1 Introduction↩︎

Throughout this paper, let us fix a positive integer \(d\) and a positive real \(T\). Let \((\Omega,\mathcal{F},\mathbb{P})\) be a complete probability space carrying a standard \(d\)-dimensional Brownian motion \((B_t)_{t\geq0}, (\mathcal{F}_t)_{t\geq0}\) the completed natural \(\sigma\)-algebra generated by \((B_t)_{t\geq0}\) and \(\mathcal{F}_T := \mathcal{F}\). We consider the following one-dimensional quadratic backward stochastic differential equation (BSDE in short): \[\begin{align} \label{BSDE1461} Y_t=\xi-\int_t^Tg(s,Y_s,Z_s){\rm d}s+\int_t^TZ_s\cdot{\rm d}B_s,~~~~t\in[0,T], \end{align}\tag{1}\] where \(T\) is the terminal time, and the terminal value \(\xi\) is an \(\mathcal{F}_T\)-measurable unbounded random variable. The generator \[g(\omega,t,y,z):\Omega\times[0,T]\times{\mathbb{R}}\times{\mathbb{R}}^d\mapsto{\mathbb{R}}\] is \((\mathcal{F}_t)\)-progressively measurable for each \((y,z)\in{\mathbb{R}}\times{\mathbb{R}}^d\), which is continuous in \((y,z)\) and has a quadratic growth in \(z\). The BSDE with parameters \((\xi,T,g)\) is usually denoted by BSDE \((\xi,T,g)\).

In 1990, the seminar paper Pardoux-Peng?, 1990? first introduced the nonlinear BSDEs and established an existence and uniqueness result for the adapted solution of BSDEs under the uniformly Lipschitz continuous condition of the generator, which laid the foundation for the theoretical and applied research of BSDEs. After that, a lot of efforts have been made in order to study the well-posedness of these equations and their practical applications. Since it appears naturally in the utility maximization problem, quadratic BSDE is a class of BSDEs that have attracted particular attention, see for example Hu?, 2005?, Rouge?, 2000?, Fan?, 2016? together with the updated papers Tian?, 2023?, Hu-Tang-Wang?, 2022?, Fan-Hu-Tang?, 2023? for more details. In particular, Kobylanski Kobylanski?, 2000? first studied quadratic BSDEs with bounded terminal values, and established a rather general existence and uniqueness result on the bounded solution. This result was further extended in Fan?, 2016? and [1] to quadratic BSDEs with finite and infinite horizons, where the generator \(g\) can satisfy a monotonicity condition with general growth in \(y\) and a locally-Lipschitz continuous condition with quadratic growth in \(z\). Roughly speaking, the monotonicity condition of the generator \(g\) in \(y\) can be read as follows: there exists a constant \(\beta\geq 0\) such that \({\rm d}\mathbb{P}\times{\rm d} t-a.e.\), for each \((y_1,y_2,y,z)\in{\mathbb{R}}\times{\mathbb{R}}\times{\mathbb{R}}\times{\mathbb{R}}^d\), \[\label{eq:1462} -{\rm{sgn}}(y_1-y_2)\big(g(t,y_1,z)-g(t,y_2,z)\big)\leq\beta|y_1-y_2|.\tag{2}\]

By applying the localization method, the authors in Briand-Hu?, 2006? obtained the first existence result for the unbounded solution of a quadratic BSDE with unbounded terminal value admitting a certain exponential moment. Based on the \(\theta\)-difference technique for convex functions, the uniqueness was established in Briand-Hu?, 2008? when the terminal value possesses every exponential moments and the generator \(g\) is uniformly Lipschitz continuous in \(y\) and convex in \(z\). By virtue of the Legendre-Fenchel transform of convex functions, this result was further strengthened in Delbaen?, 2011?, which proved that the uniqueness holds among some proper space when the terminal value admits a certain exponential moment. Furthermore, the critical case was tackled in Delbaen?, 2015?, but an additional assumption that the generator \(g\) is independent of \(y\) and strongly convex in \(z\) was required. We would like to especially mention that in Briand-Hu?, 2008? and Delbaen?, 2011?, the generator \(g\) is forced to satisfy the uniformly Lipschitz continuous condition and then the linear growth condition in the first unknown variable \(y\) in order to guarantee uniqueness of the unbounded solution.

Recently, the authors in Fan-Hu-Tang?, 2020? proved a uniqueness result for the unbounded solution of a quadratic BSDE under the assumptions that the terminal value has every exponential moments and the generator \(g\) satisfies an extended convexity condition in \((y,z)\), which admits a general growth of \(g\) in \(y\) and weakens, at the first time, the uniformly Lipschitz continuous condition of the generator \(g\) in \(y\) used in Briand-Hu?, 2008?, Delbaen?, 2011?, to the best of our knowledge. On the other hand, according to Remark 8 in Fan-Hu-Tang?, 2020? and the analysis in page 29 of [1], we can conclude that if \(g\) satisfies the extended convexity condition in \((y,z)\), then it has to satisfy the monotonicity condition 2 in \(y\) and the locally-Lipschitz continuous condition in \((y,z)\). Observe for example that the following function \[f(y):=-\mathbb{1}_{y<0} y^2+\mathbb{1}_{y\geq 0}\sqrt{y},\;\;y\in{\mathbb{R}}\] satisfies the monotonicity condition 2 with general growth, but it is not locally-Lipschitz continuous. Then, a question is naturally asked: under the basic assumptions that the generator \(g\) satisfies the monotonicity condition 2 with general growth in \(y\) and the convexity condition with quadratic growth in \(z\), without imposing the locally-Lipschitz continuous condition of \(g\) in \(y\), does the uniqueness for the unbounded solution of a quadratic BSDE hold still? The present paper gives some affirmative answers.

More specifically, the main purpose of this paper is to study the existence and uniqueness for the unbounded solution of BSDE 1 when the terminal value \(\xi^-\) admits a certain exponential moment, and the generator \(g\) may be not locally-Lipschitz continuous in \(y\), and be convex in \(z\). Our result strengthens those established in Delbaen?, 2011?, by relaxing the continuity and growth condition of \(g\) in \(y\). The paper is organized as follows. In Proposition 1 of Section 2, we prove a general existence result for the unbounded solution of BSDE (1 ) in the spirit of Fan-Hu-Tang?, 2020?, which can be compared with Corollary 2.2 in Delbaen?, 2011?. Here we work with \(\xi^-\) admitting a certain exponential moment and \(\xi^+\) admitting every exponential moments or being bounded. Section 3 is devoted to the uniqueness for the unbounded solution of BSDE (1 ) in the case of the generator \(g\) satisfying the monotonicity condition 2 with general growth in \(y\) and the convexity condition with quadratic growth in \(z\), without imposing the locally-Lipschitz continuous condition of \(g\) in \(y\). The main results are stated and proved in Theorem 2. We remark that the authors in Delbaen?, 2011? proved the uniqueness result when both \(\xi^+\) and \(\xi^-\) have a certain exponential moment. In contrast, we lift the requirement for the integrability of \(\xi^+\), but actually this requirement is not necessary, see (ii) of Remark 4 for more details. In Example 5, we provide several examples to which Theorem 2 but no existing results can apply. In Theorem 6 of Section 4, we prove the uniqueness for the unbounded solution of BSDE 1 in the case of \(g\) being convex in \((y,z)\). Actually, the assumptions used in Theorem 6 are strictly stronger than those in Delbaen?, 2011?. However, we get a more accurate conclusion in this setting, that is, \(Y_\cdot\) admits a more explicit expression (see (ii) of Remark 8 for more details). In order to prove Theorems 2 and 6, we apply some techniques and ideas used in Delbaen?, 2011? and Delbaen?, 2015?, including the Legendre-Fenchel transform of convex functions, Girsanov’s theorem, the de La Vall\(\acute{\rm{e}}\)e Poussin lemma and Fenchel’s inequality. At the same time, in order to deal with the monotonicity condition of the generator \(g\) in \(y\) as well as the general growth condition, we adopt an exponential shifting transform technique, which is an innovative idea. Finally, in Theorem 10 of Section 5, it is the first time, to the best of our knowledge, that the corresponding comparison theorems for unbounded solutions of BSDEs (1 ) are established under the assumptions used in Section 3, via Proposition 5 in Briand-Hu?, 2006?, the uniqueness stated in Theorems 2 and 6, and the infinite evolution technique combined with the localization argument. The conclusion of this paper is presented in Section 6.

Let us close this introduction with some notations that will be used later. Let \(x\cdot y\) denote the Euclidean inner product for \(x, y\in{\mathbb{R}}^d\). Denote by \(\mathbb{1}_A\) the indicator of set \(A\), and \({\rm{sgn}}(x):=\mathbb{1}_{x>0}-\mathbb{1}_{x\leq0}\). Let \(a\wedge b\) be the minimum of \(a\) and \(b\), \(a^-:=-(a\wedge0)\) and \(a^+:=(-a)^-\). Moreover, let \({\mathbb{R}}_+:=[0,+\infty)\), and \({\mathbb{R}}_-:=(-\infty,0]\). By a solution to BSDE (1 ), we mean a pair of \((\mathcal{F}_t)\)-progressively measurable processes \((Y_t,Z_t)_{t\in[0,T]}\) valued in \({\mathbb{R}}\times{\mathbb{R}}^d\) such that \(\mathbb{P}-a.s.\), the function \(t\rightarrow Y_t\) is continuous, \(t\rightarrow Z_t\) is square-integrable, \(t\rightarrow g(t,Y_t,Z_t)\) is integrable, and \((Y_\cdot,Z_\cdot)\) verifies (1 ). We denote by \(\partial f\) the subdifferential of a function \(f:{\mathbb{R}}^{d}\rightarrow{\mathbb{R}}\), and the subdifferential of \(f\) at \(z_0\) is the non-empty convex compact set of elements \(u\in{\mathbb{R}}^{1\times d}\) such that \[f(z)-f(z_0)\geq u(z-z_0),~~~~~\forall z\in{\mathbb{R}}^d.\] Moreover, for any \((\mathcal{F}_t)\)-progressively measurable process \((q_t)_{t\in[0,T]}\) such that \(\int_0^T |q_s|^2 {\rm d}s<+\infty ~\mathbb{P}-a.s.\), we denote by \(\mathcal{E}(q)\) the Dol\(\acute{e}\)ans-Dade exponential \[\Big(\exp\Big(\int_0^tq_s\cdot{\rm d}B_s-\frac{1}{2}\int_0^t |q_s|^2 {\rm d}s\Big)\Big)_{t\in[0,T]}.\] Let us recall the following Fenchel’s inequality \[xy\leq{\rm{exp}}(x)+y(\ln y-1),~~~\forall(x,y)\in{\mathbb{R}}\times{\mathbb{R}}^+.\] Then we have \[\begin{align} xy=px\frac{y}{p}\leq{\rm{exp}}(px)+\frac{y}{p}(\ln y-\ln p-1). \end{align}\] Furthermore, we define the following two process spaces.

\(\bullet\) \(\mathcal{S}^p([0,T];{\mathbb{R}})\) is the set of all \((\mathcal{F}_t)\)-progressively measurable and continuous real-valued processes \((Y_t)_{t\in[0,T]}\) satisfying \[\|Y\|_{{\mathcal{S}}^p}:=\left(\mathbb{E}\left[\sup_{t\in[0,T]} |Y_t|^p\right]\right)^{1/p}<+\infty.\] We set \(\mathcal{S}=\bigcup\limits_{p>1}\mathcal{S}^p\).

\(\bullet\) \(\mathcal{M}^p([0,T];{\mathbb{R}}^d)\) is the set of all \((\mathcal{F}_t)\)-progressively measurable \({\mathbb{R}}^d\)-valued processes \((Z_t)_{t\in[0,T]}\) satisfying \[\|Z\|_{\mathcal{M}^p}:=\left\{\mathbb{E}\left[\left(\int_0^T |Z_t|^2{\rm d}t\right)^{p/2}\right] \right\}^{1/p}<+\infty.\] Recall that an \((\mathcal{F}_t)\)-progressively measurable real-valued process \((Y_t)_{t\in[0,T]}\) belongs to class \((D)\) if the \(\{Y_\tau : \tau\in\Sigma_T\}\) is a family of uniformly integrable random variables, where \(\Sigma_T\) represents the set of all \((\mathcal{F}_t)\)-stopping times \(\tau\) valued in \([0,T]\).

In the whole paper, we are always given two \((\mathcal{F}_t)\)-progressively measurable \({\mathbb{R}}_+\)-valued processes \((\underline{\alpha}_t)_{t\in[0,T]}\) and \((\overline{\alpha}_t)_{t\in[0,T]}\) satisfying \(\int_0^T(\underline{\alpha}_t+ \overline{\alpha}_t) {\rm d}t<+\infty ~\mathbb{P}-a.s.\), and two constants \(\beta\geq0\) and \(\gamma>0\) together with a continuous nondecreasing function \(\varphi(\cdot):{\mathbb{R}}_+\rightarrow{\mathbb{R}}_+\) with \(\varphi(0)=0\).

2 An existence result↩︎

This section will put forward and prove a general existence result for unbounded solutions of BSDE 1 . We first introduce the following assumption on the generator \(g\).

(EX) \({\rm d}\mathbb{P}\times{\rm d} t-a.e.\), for each \((y,z)\in{\mathbb{R}}\times{\mathbb{R}}^d\), we have

1. \(-\mathbb{1}_{y>0}g(t,y,z)\leq \underline{\alpha}_t+\beta|y|+\frac{\gamma}{2}|z|^2\);

2. \(\mathbb{1}_{y\leq0}g(t,y,z)\leq \overline{\alpha}_t+\beta|y|+\frac{\gamma}{2}|z|^2\);

3. \(|g(t,y,z)|\leq \overline{\alpha}_t+\varphi(|y|)+\frac{\gamma}{2}|z|^2\).

The following existence proposition is the main result of this section, which can be compared with Theorem 2.1 and Corollary 2.2 in Delbaen?, 2011?.

Proof. We define, for any non-negative integrable function \(l(\cdot):[0,T]\rightarrow{\mathbb{R}}_+\) and any constants \(\kappa\geq0\) and \(\lambda>0\), the following function: \[\psi(s,x;l_\cdot,\kappa,\lambda)=\exp\Big(\lambda e^{\kappa s}x+\lambda\int_0^sl(r)e^{\kappa r}{\rm d}r\Big),~~~(s,x)\in[0,T]\times{\mathbb{R}}_+.\] (i) In view of (EX), by applying \(\rm{It\hat{o}}\)-Tanaka’s formula to \(\psi(t,Y^+_t;\underline{\alpha}_\cdot,\beta,\overline{p})\) and \(\psi(t,Y^-_t;\overline{\alpha}_\cdot,\beta,p)\) and using an identical argument as in Proposition 1 of Fan-Hu-Tang?, 2020?, we can deduce that for any bounded solution \((Y_\cdot,Z_\cdot)\) of BSDE (1 ), it holds that \(\mathbb{P}-a.s.\), for each \(t\in[0,T]\), \[\begin{align} \label{eq:2463} \overline{p} Y^+_t\leq\exp(\overline{p} Y^+_t)\leq\psi(t,Y^+_t;\underline{\alpha}_\cdot,\beta,\overline{p}) \leq\mathbb{E}\Big[\psi(T,\xi^+;\underline{\alpha}_\cdot,\beta,\overline{p})\big|\mathcal{F}_t\Big],\\ p Y^-_t\leq\exp(p Y^-_t)\leq\psi(t,Y^-_t;\overline{\alpha}_\cdot,\beta,p) \leq\mathbb{E}\Big[\psi(T,\xi^-;\overline{\alpha}_\cdot,\beta,p)\big|\mathcal{F}_t\Big].\nonumber \end{align}\tag{3}\] Then, by virtue of the last two inequalities together with the assumptions of Proposition 1, proceeding as the proof of Proposition 3 in Briand-Hu?, 2008? with a localization argument as well as Doob’s maximal inequality for martingales, we can conclude that BSDE (\(\ref{BSDE1461}\)) admits a solution \((Y_\cdot,Z_\cdot)\) satisfying (?? ), (?? ) and \(Z_\cdot\in\mathcal{M}^2\). The detailed proof is omitted here.

(ii) Since the assumptions in (ii) are stronger than those in (i), it suffices to prove that \(Y_\cdot^+\leq K\) for some constant \(K>0\). Indeed, since \(\xi^+ +\int_0^T \underline{\alpha}_t {\rm d}t\leq M\) for a constant \(M>0\), it follows from 3 that \[2\gamma Y^+_t\leq \mathbb{E}[\psi(T,\xi^+;\underline{\alpha}_\cdot,\beta,2\gamma)|\mathcal{F}_t]\leq \exp\left(2\gamma e^{\beta T}M\right),\;\;t\in [0,T],\] which is the desired result. The proof is then complete.

3 The uniqueness for the case of \(g\) satisfying monotonicity in \(y\) and convexity in \(z\)↩︎

This section is devoted to the uniqueness for the unbounded solution of BSDE (\(\ref{BSDE1461}\)) with generator \(g\) satisfying a monotonicity condition with general growth in \(y\) and a convexity condition with quadratic growth in \(z\). We use the following assumptions.

(UN1) \({\rm d}\mathbb{P}\times{\rm d} t-a.e.\), for each \((y_1,y_2,y,z)\in{\mathbb{R}}\times{\mathbb{R}}\times{\mathbb{R}}\times{\mathbb{R}}^d\), we have

1. \(-{\rm{sgn}}(y_1-y_2)\big(g(t,y_1,z)-g(t,y_2,z)\big)\leq\beta|y_1-y_2|\);

2. \(g(t,y,\cdot)\) is convex.

(UN2) \({\rm d}\mathbb{P}\times{\rm d} t-a.e.\), for each \((y,z)\in{\mathbb{R}}\times{\mathbb{R}}^d, ~\mathbb{1}_{y>0}g(t,y,z)\leq \overline{\alpha}_t+\beta|y|+\frac{\gamma}{2}|z|^2\).
(UN2’) \({\rm d}\mathbb{P}\times{\rm d} t-a.e.\), for each \((y,z)\in{\mathbb{R}}\times{\mathbb{R}}^d\), we have \[\mathbb{1}_{y>0}\big(g(t,y,z)-g(t,0,z)\big)\leq\overline{\alpha}_t+\beta|y|\] and \[|g(t,y,z)-g(t,0,z)|\leq\overline{\alpha}_t+\varphi(|y|).\]

If \(g(t,y,\cdot)\) is a convex function, then the Legendre-Fenchel transform of \(g\) can be defined as follows: \[\begin{align} \label{fdy3461} f(\omega,t,y,q):=\sup_{z\in{\mathbb{R}}^d}\big(qz-g(\omega,t,y,z)\big), ~~\forall (\omega,t,y,q)\in\Omega\times[0,T]\times{\mathbb{R}}\times{\mathbb{R}}^{1\times d}. \end{align}\tag{4}\] The following theorem presents several uniqueness results for solutions of BSDE (1 ).

The existence has been proved in Proposition 1. Now let us verify the uniqueness.
Proof of (i) of Theorem 2. According to (2) of (EX), (UN2) and (UN1), we can deduce that the generator \(f\) defined in (4 ) satisfies the following properties: \({\rm d}\mathbb{P}\times{\rm d} t-a.e.\),

•   \(\forall (y,q)\in{\mathbb{R}}\times{{\mathbb{R}}^{1\times d}}, ~f(t,y,q)\geq-\overline{\alpha}_t-\beta |y|+\frac{1}{2\gamma}|q|^2;\)

•   \(\forall q\in{{\mathbb{R}}^{1\times d}}, ~\forall(y_1,y_2)\in{{\mathbb{R}}^2},~{\rm{sgn}}(y_1-y_2)\big(f(t,y_1,q)-f(t,y_2,q)\big)\leq \beta|y_1-y_2|\);

•   \(\forall y\in{\mathbb{R}}, ~f(t,y,\cdot)\) is convex.

The last two assertions is obvious. We only need to verify the first assertion. In fact, it follows from (2) of (EX) and (UN2) that \({\rm d}\mathbb{P}\times{\rm d} t-a.e., ~\forall (y,z)\in{\mathbb{R}}\times{\mathbb{R}}^d\), \[g(t,y,z)\leq \overline{\alpha}_t+\beta|y|+\frac{\gamma}{2}|z|^2.\] Then, by (4 ) we have \({\rm d}\mathbb{P}\times{\rm d} t-a.e.\), for each \((y,q)\in{\mathbb{R}}\times{\mathbb{R}}^{1\times d}\), \[f(t,y,q)\geq -\overline{\alpha}_t-\beta|y|+\sup_{z\in{\mathbb{R}}^d}\big(qz-\frac{\gamma}{2}|z|^2\big)=-\overline{\alpha}_t-\beta|y|+\frac{1}{2\gamma}|q|^2.\] Let us proceed the proof by giving a uniform integrability result.

Proof. In view of \(q_s^*\in\partial_z g(s,Y_s,Z_s)\), we have \(g(s,Y_s,Z_s)=q_s^*Z_s-f(s,Y_s,q_s^*)\), which together with \(f(t,y,q)\geq-\overline{\alpha}_t-\beta |y|+\frac{1}{2\gamma}|q|^2\) yields \[\begin{align} g(s,Y_s,Z_s)=q_s^*Z_s-f(s,Y_s,q_s^*)&\leq \frac{1}{2}\Big(\frac{1}{2\gamma}|q_s^*|^2+2\gamma|Z_s|^2\Big)+\overline{\alpha}_s+\beta|Y_s|-\frac{1}{2\gamma}|q_s^*|^2\\ &=\gamma|Z_s|^2+\overline{\alpha}_s+\beta|Y_s|-\frac{1}{4\gamma}|q_s^*|^2; \end{align}\] \[\begin{align} \frac{1}{4\gamma}|q_s^*|^2&\leq-g(s,Y_s,Z_s)+\gamma|Z_s|^2+\overline{\alpha}_s+\beta|Y_s|. \end{align}\] It follows from the last inequality that \(\int_0^T|q_s^*|^2{\rm d}s<+\infty ~\mathbb{P}-a.s.\).

For each \(n\geq1\), let us define \[M_t:=\exp\Big(\int_{0}^{t}q_s^*{\rm d}B_s-\frac{1}{2}\int_{0}^{t}|q_s^*|^2{\rm d}s\Big), ~t\in[0,T],\] \[\tau_n:=\inf \Big\{t \in[0, T]: \int_0^t|q_s^*|^2 {\rm d}s+\int_0^t|Z_s|^2 {\rm d}s\geq n\Big\} \wedge T, \quad \frac{{\rm d}{\mathbb{Q}}_n^*}{{\rm d}\mathbb{P}}:=M_{\tau_n}.\] We set \(B_t^{q^*}:=B_t-\int_0^t q^*_s{\rm d}s, ~t\in[0,T]\). Then \((B_t^{q^*})_{t\in[0,T]}\) is a standard Brownian motion under \({\mathbb{Q}}_n^*\). Now we show that \((M_{\tau_n})_n\) is uniformly integrable.

According to (?? ), (?? ) and the Fenchel’s inequality, we have \[\begin{align} \mathbb{E}^{{\mathbb{Q}}_n^{*}}\Big[\sup_{t\in[0,T]} A_\beta(t)\Big]&=\mathbb{E}\Big[M_{\tau_n} \sup_{t\in[0,T]} A_\beta(t)\Big] \\ &\leq \mathbb{E}\Big[\exp\big(p\sup_{t\in[0,T]} A_\beta(t)\big)\Big]+\frac{1}{p}\mathbb{E}\big[M_{\tau_n}(\ln M_{\tau_n}-\ln p-1)\big]\\ &=\mathbb{E}\Big[\exp\big(p\sup_{t\in[0,T]} A_\beta(t)\big)\Big]+\frac{1}{p}\mathbb{E}[M_{\tau_n}\ln M_{\tau_n}]-\frac{1}{p}(\ln p+1). \end{align}\] Some uncomplicated calculations give \[\begin{align} \label{t} \notag\mathbb{E}[M_{\tau_n}\ln {M_{\tau_n}}]&=\mathbb{E}^{{\mathbb{Q}}_n^{*}}[\ln M_{\tau_n}]=\mathbb{E}^{{\mathbb{Q}}_n^{*}}\Big[\int_{0}^{\tau_n}q^*_s{\rm d}B_s-\frac{1}{2}\int_{0}^{\tau_n}|q^*_s|^2{\rm d}s\Big]\\ &=\mathbb{E}^{{\mathbb{Q}}_n^{*}}\Big[\int_{0}^{\tau_n}q^*_s{\rm d}B^{q^*}_s+\frac{1}{2}\int_{0}^{\tau_n}|q^*_s|^2{\rm d}s\Big]=\frac{1}{2}\mathbb{E}^{{\mathbb{Q}}_n^{*}}\Big[\int_{0}^{\tau_n}|q^*_s|^2{\rm d}s\Big]. \end{align}\tag{5}\] The calculation similar to (5 ) will be used several times later, and this calculation process will not be repeated in detail. For ease of notations, we denote the sum of constants by \(C_p\), then we have \[\begin{align} \label{3a} \mathbb{E}^{{\mathbb{Q}}_n^{*}}\Big[\sup_{t\in[0,T]} A_\beta(t)\Big]\leq C_p+\frac{1}{2p} \mathbb{E}^{{\mathbb{Q}}_n^*}\Big[\int_0^{\tau_n}|q_s^*|^2 {\rm d}s\Big], \end{align}\tag{6}\] and for each \(\overline{p}>\gamma\), in the same manner, \[\begin{align} \label{3aa} \mathbb{E}^{{\mathbb{Q}}_n^{*}}\Big[\sup_{t\in[0,T]} \overline{A}_\beta(t)\Big]\leq C_{\overline{p}}+\frac{1}{2\overline{p}} \mathbb{E}^{{\mathbb{Q}}_n^*}\Big[\int_0^{\tau_n}|q_s^*|^2 {\rm d}s\Big]. \end{align}\tag{7}\] Since \(g(s,Y_s,Z_s)=q_s^*Z_s-f(s,Y_s,q_s^*)\), by BSDE (1 ) we have \[Y_{t\wedge\tau_n}=Y_{\tau_n}+\int_{t\wedge\tau_n}^{\tau_n}f(s,Y_s,q_s^*){\rm d}s+\int_{t\wedge\tau_n}^{\tau_n} Z_s \cdot {\rm d}B_s^{q^*}, ~~t\in[0,T].\] Applying \(\rm{It\hat{o}}\)’s formula to \(e^{\beta s} Y_s\) gives \[\begin{align} {\rm d}(e^{\beta s} Y_s)=e^{\beta s}\big(\beta Y_s-f(s,Y_s,q_s^*)\big){\rm d}s-e^{\beta s}Z_s\cdot{\rm d}B_s^{q^*}, ~s\in[0,\tau_n]. \end{align}\] In view of \(f(t,y,q)\geq-\overline{\alpha}_t-\beta|y|+\frac{1}{2\gamma}|q|^2\) and \(Y_{\tau_n}\geq-Y_{\tau_n}^-\), we deduce that \[\begin{align} Y_0&=e^{\beta \tau_n} Y_{\tau_n}+\int_0^{\tau_n}e^{\beta s}\big(f(s,Y_s,q_s^*)-\beta Y_s\big){\rm d}s+\int_0^{\tau_n}e^{\beta s}Z_s\cdot{\rm d}B_s^{q^*}\\ &\geq -e^{\beta \tau_n} Y_{\tau_n}^-+\int_0^{\tau_n}e^{\beta s}\Big(-\overline{\alpha}_s-\beta|Y_s|+\frac{1}{2\gamma}|q_s^*|^2-\beta Y_s\Big){\rm d}s+\int_0^{\tau_n}e^{\beta s}Z_s\cdot{\rm d}B_s^{q^*}\\ &= -e^{\beta \tau_n} Y_{\tau_n}^-+\int_0^{\tau_n}e^{\beta s}\Big(-\overline{\alpha}_s-2\beta Y_s^++\frac{1}{2\gamma}|q_s^*|^2\Big){\rm d}s+\int_0^{\tau_n}e^{\beta s}Z_s\cdot{\rm d}B_s^{q^*}. \end{align}\] It then follows from (6 ) and (7 ) that for each \(\overline{p}>\gamma\), \[\begin{align} \label{tt} \notag Y_0&\geq \mathbb{E}^{{\mathbb{Q}}_n^*}\Big[-e^{\beta \tau_n} Y_{\tau_n}^-+\int_0^{\tau_n}e^{\beta s}\Big(-\overline{\alpha}_s-2\beta Y_s^++\frac{1}{2\gamma}|q_s^*|^2\Big){\rm d}s\Big]\\ \notag &\geq -\mathbb{E}^{{\mathbb{Q}}_n^*}\Big[e^{\beta \tau_n}\Big( Y_{\tau_n}^-+\int_0^{\tau_n}\overline{\alpha}_s{\rm d}s\Big)\Big]-2\beta \mathbb{E}^{{\mathbb{Q}}_n^*}\Big[\int_0^{\tau_n}e^{\beta s}Y_s^+{\rm d}s\Big]+\frac{1}{2\gamma}\mathbb{E}^{{\mathbb{Q}}_n^*}\Big[\int_0^{\tau_n}e^{\beta s}|q_s^*|^2{\rm d}s\Big]\\ \notag &\geq-\mathbb{E}^{{\mathbb{Q}}_n^*}\Big[A_\beta(\tau_n)\Big]-2\beta T\mathbb{E}^{{\mathbb{Q}}_n^*}\Big[\sup_{s\in[0,T]}e^{\beta s}Y_s^+\Big]+\frac{1}{2\gamma}\mathbb{E}^{{\mathbb{Q}}_n^*}\Big[\int_0^{\tau_n}|q_s^*|^2{\rm d}s\Big]\\ &\geq-C_p-2\beta T C_{\overline{p}}+\frac{1}{2}\Big(\frac{1}{\gamma}-\frac{1}{p}-\frac{2\beta T}{\overline{p}}\Big)\mathbb{E}^{{\mathbb{Q}}_n^*}\Big[\int_0^{\tau_n}|q_s^*|^2{\rm d}s\Big]. \end{align}\tag{8}\] Set \(\overline{p}>\frac{2\beta Tp\gamma}{p-\gamma}\). Then \(\frac{1}{2}\Big(\frac{1}{\gamma}-\frac{1}{p}-\frac{2\beta T}{\overline{p}}\Big)>0.\) Hence, in view of (5 ) and (8 ), we have \[\begin{align} \label{3g} 2\mathbb{E}[M_{\tau_n}\ln M_{\tau_n}]=\mathbb{E}^{{\mathbb{Q}}_n^{*}}\Big[\int_0^{\tau_n}|q_s^*|^2{\rm d}s\Big]<C_{p,\overline{p},T,\beta,\gamma}, \end{align}\tag{9}\] where \(C_{p,\overline{p},T,\beta,\gamma}\) is a positive constant independent of \(n\). Hence, \(\sup\limits_n\mathbb{E}[M_{\tau_n}\ln M_{\tau_n}]<+\infty\). Thus, \((M_{\tau_n})_n\) is uniformly integrable by the de La Vall\(\acute{\rm{e}}\)e Poussin lemma, and then \(\mathbb{E}[M_T]=1\), which means that \(\mathcal{E}(q^*)=(M_t)_{t\in[0,T]}\) is a uniformly integrable martingale and defines a probability \({\mathbb{Q}}^*\sim\mathbb{P}\). Moreover, applying Fatou’s lemma in (9 ), we obtain \[2\mathbb{E}[M_T\ln M_T]=\mathbb{E}^{{\mathbb{Q}}^*}\Big[\int_0^T|q_s^*|^2{\rm d}s\Big]\leq\liminf_n\mathbb{E}^{{\mathbb{Q}}_n^* }\Big[\int_0^{\tau_n}|q_s^*|^2{\rm d}s\Big]<+\infty.\] The proof of Proposition 3 is then complete.

In the sequel, let \((Y_\cdot, Z_\cdot)\) and \((Y'_\cdot,Z'_\cdot)\) be two solutions of BSDE (\(\ref{BSDE1461}\)) such that for each \(\overline{p}>\gamma\), \[\begin{align} \label{h1} \mathbb{E}\Big[\exp\big(p\sup_{t\in[0,T]}A_\beta(t)\big)\Big]<+\infty, ~ \mathbb{E}\Big[\exp\big(\overline{p}\sup_{t\in[0,T]}\overline{A}_\beta(t)\big)\Big]<+\infty, \end{align}\tag{10}\] \[\begin{align} \label{h2} \mathbb{E}\Big[\exp\big(p\sup_{t\in[0,T]}A'_\beta(t)\big)\Big]<+\infty, ~ \mathbb{E}\Big[\exp\big(\overline{p}\sup_{t\in[0,T]}\overline{A'}_\beta(t)\big)\Big]<+\infty, \end{align}\tag{11}\] with \(A'_\beta(t):=e^{\beta t}\big((Y'_t)^-+\int_0^t\overline{\alpha}_s{\rm d}s\big)\) and \(\overline{A'}_\beta(t):=e^{\beta t}\big((Y'_t)^++\int_0^t\underline{\alpha}_s{\rm d}s\big)\) as well as \(A_\beta(\cdot)\) and \(\overline{A}_\beta(\cdot)\) being defined respectively in (?? ) and (?? ). In order to obtain the desired uniqueness, by a symmetry argument it suffices to show that for each \(t\in[0,T], Y_t\geq Y'_t~ \mathbb{P}-a.s.\). For \(t\in[0,T)\), let us denote \(A:=\{Y_t<Y'_t\}\) and the stopping time \(\tau:=\inf\{s\geq t | Y_s\geq Y'_s\}\). Then for \(s\in[t,\tau]\), we have \(\mathbb{1}_A Y_s\leq \mathbb{1}_A Y'_s\) and \(\mathbb{1}_A Y_\tau=\mathbb{1}_A Y'_\tau ~\mathbb{P}-a.s.\) since \(t\rightarrow Y_t\) is continuous \(\mathbb{P}-a.s.\). It implies that \((\mathbb{1}_A(Y_s-Y'_s))_{s\in[t,\tau]}\) is a non-positive process. Let us consider an \((\mathcal{F}_t)\)-progressively measurable process \((q_s^*)_{s\in[0,T]}\) valued in \({\mathbb{R}}^{1\times d}\) such that \(q_s^*\in\partial_z g(s,Y_s,Z_s)\) for all \(s\in[0,T]\). Thanks to Proposition 3, we know that \((M_t)_{t\in[0,T]}\) defines a probability that we will be denoted by \({\mathbb{Q}}^*\), with \[M_t:=\exp\Big(\int_{0}^{t}q_s^*{\rm d}B_s-\frac{1}{2}\int_{0}^{t}|q_s^*|^2{\rm d}s\Big)~~\text{and}~~\frac{{\rm d}{\mathbb{Q}}^*}{{\rm d}\mathbb{P}}:=M_T.\] Then, \[{\rm d}\big(Y_s-Y'_s\big)=\Big(g(s,Y_s,Z_s)-g(s,Y'_s,Z'_s)-q_s^*(Z_s-Z'_s)\Big){\rm d}s-(Z_s-Z'_s)\cdot{\rm d}B_s^{q^*}, ~s\in[0,T].\] Applying \(\rm{It\hat{o}}\)’s formula to \(e^{\beta s}\mathbb{1}_A(Y_s-Y'_s)\), we obtain that for each \(s\in[t,\tau]\), \[\begin{align} {\rm d}\big[e^{\beta s}\mathbb{1}_A(Y_s-Y'_s)\big]=&\beta e^{\beta s}\mathbb{1}_A(Y_s-Y'_s){\rm d}s+e^{\beta s}\mathbb{1}_A\big(g(s,Y_s,Z_s)-g(s,Y_s,Z'_s)-q_s^*(Z_s-Z'_s)\\ &+g(s,Y_s,Z'_s)-g(s,Y'_s,Z'_s)\big){\rm d}s-e^{\beta s}\mathbb{1}_A(Z_s-Z'_s)\cdot{\rm d}B_s^{q^*}. \end{align}\] Noticing by \(q_s^*\in\partial_z g(s,Y_s,Z_s)\) that \(g(s,Y_s,Z_s)-g(s,Y_s,Z'_s)-q_s^*(Z_s-Z'_s)\leq0\), we have \[\begin{align} {\rm d}\big[e^{\beta s}\mathbb{1}_A(Y_s-Y'_s)\big]\leq e^{\beta s}\mathbb{1}_A\big(\beta(Y_s-Y'_s)+g(s,Y_s,Z'_s)-g(s,Y'_s,Z'_s)\big){\rm d}s-e^{\beta s}\mathbb{1}_A(Z_s-Z'_s)\cdot{\rm d}B_s^{q^*}. \end{align}\] In view of (1) of (UN1) together with the fact that \((\mathbb{1}_A(Y_s-Y'_s))_{s\in[t,\tau]}\) is a non-positive process, it is not very hard to verify that \[\mathbb{1}_A\big(\beta(Y_s-Y'_s)+g(s,Y_s,Z'_s)-g(s,Y'_s,Z'_s)\big)\leq0, ~~s\in[t,\tau].\] Then, we have \[{\rm d}\big[e^{\beta s}\mathbb{1}_A(Y_s-Y'_s)\big]\leq-e^{\beta s}\mathbb{1}_A(Z_s-Z'_s)\cdot{\rm d}B_s^{q^*}, ~~s\in[t,\tau].\] Define the following stopping time, for each \(n\geq1\), \[\sigma^t_n:=\inf\Big\{s\geq t:\int_t^s|Z_u|^2{\rm d}u+\int_t^s|Z'_u|^2{\rm d}u\geq n\Big\}\wedge\tau\] with the convention \(\inf\emptyset=+\infty\). Then, we obtain \[e^{\beta t}\mathbb{1}_A(Y_t-Y'_t)\geq e^{\beta \sigma^t_n}\mathbb{1}_A(Y_{\sigma^t_n}-Y'_{\sigma^t_n})+\int_t^{\sigma^t_n}e^{\beta s}\mathbb{1}_A(Z_s-Z'_s)\cdot{\rm d}B_s^{q^*}.\] It follows that \[\begin{align} \label{j} e^{\beta t}\mathbb{1}_A(Y_t-Y'_t)\geq \mathbb{E}^{{\mathbb{Q}}^*}\Big[e^{\beta \sigma^t_n}\mathbb{1}_A(Y_{\sigma^t_n}-Y'_{\sigma^t_n})\big|\mathcal{F}_t\Big]. \end{align}\tag{12}\] For ease of notations, let us set \(\widehat{Y}_\cdot:=Y_\cdot-Y'_\cdot\), so \(|Y_{\sigma^t_n}-Y'_{\sigma^t_n}|\leq\sup\limits_{s\in[0,T]}|\widehat{Y}_s|\). In view of (10 ) and (11 ) together with (?? ) in Proposition 3, by using an identical proof to (6 ) and (7 ), we obtain that, picking \(\overline{p}=p\), \[\mathbb{E}^{{\mathbb{Q}}^*}\Big[\sup_{s\in[0,T]}Y_s^-\Big]\leq\mathbb{E}^{{\mathbb{Q}}^*}\Big[\sup_{s\in[0,T]}A_\beta(s)\Big]\leq C_p+\frac{1}{2p} \mathbb{E}^{{\mathbb{Q}}^*}\Big[\int_0^T|q_s^*|^2 {\rm d}s\Big]=:\overline{C}_p,\] \[\begin{align} \mathbb{E}^{{\mathbb{Q}}^*}\Big[\sup\limits_{s\in[0,T]}Y_s^+\Big]\leq \overline{C}_{p}, ~ \mathbb{E}^{{\mathbb{Q}}^*}\Big[\sup\limits_{s\in[0,T]}(Y'_s)^-\Big]\leq \overline{C}_p ~~\text{and~~} \mathbb{E}^{{\mathbb{Q}}^*}\Big[\sup\limits_{s\in[0,T]}(Y'_s)^+\Big]\leq \overline{C}_{p}. \end{align}\] Then, we have \[\begin{align} \mathbb{E}^{{\mathbb{Q}}^*}\Big[\sup_{s\in[0,T]}|\widehat{Y}_s|\Big]&=\mathbb{E}^{{\mathbb{Q}}^*}\Big[\sup_{s\in[0,T]}(\widehat{Y}^+_s+\widehat{Y}^-_s)\Big]\leq4\overline{C}_p+\frac{2}{p} \mathbb{E}^{{\mathbb{Q}}^*}\Big[\int_0^T|q_s^*|^2 {\rm d}s\Big]<+\infty. \end{align}\] Subsequently, by sending \(n\rightarrow\infty\) in (12 ) and applying the Lebesgue’s dominated convergence theorem we get that \[\begin{align} e^{\beta t}\mathbb{1}_A(Y_t-Y'_t)\geq \mathbb{E}^{{\mathbb{Q}}^*}\Big[e^{\beta \tau}(\mathbb{1}_A Y_\tau-\mathbb{1}_A Y'_\tau)\big|\mathcal{F}_t\Big]=0 ~~\mathbb{P}-a.s., \end{align}\] which implies that for each \(t\in[0,T], ~\mathbb{1}_A(Y_t-Y'_t)\geq0 ~\mathbb{P}-a.s.\). Considering that \((\mathbb{1}_A(Y_s-Y'_s))_{s\in[t,\tau]}\) is a non-positive process, we know that for each \(t\in[0,T], ~\mathbb{1}_A(Y_t-Y'_t)=0 ~\mathbb{P}-a.s.\), that is to say, \(\mathbb{P}(A)=0\) and \(Y_t\geq Y'_t ~\mathbb{P}-a.s.\). The proof of (i) of Theorem 2 is then complete.

Proof of (ii) of Theorem 2. According to (2) and (3) of (EX) and (UN1), we can deduce that the function \(f\) defined in (4 ) satisfies the following properties: \({\rm d}\mathbb{P}\times{\rm d} t-a.e.\),

•   \(\forall (y,q)\in{\mathbb{R}}_+\times{{\mathbb{R}}^{1\times d}}, ~f(t,y,q)\geq-\overline{\alpha}_t-\varphi(y^+)+\frac{1}{2\gamma}|q|^2;\)

•   \(\forall (y,q)\in{\mathbb{R}}_-\times{{\mathbb{R}}^{1\times d}}, ~f(t,y,q)\geq -\overline{\alpha}_t-\beta y^-+\frac{1}{2\gamma}|q|^2;\)

•   \(\forall q\in{{\mathbb{R}}^{1\times d}}, ~\forall(y_1,y_2)\in{{\mathbb{R}}^2},~{\rm{sgn}}(y_1-y_2)\big(f(t,y_1,q)-f(t,y_2,q)\big)\leq \beta|y_1-y_2|\);

•   \(\forall y\in{\mathbb{R}}, ~f(t,y,\cdot)\) is convex.

It then follows that \({\rm d}\mathbb{P}\times{\rm d} t-a.e.\), for each \(y\leq K\) with some constant \(K>0\), we have \[\begin{align} f(t,y,q)\geq -\overline{\alpha}_t-\beta y^--\varphi(y^+)+\frac{1}{2\gamma}|q|^2 \geq-\overline{\alpha}_t-\varphi(K)-\beta y^-+\frac{1}{2\gamma}|q|^2. \end{align}\] With the above observation together with the requirement that \(Y_\cdot \leq K\), the desired uniqueness assertion can be proved in the same way as in (i) of Theorem 2.

Proof of (iii) of Theorem 2. Note by (3) of (EX) that \({\rm d}\mathbb{P}\times{\rm d} t-a.e., ~|g(t,0,z)|\leq\overline{\alpha}_t+\frac{\gamma}{2}|z|^2\) for each \(z\in{\mathbb{R}}^d\). Assumptions (EX) and (UN2’) can imply (UN2) with \(2\overline{\alpha}_\cdot\) instead of \(\overline{\alpha}_\cdot\). Therefore, the assumptions in (iii) of Theorem 2 is stronger than those in (i) of Theorem 2. Then, we can conclude that \(f\) defined in (4 ) satisfies the following properties: \({\rm d}\mathbb{P}\times{\rm d} t-a.e.\),

• \(\forall (y,q)\in{\mathbb{R}}\times{{\mathbb{R}}^{1\times d}}, ~f(t,y,q)\geq-\overline{\alpha}_t-\beta|y|+\frac{|q|^2}{2\gamma};\)

• \(\forall q\in{{\mathbb{R}}^{1\times d}},~ ~\forall(y_1,y_2)\in{{\mathbb{R}}^2},~{\rm{sgn}}(y_1-y_2)\big(f(t,y_1,q)-f(t,y_2,q)\big)\leq \beta|y_1-y_2|\);

• \(\forall y\in{\mathbb{R}}, ~f(t,y,\cdot)\) is convex.

Furthermore, it follows from (2) that \({\rm d}\mathbb{P}\times{\rm d} t-a.e.\), for each \((y,q)\in{\mathbb{R}}_-\times{{\mathbb{R}}^{1\times d}}, ~f(t,0,q)\leq f(t,y,q)+\beta|y|\). By (UN2’) and (4 ) we can deduce that \({\rm d}\mathbb{P}\times{\rm d} t-a.e.\),

• \(\forall (y,q)\in{\mathbb{R}}^+\times{{\mathbb{R}}^{1\times d}}, ~f(t,0,q)\leq f(t,y,q)+\overline{\alpha}_t+\beta|y|;\)

• \(\forall (y,q)\in{\mathbb{R}}\times{{\mathbb{R}}^{1\times d}}, ~|f(t,y,q)-f(t,0,q)|\leq \overline{\alpha}_t+\varphi(|y|).\)

Therefore, \({\rm d}\mathbb{P}\times{\rm d} t-a.e.\), for each \((y,q)\in{\mathbb{R}}\times{{\mathbb{R}}^{1\times d}}\), we have \[\begin{align} \label{x} f(t,0,q)\leq f(t,y,q)+\overline{\alpha}_t+\beta|y|. \end{align}\tag{13}\] Now, let \(q\) be in \(\mathcal{A}\), if this set is not empty. In view of (2), (5), the definition of \(\mathcal{A}\) and the Girsanov’s theorem, by Theorem 6.2 in Briand?, 2003? we know that BSDE (?? ) admits a unique solution \((Y_\cdot^q,Z_\cdot^q)\) such that \((Y_t^q)_{t\in[0,T]}\) belongs to class \((D)\) under \({\mathbb{Q}}\). Let \((Y_\cdot,Z_\cdot)\) be a solution of BSDE (1 ) satisfying (?? ) and (?? ) for any \(\overline{p}>\gamma\). We need to prove \[Y_\cdot=\mathop{\operatorname{ess\,inf}}\limits_{q\in\mathcal{A}}Y_\cdot^q,\] which consists in minimizing \(Y_\cdot^{q}\) among all the admissible controls \(q\in\mathcal{A}\). The following proof is divided into three steps.

First step. Let us start with showing that \(Y_\cdot\leq Y_\cdot^{q}\) for any \(q\in\mathcal{A}\). Firstly, we have \(Y_T=Y^{q}_T=\xi\). For each \(t\in[0,T]\) and each integer \(m\geq1\), denote the following stopping time \[\tau^t_m:=\inf\Big\{s\geq t,\int_t^s|Z_u|^2{\rm d}u+\int_t^s|Z^{q}_u|^2{\rm d}u+\int_t^s|q_u|^2{\rm d}u\geq m\Big\}\wedge T\] with the convention \(\inf\emptyset=+\infty\). Applying \(\rm{It\hat{o}}\)-Tanaka’s formula to \(e^{\beta s} (Y_s-Y_s^{q})^+\), we obtain \[\begin{align} \label{q} {\rm d}\big[e^{\beta s}(Y_s-Y_s^q)^+\big]\geq &\Big(\beta e^{\beta s}(Y_s-Y_s^q)^++e^{\beta s}\mathbb{1}_{\{Y_s-Y_s^q>0\}}\big(f(s,Y_s^q,q_s)+g(s,Y_s,Z_s)-q_sZ_s\big)\Big){\rm d}s \notag \\ &-e^{\beta s}\mathbb{1}_{\{Y_s-Y_s^q>0\}}(Z_s-Z^{q}_s)\cdot{\rm d}B_s^q, ~~~s\in[t,\tau]. \end{align}\tag{14}\] By (1) of (UN1) and (4 ), we have \[\begin{align} \mathbb{1}_{\{Y_s-Y_s^q>0\}}&\Big(f(s,Y_s^q,q_s)+g(s,Y_s,Z_s)-q_sZ_s\Big)\\ =&\mathbb{1}_{\{Y_s-Y_s^q>0\}}\Big(\big(f(s,Y_s^q,q_s)+g(s,Y_s^q,Z_s)-q_sZ_s\big)+\big(g(s,Y_s,Z_s)-g(s,Y_s^q,Z_s)\big)\Big)\\ \geq &\mathbb{1}_{\{Y_s-Y_s^q>0\}}\big(0-\beta(Y_s-Y_s^q)^+\big)\geq-\beta(Y_s-Y_s^q)^+, ~~s\in[t,\tau], \end{align}\] which together with (14 ) yields that \[\begin{align} {\rm d}\big[e^{\beta s}(Y_s-Y_s^q)^+\big] \geq -e^{\beta s}\mathbb{1}_{\{Y_s-Y_s^q>0\}}(Z_s-Z^{q}_s)\cdot{\rm d}B_s^q, ~~s\in[t,\tau]. \end{align}\] Hence, for each \(t\in[0,T]\) and \(m\geq1\), we have \[e^{\beta t} (Y_t-Y_t^{q})^+\leq e^{\beta \tau^t_m} (Y_{\tau^t_m}-Y_{\tau^t_m}^{q})^++\int_t^{\tau^t_m}e^{\beta s}\mathbb{1}_{\{Y_s-Y_s^q>0\}}(Z_s-Z^{q}_s)\cdot{\rm d}B_s^q.\] It then follows that \[\begin{align} \label{b} e^{\beta t} (Y_t-Y_t^{q})^+\leq\mathbb{E}^{{\mathbb{Q}}}\big[e^{\beta\tau^t_m} (Y_{\tau^t_m}-Y_{\tau^t_m}^{q})^+\big|\mathcal{F}_t\big]. \end{align}\tag{15}\] In view of (?? ) and (?? ), by picking \(\overline{p}=p\), the Fenchel’s inequality gives \[\begin{align} \mathbb{E}^{{\mathbb{Q}}}\Big[\sup_{s\in[0,T]} Y_s^-+\sup_{s\in[0,T]} Y_s^+\Big]&=\mathbb{E}\Big[M_T \sup_{s\in[0,T]} Y_s^-\Big]+\mathbb{E}\Big[M_T \sup_{s\in[0,T]} Y_s^+\Big]\\ &\leq \mathbb{E}\Big[\exp\Big(p\sup_{s\in[0,T]} Y_s^+\Big)\Big]+\mathbb{E}\Big[\exp\Big(p\sup_{s\in[0,T]} Y_s^-\Big)\Big]+\frac{2}{p}\mathbb{E}[M_T(\ln M_T-\ln p-1)]\\ &\leq 2C_p+\frac{1}{p}\mathbb{E}^{{\mathbb{Q}}}\Big[\int_{0}^{T}|q_s|^2{\rm d}s\Big]<+\infty. \end{align}\] Thus, in view of the last inequality and the fact that \(Y_\cdot^q\) belongs to class \((D)\) under \({\mathbb{Q}}\), by letting \(m\rightarrow\infty\) in (15 ) we obtain \[\begin{align} e^{\beta t} (Y_t-Y_t^{q})^+\leq\mathbb{E}^{{\mathbb{Q}}}\big[e^{\beta T} (Y_{T}-Y_{T}^{q})^+\big|\mathcal{F}_t\big]=0 ~~\mathbb{P}-a.s., \end{align}\] which means that for each \(t\in[0,T], ~Y_t\leq Y_t^q ~\mathbb{P}-a.s.\).

Second step. Set \(q_s^*\in\partial_{z}g(s,Y_s,Z_s)\). Then we have \[f(s,Y_s,q_s^*)=q_s^*Z_s-g(s,Y_s,Z_s), ~~s\in[0,T].\] Consequently, if \(q_\cdot^*\in\mathcal{A}\), then by BSDEs (1 ) and (?? ) together with uniqueness of the integrable solution of BSDE (?? ), we can conclude that \(Y_\cdot=Y_\cdot^{q^*}\).

Third step. We conclude the proof by verifying \(q^*\in\mathcal{A}\). Note that the assumptions of (iii) of Theorem 2 is stronger than those in (i) of Theorem 2. By Proposition 3 we know that \(\int_0^T|q_s^*|^2{\rm d}s<+\infty ~\mathbb{P}-a.s.\), \((M^*_t)_{t\in[0,T]}\) is a uniformly integrable martingale and defines a probability \({\mathbb{Q}}^*\sim\mathbb{P}\), where \[M^*_t:=\exp\Big(\int_{0}^{t}q_s^*{\rm d}B_s-\frac{1}{2}\int_{0}^{t}|q_s^*|^2{\rm d}s\Big), ~t\in[0,T]~~\text{and}~~\frac{{\rm d}{\mathbb{Q}}^*}{{\rm d}\mathbb{P}}:=M^*_T.\] Moreover, we also have that \(\mathbb{E}^{{\mathbb{Q}}^*}\big[\int_{0}^{T}|q_s^*|^2{\rm d}s\big]=2\mathbb{E}[M^*_T \ln M^*_T]<+\infty\), \[\begin{align} \label{4a} \mathbb{E}^{{\mathbb{Q}}^*}\big[\sup_{t\in[0,T]} A_\beta(t)\big]\leq C_p+\frac{1}{2p} \mathbb{E}^{{\mathbb{Q}}^*}\Big[\int_0^T|q_s^*|^2 {\rm d}s\Big]<+\infty, \end{align}\tag{16}\] and \[\begin{align} \label{4aa} \mathbb{E}^{{\mathbb{Q}}^*}\Big[\sup_{t\in[0,T]} \overline{A}_\beta(t)\Big]\leq C_p+\frac{1}{2p} \mathbb{E}^{{\mathbb{Q}}^*}\Big[\int_0^T|q_s^*|^2 {\rm d}s\Big]<+\infty. \end{align}\tag{17}\] Thus, it suffices to show that \[\begin{align} \label{320} \mathbb{E}^{{\mathbb{Q}}^*}\Big[\int_0^T|f(s,0,q_s^*)|{\rm d}s\Big]<+\infty. \end{align}\tag{18}\] To do this, we first show that \[\begin{align} \label{321} \mathbb{E}^{{\mathbb{Q}}^*}\Big[\int_0^T|f(s,Y_s,q_s^*)|{\rm d}s\Big]<+\infty. \end{align}\tag{19}\] In fact, in view of \(f(t,y,q)\geq-\overline{\alpha}_t-\beta|y|+\frac{1}{2\gamma}|q|^2\), we deduce \[f^-(s,Y_s,q_s^*)\leq\overline{\alpha}_s+\beta|Y_s|,\] so, in view of (16 ) and (17 ), \[\begin{align} \mathbb{E}^{{\mathbb{Q}}^*}\Big[\int_0^T f^-(s,Y_s,q_s^*){\rm d}s\Big]\leq\mathbb{E}^{{\mathbb{Q}}^*}\Big[\int_0^T (\overline{\alpha}_s+\beta|Y_s|){\rm d}s\Big]<+\infty. \end{align}\] Moreover, by BSDE (1 ) we have \[\begin{align} Y_t=\xi+\int_t^T f(s,Y_s,q_s^*){\rm d}s+\int_t^T Z_s\cdot{\rm d}B_s^{q^*}=\mathbb{E}^{{\mathbb{Q}}^*}\Big[\xi+\int_t^T f(s,Y_s,q_s^*){\rm d}s\big|\mathcal{F}_t\Big], ~~t\in[0,T]. \end{align}\] It then follows that \[\begin{align} Y_0=\mathbb{E}^{{\mathbb{Q}}^*}\Big[\xi+\int_0^T f(s,Y_s,q_s^*){\rm d}s\Big]\geq\mathbb{E}^{{\mathbb{Q}}^*}\Big[-|\xi|+\int_0^T f^+(s,Y_s,q_s^*){\rm d}s-\int_0^T f^-(s,Y_s,q_s^*){\rm d}s\Big]. \end{align}\] Thus, \[\mathbb{E}^{{\mathbb{Q}}^*}\Big[\int_0^T f^+(s,Y_s,q_s^*){\rm d}s\Big]\leq Y_0+\mathbb{E}^{{\mathbb{Q}}^*}\big[|\xi|\big]+\mathbb{E}^{{\mathbb{Q}}^*}\Big[\int_0^T f^-(s,Y_s,q_s^*){\rm d}s\Big]<+\infty.\] Combining the last several inequalities yields (19 ). Furthermore, in view of \(f(t,y,q)\geq-\overline{\alpha}_t-\beta|y|+\frac{1}{2\gamma}|q|^2\), we have \[f(s,0,q_s^*)\geq-\overline{\alpha}_s, ~~s\in[0,T].\] It then follows from the last inequality and (13 ) that \[\begin{align} -\overline{\alpha}_s\leq f(s,0,q_s^*)\leq f(s,Y_s,q_s^*)+\beta|Y_s|+\overline{\alpha}_s, ~~s\in[0,T], \end{align}\] which implies \[|f(s,0,q_s^*)|\leq|f(s,Y_s,q_s^*)|+\beta|Y_s|+\overline{\alpha}_s.\] Then, (18 ) follows immediately from the last inequality, (19 ), (16 ) and (17 ). Thus, we have proved the desired assertion that \(q^*\in\mathcal{A}\) is optimal for the stochastic control problem, i.e., \(Y_\cdot=Y^{q^*}_\cdot=\mathop{\operatorname{ess\,inf}}\limits_{q\in\mathcal{A}}Y^{q}_\cdot\), which naturally yields the uniqueness result. The proof is then complete.

4 The uniqueness for the case of \(g\) being convex in \((y, z)\)↩︎

This section is devoted to the uniqueness for the unbounded solution of BSDE (\(\ref{BSDE1461}\)) with generator \(g\) being convex in \((y, z)\). We use the following assumption on \(g\).

(UN3) \({\rm d}\mathbb{P}\times{\rm d} t-a.e.\), for each \((y,z)\in{\mathbb{R}}\times{\mathbb{R}}^d, ~g(t,\cdot,\cdot)\) is convex.

Suppose that \(g\) satisfies (UN3), the Legendre-Fenchel transformation of \(g\) can be defined as follows: \[\begin{align} \label{fdyc} f(\omega,t,r,q):=\sup_{(y,z)\in{\mathbb{R}}\times{\mathbb{R}}^d}\big(ry+qz-g(\omega,t,y,z)\big),~~~\forall (\omega,t,r,q)\in\Omega\times[0,T]\times{\mathbb{R}}\times{\mathbb{R}}^{1\times d}. \end{align}\tag{20}\] The \(f\) is a function valued in \({\mathbb{R}}\cup\{+\infty\}\). We define the admissible control set: \[\begin{align} \mathcal{A}:=\bigg\{&(r_t,q_t)_{t\in[0,T]}~~\text{is a pair of } (\mathcal{F}_t)\text{-progressively measurable~} {\mathbb{R}}\times{\mathbb{R}}^{1\times d} \text{-valued processes}:\\ &-\beta\leq r_\cdot\leq\beta,~~\int_{0}^{T}|q_s|^2{\rm d}s<+\infty~~ \mathbb{P}-a.s.,~~\mathbb{E}^{{\mathbb{Q}}}\Big[\int_{0}^{T}|q_s|^2{\rm d}s\Big]<+\infty,\\ &\mathbb{E}^{{\mathbb{Q}}}\Big[|\xi|+\int_{0}^{T}|f(s,r_s,q_s)|{\rm d}s\Big]<+\infty, ~\big(M_t\big)_{t\in[0,T]}~~\text{is a uniformly integrable martingale}\\ &\text{with} ~~M_t:=\exp\Big(\int_{0}^{t}q_s{\rm d}B_s-\frac{1}{2}\int_{0}^{t}|q_s|^2{\rm d}s\Big), ~t\in[0,T], ~\text{and}~\frac{{\rm d}{\mathbb{Q}}}{{\rm d}\mathbb{P}}:=M_{T}\bigg\}. \end{align}\] The following Theorem 6 gives a uniqueness result for the unbounded solution of BSDE (1 ) under assumption (UN3).

Proof. The existence result has been given in Proposition 1. Now, we show the uniqueness part. According to assumptions (EX) and (UN3), we can deduce that the function \(f\) defined in (20 ) verifies the following properties: \({\rm d}\mathbb{P}\times{\rm d} t-a.e.\),

• \(\forall (r,q)\in{\mathbb{R}}\times{{\mathbb{R}}^{1\times d}}, ~f(t,r,q)\geq-\overline{\alpha}_t+\frac{1}{2\gamma}|q|^2\);

• If \(f(t,r,q)<+\infty\) for some \((r,q)\in{\mathbb{R}}\times{\mathbb{R}}^{1\times d}\), then \(r\in[-\beta,\beta]\);

• \(f(t,\cdot,\cdot)\) is convex.

Since (3) is trivial, we need only to show (1) and (2) here. In fact, in view of (3) of (EX), we have \({\rm d}\mathbb{P}\times{\rm d} t-a.e.\), for each \(z\in{\mathbb{R}}^d, ~|g(t,0,z)|\leq\overline{\alpha}_t+\frac{\gamma}{2}|z|^2\). It then follows that \({\rm d}\mathbb{P}\times{\rm d} t-a.e.\), for each \((r,q)\in{\mathbb{R}}\times{\mathbb{R}}^{1\times d}\), \[\begin{align} f(t,r,q)&=\sup_{(y,z)\in{\mathbb{R}}\times{\mathbb{R}}^d}\big(ry+qz-g(t,y,z)\big)\geq\sup_{z\in{\mathbb{R}}^d}\big(qz-g(t,0,z)\big)\\ &\geq\sup_{z\in{\mathbb{R}}^d}\Big(qz-\overline{\alpha}_t-\frac{\gamma}{2}|z|^2\Big)=-\overline{\alpha}_t+\frac{1}{2\gamma}|q|^2. \end{align}\] Thus, (1) is true. Furthermore, thanks to (2) of (EX) and (UN2), we have \({\rm d}\mathbb{P}\times{\rm d} t-a.e.\), for each \((r,q)\in{\mathbb{R}}\times{\mathbb{R}}^{1\times d}\), \[\begin{align} f(t,r,q)&=\sup_{(y,z)\in{\mathbb{R}}\times{\mathbb{R}}^d}\big(ry+qz-g(t,y,z)\big)\geq\max\Big\{\sup_{y\leq0}\big(ry-g(t,y,0)\big), ~\sup_{y>0}\big(ry-g(t,y,0)\big)\Big\}\\ &\geq\max\Big\{\sup_{y\leq0}\big(ry-\overline{\alpha}_t+\beta y\big), ~\sup_{y>0}\big(ry-\overline{\alpha}_t-\beta y\big)\Big\}\\ &=-\overline{\alpha}_t+\max\Big\{\sup_{y\leq0}\big(r+\beta \big)y, ~\sup_{y>0}\big(r-\beta\big)y\Big\}. \end{align}\] Hence, if \(f(t,r,q)<+\infty\), then \(r+\beta\geq0\) and \(r-\beta\leq0\), i.e., \(-\beta\leq r\leq\beta\). Thus, (2) is also true.

Now, let \((r, q)\) be in \(\mathcal{A}\), if this set is not empty. Set \(B_t^q:=B_t-\int_0^t q_s{\rm d}s, t\in[0,T]\). Thanks to the Girsanov’s theorem, \((B_t^q)_{t\in[0,T]}\) is a Brownian motion under the probability \({\mathbb{Q}}\). We consider the following BSDE: \[\begin{align} \label{bsde32c} Y_t^{r,q}=\xi+\int_{t}^{T}\big(f(s,r_s,q_s)-r_sY_s^{r,q}\big){\rm d}s+\int_{t}^{T}Z_s^{r,q}\cdot{\rm d}B_s^q,~~~~t\in[0,T]. \end{align}\tag{21}\] In view of the definition of \(\mathcal{A}\), by Theorem 6.2 in Briand?, 2003? we know that BSDE (21 ) admits a unique solution \((Y_\cdot^{r,q},Z_\cdot^{r,q})\) such that \((Y_t^{r,q})_{t\in[0,T]}\) belongs to the class \((D)\) under \({\mathbb{Q}}\). Hence, the stochastic control problem consists in minimizing \(Y_\cdot^{r,q}\) among all the admissible controls \((r, q)\in\mathcal{A}\). Let \((Y_\cdot,Z_\cdot)\) be a solution of BSDE (1 ) satisfying (?? ) and (?? ) for any \(\overline{p}>\gamma\). We divide our proof into the following three steps.

First step. Let us start with showing \(Y_\cdot\leq Y_\cdot^{r,q}\) for any \((r,q)\in\mathcal{A}\). Firstly, we have \(Y_T=Y^{r,q}_T=\xi\). For each \(t\in[0,T]\) and each integer \(m\geq1\), denote the following stopping time \[\tau_m^t:=\inf\Big\{s\geq t, \int_t^s|Z_u|^2{\rm d}u+\int_t^s|Z^{r,q}_u|^2{\rm d}u+\int_t^s|q_u|^2{\rm d}u\geq m\Big\}\wedge T\] with the convention \(\inf\emptyset=+\infty\). By applying \(\rm{It\hat{o}}\)’s formula to \(e^{-\int_0^t r_s{\rm d}s} Y_t^{r,q}\) we obtain \[\begin{align} e^{-\int_0^t r_s{\rm d}s} Y_t^{r,q}=e^{-\int_0^{\tau_m^t} r_s{\rm d}s} Y_{\tau_m^t}^{r,q}+\int_t^{\tau_m^t}e^{-\int_0^s r_u{\rm d}u} f(s,r_s,q_s){\rm d}s+\int_t^{\tau_m^t}e^{-\int_0^s r_u{\rm d}u} Z_s^{r,q}\cdot{\rm d}B_s^q. \end{align}\] It then follows that \[\begin{align} Y_t^{r,q}=\mathbb{E}^{{\mathbb{Q}}}\Big[e^{-\int_t^{\tau_m^t} r_s{\rm d}s} Y_{\tau_m^t}^{r,q}+\int_t^{\tau_m^t}e^{-\int_t^s r_u{\rm d}u} f(s,r_s,q_s){\rm d}s\big|\mathcal{F}_t\Big]. \end{align}\] Since \(e^{-\int_t^{\tau_m^t} r_s{\rm d}s}\leq e^{\int_t^{\tau_m^t} \beta{\rm d}s}\leq e^{\beta T}\) and \(Y_\cdot^{r,q}\) belongs to the class \((D)\) under \({\mathbb{Q}}\), by letting \(m\rightarrow\infty\) in the last inequality, we have \[\begin{align} \label{yrq} Y_t^{r,q}=\mathbb{E}^{{\mathbb{Q}}}\Big[e^{-\int_t^T r_s{\rm d}s} \xi+\int_t^Te^{-\int_t^s r_u{\rm d}u} f(s,r_s,q_s){\rm d}s\big|\mathcal{F}_t\Big]. \end{align}\tag{22}\] Similarly, \(\rm{It\hat{o}}\)’s formula applied to \(e^{-\int_0^t r_s{\rm d}s} Y_t\) gives, in view of (20 ), \[\begin{align} e^{-\int_0^t r_s{\rm d}s} Y_t&=e^{-\int_0^{\tau_m^t} r_s{\rm d}s} Y_{\tau_m^t}+\int_t^{\tau_m^t}e^{-\int_0^s r_u{\rm d}u} (-g(s,Y_s,Z_s)+r_sY_s+q_sZ_s){\rm d}s+\int_t^{\tau_m^t}e^{-\int_0^s r_u{\rm d}u} Z_s\cdot{\rm d}B_s^q\\ &\leq e^{-\int_0^{\tau_m^t} r_s{\rm d}s} Y_{\tau_m^t}+\int_t^{\tau_m^t}e^{-\int_0^s r_u{\rm d}u} f(s,r_s,q_s){\rm d}s+\int_t^{\tau_m^t}e^{-\int_0^s r_u{\rm d}u} Z_s\cdot{\rm d}B_s^q. \end{align}\] So, we have \[\begin{align} \label{e} Y_t\leq\mathbb{E}^{{\mathbb{Q}}}\Big[e^{-\int_t^{\tau_m^t} r_s{\rm d}s} Y_{\tau_m^t}+\int_t^{\tau_m^t}e^{-\int_t^s r_u{\rm d}u} f(s,r_s,q_s){\rm d}s\big|\mathcal{F}_t\Big]. \end{align}\tag{23}\] With the fact that \(e^{-\int_t^{\tau_m^t} r_s{\rm d}s} Y_{\tau_m^t}\leq e^{\beta T} |Y_{\tau_m^t}|\) together with (?? ) and (?? ), Fenchel’s inequality gives, picking \(\overline{p}=p\), \[\begin{align} \mathbb{E}^{{\mathbb{Q}}}\Big[\sup_{s\in[0,T]} Y_s^-+\sup_{s\in[0,T]} Y_s^+\Big]&=\mathbb{E}\Big[M_T \sup_{s\in[0,T]} Y_s^-\Big]+\mathbb{E}\Big[M_T \sup_{s\in[0,T]} Y_s^+\Big]\\ &\leq \mathbb{E}\Big[\exp\Big(p\sup_{s\in[0,T]} Y_s^+\Big)\Big]+\mathbb{E}\Big[\exp\Big(p\sup_{s\in[0,T]} Y_s^-\Big)\Big]+\frac{2}{p}\mathbb{E}[M_T(\ln M_T-\ln p-1)]\\ &\leq C_p+\frac{1}{p}\mathbb{E}^{{\mathbb{Q}}}\Big[\int_{0}^{T}|q_s|^2{\rm d}s\Big]<+\infty. \end{align}\] Then, by letting \(m\rightarrow\infty\) and applying the Lebesgue’s dominated convergence theorem in (23 ), we obtain \[\begin{align} \label{y} Y_t\leq\mathbb{E}^{{\mathbb{Q}}}\Big[e^{-\int_t^T r_s{\rm d}s} \xi+\int_t^Te^{-\int_t^s r_u{\rm d}u} f(s,r_s,q_s){\rm d}s\big|\mathcal{F}_t\Big]. \end{align}\tag{24}\] Combining (22 ) and (24 ), we deduce that for each \(t\in[0,T], ~Y_t\leq Y_t^{r,q}~~\mathbb{P}-a.s.\).

Second step. Set \((r_s^*, q_s^*)\in\partial_{(y,z)}g(s,Y_s,Z_s)\). Then we have \[f(s,r_s^*,q_s^*)=r_s^*Y_s+q_s^*Z_s-g(s,Y_s,Z_s), ~~s\in[0,T].\] Consequently, if \((r^*, q^*)\in\mathcal{A}\), then by BSDEs (1 ) and (22 ) together with the above argument we can conclude that \(Y_\cdot=Y_\cdot^{r^*, q^*}\).

Third step. We conclude the proof by verifying \((r^*, q^*)\in\mathcal{A}\). Since \((r_s^*, q_s^*)\in\partial_{(y,z)}g(s,Y_s,Z_s)\), it is clear that \(f(s,r_s^*, q_s^*)<+\infty\), so we can deduce from (2) that \(-\beta\leq r_s^*\leq\beta\) and, in view of (1), \[\begin{align} g(s,Y_s,Z_s)&=r_s^*Y_s+q_s^*Z_s-f(s,r_s^*,q_s^*)\leq r_s^*Y_s+\frac{1}{2}\Big(\frac{1}{2\gamma}|q_s^*|^2+2\gamma|Z_s|^2\Big)+\overline{\alpha}_s-\frac{1}{2\gamma}|q_s^*|^2\\ &= r_s^*Y_s+\gamma|Z_s|^2+\overline{\alpha}_s-\frac{1}{4\gamma}|q_s^*|^2;\\ \frac{1}{4\gamma}|q_s^*|^2&\leq-g(s,Y_s,Z_s)+\beta |Y_s|+\gamma|Z_s|^2+\overline{\alpha}_s. \end{align}\] It follows from the last inequality that \(\int_0^T|q_s^*|^2{\rm d}s<+\infty ~\mathbb{P}-a.s.\). For each \(n\geq1\), we define \[M^*_t:=\exp\Big(\int_{0}^{t}q_s^*{\rm d}B_s-\frac{1}{2}\int_{0}^{t}|q_s^*|^2{\rm d}s\Big), ~~t\in[0,T],\] \[\tau_n:=\inf \Big\{t \in[0, T]: \int_0^t|q_s^*|^2 {\rm d}s+\int_0^t|Z_s|^2 {\rm d}s\geq n\Big\} \wedge T, \quad \frac{{\rm d}{\mathbb{Q}}_n^*}{{\rm d}\mathbb{P}}:=M^*_{\tau_n}.\] Let us further show that \(\mathbb{E}^{{\mathbb{Q}}^*}\big[\int_{0}^{T}|q_s^*|^2{\rm d}s\big]<+\infty\), and \(\big(M^*_t\big)_{t\in[0,T]}\) is a uniformly integrable martingale. To do this, we have to prove the following Lemma.

Proof of Lemma 7. According to (?? ), (?? ) and Fenchel’s inequality, we have \[\begin{align} \label{a} \mathbb{E}^{{\mathbb{Q}}_n^{*}}\Big[\sup_{t\in[0,T]} A_\beta(t)\Big]=\mathbb{E}\Big[M^*_{\tau_n} \sup_{t\in[0,T]} A_\beta(t)\Big] \leq C_p+\frac{1}{2p} \mathbb{E}^{{\mathbb{Q}}_n^*}\Big[\int_0^{\tau_n}|q_s^*|^2 {\rm d}s\Big], \end{align}\tag{25}\] and for each \(\overline{p}>\gamma\), in the same manner, \[\begin{align} \label{aa} \mathbb{E}^{{\mathbb{Q}}_n^{*}}\Big[\sup_{t\in[0,T]} \overline{A}_\beta(t)\Big]\leq C_{\overline{p}}+\frac{1}{2\overline{p}} \mathbb{E}^{{\mathbb{Q}}_n^*}\Big[\int_0^{\tau_n}|q_s^*|^2 {\rm d}s\Big]. \end{align}\tag{26}\] Since \(g(s,Y_s,Z_s)=r_s^*Y_s+q_s^*Z_s-f(s,r_s^*,q_s^*)\), and \((M^*_{t\wedge\tau_n})_{t\in[0,T]}\) is a martingale under \(\mathbb{P}\), we have \[Y_{t\wedge\tau_n}=Y_{\tau_n}+\int_{t\wedge\tau_n}^{\tau_n}\big(-r_s^*Y_s+f(s,r_s^*,q_s^*)\big){\rm d}s+\int_{t\wedge\tau_n}^{\tau_n} Z_s \cdot {\rm d}B_s^{q^*}, ~~t\in[0,T].\] Applying \(\rm{It\hat{o}}\)’s formula to \(e^{\beta s} Y_s\), we obtain \[\begin{align} {\rm d}\big(e^{\beta s} Y_s\big)=e^{\beta s}\big((\beta+r_s^*) Y_s-f(s,r_s^*,q_s^*)\big){\rm d}s-e^{\beta s}Z_s\cdot{\rm d}B_s^{q^*}, ~~s\in[0,\tau_n]. \end{align}\] Moreover, in view of \(f(t,r,q)\geq-\overline{\alpha}_t+\frac{1}{2\gamma}|q|^2\), \(Y_{\tau_n}\geq-Y_{\tau_n}^-\) and \(0\leq \beta+r_s^*\leq 2\beta\), we deduce that \[\begin{align} Y_0&=e^{\beta \tau_n} Y_{\tau_n}+\int_0^{\tau_n}e^{\beta s}\big(f(s,r_s^*,q_s^*)-(\beta+r_s^*) Y_s\big){\rm d}s+\int_0^{\tau_n}e^{\beta s}Z_s\cdot{\rm d}B_s^{q^*}\\ &\geq-e^{\beta \tau_n} Y_{\tau_n}^-+\int_0^{\tau_n}e^{\beta s}\big(-\overline{\alpha}_s-2\beta Y_s^++\frac{1}{2\gamma}|q_s^*|^2\big){\rm d}s+\int_0^{\tau_n}e^{\beta s}Z_s\cdot{\rm d}B_s^{q^*}. \end{align}\] It then follows from (25 ) and (26 ) that for each \(\overline{p}>\gamma\), similar to (8 ), \[\begin{align} \notag Y_0&\geq \mathbb{E}^{{\mathbb{Q}}_n^*}\Big[-e^{\beta \tau_n} Y_{\tau_n}^-+\int_0^{\tau_n}e^{\beta s}\big(-\overline{\alpha}_s-2\beta Y_s^++\frac{1}{2\gamma}|q_s^*|^2\big){\rm d}s\Big]\\ &\geq-C_p-2\beta TC_{\overline{p}}+\frac{1}{2}\Big(\frac{1}{\gamma}-\frac{1}{p}-\frac{2\beta T}{\overline{p}}\Big)\mathbb{E}^{{\mathbb{Q}}_n^*}\Big[\int_0^{\tau_n}|q_s^*|^2{\rm d}s\Big]. \end{align}\] Set \(\overline{p}>\frac{2\beta Tp\gamma }{p-\gamma}\). Then \(\frac{1}{2}\Big(\frac{1}{\gamma}-\frac{1}{p}-\frac{2\beta T}{\overline{p}}\Big)>0.\) Hence, \[\begin{align} \label{g} 2\mathbb{E}[M^*_{\tau_n}\ln M^*_{\tau_n}]=2\mathbb{E}^{{\mathbb{Q}}_n^{*}}[\ln M^*_{\tau_n}]=\mathbb{E}^{{\mathbb{Q}}_n^{*}}\Big[\int_0^{\tau_n}|q_s^*|^2{\rm d}s\Big]<C_{p,\overline{p},T,\beta,\gamma}, \end{align}\tag{27}\] where \(C_{p,\overline{p},T,\beta,\gamma}\) is a positive constant independent of \(n\). Hence, we obtain \(\sup\limits_n\mathbb{E}[M^*_{\tau_n}\ln M^*_{\tau_n}]<+\infty\). Then we conclude the proof of the lemma by using the de La Vall\(\acute{\rm{e}}\)e Poussin lemma.

Applying Fatou’s lemma to (27 ), we obtain \[2\mathbb{E}[M^*_T\ln M^*_T]=\mathbb{E}^{{\mathbb{Q}}^*}\Big[\int_0^T|q_s^*|^2{\rm d}s\Big]\leq\liminf_n\mathbb{E}^{{\mathbb{Q}}_n^* }\Big[\int_0^{\tau_n}|q_s^*|^2{\rm d}s\Big]<+\infty.\] Finally, it remains to show that \(\mathbb{E}^{{\mathbb{Q}}^*}\big[\int_0^T|f(s,r_s^*,q_s^*)|{\rm d}s\big]<+\infty\). In fact, in view of \(f(t,r,q)\geq-\overline{\alpha}_t+\frac{1}{2\gamma}|q|^2\), we deduce \[f^-(s,r_s^*,q_s^*)\leq\overline{\alpha}_s, ~~s\in[0,T].\] So, \[\begin{align} \mathbb{E}^{{\mathbb{Q}}^*}\Big[\int_0^T f^-(s,r_s^*,q_s^*){\rm d}s\Big]\leq\mathbb{E}^{{\mathbb{Q}}^*}\Big[\int_0^T \overline{\alpha}_s{\rm d}s\Big]<+\infty. \end{align}\] It then follows from (22 ) and \(-\beta\leq r_\cdot^*\leq\beta\) that \[\begin{align} Y_0^{r,q}&=\mathbb{E}^{{\mathbb{Q}}^*}\Big[e^{-\int_0^T r_s^*{\rm d}s} \xi+\int_0^T e^{-\int_0^s r_u^*{\rm d}u} f(s,r_s^*,q_s^*){\rm d}s\Big]\\ &\geq\mathbb{E}^{{\mathbb{Q}}^*}\Big[-e^{\beta T} |\xi|+ e^{-\beta T} \int_0^T f^+(s,r_s^*,q_s^*){\rm d}s- e^{\beta T} \int_0^T f^-(s,r_s^*,q_s^*){\rm d}s\Big]. \end{align}\] Thus, \[\mathbb{E}^{{\mathbb{Q}}^*}\Big[\int_0^T f^+(s,r_s^*,q_s^*){\rm d}s\Big]\leq e^{\beta T}Y_0^{r,q}+e^{2\beta T}\mathbb{E}^{{\mathbb{Q}}^*}\big[|\xi|\big]+e^{2\beta T}\mathbb{E}^{{\mathbb{Q}}^*}\Big[\int_0^T f^-(s,r_s^*,q_s^*){\rm d}s\Big]<+\infty.\] Combining the last several inequalities yields that \(\mathbb{E}^{{\mathbb{Q}}^*}\big[\int_0^T|f(s,r_s^*,q_s^*)|{\rm d}s\big]<+\infty\). Therefore, we show that \((r^*, q^*)\in\mathcal{A}\) is optimal, i.e., for each \(t\in[0,T], ~ Y_t=Y^{r^*, q^*}_t=\mathop{\operatorname{ess\,inf}}\limits_{(r, q)\in\mathcal{A}}Y^{r,q}_t ~\mathbb{P}-a.s.\), which naturally yields the uniqueness result. The proof of Theorem 6 is then complete.

5 Comparison theorem↩︎

In this section, we aim at establishing a general comparison theorem for unbounded solutions of BSDE (\(\ref{BSDE1461}\)) under the assumptions used in Section 3. For this purpose, we first apply Proposition 5 in Briand-Hu?, 2006? to obtain the following result.

The following comparison theorem for unbounded solutions of BSDE (\(\ref{BSDE1461}\)) is the main result of this section.

Proof. For each integer \(n\geq1\), let \[g_n(\omega,t,y,z):=\inf_{v\in{\mathbb{R}}^d}\big\{g(\omega,t,y,v)+(n+\gamma)|z-v|\big\}, ~~~(\omega,t,y,z)\in\Omega\times[0,T]\times{\mathbb{R}}\times{\mathbb{R}}^d.\] Thanks to \(g(t,y,\cdot)\) being convex and (?? ), \(g_n\) is well defined. In view of \(g\) satisfying (EX) and (UN1), it can easily be checked that \(g_n\) satisfies (EX) and (1) of (UN1). By the proof of Theorem 1 in Fan-Jiang-Tian?, 2011?, it is not hard to verify that \((g_n)_{n\geq1}\) is nondecreasing and converges to \(g\), and \(g_n\) satisfies that \({\rm d}\mathbb{P}\times{\rm d} t-a.e.\), for each \((y,z_1,z_2)\in{\mathbb{R}}\times{\mathbb{R}}^d\times{\mathbb{R}}^d\), \[\begin{align} |g_n(t,y,z_1)-g_n(t,y,z_2)|\leq(n+\gamma)|z_1-z_2|. \end{align}\]

Moreover, in view of \(g_n\leq g\), we have \[-g'(t,Y'_t,Z'_t)\leq -g(t,Y'_t,Z'_t)\leq-g_n(t,Y'_t,Z'_t).\] Thanks to Theorem 4.2 in Briand?, 2003? and Proposition 1, we can conclude that BSDE \((\xi,T,g_n)\) admits a unique solution \((Y^n_\cdot,Z^n_\cdot)\) such that (?? ) and (?? ) hold with \(Y^n_\cdot\) instead of \(Y_\cdot\) for some \(p>\gamma\) and any \(\overline{p}>\gamma\). Hence, in view of \(Y'_\cdot\in\mathcal{S}\), it is not hard to check that \((Y'_\cdot-Y^n_\cdot)^+\in\mathcal{S}\). Applying Proposition 9 gives that for each \(n\geq1\), \(Y'_\cdot\leq Y^n_\cdot\).

Similarly, since \(-g(t,Y_t,Z_t)\leq-g_n(t,Y_t,Z_t)\), we know that for each \(n\geq1\), \(Y_\cdot\leq Y^n_\cdot\). By virtue of \((g_n)_{n\geq1}\) being nondecreasing, we have that \((Y^n_t)_{n\geq1}\) is non-increasing. Hence, \(Y_\cdot\leq Y^n_\cdot\leq Y^1_\cdot\). Set \[\widehat{Y}_t:=\lim\limits_{n\rightarrow+\infty}Y^n_t,\;\;t\in [0,T].\] Since both \(Y_\cdot\) and \(Y^1_\cdot\) satisfy (?? ) and (?? ) for some \(p>\gamma\) and any \(\overline{p}>\gamma\), proceeding with a localization argument, we obtain that there exists an \((\mathcal{F}_t)\)-progressively measurable process \((\widehat{Z}_t)_{t\in[0,T]}\) such that \((\widehat{Y}_\cdot,\widehat{Z}_\cdot)\) is a solution of BSDE \((\xi,T,g)\) satisfying (?? ) and (?? ) for some \(p>\gamma\) and any \(\overline{p}>\gamma\). Furthermore, thanks to (i) of Theorem 2 together with assumptions (EX), (UN1) and (UN2) of the generator \(g\), we know that the solution of BSDE \((\xi,T,g)\) satisfying the above-mentioned conditions is unique, which implies that \(\widehat{Y}_\cdot=Y_\cdot\), i.e., \(Y_\cdot=\lim\limits_{n\rightarrow+\infty}Y^n_\cdot\). Combining with the fact that \(Y'_\cdot\leq Y^n_\cdot\), we have for each \(t\in[0,T], ~Y'_t\leq Y_t ~\mathbb{P}-a.s.\). The proof is complete.

6 Conclusions↩︎

This paper focuses on the existence and uniqueness of unbounded solutions for quadratic BSDEs, whose generator \(g\) may be not locally-Lipschitz continuous with general growth in \(y\), and be convex with quadratic growth in \(z\). Suppose that terminal value \(\xi^-\) admits a certain exponential moment and \(\xi^+\) admits every exponential moments or being bounded, and that the generator \(g\) is convex in \(z\). Under some mild growth conditions on the generator \(g\), we obtain the following results: If \(g\) satisfies the monotonicity condition with general growth in \(y\), which is strictly weaker than the uniform Lipschitz continuity condition used for example in Delbaen?, 2011?, then the quadratic BSDE admits a unique solution \((Y_\cdot,Z_\cdot)\) such that \(Y_\cdot\) satisfies certain integrability. If \(g\) is jointly convex in \((y,z)\), which is strictly stronger than the uniform Lipschitz continuity condition in \(y\) when combined with other necessary assumptions, then we can not only have existence and uniqueness of the unbounded solution, but also get an explicit expression of \(Y_\cdot\). The Legendre-Fenchel transform of convex functions, Girsanov’s theorem, the de La Vall\(\acute{\rm{e}}\)e Poussin lemma and Fenchel’s inequality play important roles in the proof of these results. In comparison to Delbaen?, 2011? and Delbaen?, 2015?, we innovatively use the exponential shifting transform with a factor \(e^{\beta t}\) or \(e^{-\int_0^t r_s{\rm d}s}\) to make fully use of the monotonicity condition of the generator \(g\) in \(y\) so that \(g\) is allowed to have a general growth rather than the usual linear growth in \(y\), which is one of the key idea of this paper. Another point that should be stressed is that the time interval partitioning method in Delbaen?, 2011? plays a crucial role in establishing the results on the uniform integrability. However, for simplicity of presentation, we do not use this method any longer, and instead lift the integrability of \(\xi^+\) so that the proof can proceed smoothly. Moreover, several examples are provided to illustrate that our results are more general than those in Delbaen?, 2011?. Finally, it is the first time, to the best of our knowledge, that the corresponding comparison theorems for unbounded solutions of the preceding BSDEs are established under general assumptions.

It is clear that \((Y_\cdot,Z_\cdot)\) is a solution of BSDE \((\xi, T, g)\) if and only if \((-Y_\cdot,-Z_\cdot)\) is a solution of BSDE \((-\xi, T, \overline{g})\), where \(\overline{g}(t,y,z):=-g(t,-y,-z)\), and that \(g\) is convex in \(z\) (resp. \((y,z)\)) if and only if \(\overline{g}\) is concave in \(z\) (resp. \((y,z)\)). Consequently, from this point of view, Theorems 2, 6 and 10 also consider quadratic BSDEs with concave generators. In addition, similar to Section 4 in Delbaen?, 2011?, we can also study the application of our Theorems 2 and 6 to quadratic PDEs, which is omitted in this paper due to the limitation of pages.

What we have seen from the above is the existence and uniqueness of unbounded solutions for quadratic BSDEs with generator \(g\) being convex (concave) in \(z\). The case of a non-convex generator was tackled in Fan-Hu-Tang?, 2020?, but more assumptions are required on the terminal value, that is, \(|\xi|\) admits every exponential moments. When \(|\xi|\) admits a certain exponential moment, the existence and uniqueness of unbounded solutions for quadratic BSDEs with non-convex generators are still unsolved, and this will be a subject of our future research.

References↩︎