In this section, our aim is to derive a necessary condition for optimal problem (2). Just as usual, in order to use the method of non-smooth analysis, we need the following assumption:

$$\begin{aligned} \text{the risk measure } \rho \text{ is Lipschitz}. \end{aligned}$$

(A.3)

The following proposition shows that \(\mathcal{E}_{0,T}^{g}\) is Lipschitz in \(L^{2}(\mathcal{F}_{T})\). For more details, see [2, 22, 23].

### Proposition 3

*Suppose that*
*g*
*satisfies* (A.1) *and* (A.2), \(\xi_{i}\in L^{2}(\mathcal {F}_{T}), i=1,2\), *and*
\((y^{i}_{t},z^{i}_{t})\)
*are the solutions of BSDE* (1) *with terminal values*
\(\xi_{i}\), *respectively*, *then there exists a constant*
\(C>0\)
*such that*

$$ E \Bigl[\sup_{t\in[0,T]} \bigl\vert y^{2}_{t}-y^{1}_{t} \bigr\vert ^{2} \Bigr]+E \biggl[ \int_{0}^{T} \bigl\vert z^{2}_{t}-z^{1}_{t} \bigr\vert ^{2}\,\mathrm{ d}t \biggr] \leq CE \bigl[ \vert \xi_{2}-\xi_{1} \vert ^{2} \bigr]. $$

(12)

### Remark 4

From Proposition 3, we know that, supposing *g* satisfies (A.1) and (A.2), then, for any \(\xi_{1}\), \(\xi_{2}\in L^{2}(\mathcal{F}_{T})\),

$$\bigl\vert \mathcal{E}_{0,T}^{g}(\xi_{1})- \mathcal{E}_{0,T}^{g}(\xi_{2}) \bigr\vert \leq C^{\frac{1}{2}} \Vert \xi _{1}-\xi_{2} \Vert $$

holds.

The following theorem is our main result.

### Theorem 1

*Under the conditions* (A.1)*–*(A.3), *if*
\(\xi^{*}\)
*is an optimal objective of problem* (2), *then there exist a non*-*negative constant*
*λ*, \(\zeta\in\partial ^{o} \rho(\xi^{*})\)
*and*
\(\eta\in\partial^{o} \mathcal{E}_{0,T}^{g}(\xi^{*})\)
*such that*

$$ \zeta+ \lambda\eta= 0 $$

(13)

*holds*.

In order to prove Theorem 1, we need the following lemma.

### Lemma 4

*Suppose that*
*g*
*satisfies* (A.1) *and* (A.2). *Then*, *for any*
\(\xi\in L^{2}(\mathcal{F}_{T})\), *we have*
\(0\notin\partial^{o} \mathcal{E}_{0,T}^{g}(\xi)\).

### Proof

For notational simplicity, we only consider the case \(d=1\). We assume that there exists \(\xi^{*}\in L^{2}(\mathcal{F}_{T})\) such that \(0\in\partial^{o} \mathcal{E}_{0,T}^{g}(\xi^{*})\). For any \(\xi\in L^{2}(\mathcal{F}_{T})\), let \(h(\xi):=\mathcal{E}_{0,T}^{g}(\xi)\), then

$$h^{o}\bigl(\xi^{*}; \eta\bigr)\geq0 $$

holds, for any \(\eta\in L^{2}(\mathcal{F}_{T})\). For each \(\xi\in L^{2}(\mathcal{F}_{T})\), suppose that \((y_{t},z_{t})\) and \((y_{t}^{r},z_{t}^{r})\) are the solutions of the following BSDEs on \([0,T]\):

$$\begin{aligned} &y_{t}=\xi+ \int_{t}^{T}g(s,y_{s},z_{s})\, \mathrm{ d}s- \int_{t}^{T}z_{s}\,\mathrm{ d}W_{s}, \\ &y_{t}^{r}=\xi+r\eta+ \int_{t}^{T}g\bigl(s,y_{s}^{r},z_{s}^{r} \bigr)\,\mathrm{ d}s- \int_{t}^{T}z_{s}^{r}\, \mathrm{ d}W_{s}, \end{aligned}$$

respectively. Thus, we have

$$\begin{aligned} y_{t}^{r}-y_{t}={}&r\eta+ \int_{t}^{T} \bigl[g\bigl(s,y_{s}^{r},z_{s}^{r} \bigr)-g(s,y_{s},z_{s}) \bigr]\,\mathrm{ d}s - \int_{t}^{T} \bigl(z_{s}^{r}-z_{s} \bigr)\,\mathrm{ d}W_{s} \\ ={}&r\eta+ \int_{t}^{T}\bigl[\alpha_{s} \bigl(y_{s}^{r}-y_{s}\bigr) +\beta_{s} \bigl(z_{s}^{r}-z_{s}\bigr)\bigr]\,\mathrm{ d}s- \int_{t}^{T}\bigl(z_{s}^{r}-z_{s} \bigr)\,\mathrm{ d}W_{s}, \end{aligned}$$

where

$$\begin{aligned} &\alpha_{s}=\frac{g(s,y_{s}^{r},z_{s})-g(s,y_{s},z_{s})}{y_{s}^{r}-y_{s}}1_{ \{ y_{s}^{r}-y_{s}\neq0 \}}, \\ &\beta_{s}=\frac{g(s,y_{s}^{r},z_{s}^{r})-g(s,y_{s}^{r},z_{s})}{z_{s}^{r}-z_{s}}1_{\{ z_{s}^{r}-z_{s}\neq0\}}, \end{aligned}$$

which imply \(\sup_{s\in[0,T]}|\alpha_{s}|\leq M\) and \(\sup_{s\in[0,T]}|\beta_{s}|\leq M\) from the Lipschitz condition (A.1). Applying Itô’s formula, we have

$$\mathcal{E}_{0,T}^{g}(\xi+r\eta)-\mathcal{E}_{0,T}^{g}( \xi)=E \biggl[r\eta\cdot \operatorname{exp} { \biggl\{ \int_{0}^{T} \biggl[\alpha_{s}- \frac{1}{2} \vert \beta_{s} \vert ^{2} \biggr]\, \mathrm{d}s+ \int_{0}^{T} \beta_{s} \, \mathrm{d}W_{s} \biggr\} } \biggr]. $$

Let \(\eta=-1\), then, by Definition 1, we can deduce that

$$h^{o}\bigl(\xi^{*};\eta\bigr)=\limsup_{\xi\rightarrow\xi^{*},r\downarrow0} \frac{\mathcal{E}_{0,T}^{g}(\xi+r\eta)-\mathcal{E}_{0,T}^{g}(\xi)}{r}< 0. $$

We get a contradiction with \(h^{o}(\xi^{*}; \eta)\geq0\). The proof of Lemma 4 is complete. □

### Proof of Theorem 1

By Lemma 4, we have \(0\notin\partial^{o} \mathcal{E}_{0,T}^{g}(\xi)\), for any \(\xi\in L^{2}(\mathcal{F}_{T})\). Hence by Definition 1, Remark 2, Definition 2 and Proposition 1, we can see that, if \(\xi^{*}\) is an optimal objective of problem (2), then the Fermat condition

$$ 0\in\partial^{o} \rho \bigl(\xi^{*} \bigr)+ N_{\mathcal{A}(x)} \bigl(\xi^{*} \bigr) $$

(14)

holds.

*Case* 1: \(\mathcal{E}_{0,T}^{g}(\xi^{*})= x\). In this case, \(\mathcal{A}(x)= \{\xi\in L^{2}(\mathcal{F}_{T})|\mathcal{E}_{0,T}^{g}(\xi )\leq\mathcal{E}_{0,T}^{g}(\xi^{*}) \}\). Thus, by Proposition 2, there exist a non-negative constant *λ*, \(\zeta\in\partial^{o} \rho(\xi^{*})\) and \(\eta\in\partial^{o} \mathcal{E}_{0,T}^{g}(\xi^{*})\) such that

$$ \zeta+ \lambda\eta= 0 $$

(15)

holds.

*Case* 2: \(\mathcal{E}_{0,T}^{g}(\xi^{*})=x_{0}< x\). In this case, denote

$$\widetilde{C}:=\bigl\{ \xi\in L^{2}(\mathcal{F}_{T})| \mathcal{E}_{0,T}^{g}(\xi)\leq x_{0}\bigr\} . $$

Since \(\xi^{*}\) is an optimal objective of problem (2), \(\xi^{*}\) is also an optimal objective of problem

$$ \min_{\xi\in\widetilde{C}}\rho(\xi). $$

(16)

Thus, by Definition 1, Remark 2, Definition 2 and Proposition 2 again, Equation (15) also holds. The proof of Theorem 1 is complete. □

From Remark 4, we know that \(\mathcal{E}_{0,T}^{g}\) is Lipschitz in \(L^{2}(\mathcal{F}_{T})\). By Definition 1, we can see that \(\partial^{o} \mathcal {E}_{0,T}^{g}(\xi)\) is not empty for any \(\xi\in L^{2}(\mathcal{F}_{T})\). Indeed, we have following result.

### Lemma 5

*Suppose that*
\(\xi\in L^{2}(\mathcal{F}_{T})\)
*and*
*g*
*satisfies* (A.1) *and* (A.2). *Let*
\((y_{t},z_{t})\)
*be the solution of BSDE* (1) *with terminal value*
*ξ*, *then for any*
\(\eta\in\partial^{o} \mathcal{E}_{0,T}^{g}(\xi)\)
*and*
\(\zeta\in L^{2}(\mathcal{F}_{T})\), *there exists*
\((\varphi_{t}, \psi_{t})\in\partial^{o} g(t,y_{t},z_{t})\)
*such that*

$$\langle\eta, \zeta\rangle=\tilde{y}_{0} $$

*holds*, *where*
\((\tilde{y}_{t}, \tilde{z}_{t})\)
*is the solution of the following BSDE on*
\([0,T]\):

$$ \tilde{y}_{t}= \zeta+ \int_{t}^{T} (\varphi_{s} \tilde{y}_{s} + \psi_{s}\tilde{z}_{s})\,\mathrm{ d}s - \int_{t}^{T}\tilde{z}_{s}\cdot\,\mathrm{ d}W_{s}. $$

(17)

### Proof

For any given \(\omega\in\varOmega\) and \(t\in[0,T]\), let \(\hat{g}(t, \hat{y}, \hat{z})=g^{o}(t, y_{t},z_{t};\hat{y}, \hat{z})\), \(\forall(\hat{y}, \hat{z})\in R\times R^{d}\). By Definition 1, we can see that \(\hat{g}(t, \hat{y}, \hat{z})\) is Lipschitz, positively homogeneous and convex in \((\hat{y}, \hat{z})\), and hence

$$ \hat{g}(t, \hat{y}, \hat{z})=\max_{(\varphi_{t}, \psi _{t})\in\partial^{o} g(t,y_{t},z_{t})} \bigl\langle (\varphi_{t}, \psi_{t}), (\hat{y}, \hat{z}) \bigr\rangle . $$

(18)

For any \(\zeta\in L^{2}(\mathcal{F}_{T})\), we consider the following BSDE on \([0,T]\):

$$ \hat{y}_{t}=\zeta+ \int_{t}^{T}\hat{g}(s, \hat{y}_{s}, \hat{z}_{s}) \,\mathrm{ d}s- \int_{t}^{T}\hat{z}_{s}\cdot\,\mathrm{ d}W_{s}. $$

(19)

Let \(\hat{f}(\zeta):=\hat{y}_{0}\). Denote

$$ D:= \bigl\{ \eta\in L^{2}(\mathcal{F}_{T})| \langle\eta, \zeta \rangle=\tilde{y}_{0}, \forall(\varphi_{t}, \psi_{t})\in\partial^{o} g(t,y_{t},z_{t}) \bigr\} , $$

(20)

where \((\tilde{y}_{t}, \tilde{z}_{t})\) is the solution of the following BSDE on \([0,T]\):

$$\tilde{y}_{t}= \zeta+ \int_{t}^{T} (\varphi_{s} \tilde{y}_{s} + \psi_{s}\tilde{z}_{s})\,\mathrm{ d}s- \int_{t}^{T}\tilde{z}_{s}\cdot\,\mathrm{ d}W_{s}. $$

By the comparison theorem of BSDE (for example, see El Karoui et al. [7]), we have \(\hat{f}(\zeta)\geq \langle\eta, \zeta\rangle\), for any \(\eta\in D\). Since *ĝ* is positively homogeneous and convex in \((\hat{y}, \hat{z})\), by Propositions 3.3 and 3.4 in Peng [16], we know that *f̂* is positively homogeneous and convex in \(L^{2}(\mathcal{F}_{T})\), and hence

$$ \hat{f}(\zeta)=\max_{\eta\in D} \langle\eta, \zeta \rangle. $$

(21)

By Definition 1, we can obtain that

$$\bigl(\mathcal{E}_{0,T}^{g}\bigr)^{o}(\xi; \zeta)= \hat{f}(\zeta)=\max_{\eta\in D} \langle\eta, \zeta\rangle. $$

This means that \(\partial^{o} \mathcal{E}_{0,T}^{g}(\xi)=D\). The proof of Lemma 5 is complete. □

In Ji and Peng [10], the authors assume that *g* is continuously differentiable in \((y,z)\). In this special case, we can get an explicit form of \(\partial^{o} \mathcal{E}_{0,T}^{g}\).

### Lemma 6

*Suppose that*
\(\xi\in L^{2}(\mathcal{F}_{T})\)
*and*
*g*
*satisfies* (A.1) *and* (A.2). *Let*
\((y_{t},z_{t})\)
*be the solution of BSDE* (1) *with terminal value*
*ξ*. *If*
*g*
*is continuously differentiable in*
\((y,z)\in R\times R^{d}\), *then we have*

$$\partial^{o} \mathcal{E}_{0,T}^{g}(\xi)= \{q_{T} \} $$

*and for any*
\(\eta\in L^{2}(\mathcal{F}_{T})\),

$$ \langle q_{T}, \eta\rangle=\tilde{y}_{0}, $$

(22)

*where*
\((\tilde{y}_{t},\tilde{z}_{t} )\)
*is the solution of the following BSDE on*
\([0,T]\):

$$\tilde{y}_{t}= \eta+ \int_{t}^{T} \bigl[g^{\prime}_{y}(s,y_{s},z_{s}) \tilde{y}_{s}+g^{\prime}_{z}(s,y_{s},z_{s}) \tilde{z}_{s} \bigr]\,\mathrm { d}s- \int_{t}^{T}\tilde{z}_{s}\cdot\,\mathrm{ d}W_{s}. $$

### Proof

For any given \(\omega\in\varOmega\) and \(t\in[0,T]\), let \(\tilde{g}(t, \tilde{y}, \tilde{z})=g^{o}(t, y_{t},z_{t};\tilde{y}, \tilde{z})\), \(\forall(\tilde{y}, \tilde{z})\in R\times R^{d}\). Since *g* satisfies (A.1), (A.2), and is continuously differentiable in \((y,z)\in R\times R^{d}\), we know that *g* is strictly differentiable in \((y,z)\in R\times R^{d}\) and

$$\tilde{g}(t, \tilde{y}, \tilde{z})=g^{\prime}_{y}(t,y_{t},z_{t}) \tilde {y}+g^{\prime}_{z}(t,y_{t},z_{t}) \tilde{z} $$

holds. For any \(\eta\in L^{2}(\mathcal{F}_{T})\), let \((\tilde{y}_{t},\tilde{z}_{t})\) be the solution of the following BSDE on \([0,T]\):

$$\tilde{y}_{t}= \eta+ \int_{t}^{T} \tilde{g}(s, \tilde{y}_{s}, \tilde{z}_{s})\,\mathrm{ d}s- \int_{t}^{T}\tilde {z}_{s}\cdot\,\mathrm{ d}W_{s}. $$

Denote \(\mathcal{E}_{0,T}^{\tilde{g}}(\eta):=\tilde{y}_{0}\). By Definition 1, we obtain

$$ \bigl(\mathcal{E}_{0,T}^{g} \bigr)^{o}(\xi; \eta)= \mathcal{E}_{0,T}^{\tilde{g}}(\eta). $$

(23)

Since *g̃* is linear in \((\tilde{y},\tilde{z})\), by Propositions 3.3 and 3.4 in Peng [16], we know that \(\mathcal{E}_{0,T}^{\tilde{g}}\) is linear in \(L^{2}(\mathcal{F}_{T})\). By Proposition 3, we can see that \(\mathcal{E}_{0,T}^{\tilde{g}}\) is continuous in \(L^{2}(\mathcal{F}_{T})\). Thus, by the Riesz representation theorem, there exists \(q_{T}\in L^{2}(\mathcal{F}_{T})\) such that

$$ \mathcal{E}_{0,T}^{\tilde{g}}(\eta)=\langle q_{T}, \eta \rangle $$

(24)

holds for any \(\eta\in L^{2}(\mathcal{F}_{T})\). From (23) and (24), we have \(\partial^{o} \mathcal{E}_{0,T}^{g}(\xi)= \{q_{T} \}\). The proof of Lemma 6 is complete. □

By Theorem 1 and Lemma 6, we immediately obtain the following theorem.

### Theorem 2

*Suppose that*
*ρ*
*satisfies* (A.3), *and*
*g*
*satisfies* (A.1), (A.2), *and is continuously differentiable in*
\((y,z)\). *Assume that*
\(\xi^{*}\)
*is an optimal objective of problem* (2), *let*
\((y^{*}_{t},z^{*}_{t})\)
*be the solution of BSDE* (1) *with terminal value*
\(\xi^{*}\). *Then there exist a non*-*negative constant*
*λ*
*and*
\(\zeta\in \partial^{o} \rho(\xi^{*})\)
*such that*

$$\zeta+ \lambda q^{*}_{T}= 0 $$

*holds*, *where*
\(q^{*}_{T}\)
*can be obtained by* (22).