag{82}\] We first prove that, for \(n\in\mathbb{N}\) and \(y \in E\) \[\begin{align} \varrho^{\nu, {\mathbf{b}}}_{p_{\mathbf{b}},(n)}( y)-&\varrho^{\nu, {\mathbf{b}}}_{p_{\mathbf{b}}}( {\mathbf{b}}(y), y)\\ &= {\mathbf{E}}_{({\mathbf{b}}(y),y)} \!\left[ e^{-{\mathfrak{q}}(\mathcal{T}^{n}_{p_{\mathbf{b}},{\mathbf{b}}})} \!\left( 1-\varrho^{\nu, {\mathbf{b}}}_{p_{\mathbf{b}}}({\mathbf{b}}( Y_{\mathcal{T}^{n}_{p_{\mathbf{b}},{\mathbf{b}}}}), Y_{\mathcal{T}^{n}_{p_{\mathbf{b}},{\mathbf{b}}}}) \right); \mathcal{T}^{n}_{p_{\mathbf{b}},{\mathbf{b}}}\leq\varkappa^{\nu,{\mathbf{b}}}_{p_{\mathbf{b}},0} \right], \label{LemB0295002} \end{align}\tag{83}\] by induction. By using 6 at \(\mathcal{T}^{1}_{p_{\mathbf{b}},{\mathbf{b}}}\) in the same way as 47 , we have, for \(y\in E\), \[\begin{align} \varrho^{\nu, {\mathbf{b}}}_{p_{\mathbf{b}}}( {\mathbf{b}}(y), y)=& {\mathbf{E}}_{({\mathbf{b}}(y),y)} \!\left[ e^{-{\mathfrak{q}}(\mathcal{T}^{1}_{p_{\mathbf{b}},{\mathbf{b}}})}\varrho^{\nu, {\mathbf{b}}}_{p_{\mathbf{b}}}({\mathbf{b}}( Y_{\mathcal{T}^{1}_{p_{\mathbf{b}},{\mathbf{b}}}}), Y_{\mathcal{T}^{1}_{p_{\mathbf{b}},{\mathbf{b}}}}); \mathcal{T}^{1}_{p_{\mathbf{b}},{\mathbf{b}}}\leq\varkappa^{\nu,{\mathbf{b}}}_{p_{\mathbf{b}},0} \right]\\ &\qquad\quad+\beta {\mathbf{E}}_{({\mathbf{b}}(y),y)} \!\left[ e^{-{\mathfrak{q}}(\varkappa^{\nu,{\mathbf{b}}}_{p_{\mathbf{b}},0})}; \varkappa^{\nu,{\mathbf{b}}}_{p_{\mathbf{b}},0}<\mathcal{T}^{1}_{p_{\mathbf{b}},{\mathbf{b}}} \right]. \label{LemB0295003} \end{align}\tag{84}\] By 81 with \(n=1\) and 84 , we have 83 with \(n=1\). We assume that 83 with \(n=k\in\mathbb{N}\) holds and prove 83 with \(n=k+1\). By 81 with \(n=k+1\) and 84 , we have, for \(y\in E\), \[\begin{align} \varrho^{\nu, {\mathbf{b}}}_{p_{\mathbf{b}},(k+1)}( y)-&\varrho^{\nu, {\mathbf{b}}}_{p_{\mathbf{b}}}( {\mathbf{b}}(y), y)\\ &= {\mathbf{E}}_{({\mathbf{b}}(y),y)} \!\left[ e^{-{\mathfrak{q}}(\mathcal{T}^{1}_{p_{\mathbf{b}},{\mathbf{b}}})} ({\varrho^{\nu, {\mathbf{b}}}_{p_{\mathbf{b}},(k)}( Y_{\mathcal{T}^{1}_{p_{\mathbf{b}},{\mathbf{b}}}})-\varrho^{\nu, {\mathbf{b}}}_{p_{\mathbf{b}}}({\mathbf{b}}( Y_{\mathcal{T}^{1}_{p_{\mathbf{b}},{\mathbf{b}}}}), Y_{\mathcal{T}^{1}_{p_{\mathbf{b}},{\mathbf{b}}}})}); \mathcal{T}^{1}_{p_{\mathbf{b}},{\mathbf{b}}}\leq\varkappa^{\nu,{\mathbf{b}}}_{p_{\mathbf{b}},0} \right]. \end{align}\] Applying 83 with \(n=k\) to the above equation and then using 6 at \(\mathcal{T}^{1}_{p_{\mathbf{b}},{\mathbf{b}}}\) in the same way as 47 , we obtain 83 with \(n=k+1\). Therefore, 83 holds for any \(n\in\mathbb{N}\) and \(y\in E\). Since \[\begin{align} \lim_{n\to\infty}\left| {\mathbf{E}}_{({\mathbf{b}}(y),y)} \!\left[ e^{-{\mathfrak{q}}(\mathcal{T}^{n}_{p_{\mathbf{b}},{\mathbf{b}}})} \!\left( 1-\varrho^{\nu, {\mathbf{b}}}_{p_{\mathbf{b}}}({\mathbf{b}}( Y_{\mathcal{T}^{n}_{p_{\mathbf{b}},{\mathbf{b}}}}), Y_{\mathcal{T}^{n}_{p_{\mathbf{b}},{\mathbf{b}}}}) \right); \mathcal{T}^{n}_{p_{\mathbf{b}},{\mathbf{b}}}\leq\varkappa^{\nu,{\mathbf{b}}}_{p_{\mathbf{b}},0} \right] \right|\qquad&\\ \leq \lim_{n\to\infty}(1+\beta){\mathbf{E}}_{({\mathbf{b}}(y),y)} \!\left[ e^{-{\mathfrak{q}}(\mathcal{T}^{n}_{p_{\mathbf{b}},{\mathbf{b}}})} ; \mathcal{T}^{n}_{p_{\mathbf{b}},{\mathbf{b}}}\leq\varkappa^{\nu,{\mathbf{b}}}_{p_{\mathbf{b}},0} \right]&=0, \end{align}\] we obtain 82 .

(3) We prove, for \(n\in\mathbb{N}\), \[\begin{align} \varrho^{\nu, {\mathbf{b}}}_{p_{\mathbf{b}},(n)}( y)=1, \qquad y\in E, \label{LemB0295005} \end{align}\tag{85}\] by induction. By 79 , 80 and 81 with \(n=1\), we have 85 with \(n=1\). If 85 with \(n=k\in\mathbb{N}\) holds, then \(\varrho^{\nu, {\mathbf{b}}}_{p_{\mathbf{b}},(k)}( Y_{\mathcal{T}^{1}_{p_{\mathbf{b}},{\mathbf{b}}}})\) in 81 with \(n=k+1\) can be replaced by \(1\). Thus, by applying 79 and 80 to the resulting expression, we obtain 85 with \(n=k+1\). Therefore, we obtain 85 for any \(n\in\mathbb{N}\).

By 82 and 85 , we obtain the identity in ?? . The proof is complete. ◻

Proof. We fix \(n\in\mathbb{N}\). We prove, for \(k\in\mathbb{N}\cup\{0\}\), \[\begin{align} U^{{\mathrm{P}}, n}_{ (k)} (x, y)\leq V^{{\mathrm{P}}, n}(x, y),\qquad (x, y)\in\mathbb{R}\times E. \label{LemC195001} \end{align}\tag{94}\] When \(k=0\), 94 holds since \(U^{{\mathrm{P}}, n}_{ (0)} (x, y)= {v_0}(x, y)\) for \((x, y)\in\mathbb{R}\times E\). We assume that 94 holds when \(k=l\in\mathbb{N}\cup\{0\}\) and prove that 94 holds when \(k=l+1\). By 19 , 6 at \(T^{{\mathrm{P}}, n+1}\) and the memoryless property of the exponential distribution, we have \[\begin{align} \mathcal{V}^{{\mathrm{P}}, n}& V^{{\mathrm{P}},n}(x, y)\\ &= {\mathbf{E}}_{(x,y)}\bigg[-\beta\int_{[0, T^{{\mathrm{P}},n+1}]}e^{-{\mathfrak{q}}(t)}{\mathrm{d}}((-\underline{X}_t\lor0)) \\ &+e^{-{\mathfrak{q}}(T^{{\mathrm{P}},n+1})}\bigg{\{}1_{\{T^{{\mathrm{P}},n+1}<T^{{\mathrm{P}},n}\}} \bigg{(}-\beta\int_{( T^{{\mathrm{P}},n+1},T^{{\mathrm{P}},n}]}e^{-{\mathfrak{q}}(t)}{\mathrm{d}}((-\underline{X}_t\lor0))\\ &+ e^{-({\mathfrak{q}}(T^{{\mathrm{P}},n})-{\mathfrak{q}}(T^{{\mathrm{P}},n+1}))}\widetilde{V}^{{\mathrm{P}},n}(X^0_{T^{{\mathrm{P}},n}}, Y_{T^{{\mathrm{P}},n}})\bigg{)} + 1_{\{T^{{\mathrm{P}},n+1}=T^{{\mathrm{P}},n}\}} \widetilde{V}^{{\mathrm{P}},n}(X^0_{T^{{\mathrm{P}},n+1}}, Y_{T^{{\mathrm{P}},n+1}}) \bigg{\}} \bigg] \\ =& {\mathbf{E}}_{(x,y)}\bigg[-\beta\int_{[0, T^{{\mathrm{P}},n+1}]}e^{-{\mathfrak{q}}(t)}{\mathrm{d}}((-\underline{X}_t\lor0)) +e^{-{\mathfrak{q}}(T^{{\mathrm{P}},n+1})}\bigg{\{}1_{\{T^{{\mathrm{P}},n+1}<T^{{\mathrm{P}},n}\}} V^{{\mathrm{P}},n}(X^0_{T^{{\mathrm{P}},n+1}}, Y_{T^{{\mathrm{P}},n+1}})\\ &+ 1_{\{T^{{\mathrm{P}},n+1}=T^{{\mathrm{P}},n}\}} \widetilde{V}^{{\mathrm{P}},n}(X^0_{T^{{\mathrm{P}},n+1}}, Y_{T^{{\mathrm{P}},n+1}}) \bigg{\}} \bigg] \\ =&\mathcal{U}^{{\mathrm{P}}, n} V^{{\mathrm{P}},n}(x, y). \label{LemC0195002} \end{align}\tag{95}\] Applying the operator \(\mathcal{U}^{{\mathrm{P}}, n}\) to both sides of 94 with \(k=l\), and using 34 with \(\nu\) replaced by \({\mathrm{P}},n\) and 95 , we have 94 with \(k=l+1\). In conclusion, we have \[\begin{align} \limsup_{k\to\infty}U^{{\mathrm{P}}, n}_{ (k)} (x, y)\leq V^{{\mathrm{P}}, n}(x, y),\qquad (x, y)\in\mathbb{R}\times E. \label{LemB0195main95001} \end{align}\tag{96}\]

Here, taking 19 into account and repeating the argument in the same manner as in 95 , with 6 applied at \(T^{{\mathrm{P}}, n+1}_1,\dots, T^{{\mathrm{P}}, n+1}_k\), we obtain the following interpretation of \(U^{{\mathrm{P}}, n}_{ (k)}\). It is the value function for dividends and capital injections over the class of strategies in \(\underline{\Pi}^{{\mathrm{P}}, n}\) under the additional constraint that dividends may be paid only at the jump times of \(\widehat{N}^{(n)}\) that coincide with one of the first \(k\) jump times of \(\widehat{N}^{(n+1)}\). We define \(\pi^{{[k]}_{(n)}}_{\underline{{\mathbf{b}}}^{{\mathrm{P}}, n}}\in\underline{\Pi}^{{\mathrm{P}}, n}\) as a strategy that uses the same control as \(\pi^{{\mathrm{P}}, n}_{\underline{{\mathbf{b}}}^{{\mathrm{P}}, n}}\) until time \(T^{{\mathrm{P}}, n+1}_k\), and does not pay dividends thereafter. Using the dominated convergence theorem as in the proof of Lemma 12, we have \[\begin{align} \lim_{k\to\infty}v_{\pi^{{[k]}_{(n)}}_{\underline{{\mathbf{b}}}^{{\mathrm{P}}, n}}}(x, y) =v_{\pi^{{\mathrm{P}}, n}_{\underline{{\mathbf{b}}}^{{\mathrm{P}}, n}}}(x, y), \qquad (x, y) \in \mathbb{R}\times E. \label{SecC002} \end{align}\tag{97}\] Also, in the proof of 26 with \(\nu\) replaced by \({\mathrm{P}}, n+1\), by replacing \(\mathcal{V}^{{\mathrm{P}}, n+1, (k)}v_0\) with \(U^{{\mathrm{P}}, n}_{(k)}\) and using the same indicator-based decomposition of expectations as in 95 , we obtain \[\begin{align} v_{\pi^{{[k]}_{(n)}}_{\underline{{\mathbf{b}}}^{{\mathrm{P}}, n}}}(x, y) \leq U^{{\mathrm{P}}, n}_{(k)}(x,y),\qquad (x, y)\in\mathbb{R} \times E. \label{SecC001} \end{align}\tag{98}\] Thus, for \(k\in\mathbb{N}\) and \((x, y)\in\mathbb{R}\times E\), we have \[\begin{align} &V^{{\mathrm{P}}, n}(x, y)-U^{{\mathrm{P}}, n}_{ (k)} (x, y)\leq V^{{\mathrm{P}}, n}(x, y)-v_{\pi^{{[k]}_{(n)}}_{\underline{{\mathbf{b}}}^{{\mathrm{P}}, n}}}(x, y)\\ &\leq V^{{\mathrm{P}}, n}(x, y)-v_{\pi^{{\mathrm{P}}, n}_{\underline{{\mathbf{b}}}^{{\mathrm{P}}, n}}}(x, y) +\left| v_{\pi^{{\mathrm{P}}, n}_{\underline{{\mathbf{b}}}^{{\mathrm{P}}, n}}}(x, y)-v_{\pi^{{[k]}_{(n)}}_{\underline{{\mathbf{b}}}^{{\mathrm{P}}, n}}}(x, y) \right| =\left| v_{\pi^{{\mathrm{P}}, n}_{\underline{{\mathbf{b}}}^{{\mathrm{P}}, n}}}(x, y)-v_{\pi^{{[k]}_{(n)}}_{\underline{{\mathbf{b}}}^{{\mathrm{P}}, n}}}(x, y) \right|, \label{Lem30995002C} \end{align}\tag{99}\] where the first inequality follows from 98 . By taking the limit of 99 as \(k\to \infty\) and by 97 , we have \[\begin{align} V^{{\mathrm{P}}, n}(x, y)-\liminf_{k\to\infty}U^{{\mathrm{P}}, n}_{ (k)} (x, y) \leq0, \qquad (x, y)\in\mathbb{R}\times E. \label{Lem30995005C} \end{align}\tag{100}\] By 96 and 100 , we obtain ?? . The proof is complete. ◻

Proof of Theorem 26. As in the proof of Theorem 23, it suffices to prove that, for \(n\in\mathbb{N}\), \[\begin{align} V^{{\mathrm{P}}, n\prime}_{+}(x, y) \leq V^{{\mathrm{P}}, {n+1}\prime}_{+}(x, y),\qquad (x,y)\in \mathbb{R}\times E, \label{barrier95result95P} \end{align}\tag{101}\] where \(V^{{\mathrm{P}}, n\prime}_+\) is the right derivative of \(V^{{\mathrm{P}}, n}\).

The functions \(x\mapsto V^{{\mathrm{P}}, n}(x, y)\), \(x\mapsto V^{{\mathrm{P}}, n+1}(x, y)\), \(x\mapsto V^{{\mathrm{P}}, n}_{(k)}(x, y)\) and \(x\mapsto U^{{\mathrm{P}}, n}_{(k)}(x, y)\) are concave for \(n, k\in\mathbb{N}\) and \(y \in E\). Hence, by Lemma 13, Lemma 27, [23] and [24], we have \(\lim_{k\to\infty}{V^{{\mathrm{P}}, n+1\prime}_{(k)+}}(x, y)={V^{{\mathrm{P}}, n+1\prime}_+}(x, y)\) and \(\lim_{k\to\infty}{U^{{\mathrm{P}}, n\prime}_{(k)+}}(x, y)={V^{{\mathrm{P}}, n\prime}_+}(x, y)\) for Lebesgue-a.e. \(x\), for any \(n\in\mathbb{N}\) and \(y \in E\). Therefore, if we show that for \(n \in \mathbb{N}\), \[\begin{align} U^{{\mathrm{P}}, n\prime}_{(k)+}(x, y) \leq V^{{\mathrm{P}}, n+1\prime}_{(k)+}(x, y),\qquad k\in\mathbb{N}\cup\{0 \},~(x,y)\in \mathbb{R}\times E, \label{barrier95convergence95P} \end{align}\tag{102}\] then 101 follows. In the following, we prove 102 by induction on \(k\).

For \(k=0\), since both \({U^{{\mathrm{P}}, n}_{(0)}}\) and \({V^{{\mathrm{P}},n+1}_{(0)}}\) equal \(v_0\), 102 holds. Assuming that 102 holds for \(k=l\in\mathbb{N}\cup\{0\}\), we show that 102 also holds for \(k=l+1\). By applying almost the same method as in the proof of 23 , we have, for \(f\in\Gamma^{{\mathrm{P}}, n+1}\) and \((x,y)\in\mathbb{R}\times E\), \[\begin{align} {(\mathcal{U}^{{\mathrm{P}}, n}f)}^\prime_+ &(x, y) = {\mathbf{E}}_{(x,y)}\bigg[e^{-{\mathfrak{q}}(T^{{\mathrm{P}},n+1}\land\tau^-_0)}\bigg{\{}1_{\{T^{{\mathrm{P}},n+1}<T^{{\mathrm{P}},n}, \underline{X}_{T^{{\mathrm{P}},n+1} }\geq0\}} f^\prime_{+}(X_{T^{{\mathrm{P}},n+1}}, Y_{T^{{\mathrm{P}},n+1}})\\ &+ 1_{\{T^{{\mathrm{P}},n+1}=T^{{\mathrm{P}},n},\underline{X}_{T^{{\mathrm{P}},n+1} }\geq0\}} \!\left( \widetilde{f}^{\prime}_{+}(X_{T^{{\mathrm{P}},n+1}}, Y_{T^{{\mathrm{P}},n+1}}) \right) +1_{\{\underline{X}_{T^{{\mathrm{P}},n+1} }<0\}}\beta \bigg{\}} \bigg]. \label{strong95Markov95P} \end{align}\tag{103}\] By 23 and 103 , we have, for \(f, g\in \Gamma^{{\mathrm{P}}, n+1}\) with \(f^\prime_+(x, y)\leq g^\prime_+(x, y)\) for \((x,y)\in\mathbb{R}\times E\), \[\begin{align} {(\mathcal{U}^{{\mathrm{P}}, n}f)}^\prime_+(x ,y )\leq {(\mathcal{V}^{{\mathrm{P}}, n+1}g)}^\prime_+(x ,y ), \qquad (x,y)\in\mathbb{R}\times E. \label{compare95periodic} \end{align}\tag{104}\] By the definitions of \(U^{{\mathrm{P}}, n}_{(l+1)}\) and \(V^{{\mathrm{P}}, n+1}_{(l+1)}\), 102 with \(k=l\) and 104 , we have 102 with \(k=l+1\). The proof is complete. ◻