*5.2. With Transaction Cost*

Now, we consider the problem given in Equation (30) with transaction cost *δ* > 0. From the previous section, we know that optimal strategies are (*c*Λ1 , *c*Λ2 )-reflected strategies with (*c*Λ1 , *c*Λ2 ) given in Proposition 2.

**Proposition 4.** *The curve* Λ → (*c*Λ1, *c*Λ2 ) *is continuous and unbounded, for* Λ ∈ [1, <sup>∞</sup>)*.*

**Proof.** From Remark 7 and the fact that *a*Λ < *c*Λ2 (by Proposition 2), we know that *c*Λ2 → ∞ as Λ → <sup>∞</sup>, so the curve is unbounded. To show the continuity of the curve, we consider two cases and use the implicit function theorem. To this end, suppose first *c*Λ1 = 0. Defining *f*(<sup>Λ</sup>, *<sup>c</sup>*2) := *<sup>G</sup>*Λ(0, *<sup>c</sup>*2) − *ζ*Λ(*<sup>c</sup>*2), we have *f*(<sup>Λ</sup>, *c*Λ2 ) = 0. Then,

$$\frac{\partial f}{\partial c\_2}(\Lambda, c\_2^{\Lambda}) = \frac{\partial G\_{\Lambda}}{\partial c\_2}(0, c\_2^{\Lambda}) - \zeta\_{\Lambda}'(c\_2^{\Lambda}) = -\zeta\_{\Lambda}'(c\_2^{\Lambda}) > 0,$$

since *c*Λ2 > *a*Λ. From here, we see that the conditions of the implicit function theorem are satisfied. Now, if *c*Λ1 > 0, define the function *f*(<sup>Λ</sup>, *c*1, *<sup>c</sup>*2)=(*f*1(<sup>Λ</sup>, *c*1, *<sup>c</sup>*2), *f*2(<sup>Λ</sup>, *c*1, *<sup>c</sup>*2)) by

$$\begin{aligned} f\_1(\Lambda, c\_1, c\_2) &:= G\_{\Lambda}(c\_1, c\_2) - \tilde{\zeta}\_{\Lambda}(c\_1), \\ f\_2(\Lambda, c\_1, c\_2) &:= G\_{\Lambda}(c\_1, c\_2) - \tilde{\zeta}\_{\Lambda}(c\_2). \end{aligned}$$

Then, *f*(<sup>Λ</sup>, *c*Λ1 , *c*Λ2 )=(0, <sup>0</sup>). Again, simple calculations show that the Jacobian determinant of this system of equations is *ζ*Λ(*c*Λ2 )*ζ*Λ(*c*Λ1 ) < 0, since *c*Λ1 < *a*Λ < *c*Λ2 . Therefore, the curve Λ → (*c*Λ1 , *c*Λ2 ) is continuous, for Λ ∈ [1, <sup>∞</sup>).

Next, we analyze the level curves of the constraint. Let <sup>Ψ</sup>*x*(*<sup>c</sup>*1, *<sup>c</sup>*2) be the expected present value of the injected capital under a (*<sup>c</sup>*1, *<sup>c</sup>*2)-reflected policy. Then, the calculations given in the proof of Lemma 1 show that

$$\begin{split} \mathbb{P}\_{\mathbf{x}}(c\_{1},c\_{2}) &:= \mathbb{E}\_{\mathbf{x}} \left[ \int\_{0}^{\infty} \mathbf{e}^{-qt} \, \mathrm{d}\mathbf{R}\_{t}^{(c\_{1},c\_{2}),0} \right] \\ &= \begin{cases} Z^{(q)}(\mathbf{x}) \frac{\overline{Z}^{(q)}(c\_{2}) - \overline{Z}^{(q)}(c\_{1})}{Z^{(q)}(c\_{2}) - Z^{(q)}(c\_{1})} - k^{(q)}(\mathbf{x}), & \text{if } 0 \le \mathbf{x} \le c\_{2}, \\\ \overline{Z}^{(q)}(c\_{2}) Z^{(q)}(c\_{1}) - \overline{Z}^{(q)}(c\_{1}) Z^{(q)}(c\_{2}) - \frac{\psi'(0+)}{q}, & \text{if } \mathbf{x} > c\_{2}. \end{cases} \end{split} \tag{39}$$

**Remark 11.** *Note that* lim *<sup>c</sup>*1→*c*2 <sup>Ψ</sup>*x*(*<sup>c</sup>*1, *<sup>c</sup>*2) = <sup>Ψ</sup>*x*(*<sup>c</sup>*2)*, where* Ψ*x is as in Equation* (32)*.*

The next lemmas describe some properties of <sup>Ψ</sup>*x*(*<sup>c</sup>*1, *<sup>c</sup>*2).

### **Lemma 8.** *Let x* ≥ 0 *be fixed.*

*1. If c*1 ≥ 0 *is fixed, then the function* <sup>Ψ</sup>*x*(*<sup>c</sup>*1, *<sup>c</sup>*2)*, given in Equation* (39)*, is strictly decreasing for all c*2 > *c*1*, and*

$$\lim\_{\varepsilon\_2 \to \infty} \overline{\Psi}\_x(\varepsilon\_1, \varepsilon\_2) = \underline{\mathbb{K}}\_{x'} \tag{40}$$

*where Kx is defined in Equation* (33)*.*

*2. If c*2 > 0 *is fixed,* <sup>Ψ</sup>*x*(*<sup>c</sup>*1, *<sup>c</sup>*2) *is strictly decreasing for all c*1 ∈ [0, *<sup>c</sup>*2)*.*

**Proof.** Let *c*1 ≥ 0 be fixed. First, assume that *c*2 ≥ *x*. To show that <sup>Ψ</sup>*x*(*<sup>c</sup>*1, *<sup>c</sup>*2) is strictly decreasing, it is sufficient to verify that

$$\frac{\overline{Z}^{(q)}(\mathbf{c}\_2) - \overline{Z}^{(q)}(\mathbf{c}\_1)}{\overline{Z}^{(q)}(\mathbf{c}\_2) - Z^{(q)}(\mathbf{c}\_1)},\tag{41}$$

is strictly decreasing, which is true if

$$\frac{\partial}{\partial c\_{2}} \left[ \frac{\overline{Z}^{(q)}(c\_{2}) - \overline{Z}^{(q)}(c\_{1})}{Z^{(q)}(c\_{2}) - Z^{(q)}(c\_{1})} \right] = \frac{Z^{(q)}(c\_{2})}{Z^{(q)}(c\_{2}) - Z^{(q)}(c\_{1})} - \frac{qW^{(q)}(c\_{2})(\overline{Z}^{(q)}(c\_{2}) - \overline{Z}^{(q)}(c\_{1}))}{[Z^{(q)}(c\_{2}) - Z^{(q)}(c\_{1})]^{2}} < 0. \tag{42}$$

Since *Z*(*q*) is a strictly log-convex function on [0, ∞) by Remark 4 (i),

$$\frac{q\mathcal{W}^{(q)}(\eta)}{Z^{(q)}(\eta)} < \frac{q\mathcal{W}^{(q)}(\emptyset)}{Z^{(q)}(\emptyset)}, \text{ for } \eta \text{ and } \emptyset \text{ such that } \eta < \emptyset.$$

Taking *ς* = *c*2 in the inequality above and integrating between *c*1 and *c*2, it follows that

$$Z^{(q)}(\mathbf{c}\_2) < \frac{q\mathcal{W}^{(q)}(\mathbf{c}\_2)[\overline{Z}^{(q)}(\mathbf{c}\_2) - \overline{Z}^{(q)}(\mathbf{c}\_1)]}{Z^{(q)}(\mathbf{c}\_2) - Z^{(q)}(\mathbf{c}\_1)}. \tag{43}$$

Then, Equation (43) yields Equation (42) and hence Equation (41) is strictly decreasing. For the case *x* > *c*2, it can be verified that

$$\begin{split} \frac{\partial}{\partial c\_{2}} \left[ Z^{(q)}(c\_{2}) \frac{\overline{Z}^{(q)}(c\_{2}) - \overline{Z}^{(q)}(c\_{1})}{Z^{(q)}(c\_{2}) - Z^{(q)}(c\_{1})} - \overline{Z}^{(q)}(c\_{2}) \right] \\ &= \frac{Z^{(q)}(c\_{1})}{Z^{(q)}(c\_{2}) - Z^{(q)}(c\_{1})} \Big[ Z^{(q)}(c\_{2}) - \frac{q W^{(q)}(c\_{2}) [\overline{Z}^{(q)}(c\_{2}) - \overline{Z}^{(q)}(c\_{1})]}{Z^{(q)}(c\_{2}) - Z^{(q)}(c\_{1})} \Big]. \end{split} \tag{44}$$

Then, using Equations (43) and (44), we obtain that <sup>Ψ</sup>*x*(*<sup>c</sup>*1, *<sup>c</sup>*2) is strictly decreasing for all *c*2 ∈ (*<sup>c</sup>*1, *<sup>x</sup>*). Similarly, we obtain Point 2 of the lemma. Now, by L'Hôpital's rule together with Exercise 8.5 (i) in Kyprianou (2014), it is not difficult to see that Equation (40) holds for any *c*1 ≥ 0.

Note that Equation (34) still holds if *K* ≥ *Kx*. On the other hand, using that *c*Λ2 → ∞ as Λ → ∞ together with Equation (18) we have that

$$\lim\_{\Lambda \to \infty} \mathbb{E}\_{\mathbf{x}} \left[ \int\_0^\infty \mathbf{e}^{-qt} \, \mathrm{d}L\_t^{(\mathbf{c}\_1^\Lambda, \mathbf{c}\_2^\Lambda)} \right] = \lim\_{\Lambda \to \infty} (c\_2^\Lambda - c\_1^\Lambda - \delta) \frac{Z^{(q)}(\mathbf{x})}{Z^{(q)}(\mathbf{c}\_2^\Lambda) - Z^{(q)}(\mathbf{c}\_1^\Lambda)} = 0,\tag{45}$$

by Remark 3 (3).

**Remark 12.** *Using the same arguments as in Lemma 7, we have that c*11 = *a*1 = 0 < *c*12 *for bounded and unbounded variation processes. Similarly, if x and K are such that K* ≥ <sup>Ψ</sup>*x*(0, *c*12) =: *Kx, then <sup>V</sup>δ*(*<sup>x</sup>*, *K*) = *<sup>V</sup>δ*,<sup>1</sup>(*x*) + *K. Note also that Kx* < *K.*

**Lemma 9.** *Let x* ≥ 0 *be fixed. Then, for each K* ∈ (*Kx*, *<sup>K</sup>*)*, there exist c* ≤ *c such that the level curve LK*(<sup>Ψ</sup>*x*) := {(*<sup>c</sup>*1, *<sup>c</sup>*2) : <sup>Ψ</sup>*x*(*<sup>c</sup>*1, *<sup>c</sup>*2) = *K*} *is continuous, contained in the set* [0, *c*] × [*c*, *c*] *and contains the points* (0, *c*) *and* (*c*, *c*)*.*

**Proof.** The continuity of the level curve is obtained as an immediate consequence of the continuity of Ψ*<sup>x</sup>*. Observe that, by Lemma 5, we know the existence of *c* > 0 such that <sup>Ψ</sup>*x*(*c*) = *K*. Meanwhile, from Lemma 8, we have that there exists *c* ∈ [*c*, ∞) such that <sup>Ψ</sup>*x*(0, *c*) = *K*. Now, the fact that the level curve *LK*(<sup>Ψ</sup>*x*) is contained in [0, *c*] × [*c*, *c*] is a consequence of Remark 11 and Lemma 8.

**Remark 13.** *Lemmas 4 and 9 yield that the parametric curve* Λ → (*c*Λ1 , *c*Λ2 ) *and the level curve LK*(<sup>Ψ</sup>*x*) *must intersect, i.e., there exists* Λ∗ *such that* <sup>Ψ</sup>*x*(*c*Λ<sup>∗</sup> 1 , *c*Λ<sup>∗</sup> 2 ) = *K, for K* ∈ (*Kx*, *Kx*]*.*

By similar arguments as in the proof of Theorem 2, by Remarks 12 and 13, and using Equation (45), we ge<sup>t</sup> the following result, whose proof is omitted.

**Theorem 3.** *Assume δ* > 0 *and let Vδ and <sup>V</sup>Dδ as in Equations* (30) *and* (31)*, respectively, then Vδ* = *<sup>V</sup>Dδ . Furthermore, if x*, *K are such that*

$$1. \quad \mathsf{K} < \underline{\mathsf{K}}\_{\mathsf{x}} \text{ then } V\_{\delta}(\mathsf{x}, \mathsf{K}) = -\infty;$$


$$V\_{\delta}(\mathbf{x}, K) = \Lambda^\* K + V\_{\delta, \Lambda^\*}(\mathbf{x}) = \mathbb{E}\_{\mathbf{x}} \left[ \int\_0^\infty e^{-qt} \, \mathbf{d} \left( L\_t^{(c\_1^{\Lambda^\*}, c\_2^{\Lambda^\*}), 0} - \delta \sum\_{0 \le s < t} \mathbf{1}\_{\left\{ \Lambda L\_s^{(c\_1^{\Lambda^\*}, c\_2^{\Lambda^\*}), 0} > 0 \right\}} \right) \right] \right]. \tag{46}$$
