Optimal Investment in an Illiquid Market with Search Frictions and Transaction Costs

Abstract

We consider an optimal investment problem to maximize expected utility of the terminal wealth, in an illiquid market with search frictions and transaction costs. In the market model, an investor’s attempt of transaction is successful only at arrival times of a Poisson process, and the investor pays proportional transaction costs when the transaction is successful. We characterize the no-trade region describing the optimal trading strategy. Our asymptotic analysis implies that the effects of the transaction costs are more pronounced (more widening effect of the no-trade region and more diminishing effect of the value function) in the market with less search frictions.

Appendices

Appendix A. Proof of Lemma 3.1

Since the parabolic type PDE (3.2) is not uniformly elliptic, we change variable as $x=h(z):=\frac{e^z}{1+e^{z}}$ and consider the PDE for v(t, h(z)).

To handle the nonlinear term in the PDE, we first consider the following map $\phi$ from $C_b([0,T]\times {{\mathbb {R}}})$ (equipped with the uniform norm) to itself:

$$\begin{aligned} \phi (f)(t,z):=\int _t^T e^{-\lambda (s-t)} {{\mathbb {E}}}\left[ K_f(s,Z_s^{(t,z)}) \right] ds, \end{aligned}$$

(A.1)

where for $(s,z)\in [t,T]\times {{\mathbb {R}}}$,

$$\begin{aligned} \begin{aligned} Z_s^{(t,z)}&:=z+ \left( \mu -r-\tfrac{\sigma ^2}{2} \right) (s-t) + \sigma (B_s-B_t),\\ K_f(s,z)&:= (\mu -r)h(z)+r-\tfrac{1}{2}\sigma ^2 h(z)^2 \\&\qquad + \lambda \sup _{\zeta \in {{\mathbb {R}}}} \left( f(s,\zeta )- \ln \left( \tfrac{1+\bar{\epsilon }h(\zeta )}{1+\bar{\epsilon }h(z)} \right) 1_{\{z<\zeta \}} -\ln \left( \tfrac{1-\underline{\epsilon }h(\zeta )}{1-\underline{\epsilon }h(z)} \right) 1_{\{z>\zeta \}} \right) . \end{aligned} \end{aligned}$$

(A.2)

We observe that for $f,g \in C_b([0,T]\times {{\mathbb {R}}})$,

$$\begin{aligned} \begin{aligned} \Vert \phi (f)-\phi (g) \Vert _\infty&\le \lambda \int _t^T e^{-\lambda (s-t)} \left( \sup _{\zeta \in {{\mathbb {R}}}} \left| f(s,\zeta )-g(s,\zeta ) \right| \right) ds\\&\le \left( 1- e^{-\lambda T} \right) \Vert f-g \Vert _\infty , \end{aligned} \end{aligned}$$

(A.3)

where the first inequality is due to the triangular inequality for supremum. Therefore, the map $\phi$ in (A.1) is a contraction map and there exists a unique ${\hat{u}}\in C_b([0,T]\times {{\mathbb {R}}})$ such that $\phi ({\hat{u}})={\hat{u}}$, by the Banach fixed point theorem.

Claim: For all $\delta \in (0,1)$, $K_{\hat{u}}\in C^{\frac{\delta }{2},\delta }([0,T]\times {{\mathbb {R}}})$.

(Proof of Claim): We first check that $K_{{\hat{u}}}$ is bounded. Since $0<h(z)<1$, we observe that

$$\begin{aligned} |K_{{\hat{u}}}(t,z) | \le |\mu |+2|r| +\tfrac{\sigma ^2}{2}+ \lambda \left( \Vert {\hat{u}} \Vert _\infty + \ln \left( \tfrac{1+\bar{\epsilon }}{1-\underline{\epsilon }} \right) \right) , \end{aligned}$$

(A.4)

i.e., $\Vert K_{{\hat{u}}}\Vert _\infty <\infty$. Therefore, to prove the claim, it is enough to check that $K_{{\hat{u}}}$ is uniformly Lipschitz with respect to z variable and $\tfrac{1}{2}$-Hölder continuous with respect to t variable. For $\Delta >0$, we obtain the following inequalities:

$$\begin{aligned} \begin{aligned}&|K_{{\hat{u}}}(t,z+\Delta )- K_{{\hat{u}}}(t,z) | \\&\quad \le (|\mu -r|+\sigma ^2)\Delta + \lambda \sup _{\zeta \in {{\mathbb {R}}}} \left| 1_{\{z+\Delta<\zeta \}}\cdot \ln \left( \tfrac{1+\bar{\epsilon }h(z+\Delta )}{1+\bar{\epsilon }h(z)}\right) \right. \\ {}&\qquad \left. + 1_{\{z<\zeta \le z+\Delta \}} \cdot \ln \left( \tfrac{1+\bar{\epsilon }h(\zeta )}{1+\bar{\epsilon }h(z)} \right) \right| \\&\quad \quad + \lambda \sup _{\zeta \in {{\mathbb {R}}}} \left| 1_{\{z+\Delta >\zeta \}} \cdot \ln \left( \tfrac{1-\underline{\epsilon }h(z+\Delta )}{1-\underline{\epsilon }h(z)}\right) - 1_{\{z\le \zeta < z+\Delta \}} \cdot \ln \left( \tfrac{1-\underline{\epsilon }h(\zeta )}{1-\underline{\epsilon }h(z)} \right) \right| \\&\quad \le \left( |\mu -r| +\sigma ^2 + \lambda \left( \bar{\epsilon }+ \tfrac{\underline{\epsilon }}{1-\underline{\epsilon }} \right) \right) \Delta , \end{aligned} \end{aligned}$$

(A.5)

where we used the mean value theorem and the bounds $0<h<1$ and $0<h'<1$. We can treat $\Delta <0$ by the same way, and conclude that $K_{\hat{u}}$ is uniformly Lipschitz with respect to z.

For $\Delta >0$, using (A.5) and the mean value theorem, we observe that

$$\begin{aligned} \begin{aligned}&{{\mathbb {E}}}\left| e^{\lambda \Delta } K_{{\hat{u}}}(s,Z_s^{(t+\Delta ,z)})-K_{{\hat{u}}}(s,Z_s^{(t,z)}) \right| \\&\quad \le \lambda e^{\lambda T} \Vert K_{{\hat{u}}}\Vert _\infty \Delta + {{\mathbb {E}}}\left| K_{{\hat{u}}}(s,Z_s^{(t+\Delta ,z)})-K_{{\hat{u}}}(s,Z_s^{(t,z)}) \right| \\&\quad \le \lambda e^{\lambda T}\Vert K_{{\hat{u}}}\Vert _\infty \Delta + \left( |\mu -r| +\sigma ^2 + \lambda \left( \bar{\epsilon }+ \tfrac{\underline{\epsilon }}{1-\underline{\epsilon }} \right) \right) {{\mathbb {E}}}\left| Z_s^{(t+\Delta ,z)}-Z_s^{(t,z)} \right| \\&\quad \le \lambda e^{\lambda T}\Vert K_{{\hat{u}}}\Vert _\infty \Delta + \left( |\mu -r| +\sigma ^2 + \lambda \left( \bar{\epsilon }+ \tfrac{\underline{\epsilon }}{1-\underline{\epsilon }} \right) \right) \\ {}&\qquad \left( \left| \mu -r-\tfrac{\sigma ^2}{2}\right| \Delta + \sigma \sqrt{\tfrac{2}{\pi }} \, \Delta ^{\frac{1}{2}} \right) , \end{aligned} \end{aligned}$$

(A.6)

where the last inequality is due to

$$\begin{aligned} {{\mathbb {E}}}\left| Z_s^{(t+\Delta ,z)}-Z_s^{(t,z)} \right|\le & {} \left| \mu -r-\tfrac{\sigma ^2}{2}\right| \Delta + \sigma \, {{\mathbb {E}}}\left| B_{t+\Delta }-B_t \right| \\ {}= & {} \left| \mu -r-\tfrac{\sigma ^2}{2}\right| \Delta + \sigma \sqrt{\tfrac{2}{\pi }} \, \Delta ^{\frac{1}{2}}. \end{aligned}$$

Using ${\hat{u}}= \phi ({\hat{u}})$ and triangular inequality, we obtain

$$\begin{aligned}&|K_{{\hat{u}}}(t+\Delta ,z)- K_{{\hat{u}}}(t,z) |\le \lambda \sup _{\zeta \in R} \left| {\hat{u}}(t+\Delta ,\zeta )-{\hat{u}}(t,\zeta ) \right| \\&\quad =\lambda \sup _{\zeta \in {{\mathbb {R}}}} \left| \int _{t+\Delta }^T e^{-\lambda (s-t-\Delta )} {{\mathbb {E}}}\left[ K_{{\hat{u}}}(s,Z_s^{(t+ \Delta ,\zeta )}) \right] ds \right. \\&\qquad \left. -\int _{t}^T e^{-\lambda (s-t)} {{\mathbb {E}}}\left[ K_{{\hat{u}}}(s,Z_s^{(t,\zeta )}) \right] ds \right| \\&\quad \le \lambda \sup _{\zeta \in {{\mathbb {R}}}} \bigg ( \int _{t+\Delta }^T e^{-\lambda (s-t)} {{\mathbb {E}}}\left| e^{\lambda \Delta } K_{{\hat{u}}}(s,Z_s^{(t+ \Delta ,\zeta )})-K_{{\hat{u}}}(s,Z_s^{(t,\zeta )}) \right| ds \\&\qquad +\int _t^{t+\Delta } e^{-\lambda (s-t)} {{\mathbb {E}}}\left| K_{{\hat{u}}}(s,Z_s^{(t,\zeta )}) \right| ds \bigg ) \le C \Delta ^{\frac{1}{2}}, \end{aligned}$$

where the generic constant C only depends on the market parameters and $\Vert {\hat{u}}\Vert _{\infty }$ and the last inequality is due to (A.4) and (A.6). We can treat $\Delta <0$ by the same way, and conclude that $K_{{\hat{u}}}$ is $\tfrac{1}{2}$-Hölder continuous with respect to t variable. (End of the proof of Claim).

Let $\delta \in (0,1)$ be fixed. Then, the above claim and Theorem 9.2.3 in [35] ensure that there exists a unique function $u\in C^{1+\frac{\delta }{2},2+\delta }((0,T)\times {{\mathbb {R}}})$ such that it satisfies the following PDE on $(t,z)\in (0,T)\times {{\mathbb {R}}}$.

$$\begin{aligned} \begin{aligned} {\left\{ \begin{array}{ll} 0=u(T,z),\\ 0= u_t + (\mu -r-\frac{\sigma ^2}{2} )u_z+\frac{1}{2} \sigma ^2 u_{zz} - \lambda u + K_{{\hat{u}}} \end{array}\right. } \end{aligned} \end{aligned}$$

(A.7)

Since u admits a unique continuous extension on $[0,T]\times {{\mathbb {R}}}$ (i.e., see chapter 8.5 in [35]), we let $u\in C^{1+\frac{\delta }{2},2+\delta }([0,T]\times {{\mathbb {R}}})$. By the Feynman-Kac formula (i.e., see Theorem 5.7.6 in [32]), the solution u of the parabolic PDE (A.7) has the stochastic representation $u=\phi ({\hat{u}})$, where $\phi$ is defined in (A.1). Since ${\hat{u}}$ is chosen as the unique fixed point of the map $\phi$, we conclude that $u={\hat{u}}$.

Our next task is to define $u(t,\pm \infty )$ for $t\in [0,T]$. Using $\lim _{z\rightarrow \infty }h(Z_s^{(t,z)})=1$ and $\lim _{z\rightarrow -\infty }h(Z_s^{(t,z)})=0$ almost surely, we obtain

$$\begin{aligned} \begin{aligned} {\left\{ \begin{array}{ll} \lim _{z\rightarrow \infty } K_u(s,Z_s^{(t,z)}) = \mu -\tfrac{1}{2}\sigma ^2 + \lambda \sup _{\zeta \in {{\mathbb {R}}}} \left( u(s,\zeta ) -\ln \left( \frac{1-\underline{\epsilon }h(\zeta )}{1-\underline{\epsilon }} \right) \right) \\ \lim _{z\rightarrow -\infty } K_u(s,Z_s^{(t,z)}) = r+ \lambda \sup _{\zeta \in {{\mathbb {R}}}} \left( u(s,\zeta )- \ln \left( 1+\bar{\epsilon }h(\zeta ) \right) \right) \end{array}\right. } a.s. \end{aligned}\nonumber \\ \end{aligned}$$

(A.8)

The above convergence and $\Vert K_u\Vert _\infty <\infty$ enable us to apply the dominated convergence theorem:

$$\begin{aligned} \begin{aligned} \lim _{z\rightarrow \pm \infty } u(t,z)&=\lim _{z\rightarrow \pm \infty } \phi (u)(t,z)=\lim _{z\rightarrow \pm \infty } \int _t^T e^{-\lambda (s-t)} {{\mathbb {E}}}\left[ K_u(s,Z_s^{(t,z)}) \right] ds\\&= {\left\{ \begin{array}{ll} \int _t^T e^{-\lambda (s-t)} \left( \mu -\tfrac{1}{2}\sigma ^2 \right. \\ \quad \left. + \lambda \sup _{\zeta \in {{\mathbb {R}}}} \left( u(s,\zeta ) -\ln \left( \frac{1-\underline{\epsilon }h(\zeta )}{1-\underline{\epsilon }} \right) \right) \right) ds, &{}\hbox { for}\ z\rightarrow \infty \\ \int _t^T e^{-\lambda (s-t)} \left( r+ \lambda \sup _{\zeta \in {{\mathbb {R}}}} \left( u(s,\zeta ) \right. \right. \\ \quad \left. \left. - \ln \left( 1+\bar{\epsilon }h(\zeta ) \right) \right) \right) ds, &{}\hbox { for}\ z\rightarrow -\infty \ \end{array}\right. }. \end{aligned} \end{aligned}$$

Therefore, we can continuously extend u to $z=\pm \infty$ and $u(t,\infty )$ and $u(t,-\infty )$ are defined by the above limit. We observe that for $z\in \{ \infty , -\infty \}$, u(t, z) satisfies

$$\begin{aligned} \begin{aligned} 0&=u_t(t,z) + (\mu -r)h(z)+r-\tfrac{1}{2}\sigma ^2 h(z)^2 - \lambda u(t,z) \\&\qquad + \lambda \sup _{\zeta \in {{\mathbb {R}}}} \left( u(t,\zeta )- \ln \left( \tfrac{1+\bar{\epsilon }h(\zeta )}{1+\bar{\epsilon }h(z)} \right) 1_{\{z<\zeta \}} -\ln \left( \tfrac{1-\underline{\epsilon }h(\zeta )}{1-\underline{\epsilon }h(z)} \right) 1_{\{z>\zeta \}} \right) , \end{aligned} \end{aligned}$$

(A.9)

where the function h is continuously extended as $h(\infty ):=1$ and $h(-\infty ):=0$.

Now we define v as $v(t,x):=u(t,h^{-1}(x))$ for $(t,x)\in [0,T]\times [0,1]$. Such v is well-defined because $h:{{\mathbb {R}}}\cup \{ \infty , -\infty \} \rightarrow [0,1]$ is bijective. We observe that for $(t,x)\in (0,T)\times (0,1)$ and $z=h^{-1}(x)$,

$$\begin{aligned} \begin{aligned} v_t(t,x)&=u_t(t,z),\\ x(1-x)v_x(t,x)&=u_z(t,z),\\ x^2(1-x)^2 v_{xx}(t,x)&=u_{zz}(t,z)-(1-2x)u_z(t,z). \end{aligned} \end{aligned}$$

(A.10)

The PDE for u in (A.7) with ${\hat{u}}$ replaced by u and the equalities in (A.10) produce the PDE for v, which is (3.2). Therefore, statement (i) is valid.

To check statement (ii), we observe that $v(t,0)=u(t,-\infty )$ and $v(t,1)=u(t,\infty )$. Then, the continuous differentiability of v(t, 0) and v(t, 1) with respect to t is followed by that of $u(t,-\infty )$ and $u(t,\infty )$, and (A.9) produces (3.3).

Finally, statement (iii) is a direct consequence of (A.10) and $u\in C^{1+\frac{\delta }{2},2+\delta }([0,T]\times {{\mathbb {R}}})$.

Appendix B. Proof of Lemma 5.3

(i) When $\epsilon =0$, Lemma 4.1 and (5.1) produce (5.2).

To prove (5.3), we first check that $Y_s^{(t,x)}$ in (4.9) safisfies

$$\begin{aligned} dY_s^{(t,x)}=Y_s^{(t,x)} (1-Y_s^{(t,x)}) \left( (\mu -r-\sigma ^2 Y_s^{(t,x)})ds + \sigma dB_s \right) . \end{aligned}$$

(B.1)

Then application of Ito’s formula produces that for $(s,x)\in [t,T) \times (0,1)$,

$$\begin{aligned} \left( Y_s^{(t,x)}-x\right) ^2&= \int _t^s Y_u^{(t,x)} (1-Y_u^{(t,x)})\left( 2(Y_u^{(t,x)}-x )( \mu -r-\sigma ^2 Y_u^{(t,x)}) \right. \\ {}&\qquad \left. + \sigma ^2 Y_u^{(t,x)} (1-Y_u^{(t,x)}) \right) du \\&\qquad + \int _t^s 2\sigma (Y_u^{(t,x)}-x ) Y_u^{(t,x)} (1-Y_u^{(t,x)}) dB_u. \end{aligned}$$

Since $0<Y_s^{(t,x)}<1$, the local martingale part above is a true martingale and we obtain

$$\begin{aligned}&\tfrac{\partial }{\partial s} \left( {{\mathbb {E}}}\left[ \left( Y_s^{(t,x)}-x\right) ^2 \right] \right) \nonumber \\ {}&\quad ={{\mathbb {E}}}\left[ Y_s^{(t,x)} (1-Y_s^{(t,x)})\left( 2(Y_s^{(t,x)}-x )( \mu -r-\sigma ^2 Y_s^{(t,x)})\right. \right. \nonumber \\ {}&\quad \left. \left. + \sigma ^2 Y_s^{(t,x)} (1-Y_s^{(t,x)}) \right) \right] =-x^2(1-x)^2 \nonumber \\ {}&\quad \cdot {{\mathbb {E}}}\left[ \tfrac{\partial }{\partial x}\left( \tfrac{Y_s^{(t,x)} \left( 1-Y_s^{(t,x)}\right) \left( \mu -r-\sigma ^2 Y_s^{(t,x)}\right) }{x(1-x)} \right) \right] , \end{aligned}$$

(B.2)

where the second equality is from elementary computations using the definition of $Y_s^{(t,x)}$ in (4.9).

For $x\in (0,1)$, we observe that

$$\begin{aligned} \tfrac{Y_s^{(t,x)}-x}{x(1-x)} = \tfrac{A^{(t,s)}-1}{A^{(t,s)} x+ (1-x)} \quad \text {with} \quad A^{(t,s)}:=e^{(\mu -r-\frac{\sigma ^2}{2})(s-t)+\sigma (B_s-B_t)}, \end{aligned}$$

(B.3)

and the above expression is decreasing in x, therefore,

$$\begin{aligned} 1- \tfrac{1}{A^{(t,s)}}<\tfrac{Y_s^{(t,x)}-x}{x(1-x)}<A^{(t,s)}-1 \quad \text {for} \quad 0<x<1. \end{aligned}$$

(B.4)

When $\epsilon =0$, the representation of $v_x$ in (4.11) becomes

$$\begin{aligned} v_x^0(t,x)=\int _t^T e^{-\lambda (s-t)} {{\mathbb {E}}}\left[ \tfrac{Y_s^{(t,x)}\left( 1-Y_s^{(t,x)}\right) \left( \mu -r-\sigma ^2 Y_s^{(t,x)} \right) }{x(1-x)} \right] ds, \end{aligned}$$

(B.5)

We take derivative with respect to x in the above expression. Then, the mean value theorem and the dominated convergence theorem, together with the inequalities (4.14) and (B.4), allow us to take derivative inside of the expectation and obtain that for $(t,x)\in [0,T)\times (0,1)$,

$$\begin{aligned} v_{xx}^0(t,x)&=\int _t^T e^{-\lambda (s-t)} {{\mathbb {E}}}\left[ \tfrac{\partial }{\partial x}\left( \tfrac{Y_s^{(t,x)} \left( 1-Y_s^{(t,x)}\right) \left( \mu -r-\sigma ^2 Y_s^{(t,x)}\right) }{x(1-x)} \right) \right] ds \end{aligned}$$

(B.6)

$$\begin{aligned}&=- \int _t^T e^{-\lambda (s-t)} \tfrac{\partial }{\partial s} \left( {{\mathbb {E}}}\left[ \left( \tfrac{Y_s^{(t,x)}-x}{x(1-x)}\right) ^2 \right] \right) \, ds \nonumber \\&=- e^{-\lambda (T-t)} \, {{\mathbb {E}}}\left[ \left( \tfrac{Y_T^{(t,x)}-x}{x(1-x)}\right) ^2 \right] - \lambda \int _t^T e^{-\lambda (s-t)} \, {{\mathbb {E}}}\left[ \left( \tfrac{Y_s^{(t,x)}-x}{x(1-x)}\right) ^2 \right] \, ds, \end{aligned}$$

(B.7)

where the second equality is due to (B.2), and the third equality is from integration by parts. Obviously (B.7) implies that $v_{xx}^0(t,x)<0$ for $(t,x)\in [0,T)\times (0,1)$.

To conclude (5.3), it only remains to check that $\lim _{x\uparrow 1} v_{xx}^0(t,x)<0$ and $\lim _{x\downarrow 0} v_{xx}^0(t,x)<0$. Indeed, (B.3) and (B.4) enable us to apply the dominated convergence theorem to (B.7) and obtain

$$\begin{aligned} \lim _{x\uparrow 1} v_{xx}^0(t,x)&= - e^{-\lambda (T-t)} \, {{\mathbb {E}}}\Big [ \left( 1- \tfrac{1}{A^{(t,T)}}\right) ^2 \Big ] - \lambda \int _t^T e^{-\lambda (s-t)} \, {{\mathbb {E}}}\Big [ \left( 1- \tfrac{1}{A^{(t,s)}}\right) ^2 \Big ] ds ,\\ \lim _{x\downarrow 0} v_{xx}^0(t,x)&= - e^{-\lambda (T-t)} \, {{\mathbb {E}}}\Big [ \left( A^{(t,T)}-1\right) ^2 \Big ] \\ {}&\quad - \lambda \int _t^T e^{-\lambda (s-t)} \, {{\mathbb {E}}}\Big [ \left( A^{(t,s)}-1\right) ^2 \Big ] ds, \end{aligned}$$

and we conclude that $\lim _{x\uparrow 1} v_{xx}^0(t,x)<0$ and $\lim _{x\downarrow 0} v_{xx}^0(t,x)<0$.

(ii) The SDE for $Y_s^{(t,x)}$ in (B.1) and $0<Y_s^{(t,x)}<1$ imply that for $(s,x)\in [t,T) \times (0,1)$,

$$\begin{aligned} \tfrac{\partial }{\partial s} \left( {{\mathbb {E}}}\left[ Y_s^{(t,x)} \right] \right) = {{\mathbb {E}}}\left[ Y_s^{(t,x)} \left( 1- Y_s^{(t,x)}\right) \left( \mu - r - \sigma ^2 Y_s^{(t,x)} \right) \right] \end{aligned}$$

(B.8)

We divide both sides of (B.8) by $x(1-x)$ and take derivative with respect to x. Then, we can put the derivative inside of the expectation as in the proof of part (i), and obtain

$$\begin{aligned} {{\mathbb {E}}}\left[ \tfrac{\partial }{\partial x} \left( \tfrac{Y_s^{(t,x)} \left( 1- Y_s^{(t,x)}\right) \left( \mu - r - \sigma ^2 Y_s^{(t,x)} \right) }{x(1-x)} \right) \right]&=\tfrac{\partial }{\partial s} \left( {{\mathbb {E}}}\left[ \tfrac{\partial }{\partial x} \left( \tfrac{Y_s^{(t,x)} }{x(1-x)}\right) \right] \right) \\&=\tfrac{\partial }{\partial s} \left( \tfrac{ {{\mathbb {E}}}\left[ \frac{\partial }{\partial x}Y_s^{(t,x)} \right] }{x(1-x)} - \tfrac{(1-2x){{\mathbb {E}}}\left[ Y_s^{(t,x)} \right] }{x^2(1-x)^2} \right) . \end{aligned}$$

We rearrange the above equation and use (B.8), (B.5), and (B.7) to obtain

$$\begin{aligned} \int _t^T e^{-\lambda (s-t)} \tfrac{\partial }{\partial s} \left( {{\mathbb {E}}}\left[ \tfrac{\partial }{\partial x} Y_s^{(t,x)} \right] \right) ds = x(1-x)v_{xx}^0(t,x) + (1-2x) v_x^0(t,x). \end{aligned}$$

(B.9)

Now we conclude that $F(t, {\hat{y}}^0(t))<1$ by the following way:

$$\begin{aligned} F(t, {\hat{y}}^0(t))&\le \lambda \int _t^T e^{-\lambda (s-t)} {{\mathbb {E}}}\left[ \tfrac{\partial }{\partial x} Y_s^{(t,x)} \right] ds \, \Big |_{x={\hat{y}}^0(t)} \\&=\left( 1- e^{-\lambda (T-t)} {{\mathbb {E}}}\left[ \tfrac{\partial }{\partial x} Y_T^{(t,x)} \right] + x(1-x)v_{xx}^0(t,x) \right. \\ {}&\quad \left. + (1-2x) v_x^0(t,x) \right) \Big |_{x={\hat{y}}^0(t)} <1, \end{aligned}$$

where the first inequality is from the definition of F in (5.4) and the positivity of $\frac{\partial }{\partial x} Y_s^{(t,x)}$ (see (4.14)), and the equality is due to integration by parts and (B.9), and the last inequality is due to the positivity of $\frac{\partial }{\partial x} Y_T^{(t,x)}$, $v_x^0(t,{\hat{y}}^0(t))=0$, and $v_{xx}^0(t,\hat{y}^0(t))<0$.

We can check that $F(t, {\hat{y}}^0(t))>-1$ by the same way as above. (iii) The expression in (4.11) produces

$$\begin{aligned} \begin{aligned} v_{x}^\epsilon (t,x_\epsilon )-v_x^0(t,x_\epsilon )&= \lambda \int _t^T e^{-\lambda (s-t)} {{\mathbb {E}}}\left[ \left( \tfrac{\partial }{\partial x} Y_s^{(t,x_\epsilon )} \right) L_y^\epsilon (s, Y_s^{(t,x_\epsilon )} ) \right] ds. \end{aligned} \end{aligned}$$

(B.10)

In the above expression, when we take limit as $\epsilon \downarrow 0$, the inequalities (4.13) and (4.14) enable us to use the dominated convergence theorem to conclude that

$$\begin{aligned} \lim _{\epsilon \downarrow 0} \left( v_{x}^\epsilon (t,x_\epsilon )-v_x^0(t,x_\epsilon ) \right) =0. \end{aligned}$$

The above limit and the continuity of $v_{x}^0$ imply (5.6).

To prove (5.7), we first observe that for $(t,x)\in [0,T)\times (0,1)$ and $(s,z)\in (t,T)\times (0,1)$, the density function for $Y_s^{(t,x)}$ is given by

$$\begin{aligned} \begin{aligned} \varphi (s,z;t,x)&:= \tfrac{\partial }{\partial z} {{\mathbb {P}}}\left( Y_s^{(t,x)} \le z \right) =\tfrac{\exp \left( -\frac{1}{2\sigma ^2 (s-t)} \left( (r-\mu +\frac{\sigma ^2}{2})(s-t) + \ln \left( \frac{z(1-x)}{(1-z)x} \right) \right) ^2 \right) }{\sigma z(1-z)\sqrt{2\pi (s-t)}}. \end{aligned}\nonumber \\ \end{aligned}$$

(B.11)

Then, the expression in (4.11) and $\frac{\partial }{\partial x} Y_s^{(t,x)}=\frac{Y_s^{(t,x)}(1-Y_s^{(t,x)})}{x(1-x)}$ imply that

$$\begin{aligned} \begin{aligned} v_x^\epsilon (t,x)-v_x^0(t,x)&= \lambda \int _t^T e^{-\lambda (s-t)} {{\mathbb {E}}}\left[ \tfrac{Y_s^{(t,x)}\left( 1-Y_s^{(t,x)}\right) }{x(1-x)} L_y^\epsilon (s, Y_s^{(t,x)} ) \right] ds\\&=\lambda \int _t^T e^{-\lambda (s-t)} \left( \int _0^1 \tfrac{z(1-z)}{x(1-x)} L_y^\epsilon (s, z) \varphi (s,z;t,x) dz \right) ds. \end{aligned} \end{aligned}$$

(B.12)

For $x\in (0,1)$, direct computations produce

$$\begin{aligned} \begin{aligned} \tfrac{\partial }{\partial x} \left( \tfrac{z(1-z)}{x(1-x)} \varphi (s,z;t,x) \right) = \tfrac{(1-z)z \left( r-\mu + (2x-\frac{1}{2})\sigma ^2 + \frac{1}{s-t} \ln \left( \frac{z(1-x)}{(1-z)x} \right) \right) \varphi (s,z;t,x) }{(1-x)^2 x^2 \sigma ^2}. \end{aligned} \end{aligned}$$

(B.13)

Assumption  implies that

$$\begin{aligned} \left| r-\mu + \tfrac{\sigma ^2}{2} \right| \le \sigma ^2, \quad \left| r-\mu + (2x-\tfrac{1}{2})\sigma ^2 \right| \le 2\sigma ^2 \quad \text {for} \quad x\in (0,1). \end{aligned}$$

(B.14)

Then, (B.11) and (B.13) produce the following:

$$\begin{aligned} \begin{aligned}&\int _0^1\left| \frac{\partial }{\partial x} \left( \tfrac{z(1-z)}{x(1-x)} \varphi (s,z;t,x) \right) \right| \, dz\\&\quad \le \int _0^1\tfrac{\left( 2\sigma ^2 + \frac{1}{s-t} \left| \ln \left( \tfrac{z(1-x)}{(1-z)x} \right) \right| \right) }{\sigma ^3(1-x)^2 x^2 \sqrt{2\pi (s-t)}}\exp \left( -\tfrac{\left( \ln \left( \frac{z(1-x)}{(1-z)x} \right) \right) ^2}{2\sigma ^2 (s-t)} + \left| \ln \left( \tfrac{z(1-x)}{(1-z)x} \right) \right| \right) \, dz \\&\quad = \int _{-\infty }^\infty \tfrac{\left( 2\sigma ^2 + \frac{1}{s-t} \left| \zeta \right| \right) e^{ -\frac{\zeta ^2}{2\sigma ^2 (s-t)} + \left| \zeta \right| }}{\sigma ^3(1-x)^2 x^2 \sqrt{2\pi (s-t)}} \cdot \tfrac{\frac{x}{1-x}e^\zeta }{(1+\frac{x}{1-x} e^\zeta )^2} \, d\zeta \\&\quad \le \int _{-\infty }^\infty \tfrac{\left( 2\sigma ^2 + \frac{1}{s-t} \left| \zeta \right| \right) e^{ -\frac{\zeta ^2}{4\sigma ^2 (s-t)} + 4\sigma ^2(s-t)}}{\sigma ^3(1-x)^3 x \sqrt{2\pi (s-t)}} \, d\zeta \\&\quad =\tfrac{2\sqrt{2} }{(1-x)^3 x}\left( 1 + \tfrac{1}{\sigma \sqrt{\pi (s-t)}} \right) e^{4\sigma ^2(s-t)}, \end{aligned} \end{aligned}$$

(B.15)

where the first inequality is due to (B.14), and the first equality is obtained by change of variables as $\zeta =\ln \left( \frac{z(1-x)}{(1-z)x} \right)$. The second inequality is due to the inequality of arithmetic and geometric means, and the second equality is obtained by direct computations.

By (B.15), we conclude that

$$\begin{aligned} \int _t^T e^{-\lambda (s-t) }\int _0^1\left| \tfrac{\partial }{\partial x} \left( \tfrac{z(1-z)}{x(1-x)} \varphi (s,z;t,x) \right) \right| \, dz \, ds <\infty , \end{aligned}$$

(B.16)

and the function $H:[0,T)\times (0,1) \rightarrow {{\mathbb {R}}}$ given by

$$\begin{aligned} H(t,x):=\lambda \int _t^T e^{-\lambda (s-t) }\int _0^1 \tfrac{\partial }{\partial x} \left( \tfrac{z(1-z)}{x(1-x)} \varphi (s,z;t,x) \right) L_y^\epsilon (s,z) \, dz \, ds \end{aligned}$$

(B.17)

is well-defined due to (B.16) and the boundedness $| L_y^\epsilon |\le \frac{\epsilon }{1-\epsilon }$ in (4.13). Then, (B.15) implies that

$$\begin{aligned} \left| H(t,x)\right| \le \tfrac{2\sqrt{2}\lambda e^{4\sigma ^2 T}}{(1-x)^3 x} \left( T + \tfrac{2\sqrt{T}}{\sigma \sqrt{\pi }} \right) \cdot \tfrac{\epsilon }{1-\epsilon } \quad \text {for} \quad (t,x)\in [0,T)\times (0,1). \end{aligned}$$

(B.18)

Now, let’s check that

$$\begin{aligned} H(t,x)=v_{xx}^\epsilon (t,x)-v_{xx}^0(t,x). \end{aligned}$$

(B.19)

Indeed, for $(t,x)\in [0,T)\times (0,1)$,

$$\begin{aligned} H(t,x)&=\lim _{\delta \rightarrow 0} \frac{1}{\delta }\int _x^{x+\delta }H(t,\eta )d\eta \\&=\lim _{\delta \rightarrow 0} \frac{1}{\delta } \,\, \lambda \int _t^T e^{-\lambda (s-t)} \int _0^1 \left( \tfrac{z(1-z)\varphi (s,z;t,x+\delta ) }{(x+\delta )(1-x-\delta )} - \tfrac{z(1-z)\varphi (s,z;t,x) }{x(1-x)} \right) \\ {}&\quad L_y^\epsilon (s,z) \, dz \, ds \\&=\lim _{\delta \rightarrow 0} \frac{1}{\delta } \left( \left( v_x^\epsilon (t,x+\delta )-v_x^0(t,x+\delta )\right) -\left( v_x^\epsilon (t,x)-v_x^0(t,x)\right) \right) \\&=v_{xx}^\epsilon (t,x)-v_{xx}^0(t,x), \end{aligned}$$

where the second equality is due to Fubini’s theorem and the fundamental theorem of calculus, and the third equality is from (B.12).

Finally, we conclude (5.7) by the following observation:

$$\begin{aligned}&\limsup _{\epsilon \downarrow 0} \left| v_{xx}^\epsilon (t,x_\epsilon )-v_{xx}^0(t,x_0)\right| \\ {}&\quad \le \limsup _{\epsilon \downarrow 0} \left| v_{xx}^\epsilon (t,x_\epsilon )-v_{xx}^0(t,x_\epsilon )\right| + \limsup _{\epsilon \downarrow 0} \left| v_{xx}^0(t,x_\epsilon )-v_{xx}^0(t,x_0)\right| \\&\quad \le \limsup _{\epsilon \downarrow 0}\tfrac{2\sqrt{2}\lambda e^{4\sigma ^2 T}}{(1-x_\epsilon )^3 x_\epsilon } \left( T + \tfrac{2\sqrt{T}}{\sigma \sqrt{\pi }} \right) \cdot \tfrac{\epsilon }{1-\epsilon } =0, \end{aligned}$$

where the second inequality is due to (B.18), (B.19), and the continuity of $v_{xx}^0$.

Appendix C. Proof of Lemma 5.10

(i) Using $\Gamma$ in (5.25), the equations (B.5) and (B.6) can be written as

$$\begin{aligned} v_x^0(t,x)=\int _0^{T-t} e^{-\lambda s} \Gamma (s,x)ds, \quad v_{xx}^0(t,x)=\int _0^{T-t} e^{-\lambda s} \Gamma _x(s,x)ds. \end{aligned}$$

(C.1)

We can do the similar argument to obtain representations for $v_{xxx}^{0}$ and partial derivatives of $v_x^{0}$ and $v_{xx}^{0}$ with respect to $\lambda$, with the observation that $\Gamma , \Gamma _x, \Gamma _{xx}$ are continuous & bounded maps on $[0,T]\times (0,1)$. The result is summarized as follows:

$$\begin{aligned} \begin{aligned}&\tfrac{\partial }{\partial \lambda } v_x^{0}(t,x)=- \int _0^{T-t} e^{-\lambda s} s \Gamma (s,x)ds, \quad \tfrac{\partial }{\partial \lambda } v_{xx}^{0}(t,x)=-\int _0^{T-t} e^{-\lambda s}s \Gamma _x(s,x)ds \\&v_{xx}^{0}(t,x)=\int _0^{T-t} e^{-\lambda s} \Gamma _x(s,x)ds,\qquad \quad v_{xxx}^{0}(t,x)=\int _0^{T-t} e^{-\lambda s} \Gamma _{xx}(s,x)ds. \end{aligned} \end{aligned}$$

(C.2)

Using $Y_t^{(t,x)}=x$, direct computations produce

$$\begin{aligned} \begin{aligned}&\Gamma (0,x)=\mu -r-\sigma ^2 x, \quad \Gamma _x(0,x) = - \sigma ^2, \quad \Gamma _{xx}(0,x)=0. \end{aligned} \end{aligned}$$

(C.3)

With (C.3) and (5.24), we apply Lemma D.2 to (C.2) and conclude (5.61).

(ii) The mean value theorem and (5.2) produce

$$\begin{aligned} \begin{aligned}&{\hat{y}}^{0,\lambda +\delta }(t) - {\hat{y}}^{0,\lambda }(t) \\ {}&\quad = - \tfrac{ v_x^{0,\lambda +\delta }(t,{\hat{y}}^{0,\lambda }(t)) - v_x^{0,\lambda }(t,\hat{y}^{0,\lambda }(t))}{v_{xx}^{0,\lambda +\delta }(t, z(\lambda ,\delta ))} \quad \text {for} z(\lambda ,\delta ) \text {between} \hat{y}^{0,\lambda }(t) \text {and} {\hat{y}}^{0,\lambda +\delta }(t), \end{aligned} \end{aligned}$$

where we specify the dependence on $\lambda$ for clarity. Since $\frac{\partial }{\partial \lambda } v_x^{0}$ exists (see (C.2)), the above equality and (5.3) ensure that ${\hat{y}}^{0,\lambda }(t)$ is differentiable with respect to $\lambda$ and

$$\begin{aligned} \begin{aligned} \tfrac{\partial }{\partial \lambda } {\hat{y}}^{0}(t) = - \frac{ \frac{\partial }{\partial \lambda } v_x^{0}(t,x)\big |_{x=\hat{y}^{0}(t)} }{ v_{xx}^{0}(t,{\hat{y}}^{0}(t))}. \end{aligned} \end{aligned}$$

(C.4)

Observe that the bounds for $\frac{\partial }{\partial x}Y_t^{(0,x)}$ and $\frac{Y_t^{(0,x)}-x}{x(1-x)}$ in (4.14) and (B.4) do not depend on the variable x. Therefore, the following convergence is uniform on $x \in (0,1)$:

$$\begin{aligned} \Gamma _x(t,x)&=-{{\mathbb {E}}}\left[ \sigma ^2 \left( \tfrac{\partial }{\partial x}Y_t^{(0,x)}\right) ^2 \right. \\ {}&\quad \left. + 2\left( \mu -r-\sigma ^2 Y_t^{(0,x)} \right) \left( \tfrac{Y_t^{(0,x)}-x}{x(1-x)}\right) \tfrac{\partial }{\partial x}Y_t^{(0,x)} \right] \xrightarrow [ t \downarrow 0]{} -\sigma ^2. \end{aligned}$$

Hence, there is a constant ${\tilde{T}}\in (0,T)$ such that $\Gamma _x(t,x)\le -\frac{\sigma ^2}{2}$ for $(t,x)\in [0,{\tilde{T}}] \times (0,1)$. This observation and the expression of $v_{xx}$ in (C.2) imply

$$\begin{aligned} \lambda v_{xx}(t,\hat{y}^0(t)) \le -\tfrac{\sigma ^2}{2} \left( 1-e^{-\lambda (T-t)} \right) \quad \text {for} \quad (t,x)\in [T-{\tilde{T}}, T) \times (0,1). \end{aligned}$$

(C.5)

We obtain $\Vert \Gamma _{xt} \Vert _\infty <\infty$ by using Ito’s formula with (B.1) and the bounds (4.14) and (B.4). Then, (C.2) and (C.3) imply

$$\begin{aligned} \left| \lambda v_{xx}(t,\hat{y}^0 (t)) + \sigma ^2 \right|&= \left| \lambda \int _0^{T-t} e^{-\lambda s} s \, \left( \tfrac{\Gamma _x(s,\hat{y}^0(t)) -\Gamma _x(0,\hat{y}^0(t))}{s} \right) ds + \sigma ^2 e^{-\lambda (T-t)} \right| \\&\le \Vert \Gamma _{xt} \Vert _\infty \left( \tfrac{1-e^{-\lambda (T-t)}}{\lambda } -(T-t) e^{-\lambda (T-t)} \right) + \sigma ^2 e^{-\lambda (T-t)}. \end{aligned}$$

This implies that there exists a constant ${\tilde{\Lambda }}$ (may depend on ${\tilde{T}}$) such that

$$\begin{aligned} \lambda v_{xx}(t,\hat{y}^0 (t)) \le -\tfrac{\sigma ^2}{2} \quad \text {for} \quad (t,\lambda )\in [0,T-{\tilde{T}}] \times [\tilde{\Lambda },\infty ). \end{aligned}$$

(C.6)

Using (C.2) and (C.3), we obtain

$$\begin{aligned}&\left| \lambda ^3 \tfrac{\partial }{\partial \lambda } v_x^{0}(t,x)\big |_{x={\hat{y}}^{0}(t)} \right| \nonumber \\&\quad \le \left| \lambda ^3 \int _0^{T-t} e^{-\lambda s} s^2 \left( \tfrac{\Gamma (s, \hat{y}^0(t))-\Gamma (0, \hat{y}^0(t))}{s}\right) ds \right| \nonumber \\ {}&\qquad + \left| \sigma ^2(y_\infty - \hat{y}^0(t))\lambda ^3 \int _0^{T-t} e^{-\lambda s} s \,ds \right| \nonumber \\&\quad \le C\left( 1-e^{-\lambda (T-t)} + \lambda (T-t)\big (1+\lambda (T-t) \big ) e^{-\lambda (T-t)} \right) \nonumber \\ {}&\quad \text {for} \quad (t,\lambda )\in [0,T]\times [1,\infty ), \end{aligned}$$

(C.7)

where the second inequality is due to $\Vert \Gamma _t \Vert _\infty <\infty$ and (5.24).

From (C.4), we obtain the boundedness of $\left| \lambda ^2 \tfrac{\partial }{\partial \lambda } {\hat{y}}^{0}(t) \right|$:

$$\begin{aligned}&\sup _{(t,\lambda )\in [0,T-{\tilde{T}}]\times [{\tilde{\Lambda }}, \infty )} \left| \lambda ^2 \tfrac{\partial }{\partial \lambda } {\hat{y}}^{0}(t) \right|<\infty \quad \text {due to} (C.6), (C.7), \sup _{x>0} x(1+x) e^{-x}<\infty , \\&\sup _{(t,\lambda )\in [0,T-{\tilde{T}}]\times [1, {\tilde{\Lambda }}]} \left| \lambda ^2 \tfrac{\partial }{\partial \lambda } {\hat{y}}^{0}(t) \right|<\infty \quad \text {due to the continuity on compact set},\\&\sup _{(t,\lambda )\in [T-{\tilde{T}},T)\times [1,\infty )} \left| \lambda ^2 \tfrac{\partial }{\partial \lambda } {\hat{y}}^{0}(t) \right|<\infty \quad \text {due to} (C.5), (C.7), \sup _{x>0}\tfrac{ x(1+x) e^{-x}}{1-e^{-x}}<\infty . \end{aligned}$$

Therefore, we conclude (5.62).

(iii) The bounds (4.14) and (5.62) enable us to use Leibniz integral rule to obtain

$$\begin{aligned} \tfrac{\partial }{\partial \lambda } {{\mathbb {E}}}\left[ \left| Y_{s}^{(t, \hat{y}^{0}(t))} - \hat{y}^{0}(s) \right| \right]&= {{\mathbb {E}}}\left[ \left( \tfrac{\partial }{\partial \lambda }\hat{y}^0(t) \cdot \tfrac{\partial }{\partial x} Y_s^{(t,x)} \Big |_{x=\hat{y}^0(t)}- \tfrac{\partial }{\partial \lambda }\hat{y}^0(s) \right) \cdot \right. \\ {}&\quad \left. {{\,\textrm{sgn}\,}}\left( Y_{s}^{(t, \hat{y}^{0}(t))} - \hat{y}^{0}(s) \right) \right] . \end{aligned}$$

The above expression, together with the bounds (4.14) and (5.62), implies (5.63).

Appendix D. Supplementary Lemmas

Lemma D.1

Let $F:[0,T]\times [0,1]^2\rightarrow {{\mathbb {R}}}$ be a continuous function. We define $f:[0,T]\times [0,1]\rightarrow [0,1]$ as

$$\begin{aligned} f(t,x):=\max \left\{ z: z\in \mathop {\textrm{argmax}}\limits _{y\in [0,1]} F(t,x,y) \right\} , \end{aligned}$$

(D.1)

then f is upper semicontinuous (which is obviously Borel-measurable).

Proof

This type of result is well-known (e.g., see p. 153 in [6]), but we give a short proof here for the sake of self-containedness.

Since F is continuous, $\mathop {\textrm{argmax}}\limits _{y\in [0,1]} F(t,x,y)$ is a nonempty closed subset of [0, 1], so the maximum element of $\mathop {\textrm{argmax}}\limits _{y\in [0,1]} F(t,x,y)$ exists and f is well-defined. Let $\{(t_n,x_n)\}_{n\in {{\mathbb {N}}}}\subset [0,T]\times [0,1]$ be a sequence converging to $(t_\infty ,x_\infty )$ such that $\lim _{n\rightarrow \infty } f(t_n,x_n)$ exists. Then, by definition of f,

$$\begin{aligned} F(t_n,x_n,f(t_\infty ,x_\infty ))\le F(t_n,x_n,f(t_n,x_n)). \end{aligned}$$

(D.2)

We let $n\rightarrow \infty$ above and using the continuity of F to obtain

$$\begin{aligned} F(t_\infty ,x_\infty ,f(t_\infty ,x_\infty )) \le F(t_\infty ,x_\infty ,\lim _{n\rightarrow \infty } f(t_n,x_n)). \end{aligned}$$

(D.3)

This implies that $\lim _{n\rightarrow \infty } f(t_n,x_n) \in \mathop {\textrm{argmax}}\limits _{y\in [0,1]} F(t_\infty ,x_\infty ,y)$, and the definition of f ensures

$$\begin{aligned} f(t_\infty ,x_\infty )\ge \lim _{n\rightarrow \infty } f(t_n,x_n). \end{aligned}$$

Therefore, f is upper semicontinuous. $\square$

Lemma D.2

Let $f(s,x): [0,t] \times (0,1) \rightarrow {{\mathbb {R}}}$ be a continuous function, and $g(\lambda ): [1,\infty ) \rightarrow (0,1)$ be a function satisfying $\lim _{\lambda \rightarrow \infty } g(\lambda )=x_\infty \in (0,1)$. Then, for $\alpha \in \{0,1,2,3,4\}$ and $t > 0$,

$$\begin{aligned} \begin{aligned}&\lim _{\lambda \rightarrow \infty } \lambda ^{\frac{\alpha }{2}+1} \int _0^{t} e^{-\lambda s}s^{\frac{\alpha }{2}} f(s,g(\lambda )) ds = c_\alpha \cdot f(0,x_\infty ), \end{aligned} \end{aligned}$$

(D.4)

where $c_0=1$, $c_1= \frac{\sqrt{\pi }}{2}$, $c_2= 1$, $c_3= \frac{3\sqrt{\pi }}{4}$, $c_4= 2$. Also, there exists a constant C such that

$$\begin{aligned} \lambda ^{\frac{\alpha }{2}+1} \int _0^t e^{-\lambda s} s^{\frac{\alpha }{2}} ds \le C, \quad \text {for} \quad (t,\lambda ,\alpha )\in [0,\infty )\times [1,\infty )\times \{0,1,2,3,4\}.\nonumber \\ \end{aligned}$$

(D.5)

Proof

Let $\eta >0$ be a given constant. The uniform continuity of f on a compact set containing the point $(0,x_\infty )$, together with $\lim _{\lambda \rightarrow \infty } g(\lambda )=x_\infty \in (0,1)$, implies that there exists $\delta >0$ such that

$$\begin{aligned} \left| f(s,g(\lambda )) - f(0,x_\infty ) \right| \le \eta \quad \text {for any} \quad (s,\lambda ) \in [0, \delta ] \times \left[ \tfrac{1}{\delta }, \infty \right) . \end{aligned}$$

(D.6)

Simple computations and (D.6) produce

$$\begin{aligned} \begin{aligned}&\limsup _{\lambda \rightarrow \infty } \Big | \lambda \int _0^t e^{-\lambda s } f(s,g(\lambda )) ds - f(0,x_\infty ) \Big | \\ {}&\quad = \limsup _{\lambda \rightarrow \infty } \bigg | \int _0^{\delta } \lambda e^{-\lambda s} (f(s,g(\lambda )) - f(0,x_\infty ))ds \\&+\int _{ \delta }^t \lambda e^{-\lambda s} (f(s,g(\lambda )) - f(0,x_\infty )) ds - e^{-\lambda t} f(0,x_\infty ) \bigg | \le \eta . \end{aligned} \end{aligned}$$

Since $\eta >0$ can be arbitrary small, we conclude (D.4) for the case of $\alpha =0$. The other cases in (D.4) can be obtained by the same way as above, using the following expressions:

$$\begin{aligned} \begin{aligned}&\lambda ^{\frac{3}{2}} \int _0^t e^{-\lambda s} s^{\frac{1}{2}} ds = -\sqrt{\lambda t} e^{-\lambda t} + \int _0^{\sqrt{\lambda t}} e^{-s^2} ds \xrightarrow [\lambda \rightarrow \infty ]{} \frac{\sqrt{\pi }}{2}, \\&\lambda ^2 \int _0^t e^{-\lambda s}s \,ds =1 -(1+\lambda t) e^{-\lambda t}\xrightarrow [\lambda \rightarrow \infty ]{} 1, \\&\lambda ^{\frac{5}{2}} \int _0^t e^{-\lambda s} s^{\frac{3}{2}} ds = -\tfrac{(3+2\lambda t)\sqrt{\lambda t} e^{-\lambda t}}{2} + \frac{3}{2}\int _0^{\sqrt{\lambda t}} e^{-s^2} ds \xrightarrow [\lambda \rightarrow \infty ]{} \frac{3\sqrt{\pi }}{4},\\&\lambda ^3 \int _0^t e^{-\lambda s}s^2 ds =2 -(2+\lambda t(2+\lambda t)) e^{-\lambda t}\xrightarrow [\lambda \rightarrow \infty ]{} 2. \end{aligned} \end{aligned}$$

(D.7)

One can easily observe that $\sup _{x>0} x^{\frac{\alpha }{2}} e^{-x} <\infty$ for $\alpha \in \{0,1,2,3,4\}$. This observation and the explicit expressions in (D.7) produce the bound (D.5). $\square$