Robust Utility Maximization Under Convex Portfolio Constraints

Abstract

We study a robust maximization problem from terminal wealth and consumption under a convex constraints on the portfolio. We state the existence and the uniqueness of the consumption–investment strategy by studying the associated quadratic backward stochastic differential equation. We characterize the optimal control by using the duality method and deriving a dynamic maximum principle.

Appendix

1.1 Proof of the Comparison Theorem

Denote by $\Delta Y_t=Y^1_t- Y_t^2$, $\Delta U_t=\check{U}^1_t- \check{U}_t^2$ and $\Delta \bar{U}_T=\bar{U}_T^1- \bar{U}_T^2$. The pair $(\Delta Y, \int (Z^1-Z^2) dW)$ is the solution of the following equation

$$\begin{aligned} d\Delta Y_t&= (\delta _t \Delta Y_t -\alpha \Delta U_t)dt +\frac{1}{2\beta } |Z^1_t|^2 dt -\frac{1}{2\beta } |Z^2_t|^2dt+(Z^1_t-Z^2_t)dW_t\\ \Delta Y_T&= \bar{\alpha }\Delta \bar{U}_T. \end{aligned}$$

which implies for any stopping time $T\ge \tau \ge t$, we have

$$\begin{aligned} \Delta Y_{\tau }-\Delta Y_{t}&= \int \limits _t^{\tau }(\delta _s \Delta Y_s -\alpha \Delta U_s)ds +\frac{1}{2\beta }\int \limits _t^{\tau } |Z_s^1|^2ds\\&- \frac{1}{2\beta } \int \limits _t^{\tau }|Z_s^2|^2ds+\int \limits _t^{\tau } (Z^1_s-Z^2_s)dW_s. \end{aligned}$$

From the inequality $\int \limits _0^t |Z_s^1|^2ds-\int \limits _0^t|Z_s^2|^2ds-2\int \limits _0^t<Z_s^1,Z_s^2>ds+2\int \limits _0^t|Z_s^2|^2ds=\int \limits _0^t|Z_s^1-Z_s^2|^2ds\ge 0$, where $<.,.>$ denotes the inner product associated with the euclidean norm, we deduce that

$$\begin{aligned} \Delta Y_{t} \!&\le \! \Delta Y_{\tau } \!-\!\int \limits _t^{\tau }(\delta _s \Delta Y_s \!-\!\alpha \Delta U_s)ds \!-\!\frac{1}{\beta }\int \limits _t^{\tau } <Z^1_s\!-\!Z^2_s,Z^{2}_s >ds \!-\!\int \limits _t^{\tau } (Z^1_s\!-\!Z^2_s)dW_s. \end{aligned}$$

We define the probability measure $Q^{*,2}$ equivalent to $P$ where its density is the $P$-martingale $Z^{*,2}_ T$ with

$$\begin{aligned} Z_T^{*,2}=\mathcal{E}_T\left( -\frac{1}{\beta }\int Z_s^2dW_s\right) . \end{aligned}$$

Since $\int \limits _0^.(Z_s^1-Z_s^2)dW_s$ is a $P$-martingale, then $\int \limits _0^.(Z_s^1-Z_s^2)dW_s+\frac{1}{\beta } \int \limits _0^.<Z_s^1-Z_s^2,Z_s^2>ds$ is a $Q^{*,2}$-local martingale. Let $(T_n)_n$ be a reducing sequence for $\int \limits _0^.(Z_s^1-Z_s^2)dW_s+\frac{1}{\beta }\int \limits _0^.<Z^1_s-Z^2_s,Z^2>ds$, then, for $n$ large enough, we have $T_n\ge t$ and so

$$\begin{aligned}&\int \limits _0^t(Z_s^1-Z_s^2)dW_s +\frac{1}{\beta } \int \limits _0^t<Z^1_s-Z^2_s,Z^2_s>ds \\&\quad = E_{Q^{*,2}}[ \int \limits _0^{\tau \wedge T_n}(Z_s^1-Z_s^2)dW_s +\frac{1}{\beta } \int \limits _0^{\tau \wedge T_n}<Z^1_s-Z^2_s,Z^2_s>ds |\mathcal{F}_t]\,\, \\&\,\text { on } \{T\ge \tau \wedge T_n \ge t\}, \end{aligned}$$

which implies

$$\begin{aligned} \Delta Y_{t}&\le E_{Q^{*,2}}[ \Delta Y_{\tau \wedge T_n } -\int \limits _t^{\tau \wedge T_n }(\delta _s \Delta Y_s -\alpha \Delta U_s)ds |\mathcal{F}_t] \,\, \text { on } \{T\ge \tau \wedge T_n \ge t\}. \end{aligned}$$

Sending $n$ to infinity, we have $\tau \wedge T_n\longrightarrow \tau$, $Q^{*,2}$ a.s. and $\Delta Y_{\tau \wedge T_n }\longrightarrow \Delta Y_{\tau }$ $Q^{*,2}$ a.s. Since $Y^1$ (resp. $Y^2$) is in $D_0^{exp}$ and $\check{U}^1$ (resp. $\check{U}^2$) is in $D_1^{exp}$, by the dominated convergence theorem, we have

$$\begin{aligned} \Delta Y_{t}&\le E_{Q^{*,2}}[ \Delta Y_{\tau } -\int \limits _t^{\tau }(\delta _s \Delta Y_s -\alpha \Delta U_s)ds |\mathcal{F}_t] \,\, \text { on } \{T\ge \tau \ge t\}. \end{aligned}$$

From the stochastic Gronwall–Bellman inequality (see Appendix C, Skiadas and Schroder [20]), we have

$$\begin{aligned} \Delta Y_{t}&\le E_{Q^{*,2}}[ \int \limits _t^{T}\alpha e^{-\int \limits _t^s \delta _s ds}\Delta U_s ds + \bar{\alpha }e^{-\int \limits _t^T \delta _s ds} \Delta \bar{U}_T |\mathcal{F}_t]. \end{aligned}$$

(6.1)

From inequalities (2.13)–(2.14), we have $\Delta Y_{t}\le 0$, $0\le t\le T$,$\,\,dt\otimes dP \text { a.s. }$ and the result follows. If $Y_0^1\!=\!Y_0^2$, then from (6.1), we have $E_{Q^{*,2}}[ \int \limits _0^{T}\alpha e^{-\int \limits _0^s \delta _s ds}\Delta U_s ds + \bar{\alpha }e^{-\int \limits _0^T \delta _s ds} \Delta \bar{U}_T]=0$, which implies $\Delta U_t\!=\!0$, $t\in [0,T]$ and $\Delta \bar{U}_T=0$. Since the BSDE (2.2)–(2.3) have a unique solution, then $Y_t^1=Y_t^2$, $t\in [0,T]$ a.e. For the last point, we argue by contradiction. If $Y_0^1=Y_0^2$, then $\Delta U_t=0$, $t\in [0,T]$ and $\Delta \bar{U}_T=0$ which contradicts our assumption and so $Y_0^1<Y_0^2$. $\Box$

1.2 Proof of the Continuity Theorem

We only prove the first statement. Since $U(c^n_t)\ge U(c_t)$, for all $0\le t\le T$ and $\bar{U}(\xi ^n)\ge \bar{U}(\xi )$, then from the comparison Theorem 2.1, the sequence $((Y^{x,c^n,\xi ^n})_{0\le t\le T})_n$ is also non-increasing and so

$$\begin{aligned} Y^{x,c^1,\xi ^1}_t\ge Y^{x,c^n,\xi ^n}_t\ge Y^{x, c,\xi }_t,\,\,\, 0\le t\le T. \end{aligned}$$

(6.2)

We define $(Y^{(\infty )}_t)_{0\le t\le T}$ as follows: $Y^{(\infty ) }_t=\displaystyle \lim _{n \longrightarrow \infty }Y^{x,c^n,\xi ^n}_t$, $0\le t\le T$. From the definition of $(Y^{x,c^n,\xi ^n}_t)_{0\le t\le T}$, we have

$$\begin{aligned} Y^{x,c^n,\xi ^n}_t=-\beta \log E_P\left[ \exp (-\frac{1}{\beta }\int \limits _t^T(\alpha U (c^n_s) -\delta _sY^{x,c^n,\xi ^n}_s)ds)-\frac{1}{\beta }\bar{\alpha }\bar{U}(\xi ^n)|\mathcal{F}_t\right] . \end{aligned}$$

From inequality (6.2) and the monotonicity property of the sequences $(\xi ^n)_n$ and $\Big ((c^n_{t})_{0\le t\le T}\Big )_n$, we have

$$\begin{aligned}&\Big |\exp (-\frac{1}{\beta }\int \limits _t^T(\alpha U(c^n_s)-\delta _s Y^{x,c^n,\xi ^n}_s)ds) -\frac{1}{\beta }\bar{\alpha }\bar{U} (\xi ^n)\Big |\\&\quad \le \exp \left( \frac{1}{\beta }\int \limits _t^T(\alpha | U(c^n_s)|+\delta _s|Y^{x,c^n,\xi ^n}_s|)ds\right) +\frac{\bar{\alpha }}{\beta }|\bar{U} (\xi ^n)|)\\&\quad \le \exp \Big (\frac{\alpha }{\beta }\int \limits _0^T(| U(c^1_s)|\!+\!|U(c_s)|)ds \!+\!\frac{\Vert \delta \Vert _{\infty }T}{\beta }\big (\mathrm {ess}\sup _{0\le t\!\le \! T}|Y^{x,c^1,\xi ^1}_t| \!+\!\mathrm ess \sup _{0\le t\le T}|Y_t^{x,c, \xi }|\big )\\&\quad +\frac{\bar{\alpha }}{\beta }(|\bar{U} (\xi ^1)|+|\bar{U}(\xi )|)\Big ):=g_T. \end{aligned}$$

From Cauchy Schwarz inequality, we have

$$\begin{aligned}&E_P[|g_T|] \le E_P\Big [\exp \Big (\frac{2\alpha }{\beta }\int \limits _0^T(| U(c^1_s)|+|U(c_s)|)ds\Big )\Big ]^{\frac{1}{2}}\nonumber \\&\quad E_P\Big [\exp \Big ( \frac{2\Vert \delta \Vert _{\infty }T}{\beta }\big (\mathrm {ess}\sup _{0\le t\le T}|Y^{x,c^1,\xi ^1}_t| +\mathrm ess \sup _{0\le t\le T}|Y_t^{x,c, \xi }|\big )+\frac{2\bar{\alpha }}{\beta }(|\bar{U} (\xi ^1)|+|\bar{U}(\xi )|)\Big )\Big ]^{\frac{1}{2}}\nonumber \\&\quad \le E_P\Big [\exp \Big (\frac{2\alpha }{\beta }\int \limits _0^T(| U(c^1_s)|+|U(c_s)|)ds\Big )\Big ]^{\frac{1}{2}}\nonumber \\&\quad E_P\Big [\exp \Big ( \frac{4\Vert \delta \Vert _{\infty }T}{\beta }\big (\mathrm {ess}\sup _{0\le t\le T}|Y^{x,c^1,\xi ^1}_t | +\mathrm ess \sup _{0\le t\le T}|Y_t^{x,c, \xi }|\big )\Big )\Big ]^{\frac{1}{4}}\nonumber \\&\quad E_P\Big [\exp \Big (\frac{4\bar{\alpha }}{\beta }(|\bar{U} (\xi ^1)|+|\bar{U}(\xi )|)\Big )\Big ]^{\frac{1}{4}}. \end{aligned}$$

(6.3)

From the boundedness on the discounting factor (H1) and since $(c,\xi )\in \mathcal{A}(x)$, $(c^1,\xi ^1)\in \mathcal{A}(x)$, $Y^{x,c, \xi }\in D_0^{exp}$ and $Y^{x,c^1, \xi ^1}\in D_0^{exp}$, we have $g_T\in L^1(P)$. By the dominated convergence theorem, we have

$$\begin{aligned} Y^{(\infty )}_t=-\beta \log E_P[\exp (-\frac{1}{\beta }\int \limits _t^T(\alpha U(c_s)-\delta _sY_s^{(\infty )})ds) -\frac{1}{\beta }\bar{\alpha }\bar{U}(\xi )|\mathcal{F}_t],\,\,0\le t\le T. \end{aligned}$$

Since there exists a unique solution to the BSDE (2.11)–(2.12), we have necessarily $Y^{(\infty )}=Y^{x,c,\xi }$ and the result follows.

1.3 Proof of the Maximum Principle

We fix $\epsilon >0$ and $\eta >0$ such that $\epsilon <\eta$.

First step: We prove that

$$\begin{aligned} \bar{\alpha } Z^{*}_TS^T_{\delta }\bar{U}^{'}(\xi ^*)&\le \lambda ^* \tilde{Z}_T \,\,dP \,\,a.s. \end{aligned}$$

(6.4)

We consider the following set

$$\begin{aligned} A_{\epsilon ,\eta } :=\Big \{Z_T^{*} S_{\delta }^T\bar{\alpha } \bar{U}'(\xi ^*)-\lambda ^* \tilde{Z}_T >0, \epsilon < \xi ^* < \eta \Big \}. \end{aligned}$$

We define $\xi _n$ as follows: $\xi _n=\xi ^* +\frac{1}{n} \mathbf{1}_{A_{\epsilon ,\eta }}$.

$\star$ We prove that $(c^*,\xi _n)\in \mathcal{A}$ : From the representation theorem under $\tilde{P}^*$, there exists a process $H_n\in \tilde{\mathcal{H}}$ such that

$$\begin{aligned} \frac{1}{n} \mathbf{1}_{A_{\epsilon ,\eta }}&= E_{\tilde{P}^*}[\frac{1}{n} \mathbf{1}_{A_{\epsilon ,\eta }}]+\int \limits _0^T{H_n}_s (diag S_s)^{-1}dS_s, \end{aligned}$$

which implies that

$$\begin{aligned} \xi _n=x+ E_{\tilde{P}^*}[\frac{1}{n} \mathbf{1}_{A_{\epsilon ,\eta }}]+ \int \limits _0^T{H_n^*}_s (diag S_s)^{-1}dS_s-\int \limits _0^Tc_s^*ds, \end{aligned}$$

where $H_n^*=H^*+H_n$. For n large enough, we have $0\le \frac{1}{n}\le \frac{\epsilon }{2}$ and so

$$\begin{aligned} {\epsilon }\le \xi _n\le \eta +\frac{\epsilon }{2}\text { on the set }\{ \epsilon <\xi ^*<\eta \}. \end{aligned}$$

From the standard assumptions on the utility functions (H4), we have

$$\begin{aligned} \bar{U}(\epsilon )\le \bar{U}(\xi _n)\le \bar{U}(\eta +\frac{\epsilon }{2}) \text { on the set }\{ \epsilon <\xi ^*<\eta \}. \end{aligned}$$

and so for $n$ large enough, $E[\exp {(\gamma |\bar{U}(\xi _n)|)}]$ is finite, which implies that $(c^*,\xi _n)\in \mathcal{A}$.

$\star$ We prove that $P(A_{\epsilon ,\eta })=0$ : From the definition of $J$ (see (4.10)) and the optimality of the strategy $(c^*,\xi ^*)$, we have

$$\begin{aligned} 0&\ge n(J(x,c^*,\xi ^{n} ,\tilde{P}^*, \lambda ^*)-J(x,c^*,\xi ^{*},\tilde{P}^*,\lambda ^*))\\&= n(Y_0^{x,c^*,\xi ^{n}}- Y_0^{x,c^*, \xi ^{*}}) -\lambda ^*E_{\tilde{P}^*}[\mathbf{1}_{A_{\epsilon ,\eta }}]\nonumber \\&\ge nE_{Q^{n}}\Big [ \bar{\alpha }S_T^\delta (\bar{U}(\xi ^{n})-\bar{U}(\xi ^*)) \Big ]-\lambda ^*E_{\tilde{P}^*}[\mathbf{1}_{A_{\epsilon ,\eta }}]\nonumber \\&= nE_{P}\Big [ Z^{Q^{n}}_T\bar{\alpha }S_T^\delta (\bar{U}(\xi ^{n})-\bar{U}(\xi ^*)) \mathbf{1}_{A_{\epsilon ,\eta }}\Big ]-\lambda ^*E_{\tilde{P}^*}[\mathbf{1}_{A_{\epsilon ,\eta }}],\nonumber \end{aligned}$$

(6.5)

where the probability measure $Q^{n}$ has a density given by the $P$-martingale $Z^{Q^{n}}=(Z_t^{ Q^{n}})_{0\le t\le T}=(\mathcal{E}_t(-\frac{1}{\beta } M^{x,c^*,\xi ^{n}}))_{0\le t\le T}$ and $M_t^{x,c^*,\xi ^{n}}=\int \limits _0^tZ_s^{x,c^*,\xi ^{n}}dW_s$.

Since there exists $\theta ^n$ between $\xi ^{n}$ and $\xi ^*$ such that $\bar{U}(\xi ^{n})-\bar{U}(\xi ^*)=\bar{U}^{'}(\theta ^n)(\xi ^{n}-\xi ^*)$, we deduce that

$$\begin{aligned} n (\bar{U}(\xi ^{n})-\bar{U}(\xi ^*))\mathbf{1}_{A_{\epsilon ,\eta }}\longrightarrow \bar{U}^{'}(\xi ^*)\mathbf{1}_{A_{\epsilon ,\eta }} \,\,dP \,\,a.s. \end{aligned}$$

(6.6)

and

$$\begin{aligned} | n(\bar{U}(\xi ^{n})-\bar{U}(\xi ^*))\mathbf{1}_{\{\epsilon <\xi ^*<\eta \}} |\le \bar{U}^{'}(\epsilon )\,\,dP \,\,a.s. \end{aligned}$$

(6.7)

From the definition of $Z_t^{Q^{n}}$, we have

$$\begin{aligned} Z_t^{Q^{n}}=\exp (-\frac{1}{\beta } M_t^{x,c^*,\xi ^{n}} -\frac{1}{2\beta ^2}<M^{x,c^*,\xi ^{n}}>_t). \end{aligned}$$

(6.8)

From the BSDE (2.11), we obtain

$$\begin{aligned} Y^{x,c^*,\xi ^{n}}_t-Y^{x,c^*,\xi ^{n}}_0&= \int \limits _0^t(\delta _s Y^{x,c^*,\xi ^{n}}_s-\alpha U(c_s^{*}))ds +\frac{1}{2\beta }\langle M^{x,c^*,\xi ^{n}}\rangle _t +M_t^{x,c^*,\xi ^{n}}.\nonumber \\ \end{aligned}$$

(6.9)

Plugging (6.9) into (6.8), we obtain

$$\begin{aligned} Z_t^{Q^{n}}= \exp \Big ( \int \limits _0^t\frac{1}{\beta }(\delta _s Y^{x,c^*,\xi ^{n}}_s-\alpha U(c_s^{*}))ds -\frac{1}{\beta } ( Y^{x,c^*,\xi ^{n}}_t -Y^{x,c^*,\xi ^{n}}_0) \Big ). \end{aligned}$$

From Proposition 2.1 (i), we have

$$\begin{aligned} \displaystyle \lim _{n \longrightarrow \infty } Z_t^{Q^{n}}&= \exp \Big ( \int \limits _0^t\frac{1}{\beta }(\delta _s Y^{x,c^*,\xi ^*}_s-\alpha U(c^{*}_s))ds -\frac{1}{\beta } ( Y^{x,c^*,\xi ^*}_t -Y^{x,c^*,\xi ^*}_0) \Big )\nonumber \\&= Z_t^*\,\,dt\otimes dP \,\,a.s. \end{aligned}$$

(6.10)

Under the boundedness on the discounting factor (H1) and since $(Y^{x,c^*,\xi ^{n}}_t)_{0\le t\le T} \in D_0^{exp}$, we have

$$\begin{aligned} \!|\!Z_t^{Q^{n}}\!|\!\!\le \! \exp \Big (\frac{T}{\beta } ||\delta ||_{\infty }\text {ess}\displaystyle \sup _{0\le t\le T}|Y_t^{x,c^*,\xi ^{n}}| \!+\!\frac{\alpha }{\beta }\int \limits _0^T |U(c^*_s)|ds\!+\!\frac{2}{\beta }\text {ess}\displaystyle \sup _{0\le t\le T}|Y_t^{x,c^*,\xi ^{n}}|\Big ).\nonumber \\ \end{aligned}$$

(6.11)

From Proposition 2.1 (i), we have $Y_t^{x,c^*,\xi ^{1}}\ge Y_t^{x,c^*,\xi ^{n}}\ge Y_t^{x,c^*, \xi ^*}$ and so

$$\begin{aligned} \text {ess}\displaystyle \sup _{0\le t\le T}|Y_t^{x,c^*,\xi ^{n}}| \le \text {ess}\displaystyle \sup _{0\le t\le T}|Y_t^{x,c^*,\xi ^{1}}| +\text {ess}\displaystyle \sup _{0\le t\le T}|Y_t^{x,c^*, \xi ^*}|. \end{aligned}$$

(6.12)

Using the inequalities (6.7), (6.11) and (6.12), we have

$$\begin{aligned}&| n(\bar{U}(\xi ^{n})-\bar{U}(\xi ^*))\mathbf{1}_{\{\epsilon < \xi ^* < \eta \}} | |Z_t^{Q^{n}}|\\&\quad \le \bar{U}^{'} (\epsilon ) \exp \Big (\frac{T}{\beta } ||\delta ||_{\infty } (\text {ess}\displaystyle \sup _{0\le t\le T}|Y_t^{x,c^*,\xi ^{1}}| +\text {ess}\displaystyle \sup _{0\le t\le T}|Y_t^{x,c^*,\xi ^*}|)\\&\quad +\frac{\alpha }{\beta }\int \limits _0^T |U(c_s^*)|ds+\frac{2}{\beta }(\text {ess}\displaystyle \sup _{0\le t\le T}|Y_t^{x,c^*,\xi ^{1}}| +\text {ess}\displaystyle \sup _{0\le t\le T}|Y_t^{x,c^*,\xi ^*}|)\Big ):=g_{T}. \end{aligned}$$

From Cauchy Schwartz inequality, we have

$$\begin{aligned} E_P[|g_{T}|] \!&\le \! \bar{U}^{'} (\epsilon ) E_P\Big [\!\exp \!\Big (\frac{2\alpha }{\beta }\int \limits _0^T|U(c_s^*)|ds\Big )\Big ]^{\frac{1}{2}}\\&E_P\Big [\exp \Big ( \frac{2(2\!+\!\Vert \delta \Vert _{\infty }T)}{\beta }\big (\mathrm ess \sup _{0\le t\le T}|Y^{x,c^*,\xi ^1}_t| \!+\!\mathrm ess \sup _{0\le t\le T}|Y_t^{x,c^*, \xi ^*}|\big ) \Big )\Big ]^{\frac{1}{2}}\nonumber . \end{aligned}$$

(6.13)

From the boundedness on the discounting factor (H1), and since $(c^*,\xi ^*)\in \mathcal{A}$, $(c^*,\xi ^{1})\in \mathcal{A}$, $Y^{x,c^*, \xi ^*}\in D_0^{exp}$ and $Y^{x,c^*, \xi ^{1}}\in D_0^{exp}$, we have $g_{T}\in L^1(P)$. By the dominated convergence theorem and substituting inequalities (6.6) and (6.10) into (6.5), we have

$$\begin{aligned} 0&\ge \displaystyle \lim _{n\longrightarrow \infty } E_{Q^{n}}\Big [\bar{\alpha } S_T^\delta n(\bar{U}(\xi ^*)-\bar{U}(\xi _n))\mathbf{1}_{\{\epsilon < \xi ^* < \eta \}} \Big ]-\lambda ^*E_{\tilde{P}^*}[\mathbf{1}_{A_{\epsilon ,\eta }}]\\&= E_{Q^{*}} \Big [ \bar{\alpha } S_T^\delta \bar{U}^{'}(\xi ^*)\mathbf{1}_{A_{\epsilon ,\eta }} \Big ]-\lambda ^*E_{\tilde{P}^*}[\mathbf{1}_{A_{\epsilon ,\eta }}]. \end{aligned}$$

which implies $P(A_{\epsilon ,\eta })=0$ for all $0<\epsilon <\eta <\infty$. Sending $\epsilon \longrightarrow 0$ and $\eta \longrightarrow \infty$, we have $A_{\epsilon ,\eta } \nearrow \Big \{Z_T^{*} S_{\delta }^T\bar{\alpha } \bar{U}'(\xi ^*)-\lambda ^* \tilde{Z}_T >0 \Big \}$ and so inequality (6.4) is proved.

Second step: We prove that

$$\begin{aligned} \bar{\alpha } Z^{*}_T S_T^\delta U^{'}(\xi ^*)&\ge \lambda ^* \tilde{Z}_T^{*} \,\,dP \,\,a.s. \end{aligned}$$

(6.14)

We consider the following set

$$\begin{aligned} B_{\epsilon ,\eta }:=\Big \{Z_T^{*}\bar{\alpha } S_T^\delta \bar{U}'(\xi ^*)-\lambda ^* \tilde{Z}_T^{*} <0, \epsilon < \xi ^* < \eta \Big \}. \end{aligned}$$

We define $\xi _n^{'}$ as follows: $\xi _n^{'}:=\xi ^* -\frac{1}{n}\mathbf{1}_{B_{\epsilon ,\eta }}$.

$\star$ We prove that $(c^*,\xi _n^{'})\in \mathcal{A}$ : As in the first step, for n large enough, we have $0\le \frac{1}{n}\le \frac{\epsilon }{2}$ and so

$$\begin{aligned} \frac{\epsilon }{2}\le \xi _n^{'}\le \eta \text { on the set }\{ \epsilon <\xi ^*<\eta \}. \end{aligned}$$

From the standard assumptions on the utility functions (H4), we have

$$\begin{aligned} \bar{U}(\frac{\epsilon }{2})\le \bar{U}(\xi _n^{'})\le \bar{U}(\eta ) \text { on the set }\{ \epsilon <\xi ^*<\eta \}. \end{aligned}$$

This shows that for $n$ large enough, $E[\exp {(\gamma |\bar{U}(\xi _n^{'})|)}]$ is finite and so $(c^*,\xi _n^{'})\in \mathcal{A}$.

$\star$ We prove that $P(B_{\epsilon ,\eta })=0$ : From the definition of $J$ (see (4.10)) and the optimality of the strategy $(c^*,\xi ^*)$, we have

$$\begin{aligned} 0&\ge n(J(x,c^*,\xi ^{'n} ,\tilde{P}^*, \lambda ^*)-J(x,c^*,\xi ^{*},\tilde{P}^*,\lambda ^*))\\&= n(Y_0^{x,c^*,\xi ^{'n}}- Y_0^{x,c^*, \xi ^{*}}) +\lambda ^*E_{\tilde{P}^*}[\mathbf{1}_{B_{\epsilon ,\eta }}]\nonumber \\&\ge nE_{Q^{'n}}\Big [ \bar{\alpha }S_T^\delta (\bar{U}(\xi ^{'n})-\bar{U}(\xi ^*)) \Big ]+\lambda ^*E_{\tilde{P}^*}[\mathbf{1}_{B_{\epsilon ,\eta }}]\nonumber \\&= nE_{P}\Big [ Z^{Q^{'n}}_T\bar{\alpha }S_T^\delta (\bar{U}(\xi ^{'n})-\bar{U}(\xi ^*)) \mathbf{1}_{B_{\epsilon ,\eta }}\Big ]+\lambda ^*E_{\tilde{P}^*}[\mathbf{1}_{B_{\epsilon ,\eta }}],\nonumber \end{aligned}$$

(6.15)

where the probability measure $Q^{'n}$ has a density given by the $P$-martingale $Z^{Q^{'n}}=(Z_t^{ Q^{'n}})_{0\le t\le T}=(\mathcal{E}_t(-\frac{1}{\beta } M^{x,c^*,\xi ^{'n}}))_{0\le t\le T}$ and $M_t^{x,c^*,\xi ^{'n}}=\int \limits _0^tZ_s^{x,c^*,\xi ^{'n}}dW_s$.

Since there exists $\theta ^n$ between $\xi ^{'n}$ and $\xi ^*$ such that $\bar{U}(\xi ^{'n})-\bar{U}(\xi ^*)=\bar{U}^{'}(\theta ^n)(\xi ^{'n}-\xi ^*)$, we deduce that

$$\begin{aligned} n (\bar{U}(\xi ^{'n})-\bar{U}(\xi ^*)) \mathbf{1}_{B_{\epsilon ,\eta }}\longrightarrow -\bar{U}^{'}(\xi ^*)\mathbf{1}_{B_{\epsilon ,\eta }} \,\,dP \,\,a.s. \end{aligned}$$

(6.16)

and

$$\begin{aligned} | n(\bar{U}(\xi ^{'n})-\bar{U}(\xi ^*))\mathbf{1}_{\{\epsilon <\xi ^*<\eta \}} |\le \bar{U}^{'}(\epsilon )\,\,dP \,\,a.s. \end{aligned}$$

(6.17)

From the definition of $Z_t^{Q^{'n}}$, we have

$$\begin{aligned} Z_t^{Q^{'n}}=\exp (-\frac{1}{\beta } M_t^{x,c^*,\xi ^{'n}} -\frac{1}{2\beta ^2}<M^{x,c^*,\xi ^{'n}}>_t). \end{aligned}$$

(6.18)

From the BSDE (2.11), we obtain

$$\begin{aligned} Y^{x,c^*,\xi ^{'n}}_t-Y^{x,c^*,\xi ^{'n}}_0&= \int \limits _0^t(\delta _s Y^{x,c^*,\xi ^{'n}}_s-\alpha U(c_s^{*}))ds +\frac{1}{2\beta }\langle M^{x,c^*,\xi ^{'n}}\rangle _t +M_t^{x,c^*,\xi ^{'n}}.\nonumber \\ \end{aligned}$$

(6.19)

Plugging (6.19) into (6.18), we obtain

$$\begin{aligned} Z_t^{Q^{'n}}= \exp \Big ( \int \limits _0^t\frac{1}{\beta }(\delta _s Y^{x,c^*,\xi ^{'n}}_s-\alpha U(c_s^{*}))ds -\frac{1}{\beta } ( Y^{x,c^*,\xi ^{'n}}_t -Y^{x,c^*,\xi ^{'n}}_0) \Big ). \end{aligned}$$

From Proposition 2.1 (ii), we have

$$\begin{aligned} \displaystyle \lim _{n \longrightarrow \infty } Z_t^{Q^{'n}}&= \exp \Big ( \int \limits _0^t\frac{1}{\beta }(\delta _s Y^{x,c^*,\xi ^*}_s-\alpha U(c^{*}_s))ds -\frac{1}{\beta } ( Y^{x,c^*,\xi ^*}_t -Y^{x,c^*,\xi ^*}_0) \Big )\nonumber \\&= Z_t^*\,\,dt\otimes dP \,\,a.s. \end{aligned}$$

(6.20)

Under the boundedness on the discounting factor (H1) and since $(Y^{x,c^*,\xi ^{'n}}_t)_{0\le t\le T} \in D_0^{exp}$, we have

$$\begin{aligned} |Z_t^{Q^{'n}}|&\le \exp \Big (\frac{T}{\beta } ||\delta ||_{\infty }\text {ess}\displaystyle \sup _{0\le t\le T}|Y_t^{x,c^*,\xi ^{'n}}| +\frac{\alpha }{\beta }\int \limits _0^T |U(c^*_s)|ds\nonumber \\&+\frac{2}{\beta }\text {ess}\displaystyle \sup _{0\le t\le T}|Y_t^{x,c^*,\xi ^{'n}}|\Big ). \end{aligned}$$

(6.21)

From Proposition 2.1 (ii), we have $Y_t^{x,c^*,\xi ^{'1}}\le Y_t^{x,c^*,\xi ^{'n}}\le Y_t^{x,c^*, \xi ^*}$ and so

$$\begin{aligned} \text {ess}\displaystyle \sup _{0\le t\le T}|Y_t^{x,c^*,\xi ^{'n}}| \le \text {ess}\displaystyle \sup _{0\le t\le T}|Y_t^{x,c^*,\xi ^{'1}}| +\text {ess}\displaystyle \sup _{0\le t\le T}|Y_t^{x,c^*, \xi ^*}|. \end{aligned}$$

(6.22)

Using the inequalities (6.17), (6.21) and (6.22), we have

$$\begin{aligned}&| n(\bar{U}(\xi ^{'n})-\bar{U}(\xi ^*))\mathbf{1}_{\{\epsilon < \xi ^* < \eta \}} | |Z_t^{Q^{'n}}|\\&\quad \le \bar{U}^{'} (\epsilon ) \exp \Big (\frac{T}{\beta } ||\delta ||_{\infty } (\text {ess}\displaystyle \sup _{0\le t\le T}|Y_t^{x,c^*,\xi ^{'1}}| +\text {ess}\displaystyle \sup _{0\le t\le T}|Y_t^{x,c^*,\xi ^*}|)\\&\quad +\frac{\alpha }{\beta }\int \limits _0^T |U(c_s^*)|ds+\frac{2}{\beta }(\text {ess}\displaystyle \sup _{0\le t\le T}|Y_t^{x,c^*,\xi ^{'1}}| \!+\!\text {ess}\displaystyle \sup _{0\le t\le T}|Y_t^{x,c^*,\xi ^*}|)\Big ):=\tilde{g}_{T}. \end{aligned}$$

From Cauchy Schwartz inequality, we have

$$\begin{aligned} E_P[|\tilde{g}_{T}|] \!&\le \! \bar{U}^{'} (\epsilon ) E_P\Big [\exp \Big (\frac{2\alpha }{\beta }\int \limits _0^T|U(c_s^*)|ds\Big )\Big ]^{\frac{1}{2}}\\&E_P\Big [\exp \Big ( \frac{2(2\!+\!\Vert \delta \Vert _{\infty }T)}{\beta }\big (\mathrm ess |Y^{x,c^*,\xi ^{'1}}_t| \!+\!\mathrm ess \sup _{0\le t\le T}|Y_t^{x,c^*, \xi ^*}|\big ) \Big )\Big ]^{\frac{1}{2}}\nonumber . \end{aligned}$$

(6.23)

From the boundedness on the discounting factor (H1), and since $(c^*,\xi ^*)\in \mathcal{A}$, $(c^*,\xi ^{'1})\in \mathcal{A}$, $Y^{x,c^*, \xi ^*}\in D_0^{exp}$ and $Y^{x,c^*, \xi ^{'1}}\in D_0^{exp}$, we have $\tilde{g}_{T}\in L^1(P)$. By the dominated convergence theorem and substituting inequalities (6.16) and (6.20) into (6.15), we have

$$\begin{aligned} 0&\ge \displaystyle \lim _{n\longrightarrow \infty } E_{Q^{'n}}\Big [\bar{\alpha } S_T^\delta n(\bar{U}(\xi ^*)-\bar{U}(\xi _n^{'}))\mathbf{1}_{\{\epsilon < \xi ^* < \eta \}} \Big ] +\lambda ^*E_{\tilde{P}^*}[\mathbf{1}_{B_{\epsilon ,\eta }}]\\&= E_{Q^{*}} \Big [ -\bar{\alpha } S_T^\delta \bar{U}^{'}(\xi ^*)\mathbf{1}_{B_{\epsilon ,\eta }} \Big ]+\lambda ^*E_{\tilde{P}^*}[\mathbf{1}_{B_{\epsilon ,\eta }}]. \end{aligned}$$

which implies $P(B_{\epsilon ,\eta })=0$ for all $0<\epsilon <\eta <\infty$. Sending $\epsilon \longrightarrow 0$ and $\eta \longrightarrow \infty$, we obtain

$$\begin{aligned} Z_T^{*} S_T^{\delta }\bar{\alpha } \bar{U}'(\xi ^*)\ge \lambda ^* \tilde{Z}_T^{*}\,\,\text { on the set }\{\xi ^* >0\}\,\, dP\,a.s. \end{aligned}$$

(6.24)

Since the utility function satisfies the Inada conditions (Assumption (H4)), we have $P(\xi ^* =0)=0$ and so $P(B_{\epsilon ,\eta })=0$ for all $0<\epsilon <\eta <\infty$.

$\star$ We prove inequality (6.14): Sending $\epsilon \longrightarrow 0$ and $\eta \longrightarrow \infty$ we have $B_{\epsilon ,\eta } \nearrow \Big \{Z_T^{*} S_{\delta }^T\bar{\alpha } \bar{U}'(\xi ^*)-\lambda ^* \tilde{Z}_T <0 \Big \}$ and so inequality (6.14) is proved.

The result follows from (6.4) and (6.14). The same argument holds for the consumption process. $\square$