Second-Order Necessary Conditions for Optimal Control with Recursive Utilities

Abstract

The necessary conditions for an optimal control of a stochastic control problem with recursive utilities are investigated. The first-order condition is the well-known Pontryagin-type maximum principle. When such a first-order necessary condition is singular in some sense, certain type of the second-order necessary condition will come in naturally. The aim of this paper is to explore such kind of conditions for our optimal control problem.

Appendix: Proof for the Claim in Sect. 4.2

We give the proof for the claim that there is a null subset $\mathcal {T}_{ij}^k \subset [0,T]$ such that, for $t \in [0,T]/\mathcal {T}_{ij}^k$,

$$\begin{aligned}&\lim _{r\rightarrow 0+} \frac{1}{r^2}\int _{t-r\beta }^{t+r\alpha }\int _{t-r\beta }^{t+r\alpha } \mathbb {E}\bigg [\gamma (u)(\delta G(t,Z_{ij}^{u_k}(u))\\&\qquad +\,\delta b^*(t,Z_{ij}^{u_k}(u))P(u))\delta b(s,Z_{ij}^{u_k}(s))I_{\{s \le u\}}\bigg ]\hbox {d}s\hbox {d}u\\&\quad =\frac{(\alpha +\beta )^2}{2}\mathbb {E}\bigg [\gamma (t)(\delta G(t,Z_{ij}^{u_k}(t))+\delta b^*(t,Z_{ij}^{u_k}(t))P(t))\delta b(t,Z_{ij}^{u_k}(t)) \bigg ]. \end{aligned}$$

Note that both $\gamma (u)(\delta G(t,Z_{ij}^{u_k}(u))+\delta b^*(t,Z_{ij}^{u_k}(u))P(u))$ and $\delta b(s,Z_{ij}^{u_k}(s))$ are square integrable processes. We shall prove a more general result that for any two processes $f,g \in L^{2}_{\mathbb {F}}(0,T)$, there exists a null subset $\mathcal {T} \subset [0,T]$ such that for $t \in [0,T]/\mathcal {T}$,

$$\begin{aligned} \lim _{r\rightarrow 0+} \frac{1}{r^2}\int _{t-r\beta }^{t+r\alpha }\int _{t-r\beta }^{t+r\alpha } \mathbb {E}[f(u)g(s)I_{\{s \le u\}}]\hbox {d}s\hbox {d}u =\frac{(\alpha +\beta )^2}{2}\mathbb {E}[f(t)g(t) ]. \end{aligned}$$

This result will prove our previous claim immediately. We divide the proof into three steps.

Step 1. We prove that, for$f \in L^{2}_{\mathbb {F}}(0,T)$, there exists a null subset$\mathcal {T} \subset [0,T]$such that, for$t \in [0,T]/\mathcal {T}$,

$$\begin{aligned} \lim _{r\rightarrow 0+} \frac{1}{r}\int _{t-r\beta }^{t+r\alpha } \mathbb {E}[ |f(t)-f(s)|^2]\hbox {d}s =0. \end{aligned}$$

Let $\zeta (t)$ be the standard mollifier (for its properties see Appendix C.4 in [25]), i.e.,

$$\begin{aligned} \zeta (t)=\left\{ \begin{aligned}\exp \left( -\frac{1}{1-t^2}\right) ,&\quad \text {if }\,|t| \le 1;\\ 0,\qquad&\quad \text {otherwise;} \end{aligned} \right. \end{aligned}$$

and $\zeta _n(t)=n\zeta (nt)$. We define $f_n(t):=f \star \zeta _n(t)$ with $\star$ representing the convolution between functions on $\mathbb {R}$. Then, we see that $f_n$ has continuous trajectories almost surely. Moreover, if we define a function g on [0, T] as $g_n(t):=\mathbb {E}[|f_n(t)-f(t)|^2]$, then $\left\| g_n\right\| _{L^1}=\int _0^T g_n(t)\hbox {d}t \longrightarrow 0$ as n goes to infinity. Define

$$\begin{aligned} T_r(f)(t):=\frac{1}{r}\int _{t-r\beta }^{t+r\alpha } \mathbb {E}[ |f(t)-f(s)|^2]\hbox {d}s, \end{aligned}$$

and $T(f)(t)=\limsup _{r \rightarrow 0}T_r(f)(t)$. It is then sufficient to prove that the set $E_{a}:=\{t \in [0,T]| T(f)(t) \ge 2a\}$ is a null set for any $a>0$. One can easily obtain that

$$\begin{aligned} T_r(f)\le & {} C\left( \frac{1}{r}\int _{t-r\beta }^{t+r\alpha }g_n(s)\hbox {d}s+T_r(f_n)(t)+g_n(t)\right) \\\le & {} C(Mg_n(t)+T_r(f_n)(t)+g_n(t)), \end{aligned}$$

for any n, with $Mg_n$ being the Hardy–Littlewood maximal function of $g_n$. Since $f_n$ has continuous paths, we have $\lim _{r \rightarrow 0} T_r(f_n)(t) =0$ for any t. Then,

$$\begin{aligned} T(f)(t) \le C (Mg_n(t)+g_n(t)), \end{aligned}$$

which implies that

$$\begin{aligned} E_a \subset \left\{ Mg_n \ge \frac{a}{2C}\right\} \cup \left\{ g_n \ge \frac{a}{2C}\right\} . \end{aligned}$$

From Hardy–Littlewood maximal inequality (see [26]),

$$\begin{aligned} \left| \left\{ Mg_n \ge \frac{a}{2C}\right\} \right| \le \frac{2C_1 C}{a}\left\| g_n\right\| _{L^1}, \end{aligned}$$

where $C_1$ is the constant from the inequality. Using Chebychev’s inequality, we also obtain

$$\begin{aligned} \left| \left\{ g_n \ge \frac{a}{2C}\right\} \right| \le \frac{2C}{a}\left\| g_n\right\| _{L^1}. \end{aligned}$$

Letting n go to infinity, we show that $|E_a|=0$.

Step 2. There exists a null subset$\mathcal {T} \subset [0,T]$such that, for$t \in [0,T]/\mathcal {T}$,

$$\begin{aligned} \lim _{r\rightarrow 0+} \frac{1}{r^2}\int _{t-r\beta }^{t+r\alpha }\int _{t-r\beta }^{t+r\alpha }\left| \mathbb {E}[f(s)g(u)]-\mathbb {E}[f(t)g(t)]\right| \hbox {d}s\hbox {d}u =0. \end{aligned}$$

Note that

$$\begin{aligned}&\left| \mathbb {E}[f(s)g(u)]-\mathbb {E}[f(t)g(t)]\right| \\&\quad \le \left| \mathbb {E}[(f(s)-f(t))g(u)]\right| +\left| \mathbb {E}[f(t)(g(u)-g(t))]\right| \\&\quad \le \mathbb {E}[|f(s)-f(t)|^2]^{\frac{1}{2}}\mathbb {E}[|g(u)|^2]^{\frac{1}{2}}+\mathbb {E}[|g(u)-g(t)|^2]^{\frac{1}{2}}\mathbb {E}[|f(t)|^2]^{\frac{1}{2}}. \end{aligned}$$

Then,

$$\begin{aligned}&\frac{1}{r^2}\int _{t-r\beta }^{t+r\alpha }\int _{t-r\beta }^{t+r\alpha } \mathbb {E}[|f(s)-f(t)|^2]^{\frac{1}{2}}\mathbb {E}[|g(u)|^2]^{\frac{1}{2}}\hbox {d}s\hbox {d}u\\&\quad =\frac{1}{r}\int _{t-r\beta }^{t+r\alpha } \mathbb {E}[|f(s)-f(t)|^2]^{\frac{1}{2}} \hbox {d}s \frac{1}{r}\int _{t-r\beta }^{t+r\alpha } E[|g(u)|^2]^{\frac{1}{2}} \hbox {d}u. \end{aligned}$$

Using Hölder inequality, we have

$$\begin{aligned} \frac{1}{r}\int _{t-r\beta }^{t+r\alpha } \mathbb {E}[|f(s)-f(t)|^2]^{\frac{1}{2}} \hbox {d}s \le (\alpha +\beta )^{\frac{1}{2}}\left( \frac{1}{r}\int _{t-r\beta }^{t+r\alpha } \mathbb {E}[|f(s)-f(t)|^2]\hbox {d}s\right) ^{\frac{1}{2}}, \end{aligned}$$

and

$$\begin{aligned} \frac{1}{r}\int _{t-r\beta }^{t+r\alpha } \mathbb {E}[|g(u)|^2]^{\frac{1}{2}} \hbox {d}u \le (\alpha +\beta )^{\frac{1}{2}}\left( \frac{1}{r}\int _{t-r\beta }^{t+r\alpha } \mathbb {E}[|g(u)|^2]\hbox {d}u\right) ^{\frac{1}{2}}. \end{aligned}$$

From Step 1, there is a null subset $\mathcal {T}^{\prime }$ such that, for $t \in [0,T]/\mathcal {T}^{\prime }$, $\frac{1}{r}\int _{t-r\beta }^{t+r\alpha } \mathbb {E}[|f(s)-f(t)|^2]^{\frac{1}{2}} \hbox {d}s \longrightarrow 0$ as r goes to 0. Since $\mathbb {E}[|g(u)|^2]$ is Lebesgue integral on [0, T], there is a null set $\mathcal {T}^{\prime \prime } \subset [0,T]$ such that, for $t \in [0,T]/\mathcal {T}^{\prime \prime }$,

$$\begin{aligned} \lim _{r \rightarrow 0+}\frac{1}{r}\int _{t-r\beta }^{t+r\alpha } \mathbb {E}[|g(u)|^2]\hbox {d}u= \mathbb {E}[|g(t)|^2]. \end{aligned}$$

Define $\mathcal {T}_1:=\mathcal {T}^{\prime } \cup \mathcal {T}^{\prime \prime }$. It is a null subset, and for $t \in [0,T]/\mathcal {T}_1$, we have

$$\begin{aligned} \lim _{r \rightarrow 0+} \frac{1}{r^2}\int _{t-r\beta }^{t+r\alpha }\int _{t-r\beta }^{t+r\alpha } \mathbb {E}[|f(s)-f(t)|^2]^{\frac{1}{2}}\mathbb {E}[|g(u)|^2]^{\frac{1}{2}}\hbox {d}s\hbox {d}u=0. \end{aligned}$$

Similarly, there is another null set $\mathcal {T}_2$ such that, for $t \in [0,T]/\mathcal {T}_2$,

$$\begin{aligned} \lim _{r \rightarrow 0+} \frac{1}{r^2}\int _{t-r\beta }^{t+r\alpha }\int _{t-r\beta }^{t+r\alpha } \mathbb {E}[|g(u)-g(t)|^2]^{\frac{1}{2}}\mathbb {E}[|f(t)|^2]^{\frac{1}{2}}\hbox {d}s\hbox {d}u=0. \end{aligned}$$

Let $\mathcal {T}:=\mathcal T_1 \cup \mathcal T_2$. This is the desired subset.

Step 3. Now we shall prove the final result.

Note that

$$\begin{aligned} \frac{(\alpha +\beta )^2}{2}\mathbb {E}[f(t)g(t) ]=\frac{1}{r^2}\int _{t-r\beta }^{t+r\alpha }\int _{t-r\beta }^{t+r\alpha } \mathbb {E}[f(t)g(t) ] I_{\{s \le u\}}\hbox {d}s\hbox {d}u. \end{aligned}$$

Thus,

$$\begin{aligned} \begin{aligned}&\left| \frac{1}{r^2}\int _{t-r\beta }^{t+r\alpha }\int _{t-r\beta }^{t+r\alpha } \mathbb {E}[f(u)g(s)I_{\{s \le u\}}]\hbox {d}s\hbox {d}u -\frac{(\alpha +\beta )^2}{2}\mathbb {E}[f(t)g(t) ]\right| \\&\quad \le \frac{1}{r^2}\int _{t-r\beta }^{t+r\alpha }\int _{t-r\beta }^{t+r\alpha }\left| \mathbb {E}[f(u)g(s)]-\mathbb {E}[f(t)g(t)]\right| I_{\{s \le u\}}\hbox {d}s\hbox {d}u\\&\quad \le \frac{1}{r^2}\int _{t-r\beta }^{t+r\alpha }\int _{t-r\beta }^{t+r\alpha }\left| \mathbb {E}[f(u)g(s)]-\mathbb {E}[f(t)g(t)]\right| \hbox {d}s\hbox {d}u. \end{aligned} \end{aligned}$$

From Step 2, we know that there is a null subset $\mathcal {T}$ such that, for $t \in [0,T]/\mathcal {T}$, the above term tends to 0, which implies that

$$\begin{aligned} \lim _{r\rightarrow 0+} \frac{1}{r^2}\int _{t-r\beta }^{t+r\alpha }\int _{t-r\beta }^{t+r\alpha } \mathbb {E}[f(u)g(s)I_{\{s \le u\}}]\hbox {d}s\hbox {d}u =\frac{(\alpha +\beta )^2}{2}\mathbb {E}[f(t)g(t) ]. \end{aligned}$$

Thus, we finish the proof. $\square$