Appendix
Proof of Lemma 1
Note that the dual value function J(y, m) can be written as follows:
$$\begin{aligned} J(y,m)= & {} \sup _{(M_t)_{t=0}^\infty \;\in {\mathcal {I}}(m)}\left\{ {\mathbb {E}}\left[ \int _0^\infty e^{-\delta t}{\tilde{u}}(y_t,m)dt\right] +{\mathbb {E}}\left[ \int _0^\infty e^{-\delta t}\int _{m}^{M_t}{\partial _m {\tilde{u}}}(y_t,\xi )d\xi dt\right] \right\} \nonumber \\= & {} \; \sup _{(M_t)_{t=0}^\infty \;\in {\mathcal {I}}(m)}\left\{ {\mathbb {E}}\left[ \int _0^\infty e^{-\delta t}\int _{m}^{M_t}{\partial _m {\tilde{u}}}(y_t,\xi )d\xi dt\right] \right\} +{\mathbb {E}}\left[ \int _0^\infty e^{-\delta t}{\tilde{u}}(y_t, m)dt\right] ,\nonumber \\ \end{aligned}$$
(54)
with \(\partial _m{\widetilde{u}}(y,\cdot )\equiv \frac{\partial {\widetilde{u}}}{\partial m}(y,\cdot )\).
Moreover, Fubini’s theorem implies that
$$\begin{aligned}&\sup _{(M_t)_{t=0}^\infty \;\in {\mathcal {I}}(m)}{\mathbb {E}}\left[ \int _0^\infty e^{-\delta t}\int _{m}^{M_t}{\partial _m {\tilde{u}}}(y_t,\xi )d\xi dt\right] \nonumber \\&\quad = \sup _{(M_t)_{t=0}^\infty \;\in {\mathcal {I}}(m)}{\mathbb {E}}\left[ \int _0^\infty e^{-\delta t}\int _{m}^{\infty }{} \mathbf{1}_{\{\xi <M_t\}}{\partial _m{\tilde{u}}}(y_t,\xi )d\xi dt\right] \nonumber \\&\quad = \int _{m}^{\infty }\sup _{ \tau _\xi }{\mathbb {E}}\left[ \int _0^\infty e^{-\delta t}{} \mathbf{1}_{\{t>\tau _\xi \}}{\partial _m{\tilde{u}}}(y_t,\xi )dt\right] d\xi \nonumber \\&\quad = \int _{m}^{\infty }\sup _{\tau _\xi }{\mathbb {E}}\left[ \int _{\tau _\xi }^\infty e^{-\delta t}{\partial _m{\tilde{u}}}(y_t,\xi )dt\right] d\xi . \end{aligned}$$
(55)
From (54) and (55), we have proved the desired result.
Proof of Lemma 2
By Feynman-Kac formula, \(J_0(y,m)\) satisfies the following ordinary differential equation (ODE):
$$\begin{aligned} \begin{aligned} {\mathcal {L}}J_0(y,m)+{\tilde{u}}(y,m)=0, \end{aligned} \end{aligned}$$
(56)
where the differential operator \({\mathcal {L}}\) is given by
$$\begin{aligned} {\mathcal {L}}= \dfrac{\theta ^2}{2}y^2\partial _{yy} + (\delta -r)y\partial _y - \delta . \end{aligned}$$
By applying the variation parameter method to the ODE (56), for any fixed \(m>0\) we obtain the following particular solution \(J_0^p(y,m)\):
$$\begin{aligned} \begin{aligned} J_0^p(y,m)\equiv \dfrac{2}{\theta ^2(n_1-n_2)}\left[ y^{n_2}\int _0^y \eta ^{-n_2-1}{\tilde{u}}(\eta ,m)d\eta +y^{n_1}\int _y^\infty \eta ^{-n_1-1}{\tilde{u}}(\eta ,m)d\eta \right] . \end{aligned} \end{aligned}$$
For \(y\ge u'(\kappa m)\),
$$\begin{aligned} \begin{aligned} \int _0^y \eta ^{-n_2-1}|{\tilde{u}}(\eta ,m)|d\eta =&\int _0^{u'(m)}\eta ^{-n_2-1}|u(m)-m\eta |d\eta \\&\quad +\int _{u'(m)}^{u'(\kappa m)}\eta ^{-n_2-1}|u(I(\eta ))-\eta I(\eta )|d\eta \\&\quad +\int _{u'(\kappa m)}^y\eta ^{-n_2-1}|u(\kappa m)-\kappa m\eta |d\eta . \end{aligned} \end{aligned}$$
(57)
Since \(u(I(y))-yI(y)\) is a continuous function in \(y\in [u'(m),u'(\kappa m)]\), there exits a constant \(C_m>0\) such that
$$\begin{aligned} |u(I(y))-yI(y)|\le C_m \;\;\text{ for }\;\;y\in [u'(m),u'(\kappa m)]. \end{aligned}$$
(58)
Moreover,
$$\begin{aligned} \begin{aligned} \int _0^{u'(m)}\eta ^{-n_2-1}|u(m)-\eta m|d\eta&\le \int _0^{u'(m)}\eta ^{-n_2-1}(|u(m)|+\eta m)d\eta \\&=\dfrac{(u'(m))^{-n_2}}{-n_2}|u(m)|+\dfrac{(u'(m))^{1-n_2}}{1-n_2}m <\infty . \end{aligned} \end{aligned}$$
(59)
By (57), (58), and (59), we deduce that for any fixed \(m>0\)
$$\begin{aligned} \int _0^y \eta ^{-n_2-1}|{\tilde{u}}(\eta ,m)|d\eta < \infty ,\;\;\text{ for } \text{ all }\;\;y>0. \end{aligned}$$
Similarly, we can obtain that for any fixed \(m>0\)
$$\begin{aligned} \int _y^\infty \eta ^{-n_1-1}|{\tilde{u}}(\eta ,m)|d\eta < \infty ,\;\;\text{ for } \text{ all }\;\;y>0, \end{aligned}$$
and thus we conclude that
$$\begin{aligned} \int _0^y \eta ^{-n_2-1}|{\tilde{u}}(\eta ,m)|d\eta +\int _y^\infty \eta ^{-n_1-1}|{\tilde{u}}(\eta ,m)|d\eta <\infty . \end{aligned}$$
It follows from Proposition 1 that
$$\begin{aligned} J_0(y,m) = J_0^p(y,m). \end{aligned}$$
This completes the proof.
Proof of Lemma 3
Since \(\partial _m{\tilde{u}}(y,m)=(u'(m)-y)\mathbf{1}_{\{y\le u'(m)\}}+\kappa (u'(\kappa m)-y)\mathbf{1}_{\{u'(\kappa m) \le y\}}\), it is easy to confirm that for any fixed \(\xi >0\)
$$\begin{aligned} \int _0^y \eta ^{-n_2-1}|\partial _m{\tilde{u}}(\eta ,\xi )|d\eta +\int _y^\infty \eta ^{-n_1-1}|\partial _m{\tilde{u}}(\eta ,\xi )|d\eta <\infty . \end{aligned}$$
It follows from Proposition 1 that
$$\begin{aligned} {\mathcal {Q}}(y,\xi ) = \dfrac{2}{\theta ^2(n_1-n_2)}\left[ y^{n_2}\int _0^y \eta ^{-n_2-1}\partial _m{\tilde{u}}(\eta ,\xi )d\eta +y^{n_1}\int _y^\infty \eta ^{-n_1-1}\partial _m {\tilde{u}}(\eta ,\xi )d\eta \right] , \end{aligned}$$
and for any fixed \(\xi >0\)
$$\begin{aligned} \liminf _{t\rightarrow \infty }e^{-\delta t}{\mathbb {E}}\left[ |{\mathcal {Q}}(y_t,\xi )|\right] =0. \end{aligned}$$
(60)
Moreover, \({\mathcal {Q}}(y,\xi )\) satisfies the following ODE:
$$\begin{aligned} \begin{aligned} {\mathcal {L}}{\mathcal {Q}}(y,\xi ) + \partial _m{\tilde{u}}(y,\xi )=0. \end{aligned} \end{aligned}$$
(61)
According to Proposition 2, we should find the solution \(H(y,\xi )\) to the variational inequality (29) satisfying the following condition:
-
1.
For any fixed \(\xi >0\) and some \(p\in [1,\infty ]\), \(H(\cdot ,\xi )\in W^{2,p}_{\text {loc}}(0,\infty )\).
-
2.
For any fixed \(\xi >0\) and some constants \(C(\xi ),\epsilon >0\),
$$\begin{aligned} |\partial _yH(y,\xi )| \le C(\xi )(y^{\epsilon }+y^{-\epsilon }),\;\;\forall y>0. \end{aligned}$$
-
3.
For any fixed \(\xi >0\), \(\liminf _{t\rightarrow \infty } e^{-\delta t}{\mathbb {E}}\left[ |H(y_t,\xi )|\right] = 0.\)
To analyze the variational inequality (29), we study properties of the following substitution:
$$\begin{aligned} P(y,\xi ) \equiv H(y,\xi )-{\mathcal {Q}}(y,\xi ). \end{aligned}$$
From (61), we transform the variation equality (29) into that of P:
$$\begin{aligned} \begin{aligned} {\left\{ \begin{array}{ll} -{\mathcal {L}}P(y,\xi )\ge -\partial _m{\tilde{u}}(y,\xi ),\;\;&{}\text{ if }\;\;P(y,\xi ) = 0,\\ -{\mathcal {L}}P(y,\xi ) = -\partial _m{\tilde{u}}(y,\xi ),\;\;&{}\text{ if }\;\;P(y,\xi ) > 0.\\ \end{array}\right. } \end{aligned} \end{aligned}$$
(62)
To solve the variational inequality (62), let us consider the following free boundary problem:
$$\begin{aligned} \begin{aligned} {\left\{ \begin{array}{ll} {\mathcal {L}}P(y,\xi ) -\partial _m {\tilde{u}}(y,\xi ) =0,\;\;&{}\text{ if }\;\;y>z(\xi ),\\ P(y,\xi )=0,\;\;&{}\text{ if }\;\;y\le z(\xi ),\\ P(z(\xi ),\xi )=\partial _y P(z(\xi ),\xi )=0. \end{array}\right. } \end{aligned} \end{aligned}$$
(63)
A general solution to the equation can be written as the sum of a general solution to the homogeneous equation and a particular solution: for \(y>z(\xi )\)
$$\begin{aligned} \begin{aligned} P(y,\xi ) =&A_1(\xi )y^{n_1} + A_2(\xi )y^{n_2}\\ -&\dfrac{2}{\theta ^2(n_1-n_2)}\left[ y^{n_2}\int _0^y \eta ^{-n_2-1}\partial _m{\tilde{u}}(\eta ,\xi )d\eta +y^{n_1}\int _y^\infty \eta ^{-n_1-1}\partial _m {\tilde{u}}(\eta ,\xi )d\eta \right] \\ =&A_1(\xi )y^{n_1} + A_2(\xi )y^{n_2}-{\mathcal {Q}}(y,\xi ). \end{aligned} \end{aligned}$$
(64)
Since \(P(y,\xi )\) should satisfy the transeversality condition \(\lim _{t\rightarrow \infty }e^{-\delta t}{\mathbb {E}}\left[ |P(y_t,\xi )|\right] =0\), we can deduce that
$$\begin{aligned} A_1(\xi ) \equiv 0. \end{aligned}$$
Thus, we can rewrite \(P(y,\xi )\) as
$$\begin{aligned} P(y,\xi ) =A(\xi )y^{n_2}-{\mathcal {Q}}(y,\xi ). \end{aligned}$$
The \({C}^1\)-condition at \(y=z(\xi )\) in (63) implies that \(P(z(\xi ),\xi )=\partial _y P(z(\xi ),\xi ) =0,\) and thus
$$\begin{aligned} \begin{aligned} \int _{z(\xi )}^\infty \eta ^{-n_1-1}\partial _m{\tilde{u}}(\eta ,\xi )d\eta =0, \end{aligned} \end{aligned}$$
and
$$\begin{aligned} \begin{aligned} A(\xi )=&\dfrac{2}{\theta ^2(n_1-n_2)}\int _0^{z(\xi )}\eta ^{-n_2-1}\partial _m{\tilde{u}}(\eta ,\xi )d\eta . \end{aligned} \end{aligned}$$
Claim 1
For any fixed \(\xi >0\), there exists a unique \(z(\xi )\) such that
$$\begin{aligned} \int _{z(\xi )}^\infty \eta ^{-n_1-1}\partial _m{\tilde{u}}(\eta ,\xi )d\eta =0. \end{aligned}$$
Moreover, \(0<z(\xi )<u'(\xi )\) for all \(\xi >0\) and \(z(\xi )\) is a strictly decreasing and continuously differentiable function in \(\xi \).
Proof of Claim 1
If \(z(\xi )\in [u'(\xi ),u'(\kappa \xi ))\) for some \(\xi \), then
$$\begin{aligned} \int _{z(\xi )}^\infty \eta ^{-n_1-1}\partial _m{\tilde{u}}(\eta ,\xi )d\eta = \kappa \int _{u'(\kappa \xi )}^\infty \eta ^{-n_1-1}(u'(\kappa \xi )-\eta )d\eta <0. \end{aligned}$$
Similarly, if \(z(\xi )\ge u'(\kappa \xi )\),
$$\begin{aligned} \int _{z(\xi )}^\infty \eta ^{-n_1-1}\partial _m{\tilde{u}}(\eta ,\xi )d\eta= & {} \kappa \int _{z(\xi )}^\infty \eta ^{-n_1-1}(u'(\kappa \xi )-\eta )d\eta \\\le & {} \kappa \int _{z(\xi )}^\infty \eta ^{-n_1-1}(z(\xi )-\eta )d\eta <0. \end{aligned}$$
Thus, it suffice to consider the case when
$$\begin{aligned} 0< z(\xi ) < u'(\xi )\;\;\text{ for } \text{ all }\;\;\xi >0. \end{aligned}$$
Now we will show that for any fixed \(\xi >0\), there exists a unique \(z(\xi )\in (0,u'(\xi ))\) such that
$$\begin{aligned} {\mathcal {W}}(z(\xi ),\xi ) =0, \end{aligned}$$
where \({\mathcal {W}}(y,\xi )\) is defined as
$$\begin{aligned} {\mathcal {W}}(y,\xi ) \equiv \int _{y}^{u'(\xi )}\eta ^{-n_1-1}(u'(\xi ) - \eta ) d\eta + \kappa \int _{u'(\kappa \xi )}^{\infty }\eta ^{-n_1 -1}(u'(\kappa \xi ) - \eta ) d\eta . \end{aligned}$$
For \(y\in (0,u'(\xi ))\), it follows from \(u'(\xi )-y>0\) that \({\mathcal {W}}(y,\xi )\) is a strictly decreasing function in \(y\in (0,u'(\xi ))\).
Moreover,
$$\begin{aligned} \begin{aligned} {\mathcal {W}}(u'(\xi ), \xi )&=-\kappa \dfrac{(u'(\kappa \xi ))^{1-n_1}}{n_1(n_1-1)}<0,\\ {\mathcal {W}}(0,\xi )&=\left[ -\frac{\eta ^{-n_1}}{n_1}u'(\xi )+\frac{\eta ^{-(n_1-1)}}{n_1-1}\right] _{\eta =0}^{u'(\xi )} -\kappa \dfrac{(u'(\kappa \xi ))^{1-n_1}}{n_1(n_1-1)}=+\infty . \end{aligned} \end{aligned}$$
The intermediate value theorem implies that for any fixed \(\xi >0\), there exists a unique \(z(\xi )\in (0,u'(\xi ))\) such that
$$\begin{aligned} {\mathcal {W}}(z(\xi ),\xi )=0. \end{aligned}$$
That is,
$$\begin{aligned} \begin{aligned} u'(\xi )\dfrac{z(\xi )^{-n_1}}{n_1}-\dfrac{z(\xi )^{1-n_1}}{n_1-1}+\dfrac{1}{n_1(n_1-1)}\left( (u'(\xi )^{1-n_1}) -\kappa (u'(\kappa \xi ))^{1-n_1}\right) =0. \end{aligned}\nonumber \\ \end{aligned}$$
(65)
By differentiating the both side of the Eq. (65) with respect to \(\xi \), we obtain that
$$\begin{aligned} \begin{aligned} \dfrac{dz}{d\xi }=\dfrac{u''(\xi )\Big ((z(\xi ))^{-n_1}-(u'(\xi ))^{-n_1}\Big )+\kappa ^2u''(\kappa \xi )\cdot (u'(\kappa \xi ))^{-n_1}}{(z(\xi ))^{-n_1-1}(u'(\xi )-z(\xi ))n_1}. \end{aligned} \end{aligned}$$
It follows from \(0<z(\xi )<u'(\xi )\), \(u''(\xi )<0\), and \(u''(\kappa \xi )<0\) that
$$\begin{aligned} \dfrac{dz}{d\xi }<0. \end{aligned}$$
Hence, we conclude that \(z(\xi )\) is a strictly decreasing and continuously differentiable function of \(\xi \).
Since \(0<z(\xi )<u'(\xi )\),
$$\begin{aligned} \begin{aligned} A(\xi )=&\dfrac{2}{\theta ^2(n_1-n_2)}\int _0^{z(\xi )}\eta ^{-n_2-1}(u'(\xi )-\eta )d\eta >0, \end{aligned} \end{aligned}$$
(66)
and thus we obtain
$$\begin{aligned} A(\xi )=\dfrac{2}{\theta ^2(n_1-n_2)}\left[ -u'(\xi )\dfrac{(z(\xi ))^{-n_2}}{n_2} + \dfrac{(z(\xi ))^{1-n_2}}{n_2-1}\right] . \end{aligned}$$
Claim 2
\(P(y,\xi )\) given by
$$\begin{aligned} \begin{aligned} P(y,\xi )= {\left\{ \begin{array}{ll} A(\xi )y^{n_2}-{\mathcal {Q}}(y,\xi ),\;\;&{}\text{ if }\;\;y>z(\xi ),\\ 0,\;\;&{}\text{ if }\;\;y\le z(\xi ), \end{array}\right. } \end{aligned} \end{aligned}$$
(67)
satisfies the variational inequality (62).
Moreover, \(P(y,\xi )\) is strictly increasing in \(y\in (z(\xi ),\infty ),\) i.e.,
$$\begin{aligned} \frac{\partial P}{\partial y}(y,\xi )>0,~~\text{ for }~~y>z(\xi ). \end{aligned}$$
Proof of Claim 2
For the case \(y>z(\xi )\), Eq. (64) implies that \(P(y,\xi )=A(\xi )y^{n_2}-{\mathcal {Q}}(y,\xi )\) satisfies
$$\begin{aligned} {\mathcal {L}}P(y,\xi ) - \partial _m{\widetilde{u}}(y,\xi )=0. \end{aligned}$$
In the other case \(y\le z(\xi )\), \(P(y,\xi )=0\) implies that \(-{\mathcal {L}}P(y,\xi )=0\), and inequality \(z(\xi )<u'(\xi )\) implies that \(\partial _m{\widetilde{u}}(y,\xi )=u'(\xi )-y>0\). Thus, we can deduce that
$$\begin{aligned} -{\mathcal {L}}P(y,\xi )\ge -\partial _m{\widetilde{u}}(y,\xi ). \end{aligned}$$
So far, we have shown the following results:
$$\begin{aligned} \begin{aligned} {\left\{ \begin{array}{ll} {\mathcal {L}}P(y,\xi ) -\partial _m {\tilde{u}}(y,\xi ) =0,\;\;&{}\text{ if }\;\;y>z(\xi ),\\ {\mathcal {L}}P(y,\xi ) -\partial _m {\tilde{u}}(y,\xi ) \le 0,\;\;&{}\text{ if }\;\;y\le z(\xi ). \end{array}\right. } \end{aligned} \end{aligned}$$
Then, we need to show the following equivalence:
$$\begin{aligned} \begin{aligned}&\{(y,\xi )~|~P(y,\xi )>0\big \}=\big \{(y,\xi )~|~y>z(\xi )\big \},\\&\{(y,\xi )~|~P(y,\xi )=0\big \}=\big \{(y,\xi )~|~y\le z(\xi )\}. \end{aligned} \end{aligned}$$
For \(y>z(\xi )\), \(P(y,\xi )\) can be represented by
$$\begin{aligned} \begin{aligned} P(y,\xi )=&A(\xi )y^{n_2}-{\mathcal {Q}}(y,\xi )\\ =&\frac{2}{\theta ^2(n_1-n_2)}\left[ -y^{n_2}\int _{z(\xi )}^y \eta ^{-n_2-1}\partial _m{\widetilde{u}}(\eta ,\xi )d\eta -y^{n_1}\int _{y}^\infty \eta ^{-n_1-1}\partial _m{\widetilde{u}}(\eta ,\xi )d\eta \right] \\ =&\frac{2}{\theta ^2(n_1-n_2)}\left[ -y^{n_2}\int _{0}^y \eta ^{-n_2-1}\partial _m{\widetilde{u}}(\eta ,\xi )\cdot \mathbf{1}_{\{\eta>z(\xi )\}}d\eta \right. \\&\left. -y^{n_1}\int _{y}^\infty \eta ^{-n_1-1}\partial _m{\widetilde{u}}(\eta ,\xi )\cdot \mathbf{1}_{\{\eta >z(\xi )\}}d\eta \right] . \end{aligned} \end{aligned}$$
Let us temporarily denote
$$\begin{aligned} \begin{aligned} g(\eta ,\xi )&\equiv \partial _m{\widetilde{u}}(\eta ,\xi )\cdot \mathbf{1}_{\{\eta >z(\xi )\}}\\&=\big (u'(\xi )-\eta \big )\mathbf{1}_{\{z(\xi )<\eta \le u'(\xi )\}}+\kappa \big (u'(\kappa \xi )-\eta \big )\mathbf{1}_{\eta \ge u'(\kappa \xi )\}}. \end{aligned} \end{aligned}$$
(68)
Since
$$\begin{aligned} y^{n_2}\dfrac{d}{dy}\left( \int _0^y\eta ^{-n_2-1}g(\eta ,\xi )d\eta \right) +y^{n_1}\dfrac{d}{dy}\left( \int _{y}^\infty \eta ^{-n_1-1}g(\eta ,\xi )d\eta \right) =0, \end{aligned}$$
we deduce that
$$\begin{aligned}&\frac{\partial P}{\partial y} =\frac{2}{\theta ^2(n_1-n_2)}\nonumber \\&\quad \left[ -n_2y^{n_2-1}\int _{0}^y \eta ^{-n_2-1}g(\eta ,\xi )d\eta -n_1y^{n_1}\int _{y}^\infty \eta ^{-n_1-1}g(\eta ,\xi )d\eta \right] . \end{aligned}$$
(69)
It follows from the integration by parts that
$$\begin{aligned} \begin{aligned} \int _{0}^y \eta ^{-n_2-1}g(\eta ,\xi )d\eta =&-\left. \frac{\eta ^{-n_2}}{n_2}g(\eta ,\xi )\right| _0^y+\int _0^y \frac{\eta ^{-n_2}}{n_2} \frac{\partial g}{\partial \eta }(\eta ,\xi )d\eta \\ =&-\frac{1}{n_2}\left( y^{-n_2}g(y,\xi )-\int _0^y \eta ^{-n_2}\frac{\partial g}{\partial \eta }(\eta ,\xi )d\eta \right) , \end{aligned} \end{aligned}$$
(70)
where \(\frac{\partial g}{\partial \eta }(\eta ,\xi )\) is given by
$$\begin{aligned} \frac{\partial g}{\partial \eta }(\eta ,\xi )= -\mathbf{1}_{\{z(\xi )<\eta \le u'(\xi )\}}-\kappa \cdot \mathbf{1}_{\{\eta \ge u'(\kappa \xi )\}}. \end{aligned}$$
(71)
Similarly, we can deduce
$$\begin{aligned} \begin{aligned} \int _{y}^\infty \eta ^{-n_1-1}g(\eta ,\xi )d\eta =&\frac{1}{n_1}\left( y^{-n_1}g(y,\xi )+\int _y^\infty \eta ^{-n_1}\frac{\partial g}{\partial \eta }(\eta ,\xi )d\eta \right) . \end{aligned} \end{aligned}$$
(72)
From Eqs. (70) and (72), we rewrite the Eq. (69) as
$$\begin{aligned} \frac{\partial P}{\partial y}=-\frac{2}{\theta ^2(n_1-n_2)}\left[ y^{n_2-1}\int _0^y \eta ^{-n_2}\frac{\partial g}{\partial \eta }(\eta ,\xi )d\eta +y^{n_1-1}\int _y^\infty \eta ^{-n_1}\frac{\partial g}{\partial \eta }(\eta ,\xi )d\eta \right] . \end{aligned}$$
Hence, we deduce that
$$\begin{aligned} \frac{\partial P}{\partial y}= {\left\{ \begin{array}{ll} \frac{2}{\theta ^2(n_1-n_2)}\left\{ \frac{1}{1-n_2}+y^{n_1-1}\left( \int _y^{u'(\xi )} \eta ^{-n_1}d\eta +\int _{u'(\kappa \xi )}^{\infty } \kappa \cdot \eta ^{-n_1} d\eta \right) \right\} ,~~&{}\text{ if }~z(\xi )<y<u'(\xi ),\\ \frac{2}{\theta ^2(n_1-n_2)}\left\{ \frac{1}{1-n_2}\left( \frac{y}{u'(\xi )}\right) ^{n_2-1}-\frac{\kappa }{1-n_1}\left( \frac{y}{u'(\kappa \xi )}\right) ^{n_1-1}\right\} ,~~&{}\text{ if }~u'(\xi )\le y < u'(\kappa \xi ),\\ \frac{2}{\theta ^2(n_1-n_2)}\left\{ y^{n_2-1}\left( \int _0^{u'(\xi )} \eta ^{-n_2}d\eta +\int _{u'(\xi )}^y \eta ^{-n_2}d\eta \right) -\frac{\kappa }{1-n_1}\right\} ,~~&{}\text{ if }~y\ge u'(\kappa \xi ). \end{array}\right. } \end{aligned}$$
and thus we conclude that \(\frac{\partial P}{\partial y}>0\) for \(y>z(\xi )\). It follows from \(P(z(\xi ), \xi )=0\) that
$$\begin{aligned} P(y,\xi )>0~~\text{ for }~~y>z(\xi ). \end{aligned}$$
In terms of \(z(\xi )\), the following relationship holds:
$$\begin{aligned} \begin{aligned}&\{(y,\xi )~|~P(y,\xi )>0\big \}=\big \{(y,\xi )~|~y>z(\xi )\big \},\\&\{(y,\xi )~|~P(y,\xi )=0\big \}=\big \{(y,\xi )~|~y\le z(\xi )\}. \end{aligned} \end{aligned}$$
By claims 1 and 2 , It follows from \(H(y,\xi )=P(y,\xi )+{\mathcal {Q}}(y,\xi )\) that
$$\begin{aligned} H(y,\xi ) =\left\{ \begin{aligned}&A(\xi ) y^{n_2}, \;&\text{ if }\;\;y >z(\xi ),\\&\dfrac{2}{\theta ^2(n_1-n_2)}\left[ y^{n_2}\int _0^y \eta ^{-n_2-1}\partial _m{\tilde{u}}(\eta ,\xi )d\eta +y^{n_1}\int _y^\infty \eta ^{-n_1-1}\partial _m {\tilde{u}}(\eta ,\xi )d\eta \right] ,&\text{ if }\;\;y\le z(\xi ), \end{aligned} \right. \end{aligned}$$
(73)
satisfies the variational inequality (29).
Claim 3
The \(H(y,\xi )\) given in (73) satisfies the following conditions 1,2 and 3.
-
1.
For any fixed \(\xi >0\) and some \(p\in [1,\infty ]\), \(H(\cdot ,\xi )\in W^{2,p}_{\text {loc}}(0,\infty )\).
-
2.
For any fixed \(\xi >0\) and some constants \(C(\xi ),\epsilon >0\),
$$\begin{aligned} |\partial _yH(y,\xi )| \le C(\xi )(y^{\epsilon }+y^{-\epsilon }),\;\;\forall y>0. \end{aligned}$$
-
3.
For any fixed \(\xi >0\), \(\liminf _{t\rightarrow \infty } e^{-\delta t}{\mathbb {E}}\left[ |H(y_t,\xi )|\right] = 0.\)
Thus, \(H(t,\xi )\) in (73) is a solution of Problem 4.
Proof of Claim 3
We will show the \(H(y,\xi )\) in (73) satisfies each conditions 1,2, and 3.
(Condition 1): From the definition of free boundary (63) and the fact that \(\partial _m {\widetilde{u}}(\cdot ,\xi )\) is differentiable almost everywhere, we can deduce that the first derivative \(\frac{\partial H}{\partial y}(y,\xi )\) of the representation (73) is differentiable almost everywhere. Morever, since \(\partial _m {\widetilde{u}}(\cdot ,\xi )\) is monotonically decreasing, we can easily deduce that the \(H(y,\xi )\) in (73) is in \(W^{2,\infty }_{\text {loc}}(0,\infty )\), i.e., \(H(\cdot ,\xi )\in W^{2,\infty }_{\text {loc}}(0,\infty )\).
(Condition 2): Consider \(\partial _y H\) derived by
$$\begin{aligned} \begin{aligned} \frac{\partial H}{\partial y}(y,\xi ) = {\left\{ \begin{array}{ll} n_2 A(\xi )y^{n_2-1}, &{}\text{ for }\;\;y >z(\xi ),\\ \displaystyle \dfrac{2}{\theta ^2(n_1-n_2)}\left[ n_2y^{n_2-1}\int _0^y \eta ^{-n_2-1}\partial _m{\tilde{u}}(\eta ,\xi )d\eta +n_1y^{n_1}\int _y^\infty \eta ^{-n_1-1}\partial _m {\tilde{u}}(\eta ,\xi )d\eta \right] , &{}\text{ for }\;\;y\le z(\xi ). \end{array}\right. } \end{aligned} \end{aligned}$$
For \(y>z(\xi )\), we can easily find a coefficient \(C_1(\xi )>0\) satisfying
$$\begin{aligned} \left| \frac{\partial H}{\partial y}(y,\xi )\right| =\left| n_2 A(\xi ) y^{n_2-1}\right| \le C_1(\xi )(y^{n_2-1}+y^{-(n_2-1)}). \end{aligned}$$
(74)
In the case of \(y\le z(\xi )\), we obtain the following inequalities:
$$\begin{aligned} \begin{aligned} \left| n_2y^{n_2-1}\int _0^y \eta ^{-n_2-1}\partial _m{\tilde{u}}(\eta ,\xi )d\eta \right|&\le \left| n_2y^{n_2-1}\right| \int _0^y \eta ^{-n_2-1}|\partial _m{\widetilde{u}}(\eta ,\xi )|d\eta \\&\le \left| n_2y^{n_2-1}\right| \int _0^y \eta ^{-n_2-1}(\eta +u'(\xi ))d\eta \\&\le C_2(\xi )(y^{-1}+y) \end{aligned} \end{aligned}$$
(75)
where \(C_2(\xi )>0\) and we have used the fact that \(|\partial _m{\widetilde{u}}(\eta ,\xi )|\le (\eta +u'(\xi ))\).
Since, we can easily obtain
$$\begin{aligned} \begin{aligned} \left| n_1y^{n_1-1}\int _y^\infty \eta ^{-n_1-1}\partial _m{\tilde{u}}(\eta ,\xi )d\eta \right|&\le C_3(\xi )(y^{-1}+y). \end{aligned} \end{aligned}$$
(76)
It follows from (75) and (76) that for \(y\le z(\xi )\)
$$\begin{aligned} \left| \frac{\partial H}{\partial y}(y,\xi )\right| \le D(\xi )\cdot (y^{-1}+y), \end{aligned}$$
(77)
where \(D(\xi )\) is a constant only depending on \(\xi \).
From (74) and (77), we can conclude that \(H(y,\xi )\) in (73) satisfies the condition 2.
(Condition 3): Since
$$\begin{aligned} (y_t)^{n_2}<z(\xi )^{n_2}\;\;\text{ for }\;\; y_t>z(\xi ),\;\;H(y_t,\xi )={\mathcal {Q}}(y_t,\xi )\;\;\text{ for }\;\; y_t \le z(\xi ), \end{aligned}$$
and
$$\begin{aligned} |H(y_t,\xi )|\le A(\xi )(y_t)^{n_2 }+ |{\mathcal {Q}}(y_t,\xi )|, \end{aligned}$$
it follows from \(A(\xi )>0\) and (60) that
$$\begin{aligned} \begin{aligned} \liminf _{t\rightarrow \infty } e^{-\delta t}{\mathbb {E}}\left[ |H(y_t,\xi )|\right]&\le \liminf _{t\rightarrow \infty }\left( e^{-\delta t}A(\xi )(z(\xi ))^{n_2}+e^{-\delta t}{\mathbb {E}}\left[ |{\mathcal {Q}}(y_t,\xi )|\right] \right) =0. \end{aligned} \end{aligned}$$
By Proposition 2, we conclude that the \(H(y,\xi )\) in (73) is a solution to the optimal stopping problem (27). Moreover, the optimal stopping time \(\tau _\xi \) to the problem (27) is given by
$$\begin{aligned} \tau _\xi =\inf \{t\ge 0 \mid H(y_t,\xi )={\mathcal {Q}}(y_t,\xi )\}= \inf \{t\ge 0 \mid y_t \le z(\xi )\}. \end{aligned}$$
Since \(z(\xi )\) is a strictly decreasing and continuous function in \(\xi \), we deduce that \(\tau _\xi \) is a increasing and continuous function in \(\xi \).
By Claims 1,2 and 3, we have proved the desired results.
Proof of Lemma 4
Since we have shown that \(z(\cdot )\) is strictly decreasing and continuously differentiable in \({\mathbb {R}}^+\), we can easily deduce the injectivity of the free boundary \(z(\cdot )\) in \({\mathbb {R}}^+\).
Since \(z(\xi )<u'(\xi )\) for all \(\xi >0\), we obtain that
$$\begin{aligned} 0\le \lim _{\xi \rightarrow \infty }z(\xi ) \le \lim _{\xi \rightarrow \infty }u'(\xi )=0 \end{aligned}$$
and thus
$$\begin{aligned} \lim _{\xi \rightarrow \infty }z(\xi )=0. \end{aligned}$$
This completes the proof.
Proof of Proposition 3
Proof of (a). It follow from Lemma 3, the free boundary \(z(\xi )\) is strictly decreasing and continuously differentiable function of \(\xi \) and the optimal stopping \(\tau _\xi \) for each \(\xi >0\) is given
$$\begin{aligned} \tau _\xi = \inf \{t\ge 0 \mid y_t \le z(\xi )\} \end{aligned}$$
(78)
Thus, we deduce that for \(\xi \ge M_{0-}\),
$$\begin{aligned} \left\{ \tau _\xi \le t \right\} =\left\{ \inf _{0\le s \le t} y_s \le z(\xi )\right\} =\left\{ I_z\left( \inf _{0\le s \le t} y_s\right) \ge \xi \right\} =\left\{ \sup _{0\le s \le t} I_z(y_s)\ge \xi \right\} . \end{aligned}$$
By the correspondence \(\left\{ \tau _\xi \le t\right\} =\left\{ M_t\ge \xi \right\} \) in Lemma 1, we obtain that for given \(y>0\) the optimal maximum process \(M_t^*(y)\) is
$$\begin{aligned} {M_t^*(y)\equiv I_z\left( \min \left\{ z(M_{0-}), \inf _{0\le s \le t} y_s \right\} \right) ,~~~~\forall ~t\ge 0}. \end{aligned}$$
Proof of (b). Since
$$\begin{aligned} H(y,\xi ) =\left\{ \begin{aligned}&A(\xi ) y^{n_2}, \;&\text{ if }\;\;y >z(\xi ),\\&{\mathcal {Q}}(y,\xi ),&\text{ if }\;\;y\le z(\xi ). \end{aligned} \right. \end{aligned}$$
(79)
It follows from the closed forms \(J_0(y,m)\) in Lemma 2 and \({\mathcal {Q}}(y,\xi )\) in Lemma 3 that
$$\begin{aligned} {\mathcal {Q}}(y,m) = \partial _m J_0 (y,m). \end{aligned}$$
(80)
From the closed form of \(H(y,\xi )\) in (73), \(J(y,M_{0-})\) is given by
$$\begin{aligned} J(y,M_{0-})&=\int _{M_{0-}}^{\infty } H(y,\xi ) d\xi +J_0(y,M_{0-})\nonumber \\&= {\left\{ \begin{array}{ll} \displaystyle \int _{M_{0-}}^\infty A(\xi )y^{n_2}d\xi + J_0(y,M_{0-}),\;\;&{}\text{ for }\;\;M_{0-} > I_z(y),\\ \displaystyle \int _{I_z(y)}^\infty A(\xi )y^{n_2}d\xi +\int _{M_{0-}}^{I_z(y)} {\mathcal {Q}}(y,\xi )d\xi + J_0(y,M_{0-}),\;\;&{}\text{ for }\;\;M_{0-} \le I_z(y). \\ \end{array}\right. } \end{aligned}$$
(81)
In the case of \(M_{0-} \le I_z(y)\), the relationship (80) implies that
$$\begin{aligned} \begin{aligned} \int _{M_{0-}}^{I_z(y)} {\mathcal {Q}}(y,\xi )d\xi + J_0(y,M_{0-}) = J_0 (y, I_z(y)). \end{aligned} \end{aligned}$$
(82)
By (81), (82), and the decomposition in (22), we obtain the semi-closed form solution.
Proof of (c).
-
(i)
\(J(y,M_{0-})\) is a \(C^{1,2}\)-function. Note that
$$\begin{aligned} \frac{\partial J}{\partial m}(y,M_{0-})=&-H(y,M_{0-})+{\partial _m J_0}(y,M_{0-})\\ =&\left\{ \begin{aligned}&-A(M_{0-})y^{n_2}+{\mathcal {Q}}(y,M_{0-})\le 0,~~&\text{ if }~~M_{0-} > I_z(y),\\&-{\mathcal {Q}}(y,M_{0-})+{\mathcal {Q}}(y,M_{0-})=0,~~&\text{ if }~~M_{0-} \le I_z(y). \end{aligned} \right. \end{aligned}$$
We easily deduce that \(A(M_{0-})y^{n_2}\) and \({\mathcal {Q}}(y,M_{0-})\) are \(C^{1,1}\) functions for \(M_{0-}>I_z(y)\). Since
$$\begin{aligned} -A(I_z(y))y^{n_2}+{\mathcal {Q}}(y,I_z(y))=0, \end{aligned}$$
we obtain that \(\partial _m J\) is a \(C^{1,1}\) function. This implies that \(J(y,M_{0-})\) is a \(C^{1,2}\)-function.
-
(ii)
\(\frac{\partial ^2 J}{\partial y^2}\) is continuous. In the case of \(M_{0-} > I_z(y)\), we easily obtain that
$$\begin{aligned} \begin{aligned} \frac{\partial ^2 J}{\partial y^2}(y,M_{0-})&=\frac{\partial ^2 J_0}{\partial y^2}(y,M_{0-})+n_2(n_2-1) y^{n_2-2}\int _{M_{0-}}^\infty A(\xi ) d\xi . \end{aligned} \end{aligned}$$
(83)
Also, in the case of \(M_{0-} \le I_z(y) \), we deduce that
$$\begin{aligned} \frac{\partial J_0}{\partial m}(y,I_z(y))&=\frac{2}{\theta ^2(n_1-n_2)}\left[ y^{n_2}\int _0^y \eta ^{-n_2-1}\partial _m {\widetilde{u}}(\eta ,I_z(y))d\eta +y^{n_1}\int _y^\infty \eta ^{-n_1-1}\partial _m {\widetilde{u}}(\eta ,I_z(y))d\eta \right] \nonumber \\&=H(y,I_z(y)), \end{aligned}$$
(84)
and
$$\begin{aligned} \begin{aligned} \frac{\partial J_1}{\partial m}(y,I_z(y))&=-y^{n_2} A(I_z(y))=-H(y,I_z(y)), \end{aligned} \end{aligned}$$
(85)
where we have used the fact that \(I_z(\cdot )\) is the inverse function of the free boundary \(z(\cdot )\). From Eqs. (84) and (85), we have
$$\begin{aligned} \begin{aligned} \frac{\partial J}{\partial y}(y,M_{0-})=&\frac{\partial J_0}{\partial y}(y,I_z(y))+\frac{\partial J_0}{\partial m}(y,I_z(y))\cdot I_z'(y) \\&\quad +n_2y^{n_2-1} \int _{I_z(y)}^{\infty }A(\xi )d\xi -y^{n_2} A(I_z(y))\cdot I_z'(y)\\ =&\frac{\partial J_0}{\partial y}(y,I_z(y))+n_2y^{n_2-1} \int _{I_z(y)}^{\infty }A(\xi )d\xi . \end{aligned} \end{aligned}$$
Similarly, we can deduce
$$\begin{aligned} \begin{aligned} \frac{\partial ^2 J_0}{\partial y\partial m}(y,I_z(y))=&\frac{2}{\theta ^2(n_1-n_2)}\left[ n_2y^{n_2-1}\int _0^y \eta ^{-n_2-1}\partial _m {\widetilde{u}}(\eta ,I_z(y))d\eta \right. \\ +&\left. n_1y^{n_1-1}\int _y^\infty \eta ^{-n_1-1}\partial _m {\widetilde{u}}(\eta ,I_z(y))d\eta \right] \\ =&\frac{\partial H}{\partial y}(y,I_z(y)), \end{aligned} \end{aligned}$$
(86)
and
$$\begin{aligned} \begin{aligned} \frac{\partial H}{\partial y}(y,I_z(y))=n_2y^{n_2-1} A(I_z(y)). \end{aligned} \end{aligned}$$
(87)
It follows from the Eqs. (86) and (87) that for \(M_{0-} \le I_z(y) \),
$$\begin{aligned} \begin{aligned} \frac{\partial ^2 J}{\partial y^2}(y,M_{0-})=&\frac{\partial ^2 J_0}{\partial y^2}(y,I_z(y))+\frac{\partial ^2 J_0}{\partial y \partial m}(y,I_z(y))\cdot I_z'(y)\\ +&n_2(n_2-1)y^{n_2-2}\int _{I_z(y)}^\infty A(\xi )d\xi -n_2y^{n_2-1} A(I_z(y)) \cdot I'_z(y)\\ =&\frac{\partial ^2 J_0}{\partial y^2}(y,I_z(y))+n_2(n_2-1)y^{n_2-2}\int _{I_z(y)}^\infty A(\xi )d\xi . \end{aligned} \end{aligned}$$
(88)
From Eqs. (83) and (88), we easily confirm that \(\frac{\partial ^2 J}{\partial y^2}\) is continuous, i.e.,
$$\begin{aligned} \frac{\partial ^2 J}{\partial y^2}(y,M_{0-})=\left\{ \begin{aligned}&\frac{\partial ^2 J_0}{\partial y^2}(y,M_{0-})+n_2(n_2-1) y^{n_2-2}\int _{M_{0-}}^\infty A(\xi ) d\xi ,~~~~\text{ if }~M_{0-} > I_z(y),\\&\frac{\partial ^2 J_0}{\partial y^2}(y,I_z(y))+n_2(n_2-1)y^{n_2-2}\int _{I_z(y)}^\infty A(\xi )d\xi ,~~~~\text{ if }~M_{0-}\le I_z(y). \end{aligned} \right. \end{aligned}$$
(89)
By (i) and (ii), we conclude tat \(J(y,M_{0-})\) is a \(C^2\)-function.
Proof of Theorem 1
We will prove the duality relationship in (40) in the following steps.
(Step 1) The dual value function \(J(y,M_{0-})\) is strictly convex in \(y>0\).
Proof of Step 1. Since \({\widetilde{u}}(y,M_{0-})\) is convex in \(y>0\) and
$$\begin{aligned} J_0(y,M_{0-}) = {\mathbb {E}}\left[ \int _0^\infty e^{-\delta t}{\widetilde{u}}(y_t,M_{0-})dt\right] , \end{aligned}$$
we deduce that \(J_0(y,M_{0-})\) is convex in \(y>0\). It follows from (66) and (89) that \(A(\xi )\) is positive for all \(\xi >0\) and thus we conclude that \(J(y,M_{0-})\) is strictly convex in \(y>0\), i.e.,
$$\begin{aligned} \partial _{yy}J(y,M_{0-})>0. \end{aligned}$$
(Step 2) For given \(M_{0-}>0\) and \(w>\kappa M_{0-}/r\), there exists unique \(y^*\) such that
$$\begin{aligned} w=-\frac{\partial J}{\partial y}(y^*,M_{0-}). \end{aligned}$$
Proof of Step 2. Since \(J(y,M_{0-})\) is strictly convex in \(y>0\), \(-\frac{\partial J}{\partial y}(y,M_{0-})\) is strictly decreasing in y.
Note that
$$\begin{aligned} -\frac{\partial J}{\partial y}(y,M_{0-})=\left\{ \begin{aligned}&-\frac{\partial J_0}{\partial y}(y,M_{0-})-n_2y^{n_2-1}\int _{M_{0-}}^\infty A(\xi )d\xi ,~~~&\text{ if }~~(y,M_{0-})\in \mathbf{NR},\\&-\frac{\partial J_0}{\partial y}(y,I_z(y))-n_2y^{n_2-1}\int _{I_z(y)}^\infty A(\xi )d\xi ,~~~&\text{ if }~~(y,M_{0-})\in \mathbf{IR}. \end{aligned} \right. \end{aligned}$$
For a sufficiently large y such that \(y\ge u'(\kappa M_{0-})\), we deduce that
$$\begin{aligned} -\frac{\partial J}{\partial y}(y,M_{0-})=&\frac{2}{\theta ^2(n_1-n_2)}\left[ -y^{n_2-1}\int _0^{u'(\kappa M_{0-})}\eta ^{-n_2}\frac{\partial {\widetilde{u}}}{\partial \eta }(\eta ,M_{0-})d\eta \right. \nonumber \\ +&\left. y^{n_2-1}\int _{u'(\kappa M_{0-})}^y\eta ^{-n_2}\cdot \kappa M_{0-}d\eta +y^{n_1-1}\int _{y}^\infty \eta ^{-n_1}\cdot \kappa M_{0-}d\eta \right] -n_2y^{n_2-1}\int _{M_{0-}}^\infty A(\xi )d\xi \nonumber \\ =&-y^{n_2-1}\left[ \frac{2}{\theta ^2(n_1-n_2)}\left( \int _0^{u'(\kappa M_{0-})}\eta ^{-n_2}\frac{\partial {\widetilde{u}}}{\partial \eta }(\eta ,M_{0-})d\eta +\frac{(u'(\kappa M_{0-}))^{1-n_2}}{1-n_2}\right) \right] \nonumber \\ -&n_2y^{n_2-1}\int _{M_{0-}}^\infty A(\xi )d\xi +\frac{\kappa M_{0-}}{r}, \end{aligned}$$
(90)
where we have used that \(\frac{\partial {\widetilde{u}}}{\partial \eta }(\eta ,M_{0-})=-\kappa M_{0-}\) for \(y\gg u'(\kappa M_{0-})\).
Thus, we obtain
$$\begin{aligned} \lim _{y\rightarrow +\infty }\left[ -\frac{\partial J}{\partial y}(y,M_{0-})\right] =\frac{\kappa M_{0-}}{r} \end{aligned}$$
For a sufficiently small y such that \(y<z(M_{0-})\),
$$\begin{aligned} \begin{aligned} -\frac{\partial J}{\partial y}(y,M_{0-})=&\frac{2}{\theta ^2(n_1-n_2)}\left[ y^{n_2-1}\int _0^{y}\eta ^{-n_2}\cdot I_z(y)d\eta +y^{n_1-1}\int _{y}^{u'(I_z(y))}\eta ^{-n_2}\cdot I_z(y)d\eta \right. \\&\left. -y^{n_1-1}\int _{u'(I_z(y))}^\infty \eta ^{-n_1}\frac{\partial {\widetilde{u}}}{\partial \eta }(\eta ,I_z(y))d\eta \right] \\&\quad -n_2y^{n_2-1}\int _{I_z(y)}^\infty A(\xi )d\xi \\ =&\frac{2}{\theta ^2(n_1-n_2)}\left[ \left( -\frac{1}{n_2-1}+\frac{1}{n_1-1}-\frac{1}{n_1-1}\left( \frac{y}{u'(I_z(y))}\right) ^{n_1-1}\right) \cdot I_z(y)\right. \\&\left. -y^{n_1-1}\int _{u'(I_z(y))}^\infty \eta ^{-n_1}\frac{\partial {\widetilde{u}}}{\partial \eta }(\eta ,I_z(y))d\eta \right] -n_2y^{n_2-1}\int _{I_z(y)}^\infty A(\xi )d\xi . \end{aligned} \end{aligned}$$
(91)
Since \(z(\xi )<u'(\xi )\) for \(\xi >0\), we can deduce that \(y<u'(I_z(y))\). This leads to
$$\begin{aligned} \left( \dfrac{y}{u'(I_z(y))}\right) ^{n_1-1}<1. \end{aligned}$$
Moreover, we know that \(\frac{\partial {\widetilde{u}}}{\partial \eta }(\eta ,I_z(y))<0\) for \(\eta >0\), and that \(A(\xi )>0\) for \(\xi >0\).
To sum up, we can deduce that
$$\begin{aligned} -\frac{\partial J}{\partial y}(y,M_{0-})>-\frac{2}{\theta ^2(n_1-n_2)}\cdot \frac{I_z(y)}{(n_2-1)}. \end{aligned}$$
From Lemma 4, we know that \(\lim _{y\rightarrow 0^+}I_z(y)=+\infty \), and this implies that
$$\begin{aligned} \lim _{y\rightarrow 0^+}\left[ -\frac{\partial J}{\partial y}(y,M_{0-})\right] =+\infty . \end{aligned}$$
Thus, we obtain that for given \(M_{0-}>0\) and \(w>\kappa M_{0-}/r\), there exists unique \(y^*\) such that
$$\begin{aligned} w=-\frac{\partial J}{\partial y}(y^*,M_{0-}). \end{aligned}$$
(Step 3) The candidate of the optimal consumption process \((c^*(y_t^*))_{t=0}^\infty \) given by
$$\begin{aligned} c^*(y_t^*)\equiv {\left\{ \begin{array}{ll} M^*(y_t^*),~~~&{}\text{ if }~~~y_t^* \le u'(M^*(y_t^*)),\\ I(y_t^*),~~~&{}\text{ if }~~~u'(M^*(y_t^*))<y_t^* < u'(\kappa M^*(y_t^*)),\\ \kappa M^*(y_t^*),~~~&{}\text{ if }~~~y_t^* \ge u'(\kappa M^*(y_t^*)) \end{array}\right. } \;\;\text{ with }\;\; y_t^*=y^* e^{\delta t}H_t \end{aligned}$$
satisfies the static budget constraint (13) with equality.
Proof of Step 3. Let us denote h(y, m) by
$$\begin{aligned} h(y,m)\equiv&-y\dfrac{\partial {\tilde{u}}}{\partial y}(y,m) \end{aligned}$$
(92)
$$\begin{aligned} =&y\left( m\cdot \mathbf{1}_{\{\eta \le u'(m)\}}+ I(y)\cdot \mathbf{1}_{\{u'(m)< y < u'(\kappa m)\}}+ \kappa m\cdot \mathbf{1}_{\{y\ge u'(m)\}}\right) \end{aligned}$$
(93)
Note that
$$\begin{aligned} h(y^*_t,M_t^*)=y_t^* c^*(y_t^*)\quad \text{ and }\quad \frac{\partial h}{\partial m}(y^*_t, M_t^*)=y_t^* \mathbf{1}_{\{y_t^* < u' (M_t^*)\}}+\kappa y_t^* \mathbf{1}_{\{y_t^* > u' (\kappa M_t^*)\}} \end{aligned}$$
with \(M^*_t=M^*_t(y^*)\).
Then, we have
$$\begin{aligned}&y^* {\mathbb {E}}\left[ \int _0^\infty H_t c^*(y_t^*) dt\right] \nonumber \\&\quad = {\mathbb {E}}\left[ \int _0^\infty e^{-\delta t} h(y^*_t,M_t^*) dt\right] \nonumber \\&\quad = {\mathbb {E}}\left[ \int _0^\infty e^{-\delta t}h(y_t^*, M_{0-})dt\right] + {\mathbb {E}}\left[ \int _0^\infty e^{-\delta t} \left( \int _{M_{0-}}^{M_t^*}\frac{\partial h}{\partial \xi }(y_t^*,\xi )d\xi \right) dt\right] ,\nonumber \\ \end{aligned}$$
(94)
As similar to the proof of Lemma 2, it is not difficult to show that
$$\begin{aligned} \int _0^y \eta ^{-n_2-1}|h(\eta ,m)|d\eta +\int _y^\infty \eta ^{-n_1-1}|h(\eta ,m)|d\eta <\infty . \end{aligned}$$
It follows from Proposition 1 and the integration by parts formula that we obtain
$$\begin{aligned}&{\mathbb {E}}\left[ \int _0^\infty e^{-\delta t}h(y_t^*, M_{0-})dt\right] \nonumber \\&\quad = -{\mathbb {E}}\left[ \int _0^\infty e^{-\delta t}y_t^*\frac{\partial {\widetilde{u}}}{\partial y}(y_t^*,M_{0-})dt\right] \nonumber \\&\quad = -\frac{2}{\theta ^2(n_1-n_2)} \left[ (y^*)^{n_2}\int _0^{y^*} \eta ^{-n_2}\frac{\partial {\widetilde{u}}}{\partial \eta }(\eta ,M_{0-})d\eta +(y^*)^{n_1}\int _{y^*}^\infty \eta ^{-n_1}\frac{\partial {\widetilde{u}}}{\partial \eta }(\eta ,M_{0-})d\eta \right] \nonumber \\&\quad = -\frac{2}{\theta ^2(n_1-n_2)} \left[ n_2(y^*)^{n_2}\int _0^{y^*} \eta ^{-n_2}{\widetilde{u}}(\eta ,M_{0-})d\eta +n_1(y^*)^{n_1}\int _{y^*}^\infty \eta ^{-n_1}{\widetilde{u}}(\eta ,M_{0-})d\eta \right] \nonumber \\&\quad = -y^* \frac{\partial J_0}{\partial y}(y^*,M_{0-}). \end{aligned}$$
(95)
It follows from the one-to-one correspondence between the process \((M_t^*)_{t=0}^\infty \) and the stopping time \((\tau _\xi )_{\xi >M_{0-}}\) developed in Sect. 3.2 that we can transform the second expectation in the last equation of (94) as
$$\begin{aligned} \begin{aligned} {\mathbb {E}}\left[ \int _0^\infty e^{-\delta t} \left( \int _{M_{0-}}^{M_t^*}\frac{\partial h}{\partial \xi }(y_t^*,\xi )d\xi \right) dt\right]&={\mathbb {E}}\left[ \int _0^\infty e^{-\delta t} \left( \int _{M_{0-}}^{\infty }\frac{\partial h}{\partial \xi }(y_t^*,\xi )\cdot \mathbf{1}_{\{\xi <M_t^*\}}d\xi \right) dt\right] \\&=\int _{M_{0-}}^{\infty }{\mathbb {E}}\left[ \int _0^\infty e^{-\delta t} \frac{\partial h}{\partial \xi }(y_t^*,\xi )\cdot \mathbf{1}_{\{t >\tau _\xi \}} dt\right] d\xi \\&=\int _{M_{0-}}^{\infty }{\mathbb {E}}\left[ e^{-\delta \tau _\xi }{\mathbb {E}}_{\tau _\xi }\left[ \int _{\tau _\xi }^\infty e^{-\delta (s-\tau _\xi )} \frac{\partial h}{\partial \xi }(y_s^*,\xi )ds \right] \right] d\xi . \end{aligned} \end{aligned}$$
Let us temporarily denote \({\widetilde{H}}(y^*,\xi )\) by
$$\begin{aligned} \begin{aligned} {\widetilde{H}}(y^*,\xi )&={\mathbb {E}}_{\tau _\xi }\left[ \int _{\tau _\xi }^\infty e^{-\delta (s-\tau _\xi )} \frac{\partial h}{\partial \xi }(y_s^*,\xi )ds \right] \\&={\mathbb {E}}\left[ e^{-\delta \tau _\xi } {\widetilde{Q}}\big (y_{\tau _\xi }^*,\xi \big )\right] , \end{aligned} \end{aligned}$$
where
$$\begin{aligned} {\widetilde{Q}}(y^*_t,\xi )\equiv {\mathbb {E}}_{t}\left[ \int _{t}^\infty e^{-\delta (s-t)} \frac{\partial h}{\partial \xi }(y_s^*,\xi )ds \right] . \end{aligned}$$
For a fixed \(\xi >0\), it is easy to show
$$\begin{aligned} \int _0^y \eta ^{-n_2-1}\left| \frac{\partial h}{\partial \xi }(\eta ,\xi )\right| d\eta +\int _y^\infty \eta ^{-n_1-1}\left| \frac{\partial h}{\partial \xi }(\eta ,\xi )\right| d\eta <\infty . \end{aligned}$$
It follows from Proposition 1 that we deduce that
$$\begin{aligned} {\widetilde{Q}}(y,\xi )= \frac{2}{\theta ^2(n_1-n_2)} \left[ y^{n_2}\int _0^{y} \eta ^{-n_2-1}\frac{\partial h}{\partial \xi }(\eta ,\xi )d\eta +y^{n_1}\int _{y}^\infty \eta ^{-n_1-1}\frac{\partial h}{\partial \xi }(\eta ,\xi )d\eta \right] . \end{aligned}$$
Since
$$\begin{aligned} \frac{\partial h}{\partial \xi } (\eta , \xi )= \frac{\partial }{\partial \xi }\left( -\eta \frac{\partial {\widetilde{u}}}{\partial \eta }(\eta ,\xi )\right) , \end{aligned}$$
it follows from (95) that
$$\begin{aligned} {\widetilde{Q}}(y,\xi )&= \frac{2}{\theta ^2(n_1-n_2)} \left[ y^{n_2}\int _0^{y} \eta ^{-n_2-1}\frac{\partial h}{\partial \xi }(\eta ,\xi )d\eta +y^{n_1}\int _{y}^\infty \eta ^{-n_1-1}\frac{\partial h}{\partial \xi }(\eta ,\xi )d\eta \right] \nonumber \\&=\dfrac{\partial }{\partial \xi }\left( -\frac{2}{\theta ^2(n_1-n_2)} \left[ n_2y^{n_2}\int _0^{y} \eta ^{-n_2-1}{\widetilde{u}}(\eta ,\xi )d\eta +n_1y^{n_1}\int _{y}^\infty \eta ^{-n_1-1}{\widetilde{u}}(\eta ,\xi )d\eta \right] \right) \\&=\dfrac{\partial }{\partial \xi }\left( -y\frac{\partial J_0}{\partial y}(y,\xi )\right) . \nonumber \end{aligned}$$
(96)
Since the stopping time \(\tau _\xi \) is characterized as
$$\begin{aligned} \tau _\xi = \inf \{t\ge 0 \mid y_t \le z(\xi )\} \end{aligned}$$
from Lemma 3, we deduce that \({\widetilde{H}}(y, \xi )\) satisfies the following ODE:
$$\begin{aligned} \begin{aligned} {\left\{ \begin{array}{ll} {\mathcal {L}}{\widetilde{H}}(y,\xi )= 0,\;\;&{}\text{ if }\;\;y>z(\xi ),\\ {\widetilde{H}}(y,\xi ) = {\widetilde{Q}}(y,\xi ),\;\;&{}\text{ if }\;\;y\le z(\xi ).\\ \end{array}\right. } \end{aligned} \end{aligned}$$
where we have characterized the boundary \(z(\xi )\) in (31).
As similar in the proof of Lemma 3, we can easily obtain the solution of \({\widetilde{H}}(y,\xi )\), i.e.,
$$\begin{aligned} {\widetilde{H}}(y,\xi ) = \left\{ \begin{aligned}&-n_2 A(\xi ) y^{n_2}\quad&\text{ if }\quad y>z(\xi ), \\&{\widetilde{Q}}(y,\xi )\quad&\text{ if }\quad y\le z(\xi ), \end{aligned} \right. \end{aligned}$$
(97)
where we have used the fact that \({\widetilde{H}}(y,\xi )\) is continuous at \(y=z(\xi )\) and the corresponding algebraic Eq. (31). Recall that \(A(\xi )\) is given by
$$\begin{aligned} A(\xi )=\dfrac{2}{\theta ^2(n_1-n_2)}\left[ -u'(\xi )\dfrac{(z(\xi ))^{-n_2}}{n_2} + \dfrac{(z(\xi ))^{1-n_2}}{n_2-1}\right] . \end{aligned}$$
It follows from (94), (95), (96), and (97) that
$$\begin{aligned} \begin{aligned}&y^* {\mathbb {E}}\left[ \int _0^\infty H_t c^*(y_t^*) dt\right] \\ =&\left\{ \begin{aligned}&-n_2(y^*)^{n_2} \int _{M_{0-}}^{\infty } A(\xi )d\xi -y^*\frac{\partial J_0}{\partial y }(y^*,M_{0-}) \quad&\text{ if }\quad z(M_{0-})<I_z(y^*) \\&-n_2(y^*)^{n_2}\int _{I_z(y^*)}^{\infty }A(\xi )d\xi -y^*\frac{\partial J_0}{\partial y }(y^*,I_z(y^*))&\text{ if }\quad z(M_{0-})\ge y^*. \end{aligned} \right. \\ =&-y^*\frac{\partial J}{\partial y}(y^*,M_{0-})\\ =&y^* w \end{aligned} \end{aligned}$$
(98)
where we have used the fact that \(y^*>0\) is a unique solution such that
$$\begin{aligned} w=-y^*\frac{\partial J}{\partial y}(y^*,M_{0-}). \end{aligned}$$
Thus, \((c^*(y_t^*))_{t=0}^\infty \) satisfies the static budget constraint (13) with equality, i.e.,
$$\begin{aligned} {\mathbb {E}}\left[ \int _0^\infty H_t c^*(y_t^*) dt\right] =w. \end{aligned}$$
(99)
(Step 4) The duality relationship (40) is established. Moreover, the candidate consumption \((c^*(y_t^*))_{t=0}^\infty \) is optimal.
Proof of Step 4. Since
$$\begin{aligned} w= {\mathbb {E}}\left[ \int _0^\infty H_t c^*(y_t^*) dt\right] = {\mathbb {E}}^{{\mathbb {Q}}}\left[ \int _0^\infty e^{-rt}c^*(y_t^*) dt\right] <\infty , \end{aligned}$$
we deduce that for given any \(T>0\),
$$\begin{aligned} \int _0^T c^*(y_t^*) dt < \infty \;\;\text{ a.s. } \end{aligned}$$
Thus, it follows from Theorem 9.4 in Karatzas and Shreve (1998) that there exists a portfolio process \((\pi ^*(y_t^*))_{t=0}^\infty \) such that \((c^*(y_t^*), \pi ^*(y_t^*))_{t=0}^\infty \) is admissible at x. Moreover, the corresponding wealth process \(W_t^{c^*,\pi ^*}\) is
$$\begin{aligned} W_t^{c^*,\pi ^*}=\dfrac{1}{H_t}{\mathbb {E}}_t\left[ \int _t^\infty H_s c^{*}(y_s^*)ds\right] \end{aligned}$$
(100)
and the dynamics of \(W_t^{c^*,\pi ^*}\) follows
$$\begin{aligned} dW_t^{c^*,\pi ^*} =[rW_t^{c^*(y_t^*),\pi ^*(y_t^*)} + (\mu -r)\pi ^*(y_t^*) - c^*(y_t^*)] dt + \sigma \pi ^*(y_t^*) dB_t. \end{aligned}$$
(101)
Recall the following weak duality stated in (39)
$$\begin{aligned} V(w,M_{0-})\le \inf _{y>0} \Big (J(y,M_{0-})+yw\Big ). \end{aligned}$$
It follows from
$$\begin{aligned} {\mathbb {E}}\left[ \int _0^\infty H_t c^*(y_t^*) dt\right] =w\;\;\text{ and }\;\;{\widetilde{u}}(y_t^*,M^*(y_t^*))=u(c^*(y_t^*))-y_t^*c^*(y_t^*), \end{aligned}$$
that
$$\begin{aligned} \begin{aligned} {\mathbb {E}}\left[ \int _0^\infty e^{-\delta t} u(c^*(y_t^*))dt\right] =&{\mathbb {E}}\left[ \int _0^\infty e^{-\delta t}{\tilde{u}}(y_t^*,M^*(y_t^*))dt\right] +{\mathbb {E}}\left[ \int _0^\infty e^{-\delta t}y^*_t c^*(y_t^*) dt\right] \\ =&J(y^*,M_{0-})+y^*w\\ =&\inf _{y>0} \Big (J(y,M_{0-})+yw\Big )\\ \ge&V(w,M_{0-}). \end{aligned} \end{aligned}$$
Since \((c^*(y_t^*), \pi ^*(y^*))_{t=0}^\infty \) is admissible, we deduce that
$$\begin{aligned} V(w,M_{0-})={\mathbb {E}}\left[ \int _0^\infty e^{-\delta t} u(c^*(y_t^*))dt\right] =\inf _{y>0} \Big (J(y,M_{0-})+yw\Big ). \end{aligned}$$
(102)
This completes the proof.
Proof of Theorem 2
Let us temporarily denote \({\mathcal {W}}(y)\) by
$$\begin{aligned} {\mathcal {W}}(y) \equiv -\dfrac{\partial J}{\partial y}(y, M^*(y))={\mathbb {E}}\left[ \int _0^\infty H_t c^*(y_t) dt\right] . \end{aligned}$$
Note that
$$\begin{aligned} {\mathcal {W}}(y^*) =w. \end{aligned}$$
Then,
$$\begin{aligned} y{\mathcal {W}}(y) = -y\dfrac{\partial J}{\partial y}(y, M^*(y))={\mathbb {E}}\left[ \int _0^\infty e^{-\delta t} y_t c^*(y_t) dt\right] . \end{aligned}$$
The Markov property implies that
$$\begin{aligned} \begin{aligned} y_t^*{\mathcal {W}}(y_t^*) =&{\mathbb {E}}_t\left[ \int _t^\infty e^{-\delta (s-t)}y_s c^*(y_s^*) ds\right] . \end{aligned} \end{aligned}$$
It follows from (100) that
$$\begin{aligned} {\mathcal {W}}(y_t^*) = \dfrac{1}{H_t}{\mathbb {E}}_t\left[ \int _t^\infty H_sc^*(y_s^*) ds\right] = W_t^{c^*,\pi ^*}. \end{aligned}$$
That is, \(W_t^{c^*,\pi ^*}={\mathcal {W}}(y_t^*)\) is the optimal wealth process and \({\mathcal {W}}(y_t^*)\) is given by
$$\begin{aligned} \begin{aligned} {\mathcal {W}}(y_t^*)=&-\frac{\partial J}{\partial y}(y_t^*,M^*(y_t^*))=-\frac{\partial J_0}{\partial y}(y_t^*,M_t^*)-n_2(y_t^*)^{n_2-1}\int _{M_t^*}^\infty A(\xi ) d\xi . \end{aligned}\nonumber \\ \end{aligned}$$
(103)
Note that \(M_t^*=M^*(y_t^*)\).
By applying the generalized Itô’s Lemma (see Harrison 1985) to \({\mathcal {W}}(y_t^*)\), we deduce
$$\begin{aligned} d{\mathcal {W}}(y_t^*)&=-\frac{\partial ^2 J}{\partial y^2}(y_t^*,M_t^*)dy_t^*-\frac{1}{2}\frac{\partial ^3 J}{\partial y^3}(y_t^*,M_t^*)(dy_t^*)^2 \nonumber \\&\quad -\frac{\partial ^2 J}{\partial y \partial m}(y_t^*,M_t^*)dM_t^*. \end{aligned}$$
(104)
It follows from (b) in Proposition 3 and (80) that
$$\begin{aligned} \frac{\partial ^2 J}{\partial y \partial m}(y_t^*,M_t^*)=&-\dfrac{\partial H}{\partial y}H(y_t^*, M_t^*)+\dfrac{\partial {\mathcal {Q}}}{\partial y}(y_t^*,M_t^*).\\ =&\left\{ \begin{aligned}&-A(M^*(y_t^*))(y^*_t)^{n_2}+{\mathcal {Q}}(y_t^*,M^*(y_t^*))\le 0,~~&\text{ if }~~y_t^*>z(M_t^*),\nonumber \\&-{\mathcal {Q}}(y^*_t,M^*(y_t^*))+{\mathcal {Q}}(y^*_t,M^*(y_t^*))=0,~~&\text{ if }~~y_t^*=z(M_t^*), \end{aligned} \right. \end{aligned}$$
(105)
Since \(H_y(y_t^*,M_t^*)={\mathcal {Q}}_y(y_t^*, M_t^*)\) only when \(y_t^*=z(M_t^*)\) or \(M_t^*\) increases, we deduce that
$$\begin{aligned} \frac{\partial ^2 J}{\partial y \partial m}(y_t^*,M_t^*) dM_t^*=0 \;\;\text{ for } \text{ all }\;\;t \ge 0. \end{aligned}$$
(106)
Note that for all \(y\ge z(m)\),
$$\begin{aligned} {\mathcal {L}}J(y,m)+{\widetilde{u}}(y,m)=0 \end{aligned}$$
By differentiating the both sides of the above equation with respect to y, we can easily obtain
$$\begin{aligned} \begin{aligned}&\frac{\theta ^2}{2}(y_t^*)^2\frac{\partial ^3 J}{\partial y^3}(y_t^*,M_t^*)+(\delta -r)y_t^*\frac{\partial ^2 J}{\partial y^2}(y_t^*,M_t^*)\\ =&r\frac{\partial J}{\partial y}(y_t^*,M_t^*)+ c_t^*-\theta ^2 y_t^*\frac{\partial J}{\partial y}(y_t^*, M_t^*) \end{aligned} \end{aligned}$$
(107)
It follows from (104), (105), (106), and (107) that we have
$$\begin{aligned} \begin{aligned} d{\mathcal {W}}(y_t^*)&=-\frac{\partial ^2 J}{\partial y^2}(y_t^*,M_t^*)dy_t^*-\frac{1}{2}\frac{\partial ^3 J}{\partial y^3}(y_t^*,M_t^*)(dy_t^*)^2\\&=\left[ r{\mathcal {W}}(y_t^*) -c_t^*+\theta ^2y_t^*\frac{\partial ^2 J}{\partial y^2}(y_t^*,M_t^*)\right] dt+\theta y_t^*\frac{\partial ^2 J}{\partial y^2}(y_t^*,M_t^*)dB_t. \end{aligned} \end{aligned}$$
(108)
By comparing the dynamics of \(e^{-rt}W_t^{c^*,\pi ^*}\) in (101) with \(e^{-rt}{\mathcal {W}}(y_t^*)\) in (108), we can obtain the optimal portfolio process \(\pi _t^*=\pi ^*(y_t^*)\) given by
$$\begin{aligned} \begin{aligned} \pi _t^*=&\frac{\theta }{\sigma }y_t^*\frac{\partial ^2 J}{\partial y^2}(y_t^*,M_t^*)\\ =&\frac{\theta }{\sigma }\left( y_t^*\frac{\partial ^2 J_0}{\partial y^2}(y_t^*,M_t^*)+n_2(n_2-1)(y_t^*)^{n_2-1}\int _{M_t^*}^\infty A(\xi ) d\xi \right) . \end{aligned} \end{aligned}$$
This completes the proof.
