Proof of Theorem 5i.. Throughout the proof, we suppress the superscript of \(V^{\text{EZ},D}\) and simply write \(V\) for notational convenience. For each \(n\in\mathbb{N}\), set \(\xi_{\tau}^n := \frac{1}{n}\vee \xi_{\tau}\). By Theorem 34, there exists a unique \(L^2\)-solution \((Y^n,Z^n)\in {\@fontswitch\relax\mathcal{P}}_+\times {\@fontswitch\relax\mathcal{P}}\) of 5 (with \(\xi_\tau\) replaced by \(\xi_{\tau}^n\)) such that \[\begin{align} Y^n_t = \mathbb{E}_t \bigg[ \int_{t\wedge \tau}^{\tau} e^{-\rho s}(1-R) (Y^n_s)^{\frac{-R}{1-R}}dD_s + \xi^n_{\tau}\bigg],\quad t\ge0. \end{align}\] Note that \(Y_t^n\) is non-increasing in \(n \in \mathbb{N}\) for every \(t\geq0\) (see Proposition 29) and \(Y^n_t > 0\) \(\mathbb{P}\)-a.s. for every \(t\geq0\) and \(n\in \mathbb{N}\). Hence \(V_t:= \lim_{n\rightarrow \infty} Y_t^n\) for \(t\geq0\) is well-defined. Moreover, the monotone convergence theorem (MCT) ensures that for \(t\geq 0\), \[\begin{align} V_t =& \lim_{n\rightarrow\infty}\mathbb{E}_t \bigg[ \int_{t\wedge \tau}^{\tau} e^{-\rho s}(1-R) (Y^n_s)^{\frac{-R}{1-R}}dD_s + \xi^n_{\tau}\bigg] \\ =& \mathbb{E}_t \bigg[ \int_{t\wedge \tau}^{\tau} e^{-\rho s}(1-R) (V_s)^{\frac{-R}{1-R}}dD_s + \xi_{\tau}\bigg]. \end{align}\] Since \(\xi_{\tau}>0\) \(\mathbb{P}\)-a.s., it holds that \(V_t \ge \mathbb{E}_t[\xi_{\tau}]>0\) \(\mathbb{P}\)-a.s. for all \(t\geq0\).

We claim that \(V\in\mathbb{UI}(g_{\text{EZ}},D)\). Since \(Y_0^n\) is non-increasing in \(n \in \mathbb{N}\) and \(V_0=\lim_{n\rightarrow \infty}Y^n_0\), it holds that \(\mathbb{E}[ \int_{0}^{\tau} e^{-\rho s}(1-R) (V_s)^{\frac{-R}{1-R}}dD_s] < V_0 \le Y_0^n<\infty\) for every \(n\in\mathbb{N}\).

Similarly, we have \(\mathbb{E}[\sup_{t\ge 0} (V_t)^{\frac{2}{1-R}}] \le \mathbb{E}[\sup_{t\ge 0} (Y^n_t)^{\frac{2}{1-R}}]<\infty\). This ensures that \(\mathbb{E}[\sup_{t\ge 0} (V_t)^2]<\infty\) by Jensen’s inequality with exponent \(\frac{1}{1-R}>1\). Hence, \(V\) is uniformly integrable and therefore is of class \(\mathbb{UI}(g_{\text{EZ}},D)\).

It remains to show that \(V\) is the unique utility process. By the martingale representation theorem, there exists \(Z\in \mathcal{P}\) such that \(\int_0^{\tau} Z_t^2 dt <\infty\) \(\mathbb{P}\)-a.s. and \((V,Z)\) solves 5 . Let \(V'\) be another utility process of class \(\mathbb{UI}(g_{\text{EZ}},D)\) and \(Z'\in \mathcal{P}\) be such that \(\int_0^{\tau} (Z'_t)^2 dt <\infty\) \(\mathbb{P}\)-a.s., and \((V',Z')\) solves 5 . Here we note that \(V_t = V'_t = \xi_{\tau}\) on \(\{t\ge \tau\}\).

Set \((\Delta V, \Delta Z):= (V-V', Z-Z')\). Then (\(\Delta V, \Delta Z)\) solves for all \(t\geq0\) \[\begin{align} \Delta V_t = \int_{t\wedge\tau}^{\tau} \frac{g_{\text{EZ}}(s,V_s) - g_{\text{EZ}}(s,V'_s)}{\Delta V_s} \Delta V_s \mathbb{1}_{\Delta V_s\neq0} dD_s - \int_{t\wedge\tau}^{\tau} \Delta Z_s dW_s. \end{align}\] Let \(\alpha=(\alpha_s)_{s\ge0}\) be defined by \(\alpha_s: = \mathbb{1}_{\Delta V_s\neq0}(g_{\text{EZ}}(s,V_s) - g_{\text{EZ}}(s,V'_s))/\Delta V_s\). Since \(y\mapsto g_{\text{EZ}}(\cdot,y)\) is decreasing, we have \(\alpha\le0\). It then follows that \[\begin{align} e^{\int_{0}^t \alpha_u du}\Delta V_t = - \int_{t\wedge\tau}^{\tau} e^{\int_{0}^s \alpha_u du} \Delta Z_s dW_s\quad for all t\geq0. \end{align}\] From the fact that \(\alpha\le 0\) and \(V,V'\in\mathbb{UI}(g_{\text{EZ}},D)\), the local martingale \(\int_0^{t}e^{\int_{0}^s \alpha_u du} \Delta Z_s dW_s\) is a martingale. Hence, \(\Delta V_t= 0\) \(\mathbb{P}\)-a.s. for all \(t\geq0\). ◻

Proof. We start with constructing \(V^{\theta,D}\) given in ?? and then show that it satisfies 10 . As \(\theta\in \Theta^D\), by Hölder’s inequality, \[\mathbb{E}[\eta_{\tau}^{\theta}e^{\int_{0}^{\tau} \frac{\theta_u^2}{2{\@fontswitch\relax\mathcal{R}}}du }(\xi_{\tau})^{\frac{1}{1-{\@fontswitch\relax\mathcal{R}}}} ]\le \mathbb{E}[(\eta_{\tau}^{\theta})^2e^{\int_{0}^{\tau} \frac{\theta^2_u}{{\@fontswitch\relax\mathcal{R}}}du } ]^{\frac{1}{2}} \mathbb{E}[(\xi_{\tau})^{\frac{2}{1-{\@fontswitch\relax\mathcal{R}}}}]^{\frac{1}{2}}<\infty.\] Furthermore, Jensen’s inequality (with exponent 2) ensures that \[\mathbb{E}\bigg[ \int_{0}^{ \tau} \eta_s^{\theta} e^{\int_{0}^{s} \frac{\theta^2_u}{2{\@fontswitch\relax\mathcal{R}}}du -\rho s} dD_s \bigg]\leq \mathbb{E}\bigg[\bigg( \int_{0}^{ \tau} \eta_s^{\theta} e^{\int_{0}^{s} \frac{\theta^2_u}{2{\@fontswitch\relax\mathcal{R}}}du -\rho s} dD_s\bigg)^2 \bigg] <\infty.\] Hence, the representation in ?? is a well-defined utility process.

By Jensen’s inequality, we have \(\mathbb{E}[\sup_{t\ge 0} (V^{\theta,D}_{t\wedge\tau})^2]<\infty\). It further implies that there exists \({\iota}\in\mathcal{P}\) such that \(\int_0^{\infty} \iota^2_t dt<\infty\), \(\mathbb{Q}^\theta\)-a.s., and \(({V}^{\theta,D},{\iota})\) solves \[\begin{align} \label{eq:thm:rob95rec:1} {V}^{\theta,D}_t = (\xi_{\tau})^{\frac{1}{1-{\@fontswitch\relax\mathcal{R}}}} + \int_{t\wedge\tau}^{\tau} e^{-\rho s} dD_s + \int_{t\wedge\tau}^{\tau} \frac{1}{2{\@fontswitch\relax\mathcal{R}}}{V}^{\theta,D}_s\theta^2_s ds - \int_{t\wedge\tau}^{\tau} {\iota}_s dW_s^{\theta}. \end{align}\tag{12}\] Since \((M_t)_{t\ge0} = (\int_{0}^{t} {\iota}_s dW_s^{\theta})_{t\ge0}\) is a local martingale under \(\mathbb{Q}^{\theta}\), we shall find an increasing sequence of stopping times \(\{T_n\}_{n\in\mathbb{N}}\) such that \(T_n\ge t\), \(\mathbb{Q}^{\theta}(\lim_{n\rightarrow\infty} T_n = \infty) = 1\) and the stopped process \((M_{t\wedge T_n})_{t\ge 0}\) is a martingale under \(\mathbb{Q}^{\theta}\). Applying \(\mathbb{E}_t^{\theta}\) to both sides of 12 gives \[\begin{align} V_{t}^{\theta,D} = \mathbb{E}^{\theta}_t\bigg[ \int_{t\wedge\tau}^{T_n\wedge\tau} e^{-\rho s} dD_s\bigg] + \frac{1}{2{\@fontswitch\relax\mathcal{R}}} \mathbb{E}^{\theta}_t\bigg[\int_{t\wedge\tau}^{T_n\wedge\tau} V^{\theta,D}_s \theta_s^2 ds\bigg]+ \mathbb{E}^{\theta}_t[V_{T_n\wedge\tau}^{\theta,D}]. \end{align}\] Then the first two terms on the right-hand side converge as \(n\to \infty\) by MCT, while the last term converges by the dominated convergence theorem (DCT). We conclude that \(V_t^{\theta}\) defined via ?? satisfies 10 . The uniqueness follows from standard localization arguments, and we omit it here. ◻

Proof. We start by proving ii., which is stated under the condition that \(\xi_{\tau}>C\) \(\mathbb{P}\)-a.s.. By Theorem 5 ii. and Itô’s lemma, \((\mathbb{Y},\mathbb{Z})\) solves \[\begin{align} \mathbb{Y}_t =& (\xi_{\tau})^{\frac{1}{1-R}} + \int_{t\wedge\tau}^{\tau} e^{-\rho s} dD_s - \frac{R}{2} \int_{t\wedge\tau}^{\tau} \frac{(\mathbb{Z}_s)^2}{\mathbb{Y}_s}ds - \int_{t\wedge\tau}^{\tau} \mathbb{Z}_s dW_s \\ =& (\xi_{\tau})^{\frac{1}{1-R}}+ \int_{t\wedge\tau}^{\tau} e^{-\rho s} dD_s- \frac{R}{2} \int_{t\wedge\tau}^{\tau} \frac{(\mathbb{Z}_s)^2}{\mathbb{Y}_s}ds- \int_{t\wedge\tau}^{\tau}\mathbb{Z}_s\theta_s ds - \int_{t\wedge\tau}^{\tau} \mathbb{Z}_s dW_s^{\theta}. \end{align}\] For any \(\theta\in \Theta^D\), denote by \(({V}^{\theta,D},{\iota}^{\theta})\) a solution of 12 . Then \[\begin{align} {V}^{\theta,D}_t -\mathbb{Y}_t &= \int_{t\wedge\tau}^{\tau}\bigg( \frac{R}{2\mathbb{Y}_s} \bigg(\mathbb{Z}_s + \frac{\mathbb{Y}_s}{R}\theta_s\bigg)^2 + \frac{\theta_s^2({V}^{\theta,D}_s -\mathbb{Y}_s )}{2R}\bigg) ds \\ &\quad - \int_{t\wedge\tau}^{\tau}({\iota}^{\theta}_s - \mathbb{Z}_s) dW_s^{\theta}. \end{align}\] Set \(\beta_{t}: = \exp(\frac{1}{{2R}}\int_0^{t}{\theta^2_s} ds)\) for \(t\geq 0\). Then using Itô’s lemma, \[\begin{align} \label{eq:pf:thm:rob95rec951} \beta_{t}( {V}^{\theta,D}_t -\mathbb{Y}_t) = \int_{t\wedge\tau}^{\tau}\beta_{s} \frac{R}{2\mathbb{Y}_s} \Big(\mathbb{Z}_s + \frac{\mathbb{Y}_s}{R}\theta_s\Big)^2ds - (M_{\tau} - M_{t\wedge \tau}), \end{align}\tag{13}\] where \(M_t := \int_0^t \beta_{s}({\iota}^{\theta}_s - \mathbb{Z}_s)dW^{\theta}_s\). Since \((M_s)_{s\geq0}\) is a \(\mathbb{Q}^{\theta}\)-local martingale, we shall find an increasing sequence of stopping times \(\{T_n\}_{n\in\mathbb{N}}\) such that \(T_n\ge t\), \(\mathbb{Q}^{\theta}(\lim_{n\rightarrow\infty} T_n = \infty) = 1\) and the stopped process \((M_{s\wedge T_n})_{s\geq t}\) is a \(\mathbb{Q}^{\theta}\)-martingale. Applying \(\mathbb{E}_t^{\theta}\) to both sides of 13 , we have for \(t\ge 0\) \[\begin{align} {V}^{\theta,D}_t -\mathbb{Y}_t &= \mathbb{E}_t^{\theta}\bigg[ \int_{t\wedge \tau}^{T_n\wedge \tau}\frac{R}{2\mathbb{Y}_s} \Big(\mathbb{Z}_s + \frac{\mathbb{Y}_s}{R}\theta_s\Big)^2 e^{\int_{t\wedge \tau}^{s}\frac{\theta^2_u}{2R} du} d s\bigg] \nonumber\\ &\quad + \mathbb{E}_t^{\theta} [e^{\int_{t\wedge \tau}^{T_n\wedge \tau}\frac{\theta^2_u}{2R} du} {V}^{\theta,D}_{T_n\wedge \tau}]-\mathbb{E}_t^{\theta} [e^{\int_{t\wedge \tau}^{T_n\wedge \tau}\frac{\theta^2_u}{2R} du}\mathbb{Y}_{T_n\wedge \tau}]\\ &=: \operatorname{I}_t^n+\operatorname{II}_t^n - \operatorname{III}_t^n. \end{align}\] We first note that \(\operatorname{I}_t^n\) converges to the second term given in ?? as \(n\to \infty\), by the MCT (as the integrand is nonnegative). Next, by using the representation given in ?? , we have that as \(n\to \infty\) \[\begin{align} \operatorname{II}_t^n = \mathbb{E}_t^{\theta}\bigg[ \int_{T_n\wedge\tau}^{\tau} e^{\int_{t\wedge \tau}^{s}\frac{\theta^2_u}{2R} du-\rho s} dD_s + e^{\int_{t\wedge \tau}^{\tau}\frac{\theta^2_u}{2R} du} (\xi_{\tau})^{\frac{1}{1-R}}\bigg] \end{align}\] goes to \(\mathbb{E}_t^{\theta}[e^{\int_{t\wedge \tau}^{\tau}\frac{\theta^2_u}{2R} du} (\xi_{\tau})^{\frac{1}{1-R}}]\), where the convergence holds by the MCT (as the integrand is nonnegative).

Lastly, we claim that the term \(\operatorname{III}_t^n\) converges to the same limit as \(\operatorname{II}_t^n\). Indeed, by the conditions that \(\theta\in\Theta^D\) and \(\xi_{\tau}\ge C\) \(\mathbb{P}\)-a.s., we have \[\begin{align} \mathbb{E}^{\theta}_t\bigg[e^{\int_{0}^{\tau}\frac{\theta^2_u}{2R} du} \sup_{s\ge t}\mathbb{Y}_{s\wedge\tau}\bigg] &\le \mathbb{E}_t\bigg[\frac{\eta^{\theta}_{\tau}}{\eta^{\theta}_{t}}e^{\int_{0}^{\tau}\frac{\theta^2_u}{2R} du} \xi_{\tau}^{\frac{1}{1-R}} \sup_{t\ge0}\Big(\mathbb{Y}_{t\wedge\tau}\xi_{\tau}^{-\frac{1}{1-R}}\Big) \bigg]\\ &\le \frac{C^{-\frac{1}{1-R}}}{\eta^{\theta}_{t}}\mathbb{E}_t [(\eta^{\theta}_{\tau})^2e^{\int_{0}^{\tau}\frac{\theta^2_u}{R} du} \xi_{\tau}^{\frac{2}{1-R}}]^{\frac{1}{2}} \mathbb{E}_t\bigg[\sup_{s\ge t} \mathbb{Y}_{s\wedge\tau}^2\bigg]^{\frac{1}{2}}<\infty, \end{align}\] where the second inequality holds by Hölder’s inequality.

Therefore, \(\mathbb{E}_t^{\theta}[e^{\int_{t\wedge \tau}^{T_n\wedge \tau}\frac{\theta^2_u}{2R} du} \mathbb{Y}_{T_n\wedge \tau}]\) is dominated by \(\mathbb{E}_t^{\theta}[e^{\int_{0}^{\tau}\frac{\theta^2_u}{2R} du} \sup_{s\ge t}\mathbb{Y}_{s\wedge\tau}]\) which is finite from the above analysis. Hence we can apply the MCT to ensure the claim to hold. Therefore, the equality ?? holds.

Moreover, ?? implies that \(V^{\theta,D}_t \ge \mathbb{Y}_t = (V^{\text{EZ},D}_t)^{\frac{1}{1-R}}\) and equality holds for \(\theta^* = -R\mathbb{Z}/\mathbb{Y}\) if \(\theta^*\in \Theta^D\). It remains to prove that \(\theta^* = -R\mathbb{Z}/\mathbb{Y}\in \Theta^D\). We first show that \((\eta^{\theta^*}_t)_{t\geq 0}\) is a martingale for \(t\ge0\). By Itô’s lemma, \[\begin{align} \label{eq:pf95thm346295logY} \ln\Big(\frac{\mathbb{Y}_{t\wedge\tau}}{\mathbb{Y}_0}\Big) = \int_{0}^{t\wedge\tau} \frac{R-1}{2} \Big(\frac{\mathbb{Z}_s}{\mathbb{Y}_s}\Big)^2 ds + \int_{0}^{t\wedge\tau} \frac{\mathbb{Z}_s}{\mathbb{Y}_s} dW_s -\int_{0}^{t\wedge\tau} \frac{e^{-\rho s}}{\mathbb{Y}_s} dD_s. \end{align}\tag{14}\] Since \((\mathbb{Y}_t)_{t\ge0}\) is continuous and never \(0\), \(1/\mathbb{Y}\) is locally bounded. Moreover, by Definition 4 i., it holds that \(\theta^* = 0\) on \(\{t\ge\tau\}\), Hence we have for\(t\ge0\), \[\begin{align} &\int_{0}^{t} \theta^*_s dW_s -\int_0^t\frac{(\theta^*_s)^2}{2}ds\\ &\quad = -R\ln\Big(\frac{\mathbb{Y}_{t\wedge\tau}}{\mathbb{Y}_0}\Big) - R\int_{0}^{t\wedge\tau}\bigg( \frac{e^{-\rho s}}{\mathbb{Y}_s} d D_s + \frac{1}{2}\bigg( \frac{\mathbb{Z}_s}{\mathbb{Y}_s}\bigg)^2ds\bigg). \end{align}\] By definition of \(\eta^{\theta^*}\) in 7 , we have that for \(t\ge0\), \(\mathbb{P}\text{-a.s.}\), \[\begin{align} \label{eq:pf95thm346295eta} \eta^{\theta^*}_t =\Big(\frac{\mathbb{Y}_0}{\mathbb{Y}_{t\wedge\tau}}\Big)^R e^{ - \int_{0}^{t\wedge\tau} \frac{R}{\mathbb{Y}_s}e^{-\rho s} d D_s -\frac{R}{2} \int_{0}^{t\wedge\tau}( \frac{\mathbb{Z}_s}{\mathbb{Y}_s})^2ds}\le \mathbb{Y}_0^R C^{-\frac{R}{1-R}}. \end{align}\tag{15}\] Here we have used the condition that \(\xi_{\tau}> C\) \(\mathbb{P}\)-a.s.. Hence \(V_t^{\text{EZ},D}\ge C\) \(\mathbb{P}\)-a.s., for \(t\geq 0\) (see Theorem 5 ii.). We know that \((\eta^{\theta^*}_t)_{t\geq 0}\) is a \(\mathbb{P}\)-local martingale. The uniform bound 15 further implies that it is a \(\mathbb{P}\)-martingale.

Next we verify \(\theta^*\in \Theta^D\) (see 11 ). For the first condition, by 14 \[\begin{align} \mathbb{E}[(\eta^{\theta^*}_{\tau})^2e^{\int_0^{\tau} \frac{(\theta^*_s)^2}{R} ds} (\xi_{\tau})^{\frac{2}{1-R}}] = \mathbb{E}[e^{ - 2R\int_{0}^{\tau} \frac{1}{\mathbb{Y}_s}e^{-\rho s} dD_s} \mathbb{Y}_{\tau}^{2(1-R)}Y_0^{2R}]<\infty, \end{align}\] where we have used the fact that \({\@fontswitch\relax\mathcal{F}}_0\) is trivial (see Section 2) in the inequality.

For the last condition, it holds that \[\begin{align} &\mathbb{E}\bigg[\bigg(\int_{0}^{\tau} e^{\int_{0}^{t} \frac{(\theta^*_s)^2}{2R}ds -\rho t}\eta_t^{\theta^*} dD_t\bigg)^2\bigg] = \mathbb{E}\bigg[\bigg(\int_{0}^{\tau} e^{ R(\ln(\frac{\mathbb{Y}_0}{\mathbb{Y}_{t}}) - \int_{0}^{t} \frac{1}{\mathbb{Y}_s}e^{-\rho s} dD_s) -\rho t} dD_t\bigg)^2\bigg]\\ &\quad \le\mathbb{Y}_0^{2R} \mathbb{E}\bigg[\bigg( \sup_{t\ge 0} \mathbb{Y}_{t\wedge\tau}^{1-R} \int_{0}^{\tau} e^{ - R \int_{0}^{t} \frac{1}{\mathbb{Y}_s}e^{-\rho s} dD_s}\frac{1}{\mathbb{Y}_t} e^{-\rho t} dD_t\bigg)^2\bigg]\\ &\quad\le \frac{\mathbb{Y}_0^{2R}}{R} \mathbb{E}\bigg[\sup_{t\ge 0} (V^{\text{EZ},D}_{t\wedge\tau})^2\bigg] <\infty, \end{align}\] as claimed. Hence, this completes the proof of ii..

Part i. follows by considering a non-increasing sequence \(\xi_{\tau}^n:= \frac{1}{n}\vee\xi_{\tau}\), \(n\in \mathbb{N}\), and then using the same arguments presented for Theorem 5i. ◻

Proof. For fixed \(t\in[0,T)\) and \(0\le k\le K\), set \(\zeta_k = \inf\{s \ge t: D_s \ge k\}\wedge T\). Then each \(\zeta_k\) is a stopping time bounded by \(T\). Moreover, since \(\{\zeta_k\le s\} = \{D_s\ge k\}\) for \(s\ge t\) and \(\zeta_{K} = T\) holds \(\mathbb{P}\)-a.s., it follows that \[\begin{align} \mathbb{E}[A_{\zeta_k}] \le& \alpha + \mathbb{E}\bigg[ \int_{\zeta_k}^{T} A_s d D_s\bigg] = \alpha + \mathbb{E}\bigg[ \int_{t}^{T} A_s \mathbb{1}_{s\ge \zeta_k}d D_s\bigg]\\ \le & \alpha + \mathbb{E}\bigg[ \int_{t}^{T} A_s \mathbb{1}_{ \zeta_k \le s \le \zeta_{K}}d D_s\bigg] = \alpha + \int_{k}^K \mathbb{E}[A_{\zeta_s}] ds. \end{align}\] Set \(u_s:= \mathbb{E}[A_{\zeta_s}]\) for \(k\in[0,K]\). Then since \(u_k \le \alpha + \int_{k}^K u_s ds\), for \(k\in[0,K]\), the standard (backward) Grönwall’s inequality ensures that \(u_k \le \alpha e^{K-k}\) for \(k\in[0,K]\). We complete the proof by letting \(k=0\). ◻

Proof. The idea is to extend the Picard iteration method given by Pardoux and Peng [74] and then to show that the corresponding sequence is Cauchy in \(\mathcal{B}_2\).

Let \(y^0 = 0\), and \(\{ (y^n_t,z^n_t)_{t\in[0,T]}\}_{n\ge1}\) be a sequence in \(\mathcal{B}_2\) defined recursively as \[\begin{align} y_t^{n+1} = \xi_T+ \int_{t}^T g(s,y_s^n)dD_s - \int_{t}^T z_s^{n} dW_s ,\quad 0\le t \le T. \end{align}\] For any \(y^n\in \mathcal{S}_2\), we construct \(y^{n+1}\) and \(z^n\) as follows. From [item:A1] and [item:A2], \[\begin{align} &\mathbb{E}\bigg[\bigg(\xi_T+\int_0^{T} g(s,y^n_s)d D_s \bigg)^2 \bigg] \le 2 \mathbb{E}\bigg[\xi_T^2+ \bigg(\int_0^{T} \big(|g(s,0)| + C_g e^{-\rho s}|y^n_s|\big)d D_s \bigg)^2 \bigg]\\ &\quad \le2\mathbb{E}[\xi_T^2] + 4 \mathbb{E}\bigg[\int_0^{T} |g(s,0)|dD_s\bigg]^2+ C_g \mathbb{E}\bigg[\sup_{0\le t\le T}|y^n_t|^2 \bigg]K^2<\infty. \end{align}\] This implies that \(\{ \mathbb{E}_t[\xi_T+\int_0^{T} g(s,y^n_s)d D_s ]\}_{t\ge0}\) is a square integrable martingale. By the martingale representation theorem, we construct a unique \(z^{n}\in\mathcal{M}_2\) and \(y^{n+1}\in\mathcal{S}_2\) such that for \(0\le t\le T\) \[\begin{align} &\int_0^t z^{n}_s dW_s = \mathbb{E}_t\bigg[\xi_T+\int_0^{T} g(s,y^n_s)d D_s \bigg] - \mathbb{E}\bigg[\xi_T+\int_0^{T} g(s,y^n_s)d D_s \bigg],\nonumber\\ &\text{ and }y^{n+1}_t: = \mathbb{E}_t\bigg[\xi_T+\int_t^{T} g(s,y^n_s)d D_s \bigg]. \end{align}\] For \(\tau \in \mathcal{T}\), applying Itô’s lemma to \((u_{s\wedge\tau}^{n+1})^2: = |y_{s\wedge\tau}^{n+1} - y_{s\wedge\tau}^{n}|^2\) gives \[\begin{align} &(u_{t\wedge\tau}^{n+1})^2 + \int_{{s\wedge\tau}}^{T\wedge\tau}| z_s^{n}- z_s^{n-1}|^2 ds \\ &= 2\int_{{s\wedge\tau}}^{T\wedge\tau} \big(g(s,y_s^n)- g(s,y_s^{n-1})\big) u_{s}^{n+1}dD_s - 2 \int_{{s\wedge\tau}}^{T\wedge\tau} u_s^{n+1}(z_s^{n}- z_s^{n-1}) dW_s. \end{align}\] It is clear by Aurand and Huang [16] that the above stochastic integral is a uniformly integrable martingale and has zero expectation. By taking expectations on both sides, together with condition [item:A1] it follows that \[\begin{align} &\mathbb{E} \big[(u_{t\wedge\tau}^{n+1})^2\big] + \mathbb{E}\bigg[\int_{t\wedge\tau}^{T\wedge\tau}| z_s^{n}- z_s^{n-1}|^2 ds\bigg]\\ &\quad \le 2\mathbb{E}\bigg[\int_{t\wedge\tau}^{T\wedge\tau} C_g u_{s}^n u_{s}^{n+1}e^{-\rho s}d{D}_s\bigg] \\ &\quad \le C_g \mathbb{E}\bigg[\int_{t\wedge\tau}^{T\wedge\tau} (u_{s}^n)^2 e^{-\rho s}d{D}_s\bigg] + C_g \mathbb{E}\bigg[\int_{t\wedge\tau}^{T\wedge\tau} (u_{s}^{n+1})^2 e^{-\rho s}d{D}_s\bigg]. \end{align}\] Lemma 26 implies \[\begin{align} \label{eq:thmA463pf951} \mathbb{E}[(u_t^{n+1})^2] \le C_g e^{C_g K} \mathbb{E}\bigg[\int_{t}^T (u_{s}^n)^2 e^{-\rho s} d{D}_s\bigg],\quad t\in[0,T]. \end{align}\tag{26}\] Consider a sequence stopping time: \(\zeta_{s}:=\inf\{t\ge0: \int_0^t e^{-\rho \tau}D_{\tau}\ge s \}\wedge T\) for \(s\in[0,K]\), which is bounded. From the proof of Lemma 26, it is obvious that \[\begin{align} \label{eq:thmA463pf952} \mathbb{E}[(u_{\zeta_t}^{n+1})^2] \le C_g e^{C_g K} \mathbb{E}\bigg[\int_{\zeta_t}^T (u_{s}^n)^2 e^{-\rho s} d{D}_s\bigg] \end{align}\tag{27}\] holds for \(t\in[0,K]\). Iterating 26 and 27 gives for \(t\in[0,T]\), \[\begin{align} \mathbb{E}[(u_t^{n+1})^2] &\le C_g e^{C_g K} \mathbb{E}\bigg[\int_{t}^T (u_{s}^n)^2 e^{-\rho s} d{D}_s\bigg]\nonumber\\ &\le C_g e^{C_g K} \mathbb{E}\bigg[\int_{0}^T (u_{s}^n)^2 e^{-\rho s} \mathbb{1}_{ \zeta_0 \le s \le \zeta_K} d{D}_s\bigg]\nonumber\\ &= C_g e^{C_g K} \mathbb{E}\bigg[\int_{0}^{K} (u_{\zeta_s}^n)^2 ds \bigg] = C_g e^{C_g K} \int_{0}^{K} \mathbb{E}[ (u_{\zeta_s}^n)^2] ds\nonumber\\ &\le (C_g e^{C_g K})^2 \int_{0}^{K} \mathbb{E}\bigg[\int_{\zeta_s}^T (u_{t}^{n-1})^2 e^{-\rho t} d{D}_t\bigg] ds\nonumber\\ &= (C_g e^{C_g K})^2 \int_{0}^{K} \int_{t_1}^{K} \mathbb{E}[ (u_{\zeta_{t_2}}^{n-1})^2] d t_2 d t_1 \nonumber\\ &\cdots\nonumber\\ &\le (C_g e^{C_g K})^n \int_{0}^{K} \int_{t_1}^{K}\cdots \int_{t_{n-1}}^K \mathbb{E}[ (u_{\zeta_{t_n}}^{1})^2] d t_{n} dt_{n-1}\cdots d t_1\nonumber\\ &\le (C_g e^{C_g K})^n \mathbb{E}\bigg[\sup_{t\in[0,T]}(u_t^1)^2\bigg] \frac{K^n}{n!}.\nonumber \end{align}\] The above estimate implies that \(\{(y^n,z^n)\}\) is a Cauchy sequence in \(\mathcal{S}_2\times\mathcal{M}_2\), due to the fact that \(y^1\in\mathcal{S}_2\) and so \(\mathbb{E}[\sup_{t\in[0,T]}(u_t^1)^2]\) is finite. It follows that \(\lim_{n\rightarrow\infty}(y^n,z^n) = (y,z) \in \mathcal{B}_2\). By standard arguments, we show that \((y,z)\) solves BSDE 5 with \((T,\xi_{T},g,D)\). The uniqueness of \((y,z)\) follows standard arguments with comparison theorem (see Proposition 29). ◻

Proof. Observe that \((\overline{Y})^{\frac{1}{1-R}}\) is the classical dividend-paying utility process, it follows directly that \((\overline{Y})^{\frac{1}{1-R}}\) is the unique solution to BSDE with parameter \((\tau, (\xi_{\tau})^{\frac{1}{1-R}}, \overline{g},D)\), with \(\overline{g}(t,v)=e^{-\rho t}\). Since \((\tau, D,\xi_{\tau})\) satisfies Condition 1, \(\mathbb{E}[\sup_{t\ge 0}\overline{Y}_t^{\frac{2}{1-R}}]<\infty\) by Doob’s maximal inequality. \(\overline{Y}\in\mathcal{S}_2\) follows as a consequence of Jensen’s inequality. ◻

Proof. Since aggregator \(\underline{g}(\cdot,y)\) satisfies condition in Theorem 27, we obtain the existence and uniqueness of the solution to BSDE 5 with parameters \((\xi,\underline{g},T,D)\). It follows from the comparison theorem (Proposition 29) that \(\underline{Y}_t \ge \tilde{Y}_t\), where \((\tilde{Y},\tilde{Z})\) is the unique solution to BSDE \[\begin{align} \tilde{Y}_t = \xi_T - \int_t^{T} e^{-\rho s} R \tilde{Y}_s dD_s - \int_{t}^{\tau} \tilde{Z}_t dW_s. \end{align}\] Moreover, since \(\tilde{Y}_t = \mathbb{E}_t[e^{-R \int_{t}^T e^{-\rho s}dD_s}\xi_T ] \ge e^{-R \|\int_0^T e^{-\rho s} dD_s \|_{\infty}} \mathbb{E}_t[\xi_T]\) \(\mathbb{P}\)-a.s., this completes the proof. ◻

Proof. For fixed \((\omega,t)\), consider the sequence \(g_n(\omega,t,y)\) associated with \(g_{\text{EZ}}\) in 3 constructed by Lemma 32. It follows by Theorem 27 that, for each \(n\ge1\), the BSDE with parameter \((T,\xi_T,g_n,D)\) has a unique solution \((y_t^n,z_t^n)_{t\in[0,\tau]} \in \mathcal{B}_2\). Lemma 32 [item:prop_agg_i] and [item:prop_agg_iii], Proposition 29 yield that for \(n\ge1\), \(y_t^1(\omega)\le y_t^{n}(\omega)\le y_t^{n+1}\), \(d\mathbb{P}\times dt\)-a.s. Note that \(y^1\) is identical to \(\underline{Y}\) defined in Lemma 31 (with current triplet \((T,D,\xi_T)\)), since \(g_1\) and \(\underline{g}\) are identical. Under the condition that \(\xi_{T}\in L^2(\mathcal{F}_T)_+\) and \(\int_0^T e^{-\rho s} dD_s\le K\) a.s., by Lemma 31, there is a real constant \(C_{\xi}\) such that \(y^1_t = \underline{Y}_t \ge e^{-R K} \mathbb{E}_t[\xi_T] \ge C_{\xi}\) for \(t\in[0,T]\) a.s.

We show that \((y^n)_{n\in \mathbb{N}}\) is bounded from above by \(\overline{Y}\) as defined in Lemma 30 (with current triplet \((T,D,\xi_T)\)). For \(n\ge1\), a direct application of Itô’s formula (cf. Jacod and Shiryaev [76]) to \(\psi:=(y^n)^{\frac{1}{1-R}}\) yields \[\begin{align} d\psi_t =& -\frac{\psi_t^{R} g_n(t,\psi_t^{1-R}) }{1-R}dD_t + \frac{(\psi_{t})^{R}z_t^n }{1-R}dW_t + \frac{R}{2(1-R)^2}(\psi_{t})^{2R-1}|z_t^n|^2 dt. \end{align}\] Note that \(\int_0^{\cdot} \frac{R}{2(1-R)^2}(\psi_{t})^{2R-1}|z_t^n|^2 dt\) is an increasing process, so \(\psi\) is a subsolution of 5 with \((T,(\xi_T)^{\frac{1}{1-R}},h,D)\), where \(h(t,y) = y^{R} g_n(t,y^{1-R})/(1-R )\). Since \(g_n\le g_{\text{EZ}}\), we have \(y^{R}g_n(t,y^{1-R})/(1-R) \le y^{R}g_{\text{EZ}}(t,y^{1-R})/(1-R) = e^{-\rho t}\). Proposition 29 yields that \(\psi \le (\overline{Y})^{\frac{1}{1-R}}\) so that \(y^n \le \overline{Y}\).

Now for each \(n\ge1\), \(C_{\xi} \le y_t^1(\omega)\le y_t^{n}(\omega)\le y_t^{n+1}\le \overline{Y}_t\), \(d\mathbb{P}\times dt\)-a.s., there exists an \(\mathcal{F}_t\)-progressively measurable process \(y\) satisfying \(\lim_{n\rightarrow\infty} y_t^n(\omega) = y_t(\omega)\), \(d\mathbb{P}\times dt\)-a.s. Hence, \(\mathbb{E}[\sup_{s\in[0,T]}|y_s|^2] \le \mathbb{E}[\sup_{s\in[0,T]}|\overline{Y}_s|^2]<\infty.\) In order to take limit of \((z^n_t)_{t\in[0,T]}\), we derive the following estimate by applying Itô’s lemma to \((y_t^n)^2\): \[\begin{align} \mathbb{E}\bigg[ \int_{0}^{T} |z^n_s|^2 ds\bigg]&= \mathbb{E}[\xi_T^2] + 2 \mathbb{E}\bigg[\int_0^{T} y_s^ng_n(s,y_s^n)d D_s \bigg] \nonumber \\ &\le \mathbb{E}[\xi_T^2] + 2 \mathbb{E}\bigg[\int_0^{T} y_s^ng_{\text{EZ}}(s,y_s^1)d D_s \bigg] \\ &\le \mathbb{E}[\xi_T^2]+ 2C_{\xi}^{\frac{-R}{1-R}} \mathbb{E}\bigg[\sup_{s\in[0,T]}|\overline{Y}_s|^2\bigg] \bigg\|\int_0^T e^{-\rho s} dD_s \bigg\|_{\infty} <\infty, \nonumber \end{align}\] where we use the fact that \(0\le g^{n}(t, y^n) \le g_{\text{EZ}}(t,y^n) \le g_{\text{EZ}}(t,y^1) \le C_{\xi}^{\frac{-R}{1-R}}e^{-\rho t}\) for any \(n\) and \(t\in[0,T]\). Therefore, there exists \(z\in\mathcal{M}_2\) and a sub sequence \((z^{n_j})_j\) of \((z^n)_n\) such that \(z^{n_j} \rightharpoonup z\) weakly in \(\mathcal{M}_2.\)

Passing to a subsequence if necessary, we can show that the whole sequence \((z^n)_n\) converges strongly to \(z\) in \(\mathcal{M}_2\) by Lebesgue’s DCT. Moreover, \(y\) is a continuous process because \(y^n\) has continuous paths by construction. We can now pass the limit in \[\begin{align} y_t^n = \xi_T + \int_{t}^T g_n(s,y^n_s)d D_s - \int_t^T z_s^n dW_s, \end{align}\] so that \((y_t,z_t)_{t\in[0,T]}\in\mathcal{B}_2\) is a solution to BSDE with \((T,\xi_T,g_{\text{EZ}},D)\). ◻

Proof. We aim to extend the result in Theorem 33 to the case with random horizon and unbounded controls.

For each \(n\in\mathbb{N}\), we construct a solution \((Y_t^n,Z_t^n)_{t\ge0} \in \mathcal{B}_2\) to the BSDE \[\begin{align} \label{eq:Y95n95construct} Y_t^n = \xi_{\tau} + \int_{t \wedge {\tau}}^{{\tau}} \mathbb{1}_{s\in[0,n\wedge\tau^n]} g_{\text{EZ}}(s,Y_s^n)dD_s - \int_{t \wedge {\tau}}^{{\tau}} Z_s^n dW_s, \quad t\ge0, \end{align}\tag{28}\] where \(\tau^n\in\mathcal{T}\) is defined as \(\tau^n: = \inf\{t\ge 0: \int_0^{t}e^{-\rho s}d D_s \ge n \}\). We denote by \(\overline{\tau}^n := {\tau}\wedge\tau^n\). Theorem 33 implies that there exits a unique solution \((y_t,z_t)_{t\in[0,n]}\in\mathcal{B}_2\) for fixed horizon \([0,n]\) BSDE \[\begin{align} y_t = \mathbb{E}_n[ \xi_{\tau}] + \int_{t}^{n} \mathbb{1}_{s\in[0,\overline{\tau}^n]} g_{\text{EZ}}(s,y_s)dD_s - \int_t^n z_s dW_s. \end{align}\] The conditions in Theorem 33 are satisfied since \(\int_{0}^n \mathbb{1}_{s\in[0,\overline{\tau}^n]} e^{-\rho s}dD_s \le n\) a.s. On the other hand, under the condition that \(\xi_{\tau}\in L^2_+(\mathcal{F}_{\tau})\), martingale representation theorem implies that there exists \((\eta_t)_{t\ge 0} \in \mathcal{M}_2\) such that \[\begin{align} \mathbb{E}_{t}[\xi_{\tau}] = \xi_{\tau}- \int_t^{{\tau}} \eta_s dW_s,\text{ on }\{t<{\tau}\};\quad \eta_t = 0 \text{ on }\{t >{\tau} \}. \end{align}\] So we construct \((Y_t^n,Z_t^n)_{t\ge0}\in \mathcal{B}_2\) as \[\begin{align} Y_t^n = y_t \mathbb{1}_{t\in[0,n\wedge\overline{\tau}^n]} + \mathbb{E}_{t}[\xi_{\tau}]\mathbb{1}_{t\in(n\wedge\overline{\tau}^n,\infty)},\quad Z_t^n = z_t \mathbb{1}_{t\in[0,n\wedge\overline{\tau}^n]} + \eta_t\mathbb{1}_{t\in(n\wedge\overline{\tau}^n,\infty)}. \end{align}\] Next, we aim to show that \((Y^n,Z^n)_{n\in\mathbb{N}}\) is Cauchy in \(\mathcal{B}_2\).

For any \(m>n\), let \(\Delta Y_t: =Y_t^m - Y_t^n\), \(\Delta Z_t : =Z_t^m-Z_t^n\). Then it suffices to show that \(|(\Delta Y, \Delta Z)|_{\mathcal{B}_2}\rightarrow 0\) as \(m,n\rightarrow \infty\).

For \(0\le t \le n\), by 28 and the fact that \(\tau^{n}(\omega)\le \tau^{m}(\omega)\) we have \[\begin{align} &\Delta Y_{t}-\Delta Y_{n\wedge {\tau}}\\ &\quad = \int_{t\wedge \overline{\tau}^n}^{n\wedge \overline{\tau}^n} \Delta g_{\text{EZ}}^{m,n}(s)dD_s + \int_{n\wedge \overline{\tau}^n}^{n\wedge \overline{\tau}^m} g_{\text{EZ}}(s,Y_s^m) dD_s- \int_{t\wedge {\tau}}^{n\wedge {\tau}} \Delta Z_s dW_s, \end{align}\] where \(\Delta g_{\text{EZ}}^{m,n}(s): =g_{\text{EZ}}(s,Y_s^m) - g_{\text{EZ}}(s,Y_s^n)\). It follows from Itô’s lemma that \[\begin{align} &|\Delta Y_{t\wedge{\tau}}|^2 + \int_{t\wedge{\tau}}^{n\wedge {\tau}} |\Delta Z_s|^2 ds = 2 \int_{t\wedge\overline{\tau}^n}^{n\wedge \overline{\tau}^n} \Delta Y_s \Delta g_{\text{EZ}}^{m,n}(s) dD_s \\ &\quad + 2 \int_{n\wedge \overline{\tau}^n}^{n\wedge \overline{\tau}^m} \Delta Y_s g_{\text{EZ}}(s,Y_s^m) dD_s- 2 \int_{t\wedge{\tau}}^{n\wedge{\tau}} \Delta Y_s \Delta Z_s dW_s +| \Delta Y_{n\wedge {\tau}}|^2 . \end{align}\] By the fact that \(g_{\text{EZ}}\) is non-increasing in its second argument, so that \(\Delta g_{\text{EZ}}^{m,n}(s)\) should have opposite sign of \(\Delta Y_s\). This, together with the non-decreasing property of \(D\), gives \(\int_{t\wedge \overline{\tau}^n}^{n\wedge \overline{\tau}^n} \Delta Y_s \Delta g_{\text{EZ}}^{m,n}(s)dD_s\le0\). \[\begin{align} &\int_{n\wedge \overline{\tau}^n}^{n\wedge \overline{\tau}^m} \Delta Y_s g_{\text{EZ}}(s,Y_s^m) dD_s = \int_{n\wedge \overline{\tau}^n}^{n\wedge \overline{\tau}^m} \Delta Y_s \big(\Delta g_{\text{EZ}}^{m,n}(s)+ g_{\text{EZ}}(s,Y^n_s)\big) dD_s \\ &\quad \le \int_{n\wedge \overline{\tau}^n}^{n\wedge \overline{\tau}^m} \Delta Y_s g_{\text{EZ}}(s,Y^n_s) dD_s = \int_{n\wedge \overline{\tau}^n}^{n\wedge \overline{\tau}^m} \Delta Y_s g_{\text{EZ}}(s,\mathbb{E}_s[\xi]) dD_s\\ &\quad \le 2 \sup_{t\in[0,{\tau}]}|\overline{Y}_t| C_4^{\frac{-R}{1-R}} \int_{n\wedge \overline{\tau}^n}^{n\wedge \overline{\tau}^m} e^{-\rho s} dD_s , \end{align}\] where the second line follows by construction that \(Y^n_s = \mathbb{E}_{s}[\xi]\) for \(s>n\wedge\tau^n\), the last line follows by the fact that \(\xi_{\tau} \ge C_4\) \(\mathbb{P}\)-a.s. for some constant \(C_4>0\), as well as the priory bound \(Y^m\), \(Y^n\le \overline{Y}\) given in Lemma 30. Hence, \[\begin{align} &|\Delta Y_{t\wedge{\tau}}|^2 + \int_{t\wedge {\tau}}^{n\wedge {\tau}} |\Delta Z_s|^2 ds\\ &\quad \le | \Delta Y_{n\wedge {\tau}}|^2 +2 \sup_{t\in[0,{\tau}]}|\overline{Y}_t| C_4^{\frac{-R}{1-R}} \int_{n\wedge \overline{\tau}^n}^{n\wedge \overline{\tau}^m} e^{-\rho s} dD_s - 2\int_{t\wedge {\tau}}^{n\wedge {\tau}} \Delta Y_s \Delta Z_s dW_s, \end{align}\] where the above local martingale is a uniformly integrable martingale. Hence, \[\begin{align} \mathbb{E}\bigg[\int_{0}^{n\wedge {\tau}} |\Delta Z_s|^2 ds \bigg] &\le \mathbb{E}[ | \Delta Y_{n\wedge {\tau}}|^2] + 2 C_4^{\frac{-R}{1-R}} \mathbb{E}\bigg[ \sup_{t\in[0,{\tau}]}|\overline{Y}_t| \int_{n\wedge \overline{\tau}^n}^{n\wedge \overline{\tau}^m} e^{-\rho s} dD_s\bigg] \nonumber \\ &=: \mathbb{E}[ | \Delta Y_{n\wedge {\tau}}|^2] + 2C_4^{\frac{-R}{1-R}} \operatorname{I}^{n,m}. \label{eq:temp95est95YD} \end{align}\tag{29}\] Moreover, it holds that \[\begin{align} \operatorname{I}^{n,m}&\leq \mathbb{E}\bigg[\sup_{t\in[0,{\tau}]}|\overline{Y}_t|^2\bigg]^{\frac{1}{2}} \mathbb{E}\bigg[\bigg( \int_{n\wedge \overline{\tau}^n}^{n\wedge \overline{\tau}^m}e^{-\rho s} dD_s \bigg)^2\bigg]^{1/2}\leq C_5(m-n).\label{eq:temp95est95YD951} \end{align}\tag{30}\] where the constant \(C_5>0\) (not depending on \(m\),\(n\),\(\Delta Y\),\(\Delta Z\)) exists because \(D\in{\@fontswitch\relax\mathcal{A}}\) has continuous path. In a similar manner, we have that \[\begin{align} &\mathbb{E}\bigg[ \sup_{t\in[0,n]} |\Delta Y_{t\wedge {\tau}}|^2 \bigg] \le \mathbb{E}[ | \Delta Y_{n\wedge {\tau} }|^2]+2C_4^{\frac{-R}{1-R}} \operatorname{I}^{n,m}+ 2\mathbb{E}\bigg[\sup_{t\in[0,n]}\int_{0}^{t\wedge {\tau}} \Delta Y_s \Delta Z_s dW_s \bigg]\nonumber\\ &\quad =: \mathbb{E}[ | \Delta Y_{n\wedge {\tau} }|^2]+2C_4^{\frac{-R}{1-R}} \operatorname{I}^{n,m}+2 \operatorname{II}^{n,m},\label{eq:temp95est95YD952} \end{align}\tag{31}\] where \(\operatorname{I}^{n,m}\) is given in 29 , and that \[\begin{align} \operatorname{II}^{n,m} &\leq C_6 \mathbb{E}\bigg[\bigg(\int_{0}^{n\wedge {\tau}} |\Delta Y_s|^2 |\Delta Z_s|^2 ds \bigg)^{1/2} \bigg]\nonumber \\ &\leq \frac{1}{4}\mathbb{E}\Big[ \sup_{t\in[0,n]} |\Delta Y_{t\wedge {\tau}}|^2 \Big] + C_6^2 \mathbb{E}\bigg[\int_{0}^{n\wedge {\tau}} |\Delta Z_s|^2 ds \bigg],\label{eq:temp95est95YD953} \end{align}\tag{32}\] where the first inequality follows from the inequality of arithmetic and geometric means, and the constant \(C_6>0\) (not depending on \(m\),\(n\),\(\Delta Y\),\(\Delta Z\)) exists by the Burkholder-Davis-Gundy’s inequality. Combining 29 , 30 , 31 , and 32 , we have some constant \(C_7>0\) (that does not depend on \(m\),\(n\)) such that \[\begin{align} \mathbb{E}\bigg[ \sup_{0\le t\le n}| \Delta Y_{t\wedge {\tau}}|^2+\int_{0}^{n\wedge {\tau}} |\Delta Z_s|^2 ds \bigg] \le C_7 \Big(\mathbb{E}\big[ | \Delta Y_{n\wedge {\tau}}|^2\big]+(m-n)\Big). \end{align}\] Next, for \(n<t \le m\), we have \[\begin{align} \Delta Y_{t} = \int_{t \wedge \overline{\tau}^n}^{m\wedge \overline{\tau}^m} g_{\text{EZ}}(s,Y_s^m)dD_s - \int_{t\wedge {\tau}}^{m\wedge {\tau}} {\Delta}Z_s dW_s, \end{align}\] where the equality holds by the fact that \(\Delta Z_s = 0\) on \(s > m\) since \(Z^m_s = Z^n_s =\eta_s\) for \(s>m\wedge{\tau} \ge m \wedge \overline{\tau}^m\).

It then follows from Itô’s lemma and similar arguments as above that \[\begin{align} &|\Delta Y_{t\wedge{\tau}}|^2 + \int_{t\wedge {\tau}}^{m\wedge{\tau}} |{\Delta}Z_s|^2 ds\\ &\quad = 2 \int_{t\wedge \overline{\tau}^m}^{m\wedge\overline{\tau}^m} \Delta Y_s g_{\text{EZ}}(s,Y^m_s) dD_s - 2\int_{t\wedge {\tau}}^{m\wedge {\tau}} \Delta Y_s {\Delta}Z_s dW_s\\ &\quad \le 2 \int_{t\wedge \overline{\tau}^m}^{m\wedge\overline{\tau}^m} |\Delta Y_s| g_{\text{EZ}}(s,\mathbb{E}_s[\xi_{\tau}]) dD_s - 2\int_{t\wedge {\tau}}^{m\wedge {\tau}} \Delta Y_s {\Delta}Z_s dW_s. \end{align}\] Note that \(n\wedge \overline{\tau}^m \le t\wedge \overline{\tau}^m \le m\wedge \overline{\tau}^m\), therefore similar argument as for the case \(0\le t \le n\) gives the existence of \(C_8>0\) independent of \(m\), \(n\) such that \[\begin{align} \mathbb{E}\bigg[ \sup_{n\le t\le m}| \Delta Y_{t\wedge {\tau}}|^2+\int_{n \wedge {\tau}}^{m \wedge {\tau}} |\Delta Z_s|^2 ds \bigg] \le C_8 (m-n). \end{align}\] Finally, \(Y_t^m = Y_t^n = \mathbb{E}_{t\wedge{\tau}}[\xi_{\tau}]\) and \(Z_t^m = Z_t^n = \eta_t\) for \(t>m\wedge{\tau}\ge m\wedge\overline{\tau}^m\), so \[\begin{align} \label{eq:mn95cauchy951} \mathbb{E}\bigg[ \sup_{t\ge 0}| \Delta Y_{t\wedge {\tau}}|^2+\int_{0}^{{\tau}} |\Delta Z_s|^2 ds \bigg] \le C_7 \mathbb{E}\big[ | \Delta Y_{n\wedge {\tau}}|^2\big] + C_9(m-n), \end{align}\tag{33}\] with \(C_9:=(C_7+C_8)\). By the DCT, we have \[\begin{align} \lim\limits_{n,m\rightarrow\infty} \mathbb{E}\big[ | \Delta Y_{n\wedge {\tau}}|^2\big] &= \mathbb{E}\Big[ \lim\limits_{n,m\rightarrow\infty}|Y^m_{n\wedge {\tau}} - Y^n_{n\wedge {\tau}}|^2 \Big] \\ &=\mathbb{E}\Big[ \lim\limits_{n\rightarrow\infty}|Y_{n\wedge\tau}-Y^n_{n\wedge {\tau}}|^2\Big] = \mathbb{E}\big[ | \xi_{\tau} - \xi_{\tau}|^2\big] = 0. \end{align}\] Since the second term in 33 vanishes as \(m,n\rightarrow\infty\), we conclude that \(\|(\Delta Y,\Delta Z)\|_{\mathcal{B}_2} \rightarrow 0\) as \(m\), \(n\rightarrow\infty\), i.e., \((Y^n,Z^n)_n\) is Cauchy in \(\mathcal{B}_2\). Since \(\mathcal{B}_2\) is complete, \(\lim_{n\rightarrow\infty}(Y^n,Z^n) = (Y,Z)\in\mathcal{B}_2\) exists. Finally, we show the limit \((Y,Z)\in\mathcal{B}_2\) solves BSDE 5 with parameter \((\tau,\xi_{\tau},g_{\text{EZ}},D)\). For each \(n\in\mathbb{N}\), \((Y_t^n,Z_t^n)_{t\ge0} \in \mathcal{B}_2\) solves the BSDE 28 , \[\begin{align} Y_t^n = \xi_{\tau} + \int_{t \wedge {\tau}}^{{\tau}} \mathbb{1}_{s\in[0,n\wedge\tau^n]} g_{\text{EZ}}(s,Y_s^n)dD_s - \int_{t \wedge {\tau}}^{{\tau}} Z_s^n dW_s, \quad t\ge0. \end{align}\] It is not difficult to pass the limit \(n\rightarrow\infty\) in the above equation and show that each term of it converges to a corresponding term in 5 for almost all \(\omega\in\Omega\) uniformly in \(t\ge0\), thus we omit the details here. ◻

Proof. Define the process \(\theta^0: = (\theta^0(t)\equiv 0 )_{t\ge0}\). It is easy to verify that \(\theta_0 \in \Theta\) and the induced measure \(\mathbb{Q}^{\theta^0}\) coincides with the reference measure \(\mathbb{P}\). It then follows by definition of \(J^*\) in 21 that \(J^*(x)\le \sup_{D\in \mathcal{A}_x} \mathbb{E}[ \int_{0}^{\tau^{x,D}} e^{-\rho t}dD_t ] + \xi_0.\) Moreover, we observe that \[\begin{align} \mathbb{E}\bigg[ \int_{0}^{\tau^{x,D}} e^{-\rho t}dD_t \bigg] = x + \mathbb{E}\bigg[\int_{0}^{\tau_N}e^{-\rho t} \big(\mu(X_t^D)-\rho X_t^D\big)dt- e^{-\rho \tau_N} X^D_{\tau_N} \bigg], \end{align}\] where \(\tau_N: = \tau^{x,D}\wedge N \wedge \inf\{t\ge0: X_t^D \ge N \}\) is an almost surely finite stopping time. Further invoking the requirement that admissible \(D\) keep the controlled process \(X^{x,D}\) non-negative, we get the following inequality of \(J^*\) \[\begin{align} J^*(x)\le x + \xi_0 + \sup_{D\in \mathcal{A}_x} \mathbb{E}\bigg[ \int_{0}^{\tau^{x,D}} e^{-\rho t}\big(\mu(X_t^D)-\rho X_t^D\big)^+ dt \bigg] \le x+ \xi_0 + \frac{\overline{\mu}}{\rho}, \end{align}\] where \((y)^+ = \max(y,0)\). The proof is complete. ◻

Proof. The proof involves a transformed process \(U_t\) of \(X_t\), which is a time-changed Brownian motion, and we prove that auxiliary versions of ?? , ?? hold with the process \(U_t\). We introduce the scale function of \(X\) in 17 as \[\begin{align} \mathfrak{s}(x): = \int_{0}^{x}\exp\Big(-2\int_0^z\frac{\mu(y)}{\sigma^2(y)}dy\Big)dz, \end{align}\] which is strictly increasing, continuously differentiable bijection of the interval \(\mathcal{X}\) onto the interval \(I=(\iota_{l},\iota_{\infty})\) with endpoints \(\iota_l:= \mathfrak{s}(\ell+)\) and \(\iota_{\infty}:= \mathfrak{s}(\infty-)\); see, e.g., Borodin and Salminen [77]. Denote by \(\mathfrak{s}^{-1}\) the inverse mapping of \(\mathfrak{s}\). We define \(U_t : = \mathfrak{s}(X_t)\), with state space \(I\), and its dynamic is given by \[\begin{align} \label{pf:lem:X95bound950} U_t = \mathfrak{s}(X_0) + \int_0^t \varsigma(U_s) dW_s, \end{align}\tag{35}\] where the dispersion function \(\varsigma(u): = (\mathfrak{s}'\cdot\sigma)(\mathfrak{s}^{-1}(u))\); see e.g., Karatzas and Shreve [78]. To further remove the dependency on the function \(\varsigma\) in 35 , we consider a time change for transformation. According to the Dambis–Dubins–Schwarz theorem (see Revuz and Yor [79]), one can rewrite the stochastic integral \(\int_{0}^{\cdot} \varsigma(U_s) dW_s\) of Brownian motion \(W\) as a time-changed Brownian motion on (an extension of) the probability space, i.e., \(\hat{W}_{\Lambda^U(\cdot)}\) that we define as follows. Define a time scale \(\Gamma^U(r)\) for \(r \in{\mathbb{R}_+}\) as \[\begin{align} \Gamma^{U}(r):=\int_0^{r} \frac{dt}{\varsigma^2\big(\mathfrak{s}(X_0)+\hat{W}_t\big)} \end{align}\] in terms of a standard Brownian motion \(\hat{W}\), and denote by \(\Lambda^{U}\) the inverse mapping of \(\Gamma^{U}\), i.e. \(\Lambda^U(t): = \inf\{r\ge 0: \Gamma^U(r)>t\}\). The representation \[\begin{align} \label{pf:lem:X95bound951} U_t = \mathfrak{s}(X_0) + \hat{W}_{\Lambda^{U}(t)} \end{align}\tag{36}\] holds. Due to the strict monotonicity of the scale function \(\mathfrak{s}\), it is sufficient to prove auxiliary properties of ?? ?? for the process \(U\) given by 36 . For convenience of presentation, we adopt the notation \[\begin{align} &\zeta^U_{y|x} :=\inf\{t\ge 0: U_t = \mathfrak{s}(y) \text{ given }U_0 =\mathfrak{s}(x)\}. \end{align}\] From the continuous path properties of standard Brownian motion together with the fact that \(\Lambda^{U}(t)>0\) holds for all \(t>0\), it is proved (see e.g.[77]) \[\begin{align} &\mathbb{P}\Big(\big \{\omega\in\Omega: \exists \epsilon(\omega)>0 \text{ s.t. } U_t \le \mathfrak{s}(x),\forall t\in[0,\epsilon(\omega)] \big \} \big| X_0 = x \Big)=0,\\ & \mathbb{P}\Big(\big \{\omega\in\Omega: \exists \epsilon(\omega)>0 \text{ s.t. } U_t \ge \mathfrak{s}(x),\forall t\in[0,\epsilon(\omega)] \big \} \big| X_0 = x \Big)=0. \end{align}\] Given the above statements, we have for any fixed \(\delta>0\), \[\begin{align} &\mathbb{P}( \zeta^U_{a|x} \ge \delta) = \mathbb{P}\Big(\inf_{t\in[0,\delta]} U_t \ge \mathfrak{s}(a) \big| X_0=x \Big) \rightarrow \mathbb{P}\Big(\inf_{t\in[0,\delta]} U_t > \mathfrak{s}(x) \big| X_0=x \Big) = 0,\\ & \mathbb{P}( \zeta^U_{b|x} \ge \delta) = \mathbb{P}\Big(\sup_{t\in[0,\delta]} U_t \ge \mathfrak{s}(b) \big| X_0=x \Big) \rightarrow \mathbb{P}\Big(\sup_{t\in[0,\delta]} U_t < \mathfrak{s}(x) \big| X_0=x \Big) = 0, \end{align}\] as \(a\uparrow x\) and \(b\downarrow x\) respectively. Hence, we conclude ?? .

To show ?? , we recognize from the time change property that \[\begin{align} \max_{0\le s\le \delta\wedge{\zeta}^U_{0|x}\wedge{\zeta}^U_{y|x}} U_s &= \max_{0\le r \le \Lambda^{U}(\delta)\wedge\Lambda^{U}(\zeta^U_{0|x})\wedge\Lambda^{U}(\zeta^U_{y|x})} \mathfrak{s}(x)+ \hat{W}_r\\ &= \max_{0\le r \le \Lambda^{U}(\delta)\wedge\hat{\zeta}_{0|x}\wedge\hat{\zeta}_{y|x}} \mathfrak{s}(x)+ \hat{W}_r, \end{align}\] where \(\hat{\zeta}_{y|x} = \inf\{ r\ge 0: B_x(r)= \mathfrak{s}(y) \}\), with \(B_x(r):= \mathfrak{s}(x)+ \hat{W}_r\).

By the monotonicity of \(\mathfrak{s}\), \(B_x(t,\omega)\) decreases as \(x\downarrow 0\). By the law of iterated logarithm (see Revuz and Yor [79]), we have \(\hat{\zeta}_{0|x}\rightarrow0\) as \(x\downarrow 0\). By definition, \(\Lambda^{\Upsilon}(\delta)\) is finite a.s. for \(\delta\in[0,\infty)\). Since \(B_x(r)\) has continuous path, considering a decreasing sequence \((x_n)_{n\in\mathbb{N}}\) with \(x_n<y\) and \(\lim_{n\rightarrow\infty} x_n=0\), we have \(\lim_{n\rightarrow\infty} \max_{0\le r \le \Lambda^{U}(\delta)\wedge\hat{\zeta}_{0|x_n}\wedge\hat{\zeta}_{y|x_n}} B_{x_n}(r) = \mathfrak{s}(0),\) and \(\max_{0\le r \le \Lambda^U(\delta)\wedge\hat{\zeta}_{y|x}}B_x(r)\le \mathfrak{s}(y)\); hence ?? holds. ◻

Proof of Proposition 13.. The boundedness of \(J^*\) follows by Lemma 35. We first show the non-decreasing property of \(J^*\). If \(y>x\), then for any admissible pair \((D,\theta)\in\mathcal{A}^x \times \Theta^D\), we can put \(\hat{D}_t = D_t + y-x\), corresponding to immediately paying dividends in the amount of \(y-x\), thereby instantly changing the initial reserve from \(y\) to \(x\) and then following strategy \(D\). Obviously, \(\hat{D}\in\mathcal{A}^y\), and that \(J^{\theta}(x;\hat{D}) = J^{\theta}(x;D) + (y-x)\), it follows that \[J^*(y) \ge J^*(x)+ (y-x),\] thus, \(J^*\) is nondecreasing.
Next, we prove ?? . Fix \(\delta>0\) and \(y>0\), in view of Lemma 36, for any \(\epsilon\in(0,1)\), choose \(x>0\) such that \[\begin{align} \mathbb{P}(\zeta_{0|x} < \delta) \ge 1-\epsilon,\qquad \mathbb{E} \bigg[\max_{0\le s\le \delta\wedge{\zeta}_{{0|x}}\wedge\zeta_{y|x}} X_s\bigg] \le \epsilon. \end{align}\] Let \(\hat{\tau} = \tau^{x,D} \wedge \zeta_{y|x} \wedge \delta\), where \(\tau^{x,D}\) defined in 19 . Since \(0\le X^{x,D}\le X\), we have \(\tau^{x,D}\le \zeta_0\) and \(\mathbb{P}(\tau^{x,D} <\zeta_{y|x} \wedge \delta) \ge \mathbb{P}(\zeta_{0|x} <\zeta_{y|x} \wedge \delta) \ge 1-\epsilon\). Due to the requirement that \(- X^D_t \ge 0\) for all \(t\ge 0\), we have and \(D_{\hat{\tau}}\le X_{\hat{\tau}} \le \max_{0\le s\le \hat{\tau}}X_s\le \max_{0\le s\le \delta\wedge{\zeta}_{{0|x}}\wedge\zeta_{y|x}} X_s\). Therefore \(\mathbb{E}_x[D_{\hat{\tau}}]\le \epsilon\). Consider the objective function \(J^{\theta^0}(x;D)\) in 20 with \(\theta^0= (\theta^0(t)\equiv 0 )_{t\ge0}\), we have \[\begin{align} &J^{\theta^0}(x;D) = \mathbb{E}\Big[\int_0^{\hat{\tau}} e^{-\rho t} dD_t + \mathbb{1}_{\hat{\tau}<\tau^{x,D}}\int_{\hat{\tau}}^{\tau^{x,D}} e^{-\rho t} dD_t + \xi e^{-\rho \tau^{x,D}} \Big]\\ &\le \mathbb{E}[D_{\hat{\tau}}] + \mathbb{E}\bigg[ \mathbb{1}_{\hat{\tau}<\tau^{x,D}} \mathbb{E}\bigg[\int_{\hat{\tau}}^{\tau^{x,D}} e^{-\rho t} dD_t \bigg| \mathcal{F}_{\hat{\tau}} \bigg]\bigg] + \xi \\ &\le \epsilon + \mathbb{E}[ \mathbb{1}_{\hat{\tau}<\tau^{x,D}} J^*(y)] + \xi \le \epsilon + J^*(y)\mathbb{P}(\hat{\tau}<\tau^{x,D})+\xi \le \xi + \big(1+J^*(y)\big)\epsilon. \end{align}\] Due to the arbitrariness of \(\epsilon\), \(\inf_{\theta}J^{\theta}(0+;\hat{D})\le J^{\theta^0}(0+;\hat{D}) \le \xi\), thus by arbitrariness of \(D\), \(J^*(0+) \le \xi\); together with \(J^*(0+) \ge \xi\) by Lemma 35, we conclude the validity of ?? .
Lastly, we prove that \(J^*\) is continuous at any \(x>0\). For arbitrary \(\theta\in\Theta\), let the measure \(\mathbb{Q}^{\theta}\) be defined in 8 and Brownian motion \(W^{\theta}\) under measure \(\mathbb{Q}^{\theta}\). Under \(\mathbb{Q}^{\theta}\), the uncontrolled surplus \(X\) satisfies \[\begin{align} dX_t = \big(\mu(X_t) + \sigma(X_t)\theta(X_t) \big)dt + \sigma(X_t) dW^{\theta}_t,\quad X_0 = x. \end{align}\] For arbitrary \(x>0\), define the stopping time \(\zeta^{\mathbb{Q}}_{0,y|x}:=\inf\{t\ge0: X_t \notin (0,y)\}\) under measure \(\mathbb{Q}\). It is well-known from the classical theory of diffusion that \[\begin{align} \label{pf:prop95v95cts951} \mathbb{Q}(X_{\zeta^{\mathbb{Q}}_{0,y|x}} = y) = \frac{\mathfrak{s}^{\mathbb{Q}}(x) - \mathfrak{s}^{\mathbb{Q}}(0)}{ \mathfrak{s}^{\mathbb{Q}}(y) - \mathfrak{s}^{\mathbb{Q}}(0) }, \end{align}\tag{37}\] where \(\mathfrak{s}^{\mathbb{Q}}\) is the scale function of \(X_t\) under measure \(\mathbb{Q}\), that is, \[\mathfrak{s}^{\mathbb{Q}}(x) : = \int_{0}^{x}\exp\bigg(-2\int_0^z\frac{\mu(y)+\sigma(y)\theta(y)}{\sigma^2(y)}dy\bigg)dz.\] 37 implies that for arbitrary \(\epsilon>0\), there exists \(y({\epsilon}) > x\) such that \[\mathbb{Q}(X_{\zeta^{\mathbb{Q}}_{0,y|x}} = y) \ge 1-\epsilon~for~x< y<y(\epsilon).\] Then, following similar arguments for proving ?? , we have for any \(\epsilon>0\), there exists \(y'(\epsilon)>x\) such that \(\mathbb{Q}(\zeta^{\mathbb{Q}}_{y|x}<\zeta^{\mathbb{Q}}_{0|x}\wedge \epsilon) \ge 1-\epsilon\) for \(x<y<y'(\epsilon)\). For arbitrary \(D\in \mathcal{A}^y\) with \(x<y<y'(\epsilon)\), we construct a dividend strategy for surplus process \(X\) with \(X_0 = x\) as follows: \(D'_t = 0\) when \(t< \zeta^{\mathbb{Q}}_{0,y|x}\wedge\epsilon\) and for \(t\ge \zeta^{\mathbb{Q}}_{0,y|x}\wedge\epsilon\), \(D'_t = X_{\epsilon}\) on \(\{\epsilon< \zeta^{\mathbb{Q}}_{0,y|x}\}\), \(D'_t = D_{t-\zeta^{\mathbb{Q}}_{y|x}}\) on \(\{\zeta^{\mathbb{Q}}_{y|x}<\epsilon\wedge\zeta^{\mathbb{Q}}_{0|x}\}\), \(D_t \equiv 0\) otherwise. We have for \(x<y<y'(\epsilon)\), \[\begin{align} &J^{\theta}(x,{D}') = \mathbb{E}^{\theta} \bigg[ \int_{0}^{\tau^{x,D'}} e^{\int_{0}^{s} \frac{\theta^2_u}{2{\@fontswitch\relax\mathcal{R}}}du -\rho s} dD'_s + e^{\int_{0}^{\tau^{x,D'}} \frac{\theta^2_u}{2{\@fontswitch\relax\mathcal{R}}}-\rho du}\xi_{\tau^{x,D'}}\bigg] \\ &\ge \mathbb{E}^{\theta} \bigg[ \mathbb{1}_{\zeta^{\mathbb{Q}}_{y|x}<\zeta^{\mathbb{Q}}_{0|x}}\bigg(\int_{0}^{\zeta^{\mathbb{Q}}_{y|x}} e^{\int_{0}^{s} \frac{\theta^2_u}{2{\@fontswitch\relax\mathcal{R}}}du -\rho s} dD_s \\ &\qquad\qquad\qquad\qquad\quad +\int_{\zeta^{\mathbb{Q}}_{y|x}}^{\zeta^{\mathbb{Q}}_{0|y}} e^{\int_{0}^{s} \frac{\theta^2_u}{2{\@fontswitch\relax\mathcal{R}}}du -\rho s} dD_s+ e^{\int_{0}^{\zeta^{\mathbb{Q}}_{y|x}} \frac{\theta^2_u}{2{\@fontswitch\relax\mathcal{R}}}du}\xi_{\tau^{x,D'}}\bigg)\bigg]\\ &\ge \mathbb{E}^{\theta} \bigg[ \mathbb{1}_{\zeta^{\mathbb{Q}}_{y|x}<\zeta^{\mathbb{Q}}_{0|x}}\mathbb{E}^{\theta}\bigg[ \int_{\zeta^{\mathbb{Q}}_{y|x}}^{\zeta^{\mathbb{Q}}_{0|y}} e^{\int_{0}^{s} \frac{\theta^2_u}{2{\@fontswitch\relax\mathcal{R}}}du -\rho s} dD_s + e^{\int_{0}^{\zeta^{\mathbb{Q}}_{y|x}} \frac{\theta^2_u}{2{\@fontswitch\relax\mathcal{R}}}du}\xi_{\tau^{x,D'}}\bigg|\mathcal{F}_{\zeta^{\mathbb{Q}}_{y|x}}\bigg] \bigg]\\ & \ge \mathbb{Q}(\zeta^{\mathbb{Q}}_{y|x}<\zeta^{\mathbb{Q}}_{0|x}\wedge \epsilon)e^{-\rho\epsilon}J^{\theta}(y,{D}) \ge (1-\epsilon)e^{-\rho\epsilon}J^{\theta}(y,{D}). \end{align}\] Hence, \[\begin{align} &\inf_{\theta} J^{\theta}(y,{D}) - J^{*}(x) \le \inf_{\theta} J^{\theta}(y,{D}) - \inf_{\theta} J^{\theta}(x;D') \\ &\quad \le \inf_{\theta} J^{\theta}(y,{D}) -(1-\epsilon)e^{-\rho\epsilon}J^{\theta}(y,{D}) \le \big(1-(1-\epsilon)e^{-\rho \epsilon}\big) J^*(h), \end{align}\] for some \(h>y\) sufficiently large. Therefore \(J^*(x)\) is continuous at \(x>0\). ◻

Proof of Theorem 18.. From Proposition 37, define \(v^*:{\mathbb{R}_+}\rightarrow \mathbb{R}\) by \[\begin{align} \label{eq:const95v} v^*(x):=\left\{ \begin{aligned} &v_{b^*}(x)\quad &&if\;\;x\in [0,b^*];\\ &x-b^*+\psi^+(b^*)\quad && if\;\;x\in (b^*,\infty). \end{aligned} \right. \end{align}\tag{43}\] The function \(v_{b^*}\) is in \(\mathcal{C}^2(\mathbb{R}_+)\) and satisfies 38 with \(\gamma = 0\). In particular, \(v_{b^*}''(b^*) = 0\) by construction. Indeed, substituting the boundary conditions \(v_{b^*}'(b^*) = 1\) and \(v_{b^*}(b^*) = \psi^+(b^*)\) into the ODE at \(x = b^*\) gives \[\frac{\sigma^2(b^*)}{2} v_{b^*}''(b^*) + \mu(b^*) - {\@fontswitch\relax\mathcal{R}}\frac{\sigma^2(b^*)}{2} \frac{1}{\psi^+(b^*)} - \rho \psi^+(b^*) = 0.\] Since \(\psi^+(b^*)\) solves the quadratic equation \(Q(\psi; b^*) = 0\), it follows that \(v_{b^*}''(b^*) = 0\). By the definition given in 43 , the function \(v^*\) is in \(\mathcal{C}^2({\mathbb{R}_+})\) and satisfies the free boundary problem 24 to the left of \(b^*\). It remains to validate for \(x\in(b^*,\infty)\), \({\@fontswitch\relax\mathcal{L}} v^*\le 0\) as \(({v^*})^{\prime}(x)=1\) is trivial.

For \(x \in (b^*, \overline{b})\), note that \(\psi^+(x) - x\) is decreasing, as stated in Assumption 15 ii.. Consequently, we have \(v^*(x) = x - b^* + \psi^+(b^*) \ge \psi^+(x)\). Hence \(\mathcal{L}v^*(x)\) is given by \[\mathcal{L}v^*(x) = -\frac{Q(v^*(x); x)}{v^*(x)} \le 0,\] where \(Q(\psi; x)\) is defined in Assumption 15 ii..

If \(\overline{b} = \infty\), the proof concludes here. Otherwise, for \(x \in [\overline{b}, \infty)\), we have \(Q(v^*(x); x) < 0\), which implies \(\mathcal{L}v^*(x) < 0\). Thus, we conclude that \(v^*(x)\) satisfies \(\mathcal{L}v^* \le 0\) on the interval \((b^*, \infty)\). This means it solves the free boundary problem 24 with the free boundary \(b^* \in (\underline{b}, \hat{b})\). This completes the proof. ◻

Proof of of i.. We first note that by 42 , the following inequality holds: \[\begin{align} \label{eq:est95g1} g_{b,\gamma}'(b)= \frac{2}{(\sigma(b)\psi^+(b))^2}Q(\psi^+(b))+\frac{2\gamma}{(\sigma(b))^2}-\frac{1}{(\psi^+(b))^2}<-\frac{1}{(\psi^+(b))^2}, \end{align}\tag{44}\] where we use that \(Q(\psi^+(b))=0\) (defined in Assumption15ii.) and \(\gamma<0\).

Define \(\phi_{b,\gamma}:\mathcal{X}\rightarrow \mathbb{R}\) as \(\phi_{b,\gamma}(x) := g_{b,\gamma}(x) - 1/\psi^+(x)\). Since \(g_{b,\gamma}(b)={1}/{\psi^+(b)}\) and \((\psi^+)'(b)\le 1\) (because \(\psi^+(x)-x\) is decreasing on \([\underline{b},\overline{b})\) and \(b\) is in \((\underline{b},\overline{b})\); see Assumption15ii.), the inequality in 44 gives \[\begin{align} \label{eq:est95g2} \phi_{b,\gamma}(b)=0,~\phi_{b,\gamma}'(b)=g'_{b,\gamma}(b) + \frac{(\psi^+)'(b)}{(\psi^+(b))^2} < \frac{(\psi^+)'(b)-1}{(\psi^+(b))^2} \le0 . \end{align}\tag{45}\]

We claim that \(y_{b,\gamma}\leq \underline{b}\). Since the case with \(y_{b,\gamma}=0\) is trivial, we can and do consider only the case with \(y_{b,\gamma}=\sup\{x\in(0,b):\phi_{b,\gamma}(x)=0\}\). Arguing by contradiction, assume that \(y_{b,\gamma}>\underline{b}\). Then by the continuity of \(\phi_{b,\gamma}\) (due to that of \(g_{b,\gamma}\) and \(\psi^+\)), we have that \(\phi_{b,\gamma}(y_{b,\gamma})= g_{b,\gamma}(y_{b,\gamma}) - {1}/{\psi^+(y_{b,\gamma})}=0.\) Furthermore, using the same arguments devoted for the second property given in 45 (along with \(y_{b,\gamma}>\underline{b}\)), we have \[(\phi_{b,\gamma})'(y_{b,\gamma})=g'_{b,\gamma}(y_{b,\gamma}) + \frac{(\psi^+)'(y_{b,\gamma})}{(\psi^+(y_{b,\gamma}))^2} <0,\] which contradicts Lemma 38ii.. Hence, the claim holds true.

Furthermore, from the definition of \(y_{b,\gamma}\) and the fact that \((\phi_{b,\gamma})'(b)=(g_{b,\gamma}- 1/\psi^+)'(b)<0\) (see 45 ), the last statement of i. is true. This completes the proof.

To that end, let \(y_{b_2,\gamma_2}\) be defined as Lemma 39i. (by assigning \(b=b_2\) and \(\gamma=\gamma_2\)). Define \(\Delta g: [0,b]\ni x\rightarrow \Delta g(x)\in \mathbb{R}\) by \(\Delta g(x):= g_{b_1,\gamma_1}(x)-g_{b_2,\gamma_2}(x)\) for \(x\in [0,b_1]\). Since \(g_{b_i,\gamma_i}\), \(i=1,2\), is the solution of 42 (when \(b=b_i\), \(\gamma=\gamma_i\)), the following holds \[\begin{align} &\dfrac{1}{2}\sigma^2(x)(\Delta g)'(x) + \Delta g(x) \Big( \mu(x)+ \frac{(1-{\@fontswitch\relax\mathcal{R}})}{2}\sigma^2(x)\left({\Delta g(x)} + 2g_{b_2,\gamma_2}(x)\right)\Big) \nonumber\\ &= \gamma_1-\gamma_2,\quadfor\;\;x\in [0,b_1]\label{eq:est95g4952} \end{align}\tag{46}\] and \(\Delta g(b_1) = 1/{\psi^+(b_1)}-g_{b_2,\gamma_2}(b_1)<0\) which follows from Lemma 39i. with fact that \(b_1\in (y_{b_2,\gamma_2},b_2)\) and \(g_{b_1,\gamma_1}(b_1)=1/{\psi^+(b_1)}\).

Here, we claim that \(\Delta g(x)<0\) on \((0,b_1)\). Arguing by contradiction, assume it does not hold; then there exists \(y_{3}:=\sup\{x\in(0,b_1): \Delta g(x) = 0\}\). Since \(\Delta g(y_3)=0\) (by its continuity) and \(y_3\in[0,b_1]\), assigning \(x=y_3\) into 46 ensures that \(\Delta g'(y_3)=2(\gamma_1-\gamma_2)/\sigma^2(y_3)>0\). This together with the fact that \(\Delta g(b_1)<0\) contradicts Lemma 38i. This completes the proof.

This proof is similar to the one given in Cohen et al. [20], so we omit the details here. ◻

Proof of i.. Let \(y_{b,\gamma}\in[0,\underline{b}]\) for every \(\gamma<0\) be defined as in Lemma 39i. Then \(y_{b,\gamma}\) increases in \(\gamma<0\). Indeed, Lemma 39ii. ensures that for any \(\gamma_1,\gamma_2\in(-\infty,0)\) such that \(\gamma_1>\gamma_2\), the following holds that \(g_{b,\gamma_1}(x) <g_{b,\gamma_2}(x)\) for every \(x\in(0,b)\). Combining this with the definition of \(y_{b,\gamma_i}\), \(i=1,2\), (see Lemma 39i.), we have \(y_{b,\gamma_1}>y_{b,\gamma_2}\). This ensures that the (increasing) monotonicity of \(y_{b,\gamma}\) with resepect to \(\gamma<0\).

Consider an increasing sequence \((\gamma_n)_{n\in \mathbb{N}}\subseteq (-\infty,0)\) such that \(\gamma_n\uparrow 0\) as \(n\rightarrow\infty\). Then the monotonicity of \((y_{b,\gamma_n})_{n\in\mathbb{N}}\) together with uniformly boundedness within \([0,\underline{b}]\) ensures that there exists \(y_{b}^*:=\lim_{n\rightarrow\infty}y_{b,\gamma_n}\) satisfying \(y_{b}^*\in[0,\underline{b}]\) and \(y_{b}^*\geq y_{b,\gamma_n}\) for every \(n\in \mathbb{N}\). Combined with Lemma 39i., this ensures that for every \(n\in\mathbb{N}\) and every \(x\in[y_{b}^*,b)\), \(g_{b,\gamma_n}(x)\geq 1/\psi^+(x).\) Therefore combined with Lemma 39iii. and letting \(n\rightarrow \infty\), this completes the proof.

Lemma 39ii. ensures that for every \(\gamma<0\) and every \(x\in(0,b_1]\), \(g_{b_1,0}(x)=g_{b_1}(x)< g_{b_2,\gamma}(x).\) Combined with Lemma 39iii. and letting \(\gamma\uparrow 0\), this completes the proof.

We establish the proof for the case where \(\gamma = 0\), as the analysis can be readily extended to accommodate the case where \(\gamma\neq 0\). We start with proving the first convergence therein. Let \(\varepsilon\in [0,1\wedge(\overline{b}-b))\). Since \(g_{b+\varepsilon}\in C^2(\mathbb{R}_+)\) (resp. \(g_{b}\in C^2(\mathbb{R}_+)\)) is the solution of 42 (when \(\gamma=0\) with \(b+\varepsilon\) (resp. \(b\))), the following holds: for every \(x\in\mathbb{R}_+\), \[\begin{align} &g_{b+\varepsilon}(x) - g_{b}(x) -\big(g_{b+\varepsilon}(b) - g_{b}(b)\big)= - \int_{x}^b \big(g'_{b+\varepsilon}(y) - g'_{b}(y) \big) dy \nonumber\\ &= \int_{x}^b \Big( (1-{\@fontswitch\relax\mathcal{R}})\sigma^2(x)\big(g_{b+\varepsilon}(y) +g_b(y)\big) + \frac{2\mu(x)}{\sigma^2(x)} \Big)\big(g_{b+\varepsilon}(y) - g_b(y)\big) dy. \label{eq:est95g5} \end{align}\tag{47}\]

Lemma 40ii. ensures that \(g_b(x)\le g_{b+\varepsilon}(x)\le g_{b+1}(x)\le C_0\) for every \(x\in[0,b]\), where \(C_0>0\) is a constant (that depends on \(1\wedge (\overline{b}-b)\) but not on \(\varepsilon\)). Furthermore, \(\sigma\) and \(\mu\) are uniformly bounded on \([0,b]\) (see Assumption 15). Hence, combined with 47 these properties ensure that there is a constant \(C_1>0\) (that depends on \(C_0\) but not on \(\varepsilon\)) such that for every \(x\in[0,b)\), \(0\leq g_{b+\varepsilon}(x) - g_{b}(x) \leq g_{b+\varepsilon}(b) - g_{b}(b) + C_1 \int_b^x \big(g_{b+\varepsilon}(y) - g_b(y)\big) dy.\) An application of Grönwall’s inequality gives the existence of a constant \(C_2>0\) satisfying \[\begin{align} \label{eq:est95g6} 0\leq \sup_{x\in[0,b]}(g_{b+\varepsilon}(x) - g_{b}(x)) \leq C_2 (g_{b+\varepsilon}(b) - g_{b}(b)). \end{align}\tag{48}\]

Furthermore, we claim that there is a constant \(C_3\) (that depends on \(1\wedge (\overline{b}-b)\) but not on \(\varepsilon\)) such that \(0\leq g_{b+\varepsilon}(b) - g_{b}(b) \leq C_3 \varepsilon.\) Indeed, \[\begin{align} \left|g_{b+\varepsilon}(b) - g_{b}(b) \right| \le |g_{b+\varepsilon}(b+\varepsilon) -g_{b}(b) | + |g_{b+\varepsilon}(b+\varepsilon) - g_{b+\varepsilon}(b)|: = \operatorname{I} + \operatorname{II}. \end{align}\] \(\operatorname{I}\) is bounded by \(\varepsilon \sup_{x\in[b,b+\varepsilon]}|{(\psi^+)'(x)}/{(\psi^+)^2(x)}|\), which is of \(O(\varepsilon)\) by the fact that \(g_{b+\varepsilon}(b+\varepsilon)=1/\psi^+(b+\varepsilon)\), \(g_{b}(b)=1/\psi^+(b)\), and \(\psi^+\in C^1([0,\overline{b}])\). \(\operatorname{II}\) is of \(O(\varepsilon)\) with similar arguments devoted for 47 .

Combined with 48 , this ensures that \[\begin{align} \lim_{\varepsilon\downarrow 0}\sup_{x\in[0,b]}\Big(g_{b+\varepsilon}(x) - g_{b}(x)\Big) =0. \end{align}\] ?? holds because \(v_{b}(x) = \psi^+(b)\exp(\int_{b}^x g_{b}(u) du )\) and \(v_b'(x) = v_b(x)g_b(x)\).

The proof in the case \(\varepsilon\uparrow 0\) is similar therefore we omit it. ◻

Proof. Let \(\gamma>0\) and \(v_{b,\gamma}\) be the solution of 38 . Since \(v_{b,\gamma}'(b)=1\) and \(v_{b,\gamma}(b)= \psi^+(b)\), the nonlinear ODE given in 38 ensures that \(v''_{b,\gamma}(b) = 2\gamma\psi^+(b)/\sigma^2(b)>0\).

We claim that for every \(x\in(0,b)\), \(v'_{b,\gamma}(x)<1\). Arguing by contradiction, we assume that it does not hold. Then together with \(v_{b,\gamma}'(b)=1\), the following supremum is attained: \(x_1:=\sup\{x\in(0,b): v'_{b,\gamma}(x) = 1 \}\). By definition of this, it is clear that for every \(x\in[x_1,b]\), \(v'_{b,\gamma}(x)\le1,\) which ensures that \(v_{b,\gamma}(x) - v_{b,\gamma}(b)\geq x-b\) for every \(x\in[x_1,b]\).

Furthermore, from Assumption 15ii. (with \(b<\underline{b}\)) and \(v_{b,\gamma}=\psi^+(b)\), it follows that for every \(x\in[x_1,b]\), \(v_{b,\gamma}(x_1) \ge \psi^+(b)-(b-x_1) \ge \psi^+(x_1)\). This implies that \(Q(v_{b,\gamma}(x_1)) \ge Q(\psi^+(x_1))=0\) (see Assumption 15ii.). From the nonlinear ODE 38 , it follows that \[\begin{align} \frac{1}{2}\sigma^2(x_1) v''_{b,\gamma}(x_1)=Q\big(v_{b,\gamma}(x_1)\big)\frac{1}{v_{b,\gamma}(x_1)}+\gamma v_{b,\gamma}(x_1) \ge \gamma \psi^+(x_1)>0. \end{align}\] Combining this with the fact that \(v''_{b,\gamma}(b)>0\) contradicts Lemma 38ii. Hence we have shown that the claim holds.
Using Lemma 39iii. and letting \(\gamma\downarrow0\), we complete the proof. ◻

Proof of i.. Arguing by contradiction, we assume that it does not hold. Recalling the one-to-one correspondence \(h_{b,\gamma}=(v_{b,\gamma})^{1-{\@fontswitch\relax\mathcal{R}}}\) (see 39 ), we obtain that \[\max_{x\in[0,b]}\left\{ (x+\xi_0)^{1-{\@fontswitch\relax\mathcal{R}}} - h_{b,\gamma}(x) \right\} > 0.\] We note that from Assumption15iii. (with \(b\leq \hat{b}\)) and the condition \(v_{b,\gamma}(0) \ge \xi_0\), it follows that \(h_{b,\gamma}(x)\ge(x+\xi_0)^{1-{\@fontswitch\relax\mathcal{R}}}\) when \(x=0\) and \(x=b\). Hence this implies that \((x+\xi_0)^{1-\epsilon} - h_{b,\gamma}(x)\) takes its maximum at \(x_0\in(0,b)\). We have \[\begin{align} \label{pf:lem95comp951} -{\@fontswitch\relax\mathcal{R}}(1-{\@fontswitch\relax\mathcal{R}})(x_0+\xi)^{-1-{\@fontswitch\relax\mathcal{R}}} \le h_{b,\gamma}''(x_0),~(1-{\@fontswitch\relax\mathcal{R}})(x_0+\xi)^{-{\@fontswitch\relax\mathcal{R}}} =h_{b,\gamma}'(x_0). \end{align}\tag{49}\] From the linear ODE given in 40 , we get to a contradiction \[\begin{align} &0 = \dfrac{1}{2}\sigma^2(x_0)h_{b,\gamma}''(x_0)+ \mu(x_0)h_{b,\gamma}'(x_0) - (1-{\@fontswitch\relax\mathcal{R}})(\rho+\gamma) h_{b,\gamma}(x_0)\\ > & (1-{\@fontswitch\relax\mathcal{R}})(x_0+\xi_0)^{-1-{\@fontswitch\relax\mathcal{R}}}\Big(-\frac{{\@fontswitch\relax\mathcal{R}}}{2}\sigma^2(x_0) + \mu(x_0)(x_0+\xi) -(\rho+\gamma)(x_0+\xi)^2\Big)\ge0, \end{align}\] where the first inequality follows from 49 and that \(h_{b,\gamma}(x_0) < (x_0+\xi)^{1-{\@fontswitch\relax\mathcal{R}}}\), and the second inequality follows from Assumption15ii. and \(\gamma\le 0\). This completes the proof of i.

This statement is proved in three steps.

 We claim that \(v_{\underline{b},\gamma}(0)> \xi_0\) and \(v_{\hat{b},\gamma}(0)<\xi_0\).

Lemma 41 (when \(b=\underline{b}\)) ensures that \(v_{\underline{b},\gamma}(0)\geq v_{\underline{b},\gamma}(\underline{b})-(\underline{b}-x)\) for every \(x\in[0,\underline{b}]\). In particular, \(v_{\underline{b},\gamma}(0)>v_{\underline{b},\gamma}(\underline{b})-\underline{b}\) because it cannot be the case that \(v_{\underline{b},\gamma}'(x)\equiv 0\) for \(x\in[0,\underline{b}]\). Furthermore, from the fact that \(v_{\underline{b},\gamma}(\underline{b})=\psi^+(\underline{b})\) and Assumption 15ii.iii. (with \(\underline{b}<\hat{b}\)), it follows that \(v_{\underline{b},\gamma}(0) > \psi^+(\underline{b})-\underline{b}> \psi^+(\hat{b})-\hat{b} = \xi_0\).

It remains to show \(v_{\hat{b},\gamma}(0)<\xi_0\). Arguing by contradiction, we assume that it does not hold. From Lemma 42i., it follows that \(v_{\hat{b},\gamma}(x) \ge x + \xi_0\) for every \(x\in[0,\hat{b}]\). Since \(v_{\hat{b},\gamma}\) is the solution of 38 , assigning \(b=\hat{b}\) into 38 ensures that \(v''_{\hat{b},\gamma}(\hat{b}) = 2\gamma\sigma^2(\hat{b})\psi^+(\hat{b})<0\).

By using a proof by contradiction, standard arguments shows that \(v_{\hat{b},\gamma}'(x)>1\), for \(x\in(\underline{b},\hat{b})\). From the claim, the fact that \(v_{\hat{b},\gamma}(\hat{b})=\psi^+(\hat{b})\), and Assumption 15ii. (with \(\hat{b}\in[\underline{b},\overline{b}]\)), it follows that \(v_{\hat{b},\gamma}(\underline{b}) < \psi^+(\hat{b}) - (\hat{b}-\underline{b}) = \xi_0+\underline{b}\). This contradicts with Lemma 42i. Therefore, we conclude that \(v_{\hat{b},\gamma}(0)< \xi_0\).
We claim that \(b^*(\gamma): = \inf\{b\in(\underline{b},\hat{b}): v_{b,\gamma}(0) = \xi_0\}\) is well-defined and that \(v_{b,\gamma}(0) < \xi_0\) for every \(b\in( b^*(\gamma), \hat{b})\).

By Lemma 40iii., we have the continuity of \(b\mapsto v_{b,\gamma}(0)\) with fixed \(\gamma\le 0\). There \(b^*(\gamma): = \inf\{b\in(\underline{b},\hat{b}): v_{b,\gamma}(0) = \xi_0\}\) is well-defined.

Now we claim that \(v_{b,\gamma}(0) < \xi_0\) for every \(b\in( b^*(\gamma), \hat{b})\). From Lemma 40i. together with \(b^*(\gamma)>\underline{b}\), it follows that \(g_{b,\gamma}(x) \ge 1/\psi^+(x)\) for every \(x\in [b^*(\gamma),b]\). Combining this with the fact that \((\psi^+)'(x) < 1\) for every \(x\in[b^*(\gamma),b]\) (see Assumption 15ii.) ensures that \(g_{b,\gamma}(x) > (\psi^+)'(x)\psi^+(x)\) for every \(x\in [b^*(\gamma),b]\).

Furthermore, as \(b>b^*(\gamma)\), Lemma 40ii. ensures that \(g_{b,\gamma}(x) > g_{b^*(\gamma),\gamma}(x)\) for \(x\in(0,b^*(\gamma)]\). Hence we conclude that \[\begin{align} v_{b,\gamma}(0) &=\psi^+(b) \exp\bigg({\int_{b}^{b^*(\gamma)}g_{b,\gamma}(y)dy + \int_{b^*(\gamma)}^{0}g_{b,\gamma}(y)dy }\bigg)\\ & <\psi^+(b^*(\gamma)) \exp \bigg(\int_{b^*(\gamma)}^{0}g_{b^*(\gamma),\gamma}(y)dy \bigg) = v_{b^*(\gamma),\gamma}(0) = \xi_0. \end{align}\] . We claim that \(b^*(\gamma)\) increases in \(\gamma<0\), which ensures that \(b^*\) defined in ?? is well-defined by monotone convergence. Furthermore, we claim that \(b^*\in (\underline{b},\hat{b})\) and \(v_{b^*}(0) = \xi_0.\)

Let \(b\in(\underline{b},\overline{b})\) and \(\gamma_1,\gamma_2\in(-\infty,0]\) with \(\gamma_1\ge\gamma_2\). Then Lemma 39ii. ensures that \(g_{b,\gamma_1}(x) < g_{b,\gamma_2}(x)\) for every \([0,b]\). From this and the relationship given in 41 , it follows that \(v_{b,\gamma_1}(x) \ge v_{b,\gamma_2}(x)\) on \([0,b]\). In particular, since it cannot be the case that \(g_{b,\gamma_1} \equiv v_{b,\gamma_2}\) for \(x\in[0,b]\), hence \(v_{b,\gamma_1}(0) > v_{b,\gamma_2}(0)\).

From this and the fact that \(b^*(\gamma_1)=\inf\{b\in(\underline{b},\hat{b}): v_{b,\gamma_1}(0) = \xi_0\}\), it follows that \(\xi_0 = v_{b^*(\gamma_1),\gamma_1}(0) > v_{b^*(\gamma_1),\gamma_2}(0)\). Furthermore, Step 2 ensures that \(v_{b,\gamma_2}(0) < \xi_0\) for every \(b\in( b^*(\gamma_2), \hat{b})\). We hence have \(b^*(\gamma_1) > b^*(\gamma_2)\).

By the monotonicity of \(b^*(\gamma)\) and its uniform boundedness, we can apply the MCT to have the existence of \(b^*\) in ?? .

It remains to show that \(v_{b^*}(0) = \xi_0\) and \(b^*\in (\underline{b},\hat{b})\). To that end, consider an increasing sequence \((\gamma_n)_{n\in \mathbb{N}}\) such that \(\gamma_n\in (-\infty,0)\) for every \(n\in\mathbb{N}\) and \(\lim_{n\rightarrow \infty } \gamma_n = 0\).

Since \(\lim_{n\rightarrow \infty} b^*(\gamma_n) = b^*\), Lemma 40iii. ensures that for any \(\varepsilon>0\), there exists \(N_\varepsilon \in \mathbb{N}\) satisfying that for every \(n\geq N_\varepsilon\), \(|v_{b^*}(0) - v_{b^*(\gamma_n)}(0)|<\varepsilon.\) Furthermore Lemma 39iii. ensures that there exists \(\widetilde{N}_\varepsilon\in \mathbb{N}\) satisfying that for every \(n\geq \widetilde{N}_\varepsilon\), \(|v_{b^*(\gamma_n),\gamma_n}(0) - v_{b^*(\gamma_n)}(0)|<\varepsilon.\) Since \(v_{b^*(\gamma_n),\gamma_n}(0) = \xi_0\) for every \(n\in\mathbb{N}\) (see Step 2.), we hence obtain that for every \(n\geq \widetilde{N}_\varepsilon \vee N_\varepsilon\), \[\begin{align} |v_{b^*}(0) - \xi_0 |\leq |v_{b^*}(0) - v_{b^*(\gamma_n)}(0)|+|v_{b^*(\gamma_n),\gamma_n}(0) - v_{b^*(\gamma_n)}(0)| < 2\varepsilon. \end{align}\] By letting \(\varepsilon\downarrow 0\), we show \(v_{b^*}(0) = \xi_0\). Finally, we know that \(\underline{b}<b^*(\gamma)< \hat{b}\) for every \(\gamma<0\) by Step 1, therefore \(\underline{b}\le \lim_{\gamma\uparrow 0}b^*(\gamma) = b^*\le \hat{b}\). The strict inequality \(\underline{b}< b^*\) follows by the increasing monotonicity of \(b^*(\gamma)\). ◻

Proof. Given the existence of \(b^*\in(\underline{b},\hat{b}]\) satisfying \(v_{b^*}(0)=\xi_0\) by Lemma 42ii., we first prove that \(v_{b^*}'(x)\geq 1\) for every \(x\in[0,b^*]\).

From Lemma 42ii., define \(b^{\gamma}\) for every \(\gamma<0\) by \[\begin{align} \label{eq:quad95perturb950} b^{\gamma}:=b^*(\gamma)= \inf\{b\in(\underline{b},\hat{b}): v_{b,\gamma}(0) = \xi_0\}. \end{align}\tag{50}\] Furthermore, denote by \(v_{b^{\gamma},\gamma}\in C^2((0,b^{\gamma})\cup(b^{\gamma},\infty))\cap C^1(\mathbb{R}_+)\) the solution of 38 (when \(b=b^{\gamma}\)).
The proof is achieved in two steps.
We star with perturbation analysis with a fixed \(\gamma\in(-\rho,0)\). We claim \[\begin{align} \label{eq:claim95195shooting} v_{b^\gamma,\gamma}'(b^\gamma)(x)\ge 1,\forall x\in(0,b^{\gamma}]. \end{align}\tag{51}\] From the boundary conditions \(v_{b^{\gamma},\gamma}'(b^{\gamma})=1\) and \(v_{b^{\gamma},\gamma}(b^{\gamma})= \psi^+(b^{\gamma})\), the \(C^2\) property of \(v_{b^{\gamma},\gamma}\) and nonlinear ODE 38 gives \(v''_{b^{\gamma},\gamma}(b^{\gamma}) = 2\gamma\sigma^2(b^{\gamma})\psi^+(b^{\gamma})<0.\) Argue by contradiction, assume 51 doesn’t hold. Then combined with the fact that \(v_{b^\gamma,\gamma}'(b^\gamma)=1\), we have the existence of \[x_1:=\sup\{ x\in(0,b^\gamma): v'_{b^\gamma,\gamma}(x) = 1 \}.\] In the following, we prove that the existence of \(x_1\) leads to a contradiction. i. Case \(x_1 \in (\underline{b},{b}^\gamma)\). Since \(v_{b^\gamma,\gamma}'(x_1)= 1\) (noting that \(v_{b^\gamma,\gamma}\in C^1(\mathbb{R}_+)\)), it follows that \(v_{b^\gamma,\gamma}'(x) > 1\) for every \(x\in(x_1,b^\gamma)\). Furthermore, since \(\psi^+(x)-x\) decreases on \([x_1,b^\gamma]\subset [\underline{b},\hat{b}]\) (see Assumption 15ii. and \(x_1 \in (\underline{b},{b}^\gamma)\)) and \(v_{b^\gamma,\gamma}(b^\gamma)=\psi^+(b^\gamma)\), hence \[\begin{align} v_{b^\gamma,\gamma}(x_1) \le \psi^+(b^\gamma)-(b^\gamma-x_1)\le \psi^+(x_1). \end{align}\] Moreover, since \(x_1 \in (\underline{b},\hat{b})\) and \(v_{b^\gamma,\gamma}(0)= \xi_0\) (by definition of \(b^\gamma\)), Lemma42i. ensures that \(v_{b^\gamma,\gamma}(x_1) \ge x_1 +\xi_0 \ge \psi^-(x_1)\), so that \(Q(v_{b^\gamma,\gamma}(x_1);x_1)\le 0\) by recalling that \(Q(\cdot;x_1)\) defined in Assumption 15ii. This implies that \[\begin{align} v''_{b^\gamma,\gamma}(x_1) = \frac{2}{\sigma^2(x_1)}\Big(\gamma v_{b^\gamma,\gamma}(x_1) +\frac{Q(v_{b^\gamma,\gamma}(x_1);x_1)}{v_{b^\gamma,\gamma}(x_1)}\Big)\le \frac{2}{\sigma^2(x_1)}\gamma (x_1+\xi_0)<0. \end{align}\] This together with the fact that \(v''_{b^{\gamma},\gamma}(b^{\gamma})<0\) contradicts Lemma 38ii.

ii. Case \(x_1\in(0,\underline{b}]\). Recalling \(x_1=\sup\{ x\in(0,b^\gamma): v'_{b^\gamma,\gamma}(x) = 1 \}\) and the fact that \(v_{b^\gamma,\gamma}'(b^\gamma)=1\) and \(v_{b^\gamma,\gamma}''(b^\gamma)<0\), we employ Lemma 38ii. to obtain that \(v''_{b^\gamma,\gamma}(x_1)\geq 0\), which can be rewritten by \[\begin{align} \label{eq:quad95perturb} v''_{b^\gamma,\gamma}(x_1)= \frac{2}{\sigma^2(x_1)}\frac{1}{v_{b^\gamma,\gamma}(x_1)}\widetilde{Q}_{\gamma}(v_{b^\gamma,\gamma}(x_1);x_1)\geq 0, \end{align}\tag{52}\] where \(\widetilde{Q}_\gamma(\psi;x_1):=(\rho+\gamma) \psi^2-\mu(x_1)\psi + \frac{\@fontswitch\relax\mathcal{R}}{2}\sigma^2(x_1)\) (which is well-defined by Assumption 15ii. with the fact that \(\rho>\rho+\gamma>0\)) and we have used the nonlinear ODE of \(v_{b^\gamma,\gamma}\) given in 38 .

Lemma 42i. together with Assumption 15iii. (and \(x_1\in(0,\underline{b}]\subset [0,\hat{b}]\)) ensures that \(v_{b^\gamma,\gamma}(x_1) \geq x_1+\xi_0\geq \psi^-(x_1)\). Furthermore, recalling that \(\gamma\) satisfies \(\rho>\rho+\gamma>0\), we have \[\begin{align} v_{b^\gamma,\gamma}(x_1) \geq& \frac{\mu(x_1) - \sqrt{\mu^2(x_1)-2{\@fontswitch\relax\mathcal{R}}\rho \sigma^2(x_1)}}{2\rho}\tag{53} \\ >&\frac{\mu(x_1) - \sqrt{\mu^2(x_1)-2{\@fontswitch\relax\mathcal{R}}(\rho+\gamma) \sigma^2(x_1)}}{2(\rho+\gamma)} =:\widetilde{\psi^-}(x_1),\tag{54} \end{align}\] where we note that RHS of 53 equals \(\psi^-(x_1)\) and that the last term \(\widetilde{\psi^-}(x_1)\) is the smaller one among two roots of the quadratic function \(\widetilde{Q}_\gamma(\cdot;x_1) = 0\).

Since \(\widetilde{Q}_{\gamma}(v_{b^\gamma,\gamma}(x_1);x_1)\geq0\) (see 52 ), from 54 it follows that \[\begin{align} \label{eq:quad95perturb3} v_{b^\gamma,\gamma}(x_1)\geq \frac{\mu(x_1) + \sqrt{\mu^2(x_1)-2{\@fontswitch\relax\mathcal{R}}(\rho+\gamma) \sigma^2(x_1)}}{2(\rho+\gamma)}=:\widetilde{\psi^+}(x_1), \end{align}\tag{55}\] where the right-hand side \(\widetilde{\psi^+}(x_1)\) is the other (larger) root of \(\widetilde{Q}_\gamma(\cdot;x_1)\).

Now we claim that for \(x\) fixed, if \((\psi^+)'(x) \ge 1\) then \((\widetilde{\psi^+})'(x) \ge 1\). By taking derivative of \(\psi^+(x)\) w.r.t. \(x\), we have \[\begin{align} (\psi^{+})'(x) = \frac{1}{2\rho}\mu'(x) + \frac{\mu(x)\mu'(x)}{2\rho\sqrt{\mu(x)^2-2{\@fontswitch\relax\mathcal{R}}\rho\sigma^2(x)}} - \frac{{\@fontswitch\relax\mathcal{R}}\sigma(x)\sigma'(x)}{\sqrt{\mu(x)^2-2{\@fontswitch\relax\mathcal{R}}\rho\sigma^2(x)}}. \end{align}\] When \((\psi^+)'(x) \ge 1\), it is clear that \(\mu'(x)>0\) since \(\sigma'(x)\ge 0\) by Assumption 15. Therefore the first term and the last term in above equation are decreasing w.r.t \(\rho\). It remains to show the same monotonicity for the second term in \((\psi^{+})'(x)\). Take derivative of the function \(h(\rho;x): = \rho\sqrt{\mu(x)^2-2{\@fontswitch\relax\mathcal{R}}\rho\sigma^2(x)}\) w.r.t \(\rho\), we have \[\begin{align} \partial_{\rho} h(\rho;x) = \sqrt{\mu(x)^2-2{\@fontswitch\relax\mathcal{R}}\rho\sigma^2(x)} - \frac{{\@fontswitch\relax\mathcal{R}}\rho \sigma^2(x)}{\sqrt{\mu(x)^2-2{\@fontswitch\relax\mathcal{R}}\rho\sigma^2(x)}} \ge 0, \end{align}\] where the inequality is followed by Assumption 15i. with \((x_2,x_1)\subset [0,\underline{b}]\). As a result, we show that \(1 \le (\psi^{+})'(x)\le (\widetilde{\psi^+})'(x)\) since \(\rho+\gamma < \rho\) for \(\gamma<0\). Combined with Assumption  15ii., which says that \(\psi^+(x)-x\) increases on \([0,\underline{b}]\), we have \(\widetilde{\psi^+}(x)-x\) increases on \([0,\underline{b}]\).

To proceed, recalling that \(v'_{b^\gamma,\gamma}(x_1) = 1\) and \(v''_{b^\gamma,\gamma}(x_1)\geq 0\) (see 52 ), standard arguments by using a proof by contradiction shows that there exists \(x_2\) defined as \(x_2:=\sup\{ x\in(0,x_1): v'_{b^\gamma,\gamma}(x) = 1 \}\). Since \(v'_{b^\gamma,\gamma}(x)<1\) for every \(x\in(x_2,x_1)\) (from the definition of \(x_2\) and the fact that \(v'_{b^\gamma,\gamma}(x_1) = 1\) and \(v''_{b^\gamma,\gamma}(x_1)\geq 0\)), the following hold: \(v'_{b^\gamma,\gamma}(x_2)=1\) and \[\begin{align} v_{b^\gamma,\gamma}(x_2)&>v_{b^\gamma,\gamma}(x_1)-(x_1-x_2)\nonumber\\ &\geq \frac{\mu(x_1) + \sqrt{\mu^2(x_1)-2{\@fontswitch\relax\mathcal{R}}(\rho+\gamma) \sigma^2(x_1)}}{2(\rho+\gamma)}-(x_1-x_2)\nonumber\\ &\geq \frac{\mu(x_2) + \sqrt{\mu^2(x_2)-2{\@fontswitch\relax\mathcal{R}}(\rho+\gamma) \sigma^2(x_2)}}{2(\rho+\gamma)}, \label{eq:quad95perturb4} \end{align}\tag{56}\] where the second inequality follows from 55 and the last inequality follows from the fact that \(\widetilde{\psi^+}(x)-x\) increases on \((x_2,x_1)\) since \((x_2,x_1)\subset [0,\underline{b}]\).

Recalling the nonlinear ODE 38 , the inequality given in 56 ensures that \[v''_{b^\gamma,\gamma}(x_2)=\frac{2}{\sigma^2(x_2)}\frac{1}{v_{b^\gamma,\gamma}(x_2)}\widetilde{Q}_{\gamma}(v_{b^\gamma,\gamma}(x_2);x_2)>0.\] This together with 52 contradicts Lemma 38. (Lemma 2.2ii. requires \(v''_{b^\gamma,\gamma}(x_1)>0\), but we can see it cannot be the case that \(v''_{b^\gamma,\gamma}(x_2)>0\), \(v''_{b^\gamma,\gamma}(x_1)=0\) and \(v''_{b^\gamma,\gamma}(b^{\gamma})<0\)).

By deriving contradictions for the above two cases, we have 51 .

Recall \(b^\gamma\) for every \(\gamma<0\) given in 50 and let \(b^*\) be defined by \(b^*=\lim_{\gamma\uparrow 0} b^\gamma\) (see Lemma 42ii.). We claim that \(v'_{b^*}(x)\ge 1\) for every \(x\in (0,b^*]\).

To that end, let us consider an increasing sequence \((\gamma_n)_{n\in\mathbb{N}}\) such that \(-\rho <\gamma_n<0\) for every \(n\in\mathbb{N}\) and \(\lim_{n\rightarrow \infty } \gamma_n = 0\). Since \(\lim_{j\rightarrow \infty} b^{\gamma_n} = b^*\) and \((b^{\gamma_n})_{n\in\mathbb{N}}\) is a increasing sequence (see Lemma 42ii.), the following hold: \(b^*\geq b^{\gamma_n} \geq b^{\gamma_0}\) for every \(n\in \mathbb{N}\).

Combining this with Step 1. (along with \(v_{b^{\gamma_n},\gamma_n}\in C^1(\mathbb{R}_+)\) for every \(n\in\mathbb{N}\)) ensures that for every \(n\in\mathbb{N}\) and every \(x\in (0,b^*]\), \[\begin{align} \label{eq:quad95perturb5} v_{b^{\gamma_n},\gamma_n}'(x) \ge 1. \end{align}\tag{57}\]

Fix \(y\in(0,b^{\gamma_0}]\subset (0,b^*]\). For arbitrary \(\varepsilon>0\), then Lemma 39iii. ensures that there exists \({N}_\varepsilon\in \mathbb{N}\) such that \(|v'_{b^{\gamma_n},\gamma_n}(x) - v'_{b^{\gamma_n}}(x)|<\varepsilon\) for every \(n\geq {N}_\varepsilon\). Furthermore, Lemma 40iii. ensures that there exists \(\widetilde{N}_\varepsilon\in \mathbb{N}\) such that \(|v'_{b^*}(x) - v'_{b^{\gamma_n}}(x)|<\varepsilon\) for every \(n\geq \widetilde{N}_\varepsilon\). Therefore, from 57 , it follows that for every \(n\geq \widetilde{N}_\varepsilon\vee N_\varepsilon\), \(v'_{b^*}(x) >v'_{b^{\gamma_n}}(x) -\varepsilon> v'_{b^{\gamma_n},\gamma_n}(x)- 2\varepsilon\geq 1- 2\varepsilon.\) By letting \(\varepsilon\rightarrow0\), we conclude that \(v_{b^*}'(x)\ge1\) for \(y\in (0, b^{\gamma_0}]\).

It remains to show \(v_{b^*}'(y)\ge1\) on \((b^{\gamma_0},b^*]\).
Assume the contrary: \(\min_{y\in[b^{\gamma_0},b^*]} v_{b^*}'(y) <1\). Then \(v_{b^*}'(y) -1\) takes its minimum at \(x_0\in(b^{\gamma_0},b^*)\) because \(v'_{b^*}(b^*) = 1\) and \(v'_{b^*}(b^{\gamma_0})\ge1\) by above analysis. Then we have that \(v_{b^*}''(x_0) = 0\) and \(v'_{b^*}(x_0)<1\). By the continuity of \(v'_{b^*}\), let \(\epsilon=(1-v'_{b^*}(x_0))/2\) there exists \(x_0'\) in a local neighborhood of \(x_0\) such that \(v_{b^*}''(x_0') > 0\) and \(v_{b^*}'(x_0')< v'_{b^*}(x_0) + \epsilon<1\). Hence \[g_{b^*}(x_0') = \dfrac{v'_{b^*}(x_0')}{v_{b^*}(x_0')} < \frac{1}{x_0'+\xi} < \frac{1}{\psi^{-}(x_0')},\] where the first inequality follows Lemma 42i. that \(v_{b^*}(x_0')\ge x_0'+\xi\) since \(v_{b^*}(0)=\xi\), while the last inequality follows Assumption 15iii. Recall the ODE 38 of \(v_{b^*}\), we have \[\begin{align} \frac{1}{2}\sigma^2(x_0) v''_{b^*}(x_0') = \frac{1}{v_{b^*}(x_0')}Q(v_{b^*}(x_0');x_0') > 0. \end{align}\] Combined with the above arguments, we conclude that \(g_{b^*}(x_0') < 1/\psi^+(x_0')\). This contradicts with Lemma 40i. which gives \(g_{b^*}(x_0')\ge{1}/{\psi^+(x_0')}\) as \(x_0>b^{\gamma_0}>\underline{b}\). As a result, we see \(v_{b^*}'(y)\ge1\) on \((b^{\gamma_0},b^*]\).

Finally, we show that \(b^*<\hat{b}\), where \(\hat{b}\) defined in Assumption 15iii. We only need to show \(b^* \neq \hat{b}\) since \(b^*\le \hat{b}\) by Lemma 42ii. When \(b^* = \hat{b}\), by definition of \(\hat{b}\), we have \(v_{b^*}(0)=\xi_0\) while \(v_{b^*}(b^*) = \psi^+(b^*) = b^* + \xi_0\). Together with our previous result that \(v_{b^*}'(x)\ge 1\) on \((0,b^*]\), we conclude \(v_{b^*}'(x) \equiv 1\) on \((0,b^*]\), which cannot be the case due to the structure of ODE 38 . Therefore, we have \(b^*<\hat{b}\). ◻

Proof. Theorem 19 is proved in two steps. As the proof for the case \(x=0\) is obvious, we assume \(x\in{\mathbb{R}_+}\). For simplicity, we omit the superscript \(x\) in \(X^{x,D}\) and \(\tau^{x,D}\), as defined in 18 and 19 .

Recall that \(J^*\) is defined in 21 , we claim that for every \(x\in{\mathbb{R}_+}\), \[\begin{align} \tag{58} v^*(x)\geq& \sup_{D\in {\@fontswitch\relax\mathcal{A}}^x}\inf_{\theta \in \Theta_D}\mathbb{E}^{\theta}\bigg[\int_{0}^{\tau^{D}}e^{-\rho t} \Big( dD_t + \frac{\theta_t^2}{2{\@fontswitch\relax\mathcal{R}}} v^*(X_t^{D}) dt\Big) + e^{-\rho \tau^{D}}\xi_0 \bigg],\\ v^*(x) \ge& J^*(x).\tag{59} \end{align}\]

Set an arbitrary \(D\in {\@fontswitch\relax\mathcal{A}}^x\) and define the stopping times \((\hat{\tau}_n)_{n\in \mathbb{N}}\) by \[\begin{align} \label{eq:stop95bdd95X} \hat{\tau}_n:= \inf\{t\ge 0: X^{D}_t\notin [1/n,n] \}\wedge n. \end{align}\tag{60}\] Recall the structure of the operator \(\mathcal{L}\) from 22 , and define the measure \(\mathbb{Q}^*\) via 8 with the Girsanov kernel \(\theta^*:=(\theta^*(D)_s)_{s\geq 0}\) given by \[\begin{align} \theta^*_s:= -{\@fontswitch\relax\mathcal{R}}\frac{\sigma(X_s^{D})(v^*)'(X_s^{D})}{v^*(X_s^{D})}. \label{eq:worst95kernel} \end{align}\tag{61}\] Note that by 43 and Proposition 37, \((v^*)'\) is bounded on \([0,b^*]\) and is identical to \(1\) on \((b^*,\infty)\) while \(v^*\) is bounded below by \(\xi_0\) on \({\mathbb{R}_+}\). Following similar arguments of in Theorem 9, it follows that \(\theta^*\in \Theta^D\). From this, we can denote by \(\mathbb{Q}^*\) the corresponding probability measure obtained from \(\theta^*\).

An application of Dynkin formula into the process \(e^{-\rho t} v^*(X_t^D)\) on the random time interval \([0,\tau^D\wedge\hat{\tau}_n]\) under \(\mathbb{Q}^*\) (corresponding to \(\theta^*\)) ensures that \[\begin{align} &e^{-\rho (\tau^D\wedge \hat{\tau}_n )} v^*(X_{\tau^D\wedge \hat{\tau}_n }^D) -v^*(x) =\int_0^{\tau^D\wedge \hat{\tau}_n } e^{-\rho s} \; \hat{\@fontswitch\relax\mathcal{L}}^{\theta^*}v^*(X_s^D)ds\nonumber\\ & +\int_0^{\tau^D\wedge \hat{\tau}_n } e^{-\rho s} \sigma(X_{s}^D)(v^*)'(X_{s}^D) d W^{\mathbb{Q}^*}_s- \int_0^{\tau^D\wedge \hat{\tau}_n }e^{-\rho s} (v^*)'(X_{s}^D) dD_s,\label{eq:thm95veri95pf951} \end{align}\tag{62}\] where \(\hat{\@fontswitch\relax\mathcal{L}}^{\theta^*}\) is the infinitesimal operator under \(\mathbb{Q}^*\), i.e., \[\begin{align} \hat{\@fontswitch\relax\mathcal{L}}^{\theta^*}v^*(X_s^D):=\Big(\frac{\sigma^2}{2}(v^*)''+ (\mu+\sigma\theta_s^*)(v^*)' -\rho v^*\Big)(X_{s}^D). \end{align}\] Based on the definitions of the nonlinear operator \({\@fontswitch\relax\mathcal{L}}\) in 22 , the process \(\theta^*\) in 61 , and given that \(v^*(X_s^D)>0\) for \(s\in[0,\tau^D)\) and \(v^*\) solves 24 , it follows that for every \(s\in[0,\tau^D)\): \[\begin{align} \hat{\@fontswitch\relax\mathcal{L}}^{\theta^*}v^*(X_s^D)= {\@fontswitch\relax\mathcal{L}} v^*(X_s^D)-\frac{(\theta_s^*)^2}{2{\@fontswitch\relax\mathcal{R}}}v^*(X_s^D)\leq -\frac{(\theta_s^*)^2}{2{\@fontswitch\relax\mathcal{R}}}v^*(X_s^D). \end{align}\] Furthermore, Since \(v^*\in C^2({\mathbb{R}_+})\) (see Theorem 18), \(\sigma\in C^1(\mathbb{R}_+)\) (see Assumption 15i.) and \(X^{x,D}\) is bounded in \([0,\tau^D\wedge \hat{\tau}_n ]\) (see 60 ), the Brownian local martingale term given in 62 is martingales. Lastly, since \((v^*)'(x)\geq 1\) for \(x\in {\mathbb{R}_+}\) and \(v^*(0)=\xi_0\), it follows that \(v^*(X_{\tau_D\wedge \hat{\tau}_n }^D)\geq \xi_0\). Taking the expectation over 62 under \(\mathbb{Q}^*\) gives \[\begin{align} v^*(x) \geq & \mathbb{E}^{\theta^*}\bigg[\int_0^{(\tau_D\wedge \hat{\tau}_n )} e^{-\rho s}d D_s +\int_0^{(\tau_D\wedge \hat{\tau}_n )}e^{-\rho s}\frac{(\theta^*_s)^2}{2{\@fontswitch\relax\mathcal{R}}}v^*(X^D_s) ds +e^{-\rho (\tau_D\wedge \hat{\tau}_n )} \xi_0 \bigg]. \end{align}\] Since \(\tau_D\wedge \hat{\tau}_n \rightarrow \tau_D\) as \(n\rightarrow \infty\) (see 60 ), an application of Fatou’s lemma together with \(\theta^* \in \Theta^D\) (see 61 ) ensures that \[\begin{align} v^*(x) \geq & \mathbb{E}^{\theta^*}\bigg[\int_0^{\tau_D} e^{-\rho s}d D_s+\int_0^{\tau_D} e^{-\rho s}\frac{(\theta_s^*)^2}{2{\@fontswitch\relax\mathcal{R}}}v^*(X^D_s) ds+e^{-\rho \tau_D} \xi_0 \bigg] \\ \geq & \inf_{\theta \in \Theta_D} \mathbb{E}^{\theta}\bigg[\int_0^{\tau_D} e^{-\rho s}d D_s+\int_0^{\tau_D} e^{-\rho s}\frac{\theta_s^2}{2{\@fontswitch\relax\mathcal{R}}}v^*(X^D_s) ds+e^{-\rho \tau_D} \xi_0 \bigg]. \nonumber \end{align}\] As the inequality holds for every \(D\in {\@fontswitch\relax\mathcal{A}}^x\), the claim 58 holds. To show 59 , the proof proceeds similarly by applying Dynkin formula to the process \(e^{-\rho t + \int_0^{t}({\theta}^*_s)^2/2{\@fontswitch\relax\mathcal{R}}ds} v^*(X_t^D)\) on the random time interval \([0,\tau^D\wedge\hat{\tau}_n]\) and take expectation. We obtain that \[\begin{align} v^*(x) &\geq \mathbb{E}^{\theta^*}\bigg[\int_0^{\tau_D} e^{-\rho s+\frac{1}{2{\@fontswitch\relax\mathcal{R}}}\int_0^{s} (\theta_u^*)^2 du}d D_s+e^{-\rho \tau_D+\frac{1}{2{\@fontswitch\relax\mathcal{R}}}\int_0^{\tau_D} (\theta_u^*)^2 du} \xi_0 \bigg]\geq J^{\theta^*}(x;D), \end{align}\] where \(J^{\theta}\) is defined in 20 . The last term is greater than \(\inf_{\theta\in \Theta^D} J^{\theta}(x;D)\). Taking supreme over \({D\in\mathcal{A}^x}\) on both sides of the above equation gives 59 .

Let \(D^*=D(b^*)\) be the \(b^*\)-threshold strategy. We show that \[\begin{align} \tag{63} v^*(x)\leq& \inf_{\theta \in \Theta_{D^*}}\mathbb{E}^{\theta}\bigg[\int_{0}^{\tau^{D^*}}e^{-\rho t} \Big(dD^*_t + \frac{\theta_t^2}{2{\@fontswitch\relax\mathcal{R}}} v^*(X_t^{D^*}) dt \Big) + e^{-\rho \tau^{D^*}}\xi_0 \bigg],\\ \tag{64} v^*(x) \le& J^*(x). \end{align}\] The analysis is divided into two cases, i.e., \(x\in(0,b^*]\) or \(x\in(b^*,\infty)\). Before starting the proof, we first note that from the definition of \(b^*\)-threshold strategy (see Definition 14), the following properties hold for every \(s\in(0,\tau^{D^*}]\): \[\begin{align} \label{eq:skorokhod95reflect} X^{D^{*}}_s \in [0, b^*],\;\;\quad dD^*_s>0\quad only when X_s^{D^*}=b^*. \end{align}\tag{65}\] Furthermore, if \(x\in(0,b^*]\), the above properties hold for every \(s\in[0,\tau^{D^*}]\). We first consider the case \(x\in(0,b^*]\). Let \(\theta \in \Theta({D^*})\) and \(\mathbb{Q}\) be the corresponding probability measure induced by the kernel \(\theta\). Furthermore, as in 60 , we define for every \(n\in \mathbb{N}\) by \(\hat{\tau}_n:= \inf\{t\ge 0: X^{D^*}_t\notin [1/n,n] \}\wedge\tau^{D^*}.\) An application of Dynkin formula to the process \(e^{-\rho t} v^*(X_t^{D^*})\) on the random time interval \([0,\hat{\tau}_n]\) under \(\mathbb{Q}\) gives that \[\begin{align} \label{eq:thm95veri95pf9510} \begin{aligned} v^*(x) &=\mathbb{E}^\theta \bigg[e^{-\rho (\hat{\tau}_n)} v^*(X_{\hat{\tau}_n}^{D^*})-\int_0^{\hat{\tau}_n} e^{-\rho s} (\hat{\@fontswitch\relax\mathcal{L}}^{\theta}v^*(X_s^{D^*})ds-dD^*_s ) \bigg], \end{aligned} \end{align}\tag{66}\] where \(\hat{\@fontswitch\relax\mathcal{L}}^{\theta}\) is the infinitesimal operator under \(\mathbb{Q}^*\), i.e., \[\begin{align} \hat{\@fontswitch\relax\mathcal{L}}^{\theta}v^*(X_s^{D^*}):=\Big(\frac{1}{2}\sigma^2(v^*)''(X_{s}^{D^*}) + (\mu+\sigma\theta_s)(v^*)' -\rho v^*\Big)(X_{s}^{D^*}), \end{align}\] and we use the fact that \(\mathbb{E}^{\theta}[\int_0^{\tau^{D^*}\wedge \hat{\tau}_n } e^{-\rho s} \sigma(X_{s}^{D^*})(v^*)'(X_{s}^{D^*}) d W^{\mathbb{Q}}_s]=0\) and that \(dD^*_s>0\) only when \((v^*)'(X_{s}^{D^*})=1\) for \(s\in[0,\tau^{D^*}]\). From 65 and the fact that \({\@fontswitch\relax\mathcal{L}}v^*(x)=0\) for every \(x\in(0,b^*]\) (see 24 with the free boundary \(b^*\)), it follows that for every \(s\in [0,\tau^{D^*}]\), \({\@fontswitch\relax\mathcal{L}}v^*(X_s^{D^*})=0\). Recall the definition of \({\@fontswitch\relax\mathcal{L}}\) in 22 , we obtain that for every \(s\in [0,\tau^{D^*}]\) \[\begin{align} \label{eq:claim3} \hat{\@fontswitch\relax\mathcal{L}}^{\theta}v^*(X_s^{D^*})\geq {\@fontswitch\relax\mathcal{L}}v^*(X_s^{D^*})- \frac{(\theta_s)^2}{2{\@fontswitch\relax\mathcal{R}}} v^*(X_s^{D^*}) = - \frac{(\theta_s)^2}{2{\@fontswitch\relax\mathcal{R}}} v^*(X_s^{D^*}). \end{align}\tag{67}\] By 67 and 66 , we hence obtain that \[\begin{align} \label{eq:thm95veri95pf952} \begin{aligned} v^*(x)\leq \mathbb{E}^\theta \bigg[\int_0^{\hat{\tau}_n} e^{-\rho s} \left(\frac{(\theta_s)^2}{2{\@fontswitch\relax\mathcal{R}}} v^*(X_s^{D^*})ds+dD^*_s \right)+e^{-\rho (\hat{\tau}_n)} v^*(X_{\hat{\tau}_n}^{D^*}) \bigg]. \end{aligned} \end{align}\tag{68}\] From \(v^*\in C^2({\mathbb{R}_+})\) and the fact that \(X^{D^{*}}_s \in [0, b^*]\) for every \(s\in [0,\tau^{D^*}]\) (see 65 ), it follows that \(\sup_{s\in [0,\tau^{D^*}]}|v^*(X_s^{D^*})|<\infty.\) Hence, an application of the DCT (applicable to the last term of right-hand side of 68 ) and monotone convergence theorem (applicable to the first term of right-hand side of 68 ), together with \(\tau^{D^*}\wedge \hat{\tau}_n\rightarrow \tau^{D^*}\) as \(n\rightarrow \infty\) into 68 ensures that \[\begin{align} \label{eq:thm95veri95pf953} v^*(x) &\leq \mathbb{E}^\theta \bigg[\int_0^{\tau^{D^*}} e^{-\rho s} \left(\frac{(\theta_s)^2}{2{\@fontswitch\relax\mathcal{R}}} v^*(X_s^{D^*})ds+dD^*_s \right)+e^{-\rho \tau^{D^*}} \xi_0 \bigg], \end{align}\tag{69}\] where we have used that \(v^*(X_{\tau^{D^*}}^{D^*})=v^*(0)=\xi_0\) (see 24 ). As the inequality holds for every \(\theta \in \Theta_{D^*}\), the claim 63 holds for every \(x\in(0,b^*]\).

It remains to prove the case \(x\in (b^*,\infty)\). Note that for every \(x\in (b^*,\infty)\), \(v^*(x)=x-b^*+\psi^+(b^*)\) (see 43 ). Furthermore, since the strategy \(D^*=D(b^*)\) starts with an instantaneous increase with amount \(x-b^*\), it follows that \[\begin{align} &\mathbb{E}^\theta \bigg[\int_0^{\tau^{D^*}} e^{-\rho s} \Big(\frac{(\theta_s)^2}{2{\@fontswitch\relax\mathcal{R}}} v^*(X_s^{x,D^{*}})ds+dD^*_s \Big)+e^{-\rho \tau^{D^*}} \xi_0 \bigg]\geq v^*(x) \end{align}\] holds for every \(\theta \in \Theta_{D^*}\), where we have used the inequality 69 (since it holds when \(x=b^*\)) and \(v^*(b^*)=\psi^+(b^*)\).

To show 64 , similar to the analysis in Step 1, an application of Dynkin formula to the process \(e^{-\rho t \int_0^{t}({\theta}_s)^2/(2{\@fontswitch\relax\mathcal{R}}) ds} v^*(X_t^{D^*})\) gives \[\begin{align} v^*(x) \le \inf_{\theta\in{\Theta_{D}^*}} J^{\theta}(x;D^*) \le \sup_{D\in\mathcal{A}^x} \inf_{\theta\in{\Theta_{D}}} J^{\theta}(x;D) \le J^*(x),\quad x\in{\mathbb{R}_+}. \end{align}\] This completes the proof. ◻

Proof. Part i. follows from Theorem 5 ii., while Part ii. follows from Theorem 9 ii.. Hence, we start with proving Part iii..

Let \(t\ge0\). For any \(\theta\in\Theta^D({\@fontswitch\relax\mathcal{R}})\), we have \(V_t^{\theta,D}({\@fontswitch\relax\mathcal{R}})\ge V_t^{\text{rob},D}({\@fontswitch\relax\mathcal{R}})\). By the definition 9 of \(V^{\text{rob},D}({\@fontswitch\relax\mathcal{R}})\), \[\begin{align} V^{\text{rob}}_t({\@fontswitch\relax\mathcal{R}}) \geq \mathop{\mathrm{ess\,inf}}_{\theta\in\Theta^D({\@fontswitch\relax\mathcal{R}})} \mathbb{E}^{\theta}_t\bigg[ \int_{t\wedge\tau}^{\tau}\bigg( e^{-\rho s} dD_s+ \frac{V^{\text{rob},D}_s({\@fontswitch\relax\mathcal{R}}) \theta_s^2}{2{\@fontswitch\relax\mathcal{R}}} ds \bigg) + \xi^0_{\tau}\bigg]. \end{align}\] We get the inequality in the \(\ge\) direction of ?? by noticing that \(\Theta^D({\@fontswitch\relax\mathcal{R}})\subset\Theta\). For the reverse inequality, we note that for every \(\theta \in \Theta\) \[\begin{align} -\frac{{\@fontswitch\relax\mathcal{R}}}{2} \frac{\mathbb{Z}^D({\@fontswitch\relax\mathcal{R}})^2}{V^{\text{rob},D}({\@fontswitch\relax\mathcal{R}})} = \inf_{\theta } \Big\{ \frac{V^{\text{rob},D}({\@fontswitch\relax\mathcal{R}})\theta^2}{2{\@fontswitch\relax\mathcal{R}}} + \mathbb{Z}^D({\@fontswitch\relax\mathcal{R}})\theta \Big\}\leq \frac{V^{\text{rob},D}({\@fontswitch\relax\mathcal{R}})\theta^2}{2{\@fontswitch\relax\mathcal{R}}} + \mathbb{Z}^D({\@fontswitch\relax\mathcal{R}})\theta. \end{align}\]

Substituting this into ?? , taking expectations under \(\mathbb{Q}_t^{\theta}\) for each \(\theta \in \Theta\), and then taking the essential infimum over \(\theta\in\Theta\) yield the inequality in the \(\le\) direction of ?? . This completes the argument.

We now prove Part iv. Let \(t\ge0\). By definitions 4 and 10 , we have \[\begin{align} &K_t^D - V^{\text{EZ},D}_t({\@fontswitch\relax\mathcal{R}})\nonumber\\ &\quad = \mathbb{E}_t\bigg[\xi^0_{\tau} - (\xi^0_{\tau})^{1-{\@fontswitch\relax\mathcal{R}}} + \int_{t\wedge\tau}^{\tau} e^{-\rho s} \big( 1 - (1-{\@fontswitch\relax\mathcal{R}}) (V^{\text{EZ},D}_s({\@fontswitch\relax\mathcal{R}}))^{\frac{-{\@fontswitch\relax\mathcal{R}}}{1-{\@fontswitch\relax\mathcal{R}}}}\big) dD_s \bigg] \nonumber\\ &\quad \le \mathbb{E}_t\bigg[\xi^0_{\tau} - (\xi^0_{\tau})^{1-{\@fontswitch\relax\mathcal{R}}} + \int_{t\wedge\tau}^{\tau} e^{-\rho s} {\@fontswitch\relax\mathcal{R}}V^{\text{EZ},D}_s({\@fontswitch\relax\mathcal{R}}) \, dD_s \bigg] \nonumber\\ &\quad \le {\@fontswitch\relax\mathcal{R}}\bigg( \mathbb{E}_t[(\xi^0_{\tau})^2] + \mathbb{E}_t\bigg[\sup_{s \ge t} (V_s^{\text{EZ},D}({\@fontswitch\relax\mathcal{R}}))^2\bigg] \mathbb{E}_t\bigg[ \bigg(\int_0^{\tau} e^{-\rho s} dD_s \bigg)^2 \bigg] \bigg). \end{align}\] The first inequality uses the concavity of \(f(x) = 1 - (1-{\@fontswitch\relax\mathcal{R}})x^{\frac{-{\@fontswitch\relax\mathcal{R}}}{1-{\@fontswitch\relax\mathcal{R}}}}\), which implies \(f(x) \le f(1) + f'(1)(x-1) = {\@fontswitch\relax\mathcal{R}}+ {\@fontswitch\relax\mathcal{R}}(x-1)\). The second inequality follows from \(x - x^{1-{\@fontswitch\relax\mathcal{R}}} \le {\@fontswitch\relax\mathcal{R}}x^2\) for \(x > 0\) and \({\@fontswitch\relax\mathcal{R}}\in (0,1)\). By assumption, the RHS is finite \(\mathbb{P}\)-a.s., so there exists some \({\@fontswitch\relax\mathcal{F}}_t\)-measurable \(\overline{c}_t\) (not depending on \({\@fontswitch\relax\mathcal{R}}\)) such that \(K_t^D - V_t^{\text{EZ},D}({\@fontswitch\relax\mathcal{R}}) \le {\@fontswitch\relax\mathcal{R}}\overline{c}_t,\) \(\mathbb{P}\)-a.s.. Furthermore, \(\overline{c}_t\) can be chosen so that \(\mathbb{E}[\sup_{t\ge0} \overline{c}_t] < \infty\). Indeed, it holds that \(\mathbb{E}[\sup_{t\ge0} |K^D_t - V^{\text{EZ},D}_t({\@fontswitch\relax\mathcal{R}})|] < \infty\) (by the bounds \(\mathbb{E}[\sup_{t\ge0} |K^D_{t \wedge \tau}|^2] < \infty\) and \(\mathbb{E}[\sup_{t\ge0} |V^{\text{EZ},D}_{t \wedge \tau}({\@fontswitch\relax\mathcal{R}})|^2] < \infty\), as well as \(K^D_t=V^{\text{EZ},D}_t({\@fontswitch\relax\mathcal{R}})=\xi_{\tau}^0\) on \(\{t\ge\tau\}\).)

Next, using \(V^{\text{EZ},D}_t = (V^{\text{rob},D}_t({\@fontswitch\relax\mathcal{R}}))^{1-{\@fontswitch\relax\mathcal{R}}}\), we obtain for any \(\lambda \in (0,1)\), \[\begin{align} &\lambda \ln(K_t^D) + (1-\lambda)(1-{\@fontswitch\relax\mathcal{R}}) \ln\big(V^{\text{rob},D}_t({\@fontswitch\relax\mathcal{R}})\big)\\ &\quad\le \ln\Big( \lambda K^D_t + (1-\lambda) \big(V^{\text{rob},D}_t({\@fontswitch\relax\mathcal{R}})\big)^{1-{\@fontswitch\relax\mathcal{R}}} \Big). \end{align}\] Applying Taylor’s expansion and taking the limit as \(\lambda \downarrow 0\), we deduce \[\ln(K_t^D) - \ln(V^{\text{rob},D}_t({\@fontswitch\relax\mathcal{R}})) \le \frac{{\@fontswitch\relax\mathcal{R}}\overline{c}_t}{V_t^{\text{EZ},D}({\@fontswitch\relax\mathcal{R}})} - {\@fontswitch\relax\mathcal{R}}\ln\big(V^{\text{rob},D}_t({\@fontswitch\relax\mathcal{R}})\big),\quad \text{ \mathbb{P}-a.s.}\] Setting \(\overline{c}'_t = \overline{c}_t/C\) (with the lower bound \(C\) of \(V_t^{\text{EZ},D}({\@fontswitch\relax\mathcal{R}})\)), we have ?? . ◻

Proof of Theorem 21.. We first prove the theorem under the additional assumption that \(\xi^0_{\tau} > C\) \(\mathbb{P}\)-a.s. with \(C > 0\) (as imposed in Lemma 43). The general case follows by considering the truncated sequence \(\tilde{\xi}_{\tau}^n = \frac{1}{n} \vee \xi^0_{\tau}\) for each \(n \in \mathbb{N}\), and then letting \(n \to \infty\). We present the proof in two steps.

Let \(0 < {\@fontswitch\relax\mathcal{R}}_1 < {\@fontswitch\relax\mathcal{R}}_2 < 1\). For any \(\theta \in \Theta\), \[\frac{1}{2{\@fontswitch\relax\mathcal{R}}_1} \int_0^t \theta_s^2 \, ds \ge \frac{1}{2{\@fontswitch\relax\mathcal{R}}_2} \int_0^t \theta_s^2 \, ds, \quad \forall t \ge 0.\] By the representation of \(V^{\theta,D}\) in ?? (see Proposition 8), the inequality \(V^{\theta,D}_t({\@fontswitch\relax\mathcal{R}}_1) \ge V^{\theta,D}_t({\@fontswitch\relax\mathcal{R}}_2),\) \(t\ge0,\) holds for all \(\theta\in\Theta\) (even if not in the admissible set \(\Theta^D({\@fontswitch\relax\mathcal{R}}_1)\) of \({\@fontswitch\relax\mathcal{R}}_1\)). Therefore, \[\begin{align} V^{\text{rob},D}_t({\@fontswitch\relax\mathcal{R}}_1) =& \mathop{\mathrm{ess\,inf}}_{\theta\in\Theta^D({\@fontswitch\relax\mathcal{R}}_1)} V_t^{\theta,D}({\@fontswitch\relax\mathcal{R}}_1) \\ \ge& \mathop{\mathrm{ess\,inf}}_{\theta\in\Theta^D({\@fontswitch\relax\mathcal{R}}_2)} V_t^{\theta,D}({\@fontswitch\relax\mathcal{R}}_1) \ge \mathop{\mathrm{ess\,inf}}_{\theta\in\Theta^D({\@fontswitch\relax\mathcal{R}}_2)} V_t^{\theta,D}({\@fontswitch\relax\mathcal{R}}_2) = V_t^{\text{rob},D}({\@fontswitch\relax\mathcal{R}}_2),~t\ge0. \end{align}\] The first inequality holds because \(\Theta^D({\@fontswitch\relax\mathcal{R}}_1) \subseteq \Theta^D({\@fontswitch\relax\mathcal{R}}_2)\); as \({\@fontswitch\relax\mathcal{R}}\) increases, the integrability conditions in 11 become less restrictive. On the other hand, note that \(\theta^0 := (\theta^0_t)_{t \ge 0} \equiv 0 \in \Theta^D({\@fontswitch\relax\mathcal{R}})\), for any \({\@fontswitch\relax\mathcal{R}}\in(0,1)\). Thus, \[V^{\text{rob},D}_t({\@fontswitch\relax\mathcal{R}}) = \inf_{\theta\in \Theta^D({\@fontswitch\relax\mathcal{R}})} V_t^{\theta,D}({\@fontswitch\relax\mathcal{R}}) \le V_t^{\theta^0,D}({\@fontswitch\relax\mathcal{R}}) \equiv K_t^D, \quad t\ge0,\;{\@fontswitch\relax\mathcal{R}}\in(0,1).\]

Let \(0 < {\@fontswitch\relax\mathcal{R}}_1 < {\@fontswitch\relax\mathcal{R}}_2 < 1\) and let \(\theta^{*,2} \in \Theta^D({\@fontswitch\relax\mathcal{R}}_2) \subset \Theta\) denote the worst-case kernel for \({\@fontswitch\relax\mathcal{R}}_2\) so that \(V^{\text{rob},D}({\@fontswitch\relax\mathcal{R}}_2) = V^{\theta^{*,2},D}({\@fontswitch\relax\mathcal{R}}_2)\). Specifically, \[\begin{align} \label{eq:theta422} \theta^{*,2}_t := - {\@fontswitch\relax\mathcal{R}}_2 \frac{{\mathbb{Z}}^D_t({\@fontswitch\relax\mathcal{R}}_2)}{V^{\text{rob},D}_t({\@fontswitch\relax\mathcal{R}}_2)} = -\frac{{\@fontswitch\relax\mathcal{R}}_2}{1-{\@fontswitch\relax\mathcal{R}}_2} \frac{Z^D_t({\@fontswitch\relax\mathcal{R}}_2)}{V^{\text{EZ},D}_t({\@fontswitch\relax\mathcal{R}}_2)}, \quad t\ge0. \end{align}\tag{70}\] Let \(\eta^{\theta^{*,2}}\) be defined as in 7 with \(\theta=\theta^{*,2}\). By ?? in Lemma 43iii, \[\begin{align} V^{\text{rob},D}_t({\@fontswitch\relax\mathcal{R}}_1) \le & \mathbb{E}_t^{\theta^{*,2}}\bigg[ \int_{t\wedge\tau}^{ \tau} e^{-\rho s} dD_s + \xi^0_{\tau}+ \frac{1}{2{\@fontswitch\relax\mathcal{R}}_1} \int_{t\wedge\tau}^{\tau} V^{\text{rob},D}_s({\@fontswitch\relax\mathcal{R}}_1) (\theta^{*,2}_s)^2 ds\bigg]\\ = &V^{\text{rob},D}_t({\@fontswitch\relax\mathcal{R}}_2) +\frac{1}{2}\Big(\frac{1}{{\@fontswitch\relax\mathcal{R}}_1}-\frac{1}{{\@fontswitch\relax\mathcal{R}}_2}\Big) \mathbb{E}_t^{\theta^{*,2}}\bigg[ \int_{t\wedge\tau}^{\tau} (\theta^{*,2}_s)^2 V^{\text{rob},D}_s({\@fontswitch\relax\mathcal{R}}_1) ds \bigg]\\ &+ \mathbb{E}_t^{\theta^{*,2}}\bigg[ \int_{t\wedge\tau}^{\tau} \frac{(\theta^{*,2}_s)^2}{2{\@fontswitch\relax\mathcal{R}}_2} \big( V^{\text{rob},D}_s({\@fontswitch\relax\mathcal{R}}_1)-V^{\text{rob},D}_s({\@fontswitch\relax\mathcal{R}}_2)\big)ds \bigg]. \end{align}\] Since \(V_t^{\text{rob},D}({\@fontswitch\relax\mathcal{R}}_1)\ge V_t^{\text{rob},D}({\@fontswitch\relax\mathcal{R}}_2)\) by monotonicity, Grönwall’s inequality gives \[\begin{align} &V^{\text{rob},D}_t({\@fontswitch\relax\mathcal{R}}_1)-V^{\text{rob},D}_t({\@fontswitch\relax\mathcal{R}}_2) \\ &\quad\le \frac{{\@fontswitch\relax\mathcal{R}}_2-{\@fontswitch\relax\mathcal{R}}_1}{2{\@fontswitch\relax\mathcal{R}}_1{\@fontswitch\relax\mathcal{R}}_2}\mathbb{E}_t^{\theta^{*,2}}\bigg[\int_{t\wedge\tau}^{\tau} e^{\int_{t\wedge\tau}^{s}\frac{(\theta^{*,2}_u)^2}{2{\@fontswitch\relax\mathcal{R}}_2}du} (\theta^{*,2}_s)^2 V^{\text{rob},D}_s({\@fontswitch\relax\mathcal{R}}_1) ds \bigg]\\ &\quad = \frac{{\@fontswitch\relax\mathcal{R}}_2-{\@fontswitch\relax\mathcal{R}}_1}{2{\@fontswitch\relax\mathcal{R}}_1{\@fontswitch\relax\mathcal{R}}_2}\mathbb{E}_t\bigg[\int_{t\wedge\tau}^{\tau} \eta_s^{\theta^{*,2}} e^{\int_{t\wedge\tau}^s\frac{(\theta^{*,2}_u)^2}{2{\@fontswitch\relax\mathcal{R}}_2}du} (\theta^{*,2}_s)^2 V^{\text{rob},D}_s({\@fontswitch\relax\mathcal{R}}_1) ds \bigg]:=\mathrm{I}_t. \end{align}\] Using Itô’s formula for \(\ln V^{\text{rob},D}_s({\@fontswitch\relax\mathcal{R}}_2)\), direct calculation gives \[\begin{align} \eta_s^{\theta^{*,2}} e^{\int_{t}^s\frac{(\theta^{*,2}_u)^2}{2{\@fontswitch\relax\mathcal{R}}_2}du} &= \exp\bigg\{{\@fontswitch\relax\mathcal{R}}_2\bigg(\ln\Big(\frac{V^{\text{rob},D}_t({\@fontswitch\relax\mathcal{R}}_2)}{V^{\text{rob},D}_s({\@fontswitch\relax\mathcal{R}}_2)}\Big)-\int_{t}^s\frac{e^{-\rho u}}{V^{\text{rob},D}_u({\@fontswitch\relax\mathcal{R}}_2)}dD_u \bigg)\bigg\}\\ &\le \Big(\frac{ V_{t}^{\text{rob},D}({\@fontswitch\relax\mathcal{R}}_2) }{ V_s^{\text{rob},D}({\@fontswitch\relax\mathcal{R}}_2)}\Big)^{{\@fontswitch\relax\mathcal{R}}_2}. \end{align}\] Hence, using the explicit form of \(\theta^{*,2}\) in 70 , we have \[\begin{align} \mathrm{I}_t\le& \frac{{\@fontswitch\relax\mathcal{R}}_2-{\@fontswitch\relax\mathcal{R}}_1}{2{\@fontswitch\relax\mathcal{R}}_1{\@fontswitch\relax\mathcal{R}}_2} \mathbb{E}_t\bigg[\int_{t\wedge\tau}^{\tau}\bigg( \frac{{\@fontswitch\relax\mathcal{R}}_2^2}{(1-{\@fontswitch\relax\mathcal{R}}_2)^2}\frac{|Z_s^D({\@fontswitch\relax\mathcal{R}}_2)|^2}{V_s^{\text{EZ},D}({\@fontswitch\relax\mathcal{R}}_2) \big(V^{\text{rob},D}_{s}({\@fontswitch\relax\mathcal{R}}_2)\big)^{1-{\@fontswitch\relax\mathcal{R}}_2}} \\ &\quad\quad\quad\quad\quad\quad\quad\quad\quad \times \Big(\frac{ V_{t}^{\text{rob},D}({\@fontswitch\relax\mathcal{R}}_2) }{ V_s^{\text{rob},D}({\@fontswitch\relax\mathcal{R}}_2)}\Big)^{{\@fontswitch\relax\mathcal{R}}_2} V^{\text{rob},D}_s({\@fontswitch\relax\mathcal{R}}_1)\bigg) ds \bigg]\\ \le&\frac{({\@fontswitch\relax\mathcal{R}}_2-{\@fontswitch\relax\mathcal{R}}_1){\@fontswitch\relax\mathcal{R}}_2}{2{\@fontswitch\relax\mathcal{R}}_1(1-{\@fontswitch\relax\mathcal{R}}_2)^2} \frac{\big(V_t^{\text{rob},D}({\@fontswitch\relax\mathcal{R}}_2)\big)^{{\@fontswitch\relax\mathcal{R}}_2}}{C^{\frac{1}{1-{\@fontswitch\relax\mathcal{R}}_2}}} \mathbb{E}_t\bigg[ \sup_{t\ge 0} \frac{ K^D_t}{ V^{\text{EZ},D}_t({\@fontswitch\relax\mathcal{R}}_2)} \int_{t\wedge\tau}^{\tau} |Z^D_s({\@fontswitch\relax\mathcal{R}}_2)|^2 ds \bigg], \end{align}\] where \(C^{\frac{1}{1-{\@fontswitch\relax\mathcal{R}}_2}}\) is a lower bound for \(V_s^{\text{rob}}({\@fontswitch\relax\mathcal{R}}_2)\), and we have used the fact that \(V^{\text{rob},D}_t({\@fontswitch\relax\mathcal{R}}_1) \le K^D_t\). To bound the expectation, we use ?? , which yields \[\begin{align} \sup_{t\ge0} \frac{ K_t^D}{ V^{\text{EZ},D}_t({\@fontswitch\relax\mathcal{R}}_2)} \le \sup_{t\ge0} \Big(1+{\@fontswitch\relax\mathcal{R}}_2 \frac{\overline{c}_t}{C}\Big) =\sup_{t\ge0} (1+{\@fontswitch\relax\mathcal{R}}_2 \overline{c}_t')<\infty, \text{ {\mathbb{P}}-a.s.,} \end{align}\] where \(\overline{c}_t'\) is the constant from ?? , and we use \(\mathbb{E}[\sup_{t\ge0} \overline{c}_t'] < \infty\). Together with \(\int_0^{\tau} |Z_t^D({\@fontswitch\relax\mathcal{R}}_2)|^2 dt \in L^2({\@fontswitch\relax\mathcal{F}}_{\tau})\), there exists a constant \(C'_t>0\) such that \[\begin{align} 0 \le V^{\text{rob},D}_t({\@fontswitch\relax\mathcal{R}}_1) - V^{\text{rob},D}_t({\@fontswitch\relax\mathcal{R}}_2) \le C'_t ({\@fontswitch\relax\mathcal{R}}_2 - {\@fontswitch\relax\mathcal{R}}_1), \text{ {\mathbb{P}}-a.s.} \end{align}\] This establishes the continuity of the mapping \((0,1) \ni {\@fontswitch\relax\mathcal{R}}\mapsto V_t^{\text{rob},D}({\@fontswitch\relax\mathcal{R}})\). In addition, \(\lim_{{\@fontswitch\relax\mathcal{R}}\downarrow 0} V_t^{\text{rob},D}({\@fontswitch\relax\mathcal{R}}) = K^D_t\) follows from ?? , so ?? holds. ◻

Proof of Corollary 23.. Since \(J^*(x;{\@fontswitch\relax\mathcal{R}}) = \sup_{D\in {\@fontswitch\relax\mathcal{A}}^x} V^{\text{rob},D}_0({\@fontswitch\relax\mathcal{R}})\), the monotonicity and lower semi-continuity from left follows directly from Theorem 21: \[\liminf_{{\@fontswitch\relax\mathcal{R}}\uparrow \tilde{{\@fontswitch\relax\mathcal{R}}}} J^*(x;{\@fontswitch\relax\mathcal{R}}) = \liminf_{{\@fontswitch\relax\mathcal{R}}\uparrow \tilde{{\@fontswitch\relax\mathcal{R}}}} \sup_{D\in {\@fontswitch\relax\mathcal{A}}^x} V^{\text{rob},D}_0 ({\@fontswitch\relax\mathcal{R}}) \ge \sup_{D\in {\@fontswitch\relax\mathcal{A}}^x} V^{\text{rob},D}_0 (\tilde{{\@fontswitch\relax\mathcal{R}}}) = J^*(x;\tilde{{\@fontswitch\relax\mathcal{R}}}).\] For the right continuity, we observe that \(\lim_{{\@fontswitch\relax\mathcal{R}}\downarrow \tilde{{\@fontswitch\relax\mathcal{R}}}} J^*(x;{\@fontswitch\relax\mathcal{R}}) =\sup_{{\@fontswitch\relax\mathcal{R}}\le \tilde{{\@fontswitch\relax\mathcal{R}}} }J^*(x;{\@fontswitch\relax\mathcal{R}})\), and the interchange of supremum arguments yields the desired result. ◻

Proof of Theorem 24.. We denote by \(\underline{b}_{{\@fontswitch\relax\mathcal{R}}}\), \(\overline{b}_{{\@fontswitch\relax\mathcal{R}}}\), and \(\hat{b}_{{\@fontswitch\relax\mathcal{R}}}\) the constants from Assumption 15, emphasizing their dependence on \({\@fontswitch\relax\mathcal{R}}\). We set \(\hat{\boldsymbol{b}} := \max_{{\@fontswitch\relax\mathcal{R}}\in {\@fontswitch\relax\mathcal{I}}} \hat{b}_{{\@fontswitch\relax\mathcal{R}}}\). By Theorem 19, \(J^*(x;{\@fontswitch\relax\mathcal{R}})=v^*(x;{\@fontswitch\relax\mathcal{R}})\) for all \(x \in \mathbb{R}_+\), where \(v^*(x;{\@fontswitch\relax\mathcal{R}})\) is constructed in 43 (by rewriting \(\psi^+(b^*)\) as \(\psi^+(b^*_{{\@fontswitch\relax\mathcal{R}}};{\@fontswitch\relax\mathcal{R}})\) for each \({\@fontswitch\relax\mathcal{R}}\) therein; see also Assumption 15ii.). Throughout, let \((v^*)'(\cdot;{\@fontswitch\relax\mathcal{R}})\) denote the derivative of \(v^*(\cdot;{\@fontswitch\relax\mathcal{R}})\) w.r.t. \(x\). For clarity, we organize the proof into following steps.

Consider \({\@fontswitch\relax\mathcal{R}}_1<{\@fontswitch\relax\mathcal R}_2\in{\@fontswitch\relax\mathcal{I}}\).

Let \(D_1^*\) and \(\theta^{*,2}\) denote the optimal dividend for \({\@fontswitch\relax\mathcal{R}}_1\) and the worst-case kernel for \({\@fontswitch\relax\mathcal{R}}_2\) respectively. From the proof of Theorem 19 in Section 9, we have \[\begin{align} & v^*(x ; {\@fontswitch\relax\mathcal{R}}_1) \leq J^{\theta^{*, 2}}(x; D_1^*, v^*(\cdot; {\@fontswitch\relax\mathcal{R}}_1), {\@fontswitch\relax\mathcal{R}}_1),\\ &v^*(x,{\@fontswitch\relax\mathcal{R}}_2) \geqslant J^{\theta^{*, 2}}(x; D_1^*,v^*(\cdot; {\@fontswitch\relax\mathcal{R}}_2),{\@fontswitch\relax\mathcal{R}}_2), \end{align}\] where for any Borel measurable \(f:\mathbb{R}_+\to\mathbb{R}\), \[J^{\theta^{*, 2}}(x; D, f, {\@fontswitch\relax\mathcal{R}}):=\mathbb{E}^{\theta^{*, 2}}\bigg[\int_0^{\tau^{x,D}} e^{-\rho s}\Big(d D_s+\frac{({\theta_s^{*, 2}})^2}{2 {\@fontswitch\relax\mathcal{R}}}f(X_s^{D}) d s\Big)+e^{-\rho\tau^{x,D}} \xi_0\bigg].\] Following similar arguments as in Theorem 21 (Step 2), we obtain \[\begin{align} &0\le v^*(x;{\@fontswitch\relax\mathcal{R}}_1) - v^*(x;{\@fontswitch\relax\mathcal{R}}_2)\\ &\le\frac{({\@fontswitch\relax\mathcal{R}}_2-{\@fontswitch\relax\mathcal{R}}_1){\@fontswitch\relax\mathcal{R}}_2}{2{\@fontswitch\relax\mathcal{R}}_1}v^*(x;{\@fontswitch\relax\mathcal{R}}_2)^{{\@fontswitch\relax\mathcal{R}}_2}\mathbb{E}\bigg[\int_0^{\tau^{x,D_1^*}} e^{-\rho s}\tilde{v}_{{\@fontswitch\relax\mathcal{R}}_1,{\@fontswitch\relax\mathcal{R}}_2}(X_s^{D_1^*}) d s\bigg]\le C_{\@fontswitch\relax\mathcal{I}}({\@fontswitch\relax\mathcal{R}}_2-{\@fontswitch\relax\mathcal{R}}_1), \end{align}\] with \(\tilde{v}_{{\@fontswitch\relax\mathcal{R}}_1,{\@fontswitch\relax\mathcal{R}}_2}(\cdot):=((v^*)'(\cdot;{\@fontswitch\relax\mathcal{R}}_1)\sigma(\cdot))^2\frac{v^*(\cdot;{\@fontswitch\relax\mathcal{R}}_1)}{v^*(\cdot;{\@fontswitch\relax\mathcal{R}}_2)}(v^*(\cdot;{\@fontswitch\relax\mathcal{R}}_2))^{-(1-{\@fontswitch\relax\mathcal{R}}_2)}\) and some constant \(C_{\@fontswitch\relax\mathcal{I}}>0\) (not depending on \({\@fontswitch\relax\mathcal{R}}_1\) and \({\@fontswitch\relax\mathcal{R}}_2\)). This uniform bound holds since \(X^{D_{{\@fontswitch\relax\mathcal{R}}}^*}\) is valued in \([0, \hat{\boldsymbol{b}}]\) and \(\xi_0 + x \leq v^*(x; {\@fontswitch\relax\mathcal{R}}) \leq \xi_0 + x + \overline{\mu}/\rho\) for all \({\@fontswitch\relax\mathcal{R}}\in {\@fontswitch\relax\mathcal{I}}\). Therefore, \(v^*(x; {\@fontswitch\relax\mathcal{R}})\) is continuous in \({\@fontswitch\relax\mathcal{R}}\).

Note that \(v^*(0;{\@fontswitch\relax\mathcal{R}}) = \xi_0\), and by Theorem 21, the mapping \({\@fontswitch\relax\mathcal{R}}\mapsto v^*(x;{\@fontswitch\relax\mathcal{R}})\) is decreasing. Consequently, for any \({\@fontswitch\relax\mathcal{R}}_1 < {\@fontswitch\relax\mathcal{R}}_2\in{\@fontswitch\relax\mathcal{I}}\), we have \((v^*)'(0;{\@fontswitch\relax\mathcal{R}}_1) \geq (v^*)'(0;{\@fontswitch\relax\mathcal{R}}_2)\). Suppose, to the contrary, that \({\@fontswitch\relax\mathcal{R}}\mapsto (v^*)^{\prime}(0;{\@fontswitch\relax\mathcal{R}})\) is not continuous. Then there exists \(\tilde{{\@fontswitch\relax\mathcal{R}}} \in \operatorname{int}{\@fontswitch\relax\mathcal{I}}\) such that \[\begin{align} \alpha_1 := \liminf_{\epsilon \rightarrow 0+} (v^*)^{\prime}(0;\tilde{{\@fontswitch\relax\mathcal{R}}} - \epsilon) > \limsup_{\epsilon \rightarrow 0+} (v^*)^{\prime}(0;\tilde{{\@fontswitch\relax\mathcal{R}}} + \epsilon) := \alpha_2 \ge 1, \end{align}\] where the inequality \(\alpha_2 \ge 1\) follows from the fact that \((v^*)^{\prime}(0;{\@fontswitch\relax\mathcal{R}}) \ge 1\), as it satisfies the HJB-VI 23 . For a sufficiently small \(\delta_e>0\), we have \[\begin{align} (v^*)^{\prime}(0;\tilde{{\@fontswitch\relax\mathcal{R}}} - \epsilon) > (v^*)^{\prime}(0;\tilde{{\@fontswitch\relax\mathcal{R}}} + \epsilon) + \frac{\alpha_1 - \alpha_2}{2},\quad for all \epsilon\in(0,\delta_e). \end{align}\] Since \(v^*(x;{\@fontswitch\relax\mathcal{R}})\) is twice continuously differentiable w.r.t. \(x\) on \(\mathbb{R}_+\), the derivative \((v^*)'(x;{\@fontswitch\relax\mathcal{R}})\) is continuous in \(x\). Thus, for a sufficiently small \(\delta_x > 0\), \[\begin{align} (v^*)'(x;\tilde{{\@fontswitch\relax\mathcal{R}}} - \epsilon) > (v^*)'(x;\tilde{{\@fontswitch\relax\mathcal{R}}} + \epsilon) + \frac{\alpha_1 - \alpha_2}{2},\quad for all x \in [0, \delta_x], \epsilon \in (0, \delta_e). \end{align}\] Integrating both sides from \(x = 0\) to \(x = \delta_x\) yields \[\begin{align} v^*(\delta_x;\tilde{{\@fontswitch\relax\mathcal{R}}} - \epsilon) > v^*(\delta_x;\tilde{{\@fontswitch\relax\mathcal{R}}} + \epsilon) + \frac{\alpha_1 - \alpha_2}{2} \delta_x, \quad for all\epsilon \in (0, \delta_{e}). \end{align}\] Taking the limit as \(\epsilon \rightarrow 0^+\), we obtain \(v^*(\delta_x;\tilde{{\@fontswitch\relax\mathcal{R}}}^{-}) > v^*(\delta_x;\tilde{{\@fontswitch\relax\mathcal{R}}}^{+})\), which contradicts to the continuity of \({\@fontswitch\relax\mathcal{R}}\mapsto v^*(x;{\@fontswitch\relax\mathcal{R}})\) in Corollary 23. We conclude that \({\@fontswitch\relax\mathcal{I}} \ni {\@fontswitch\relax\mathcal{R}}\mapsto (v^*)^{\prime}(0;{\@fontswitch\relax\mathcal{R}})\) must be continuous.

Recall \(g_{b^*_{{\@fontswitch\relax\mathcal{R}}}}(x) \equiv g_{b^*_{{\@fontswitch\relax\mathcal{R}}},0}(x)\) satisfying \[\begin{align} \left\{ \begin{aligned} &\frac{\sigma^2(x)}{2} (g_{b^*_{{\@fontswitch\relax\mathcal{R}}}})^{\prime}(x) + \mu(x) g_{b^*_{{\@fontswitch\relax\mathcal{R}}}}(x) + \frac{(1-{\@fontswitch\relax\mathcal{R}})}{2} \sigma^2(x) (g_{b^*_{{\@fontswitch\relax\mathcal{R}}}})^2(x) = \rho, \\ &g_{b^*_{{\@fontswitch\relax\mathcal{R}}}}(b^*_{{\@fontswitch\relax\mathcal{R}}}) = 1 / \psi^+(b^*_{{\@fontswitch\relax\mathcal{R}}}), \end{aligned} \right. \end{align}\] as in 42 . Fix \({\@fontswitch\relax\mathcal{R}}_1,{\@fontswitch\relax\mathcal{R}}_2\in{\@fontswitch\relax\mathcal{I}}\), for notational simplicity, let \(g_i\) denote \(g_{b^*_{{\@fontswitch\relax\mathcal{R}}_i}}\) for \(i=1,2\). Then, by 42 , \[\begin{align} g_i(x) &= g_i(0) - \int_0^{x} g'_i(y) \, dy = \frac{(v^*)^{\prime}(0;{\@fontswitch\relax\mathcal{R}}_i)}{v^*(0;{\@fontswitch\relax\mathcal{R}}_i)} - \int_0^{x} g'_i(y) \, dy \\ &= \frac{(v^*)^{\prime}(0;{\@fontswitch\relax\mathcal{R}}_i)}{\xi_0} + \int_0^{x} \left( \frac{2\mu(y)}{\sigma^2(y)} g_i(y) + (1-{\@fontswitch\relax\mathcal{R}}_1) g_i^2(y) - \rho \right) dy. \end{align}\] Therefore, for \(x \in [0, \hat{\boldsymbol{b}}]\), \[\begin{align} |g_1(x) - g_2(x)| \le & \frac{1}{\xi_0} \left| (v^*)^{\prime}(0;{\@fontswitch\relax\mathcal{R}}_1) - (v^*)^{\prime}(0;{\@fontswitch\relax\mathcal{R}}_2) \right| + C_{\hat{\boldsymbol{b}}} \int_{0}^{x} |g_1(y) - g_2(y)| dy \\ &+ \int_{0}^{x} ({\@fontswitch\relax\mathcal{R}}_2 - {\@fontswitch\relax\mathcal{R}}_1) g_2^2(y) dy, \end{align}\] where \(C_{\hat{\boldsymbol{b}}}:= \sup_{0 \le y \le \hat{\boldsymbol{b}}} \left\{ \frac{2\mu(y)}{\sigma^2(y)} + (1-{\@fontswitch\relax\mathcal{R}}_1) |g_1(y) + g_2(y)| \right\}\), for which we recall that \(\hat{\boldsymbol{b}}\) is a uniform upper bound for \(b^*_{{\@fontswitch\relax\mathcal{R}}}\) by Theorem 18. By Grönwall’s inequality, there is some constant \(c>0\) such that \[\begin{align} \sup_{0 \le x \le \hat{\boldsymbol{b}}} |g_1(x) - g_2(x)| \le c e^{C_{\hat{\boldsymbol{b}}}} \left( |(v^*)^{\prime}(0;{\@fontswitch\relax\mathcal{R}}_1) - (v^*)^{\prime}(0;{\@fontswitch\relax\mathcal{R}}_2)| + |{\@fontswitch\relax\mathcal{R}}_2 - {\@fontswitch\relax\mathcal{R}}_1| \right). \end{align}\] Since \({\@fontswitch\relax\mathcal{I}}\ni{\@fontswitch\relax\mathcal{R}}\mapsto (v^*)^{\prime}(0;{\@fontswitch\relax\mathcal{R}})\) is continuous (by Step 1), we have \[\begin{align} \label{eq:g95cts} \lim_{\substack{|{\@fontswitch\relax\mathcal{R}}_2 - {\@fontswitch\relax\mathcal{R}}_1| \to 0^+ \\ {\@fontswitch\relax\mathcal{R}}_1,\,{\@fontswitch\relax\mathcal{R}}_2 \in {\@fontswitch\relax\mathcal{I}}}} \sup_{0 \le x \le \hat{\boldsymbol{b}}} |g_1(x) - g_2(x)| = 0. \end{align}\tag{71}\] We thus show that \({\@fontswitch\relax\mathcal{I}}\ni {\@fontswitch\relax\mathcal{R}}\mapsto g_{b^*_{{\@fontswitch\relax\mathcal{R}}}}(x)\) is continuous, uniformly in \(x \in [0, \hat{\boldsymbol{b}}]\).

Recall the definition of \(b^*_{{\@fontswitch\relax\mathcal{R}}}\) in Proposition 37, and the relationship between \(v^*(x;{\@fontswitch\relax\mathcal{R}})\) and \(g_{b^*_{{\@fontswitch\relax\mathcal{R}}}}(x)\) as given in 41 and 43 . Without loss of generality, fix \({\@fontswitch\relax\mathcal{R}}_1<{\@fontswitch\relax\mathcal R}_2\in{\@fontswitch\relax\mathcal{I}}\). We have \[\begin{align} \label{eq:gR95relation} \frac{ \psi^+\big(b^*_{{\@fontswitch\relax\mathcal{R}}_1};{\@fontswitch\relax\mathcal{R}}_1\big) e^{-\int_0^{b^*_{{\@fontswitch\relax\mathcal{R}}_1}} g_1(y)dy} }{\psi^+\big(b^*_{{\@fontswitch\relax\mathcal{R}}_2};{\@fontswitch\relax\mathcal{R}}_2\big) e^{-\int_0^{b^*_{{\@fontswitch\relax\mathcal{R}}_2}} g_2(y)dy}} = \frac{v^*(0;{\@fontswitch\relax\mathcal{R}}_1)}{v^*(0;{\@fontswitch\relax\mathcal{R}}_2)} = 1. \end{align}\tag{72}\] To proceed, we recall several properties established in Section 4. By the definition of \(\psi^+\) in Assumption 15, the mapping \({\@fontswitch\relax\mathcal{R}}\mapsto \psi^+(x;{\@fontswitch\relax\mathcal{R}})\) is continuous and uniformly decreasing for \(x \geq 0\). Moreover, since \(\psi^+(x;{\@fontswitch\relax\mathcal{R}}) - x\) is decreasing for \(x > \underline{b}\), it follows that \((\psi^+)^{\prime}(x;{\@fontswitch\relax\mathcal{R}}) \leq 1\) for \(x > \underline{b}\). In addition, Lemma 40 i. asserts that \(g_i(x) \geq 1 / \psi^+(x;{\@fontswitch\relax\mathcal{R}}_i)\) for \(x > \underline{b}_{{\@fontswitch\relax\mathcal{R}}_i}\).

Taking logarithms on both sides of 72 , we consider the cases separately.

Case 1: \(b^*_{{\@fontswitch\relax\mathcal{R}}_1} < b^*_{{\@fontswitch\relax\mathcal{R}}_2}\). We have \[\begin{align} \ln \psi^+\big(b^*_{{\@fontswitch\relax\mathcal{R}}_1};{\@fontswitch\relax\mathcal{R}}_1\big) - \ln \psi^+\big(b^*_{{\@fontswitch\relax\mathcal{R}}_2};{\@fontswitch\relax\mathcal{R}}_1\big) + \int_{b^*_{{\@fontswitch\relax\mathcal{R}}_1}}^{b^*_{{\@fontswitch\relax\mathcal{R}}_2}} g_1 (y) \, dy &\le \int_0^{b^*_{{\@fontswitch\relax\mathcal{R}}_2}} \big(g_1(y)-g_2(y)\big) \, dy, \\ \int_{b^*_{{\@fontswitch\relax\mathcal{R}}_1}}^{b^*_{{\@fontswitch\relax\mathcal{R}}_2}} \left( g_1 (y) - \frac{(\psi^+)^{\prime}(y;{\@fontswitch\relax\mathcal{R}}_1)}{\psi^+(y;{\@fontswitch\relax\mathcal{R}}_1)} \right) \, dy &\le \int_0^{b^*_{{\@fontswitch\relax\mathcal{R}}_2}} |g_1(y)-g_2(y)| \, dy, \\ 0 \le \int_{b^*_{{\@fontswitch\relax\mathcal{R}}_1}}^{b^*_{{\@fontswitch\relax\mathcal{R}}_2}}\left( g_1 (y) -\frac{1}{\psi^+(y;{\@fontswitch\relax\mathcal{R}}_1)} \right) \, dy &\le \int_0^{b^*_{{\@fontswitch\relax\mathcal{R}}_2}} |g_1(y)-g_2(y)| \, dy, \end{align}\] where the first inequality follows from the monotonicity of \({\@fontswitch\relax\mathcal{R}}\mapsto \psi^+(x;{\@fontswitch\relax\mathcal{R}})\), and in the last line, we use \((\psi^+)^{\prime}(x;{\@fontswitch\relax\mathcal{R}}_1) \le 1\) and \(g_1(x) \ge 1/ \psi^+(x;{\@fontswitch\relax\mathcal{R}}_1)\) for \(x > b^*_{{\@fontswitch\relax\mathcal{R}}_1}>\underline{b}_{{\@fontswitch\relax\mathcal{R}}_1}\).

Case 2: \(b^*_{{\@fontswitch\relax\mathcal{R}}_1} > b^*_{{\@fontswitch\relax\mathcal{R}}_2}\). By analogous reasoning, \[\begin{align} 0 &< \int_{b^*_{{\@fontswitch\relax\mathcal{R}}_2}}^{b^*_{{\@fontswitch\relax\mathcal{R}}_1}} g_2 (y) - \frac{1}{\psi^+(y;{\@fontswitch\relax\mathcal{R}}_1)} \, dy \\ &\le \left| \ln \psi^+\big(b^*_{{\@fontswitch\relax\mathcal{R}}_2};{\@fontswitch\relax\mathcal{R}}_1\big) - \ln \psi^+\big(b^*_{{\@fontswitch\relax\mathcal{R}}_2};{\@fontswitch\relax\mathcal{R}}_2\big) \right| + \int_0^{b^*_{{\@fontswitch\relax\mathcal{R}}_1}} |g_2(y)-g_1(y)| \, dy, \end{align}\] where the strict inequality holds because \(\psi^+(x;{\@fontswitch\relax\mathcal{R}}_2) < \psi^+(x;{\@fontswitch\relax\mathcal{R}}_1)\), which implies \(g_2(x) \ge 1/\psi^+(x;{\@fontswitch\relax\mathcal{R}}_2) > 1/\psi^+(x;{\@fontswitch\relax\mathcal{R}}_1)\) for \(\underline{b}_{{\@fontswitch\relax\mathcal{R}}_2} < b^*_{{\@fontswitch\relax\mathcal{R}}_2} < x\).

For both cases, by 71 and the continuity of \({\@fontswitch\relax\mathcal{R}}\mapsto \psi^+(x;{\@fontswitch\relax\mathcal{R}})\), we can have the desired continuity of \(b^*_{\@fontswitch\relax\mathcal{R}}\). ◻