Proof. For any $0\le y< x$, we consider an $\epsilon$-optimal strategy $\theta _\epsilon$ for initial surplus $x$ such that $V(x)\le V_{\theta _\epsilon }(x)+ \epsilon$; then for initial surplus $y$, consider the admissible strategy $\theta _y$ with initial capital injection $x-y$ followed by applying strategy $\theta _\epsilon$; hence we obtain

$$V(y)\ge V_{\theta_y} =V_{\theta_\epsilon}(x) - \phi(x-y) -k \ge V(x) - \phi(x-y) -k -\epsilon.$$

For the other inequality in (2.4), the proof is similar by consider an $\epsilon$-optimal strategy for initial surplus $y$ and an admissible strategy for initial surplus $y$ with immediate dividend payment $x-y$. The increasing property is a direct consequence of (2.4). In addition, for $x\ge 0$, we consider a special dividend strategy $d$ where the initial surplus is paid as lump sum dividend at time zero and premium income are paid continuously as dividend for all $t\ge 0$, i.e. $L^{d}_t = x + ct$; hence, we obtain an upper bound $\int _{0}^{\infty }e^{-\delta t} {{\rm d}}L^{d}_t$ for the performance function under any admissible strategy in $\Theta$, that is

$$\int_{0}^{\infty} e^{-\delta t} \,{{\rm d}}L^{d}_t = x+ \frac{c}{\delta}<\infty,$$

which is exactly the right-hand side of (2.5). The linear lower bound can be obtained by considering the admissible strategy $\theta$ with dividend strategy follows $L^{d}_t = x + ct$ and no capital injection, then ruin will occur at the arrival time of the first claim, hence

$$V(x)\ge V_\theta(x) = \mathbb{E}_{x}\left[\int_{0}^{T_1}e^{-\delta t}\,{{\rm d}}L^{d}_t + e^{-\delta T_1}\pi({-}Y_1) \right] = x + \frac{c+ \lambda(K-\Phi\mu)}{\lambda+\delta},$$

where $T_1$ and $Y_1$ are the arrival time and amount of the first claim, respectively; $K<0$ and $\Phi \in [0,1]$ are the aforementioned parameters in the penalty function $\pi$. To prove the locally Lipschitz continuity, we consider initial surplus $y\ge 0$ and any $\epsilon >0$, let $\theta _x$ denotes the $\epsilon$-optimal strategy for any $x>y$, i.e. $V_{\theta _x}(x)\ge V(x)-\epsilon$. Then, consider another strategy $\theta _y$ with initial surplus $y$ that pays no dividends and no capital injections if $U^{\theta _y}_t< x$ and follows strategy $\theta _x$ if $U^{\theta _y}_t$ reaches $x$; then, $\theta _y$ is obviously an admissible strategy. Hence, we have

$$V(y)\ge V_{\theta_y}(y)\ge V_{\theta_x}(x)e^{-(\lambda+\delta) ({(x-y)}/{c})} \ge (V(x)-\epsilon)e^{-(\lambda+\delta) ({(x-y)}/{c})},$$

then, we obtain

$$V(x)-V(y)\le (e^{(\lambda+\delta)({(x-y)}/{c})} -1)V(x).$$

Finally, the upper semi-continuity at 0 can be derived by including the take-the-money-and-run strategy at 0 into our admissible set, which is reasonable since it should have a higher expected return to run the business rather than simply declare to ruin when surplus is at 0 (see e.g. [Reference Xu and Woo18]). Hence, we have $V(0)\ge \pi (0)$.

Proof. We only prove that $\mathcal {M}\varphi (x)$ is linearly bounded here, for the proof of increasing, Lipschitz continuous can refer to Xu and Woo [Reference Xu and Woo18] Lemma 5.1. Note that,

\begin{align*} \mathcal{M}\varphi(x)& = \sup_{y\ge 0}\{\varphi(x+y) - (k+ \phi y) \}\\ & \le\sup_{y\ge 0}\{x+y + m - (k+ \phi y) \}\\ & = x + m-k + \sup_{y\ge 0}\{ (1-\phi) y \}= x+m-k, \end{align*}

where the last equation holds since $\phi \ge 1$.

Proof. The proof follows easily from the $\sup$ manipulations (see e.g. [Reference Seydel15]), i.e.

$$\sup_x(f(x)+ g(x)) \le \sup_x(f(x)) + \sup_x(g(x)),$$

and

$$\sup_x(f(x)+ g(x)) \ge \sup_x(f(x)) + \inf_x(g(x)).$$

Hence, for any $x\ge 0$,

\begin{align*} \mathcal{M}(h f + (1-h)g)(x) & = \sup_{y\ge 0}\{h f(x+y) + (1-h)g(x+y) -\phi y-k \}\\ & =\sup_{y\ge 0}\{h (f(x+y)-\phi y - k) + (1-h)(g(x+y) -\phi y-k) \}\\ & \le h \sup_{y\ge 0}\{f(x+y)-\phi y - k \} +(1-h)\sup_{y\ge 0}\{g(x+y)-\phi y - k\}\\ & = h\mathcal{M}f(x) + (1-h)\mathcal{M}g(x). \end{align*}

Similarly,

\begin{align*} \mathcal{M}({-}h f + (1+h)g) & = \sup_{y\ge 0}\{{-}h f(x+y) + (1+h)g(x+y) -\phi y-k \}\\ & =\sup_{y\ge 0}\{{-}h (f(x+y)-\phi y - k) + (1+h)(g(x+y) -\phi y-k) \}\\ & \ge -h \sup_{y\ge 0}\{f(x+y)-\phi y - k \} +(1+h)\sup_{y\ge 0}\{g(x+y)-\phi y - k\}\\ & ={-}h\mathcal{M}f(x) + (1+h)\mathcal{M}g(x). \end{align*}

Proof. For the proof of (3.1), we follow the analysis in Schmidli [Reference Schmidli14] Lemma 2.38. Note that we work under the assumptions that dividend payments and capital injections cannot occur simultaneously, and any dividend payment should not result in capital injection and vice versa. Since $\Theta ^{r}_1 \subset \Theta ^{r}_2 \subset \Theta$, where $\Theta ^{r}_1$ and $\Theta ^{r}_2$ are the corresponding sets of admissible strategies with ceiling dividend rates $n_1$ and $n_2$ with $n_1 < n_2$, respectively (see e.g. [Reference Xu and Woo18]). Then, one has $V^{n}(x)$ is an increasing sequence of $n$ and $\limsup _{n\to \infty }V^{n}(x) \le V(x)$. Next, to show that $V(x) \le \liminf _{n\to \infty } V^{n}(x)$, we consider, for each $\epsilon > 0$, a dividend strategy $d_j$ with pure jump of size that is greater or equal to $\epsilon$, and combine with an admissible capital injection strategy $v$ such that $V_{(d_j,v)}(x) \ge V(x) - 2\epsilon$. On the other hand, we construct another dividend strategy $\tilde {d}_j$ with absolutely continuous dividend density that is bounded by $n$. To be specific, under strategy $\tilde {d}_j$, dividends will start to be paid at rate $n$ when a lump sum dividend payment occurs in strategy $d_j$ until the accumulated amount of dividend meet with the lump sum in strategy $d_j$. Hence, with sufficiently large $n$, the difference between the performance function of these two strategies is bounded by $\epsilon$. Therefore, we have $V(x) - V_{(\tilde {d},v)}(x) \le 3\epsilon$. Then, by letting $\epsilon \to 0$, we obtain that $V(x) \le \liminf _{n\to \infty } V^{n}(x)$. Then, (3.1) holds true.

For (3.2), since $V^{n}(x)$ is increasing in $n$ and absolutely continuous with respective to $x$, then there must exist $n^{*} > n$ and $y^{*} \ge 0$ such that

$$\sup_{y\ge 0} \{V^{n}(x+y) - (k+\phi y) \} \le V^{n^{*}}(x+y^{*}) - (k+\phi y^{*}),$$

let $n,n^{*} \to \infty$, one arrives at

\begin{align*} & \lim_{n\to \infty} \sup_{y\ge 0}\{V^{n}(x+y) - (k+\phi y) \}\\ & \quad \le V(x+y^{*}) -(k+\phi y^{*}) \le \sup_{y\ge 0}\{V(x+y) - (k+\phi y) \}. \end{align*}

Meanwhile, for each $y^{\prime } \ge 0$,

$$\sup_{y\ge 0} \{V^{n}(x+y) - (k+\phi y) \} \ge V^{n}(x+y^{\prime}) - (k+\phi y^{\prime})$$

then, let $n \to \infty$, one arrives at

$$\lim_{n\to \infty} \sup_{y\ge 0}\{V^{n}(x+y) - (k+\phi y) \} \ge V(x+y^{\prime}) -(k+\phi y^{\prime}),$$

since $y^{\prime }$ is arbitrary, one has

$$\lim_{n\to \infty} \sup_{y\ge 0}\{V^{n}(x+y) - (k+\phi y) \} \ge \sup_{y\ge 0}\{ V(x+y) -(k+\phi y) \}.$$

Then, one completes the proof.

Proof. Note that according to Lemma 5.2 in Xu and Woo [Reference Xu and Woo18] and Proposition 3.1, one directly has $V(x)\ge \mathcal {M}V(x)$ for $x\ge 0$. And for $x<0$, by definition of the value function, $V(x) = \pi (x)$.

1. (i) $V$ is subsolution: For any $x\in [0,\infty )$ and $h\in C^{1}(0,\infty )$ such that $h\ge V$ on $[0,\infty )$ and $h(x) = V(x)$, we need to show that

   (3.5) \begin{equation} \max\{(\mathcal{A}_\pi-\delta)h(x), 1-h^{\prime}(x), \mathcal{M}h(x) -h(x) \}\ge 0. \end{equation}
   When $h(x) = V(x) = \mathcal {M}V(x) = \mathcal {M}h(x)$, (3.5) holds trivially, hence we focus on the case $V(x)>\mathcal {M}V(x)$. If $1-h^{\prime }(x) \ge 0$, then (3.5) holds true obviously. Finally, if $1-h^{\prime }(x) < 0$, we consider a sequence $d_n\uparrow \infty$ as $n\to \infty$, which corresponds to the ceiling rate for the value function $V^{d_n}$ with bounded dividend density Xu and Woo [Reference Xu and Woo18] Eq. (3.1), such that there exists an associated sequence of functions $h_n\in C^{1}(0,\infty )$ with $h_n$ converges to $h$ uniformly on compact sets and $h^{\prime }_n(x) \to h^{\prime }(x)$ when $n\to \infty$, and $h_n\ge V^{d_n}$ on $[0,\infty )$ with $h_n(x) = V^{d_n}(x)$. Then according to Xu and Woo [Reference Xu and Woo18] Proposition 5.1 and Proposition 3.1, for sufficiently large $n$, we have $V^{d_n}(x)>\mathcal {M}V^{d_n}(x)$, then one must have,
   (3.6) \begin{equation} \sup_{d\in [0,d_n]}\{(\mathcal{A}_\pi -\delta)h_n(x) + (1- h^{\prime}_n(x))d \}\ge 0. \end{equation}
   Since $1-h^{\prime }(x) < 0$, there exists a sufficiently large $\bar {n}$ such that for all $n>\bar {n}$, one has $1-h^{\prime }_n(x) < 0$, and (3.6) becomes $(\mathcal {A}_\pi -\delta )h_n(x) \ge 0$, then, by letting $n\to \infty$, one arrives at $(\mathcal {A}_\pi -\delta )h(x) \ge 0$. Hence, (3.5) holds.

2. (ii) $V$ is supsolution: For any $x\in [0,\infty )$, we have $V(x)\ge \mathcal {M}V(x)$. Then, it remains to show that for any $h\in C^{1}(0,\infty )$ with $h\le V$ on $[0,\infty )$ and $h(x) = V(x)$, one has

   $$\max\{(\mathcal{A}_\pi - \delta)h(x), 1-h^{\prime}(x) \} \le 0.$$
   Similarly, we consider a sequence $d_n \uparrow \infty$ as $n\to \infty$ such that there exists an associated sequence of test function $h_n\in C^{1}(0,\infty )$ with $h_n\le V^{d_n}$ on $[0,\infty )$, $h_n(x)=V^{d_n}(x)$, and $h_n$ converges uniformly to $h$ on compact sets, $h^{\prime }_n(x)\to h^{\prime }(x)$ when $n\to \infty$. Then, from Xu and Woo [Reference Xu and Woo18] Proposition 5.1, one has
   (3.7) \begin{equation} \sup_{d\in[0,d_{n}]}\{(\mathcal{A}_\pi -\delta )h_n(x) + (1-h_n^{\prime}(x))d \} \le 0,\quad \text{for all }n. \end{equation}
   Then, one has $1-h_n^{\prime }(x)\le 0$ for sufficiently large $n$, therefore $1-h^{\prime }(x)\le 0$. In addition, for sufficiently large $n$ when $1-h_n^{\prime }(x)\le 0$, (3.7) reduces to $(\mathcal {A}_\pi -\delta )h_n(x)\le 0$, and by letting $n\to \infty$, one has $(\mathcal {A}_\pi -\delta )h(x)\le 0$. Then, we complete the proof.

Proof. Let $\bar {h}\in \mathcal {LB}^{\pi }(\mathbb {R})$ be a viscosity supersolution of (3.3). According to Lemma A.1, there exists a sequence of continuously differentiable functions $h_n$ on $\mathbb {R}$ satisfy the condition (iv) in $\mathcal {LB}^{\pi }(\mathbb {R})$ with $h_n(x) = \pi (x)$ for $x<0$, such that $h_n \le \bar {h}$ on $[0,\infty )$, and when $h_n(x) =\bar {h}(x)$ we have $(\mathcal {A}_\pi - \delta )h_n(x) \le 0$ and $h^{\prime }_n(x) \ge 1$ for $x\ge 0$. In addition, $h_n$ converges uniformly to $\bar {h}$ on compact sets and $h^{\prime }_n(x)$ converges to $\bar {h}^{\prime }(x)$ ${\rm a.e.}$ Then, let us consider the controlled process $U^{\theta }$ with an arbitrary admissible strategy $\theta =(d,v)$. Denote the cumulative dividend process as

$$L^{d}_t = \int_{0}^{t}{{\rm d}} \tilde{L}^{d}_t + \sum_{L^{d}_{s+}\neq L^{d}_s}(L^{d}_{s+} - L^{d}_{s}),$$

where $\tilde {L}^{d}_t$ denotes the continuous part of the dividend process, and $L^{d}_{s+} - L^{d}_{s}$ denotes the corresponding jump components. In addition, let capital injection strategy be an impulse strategy $v= (\omega _1,\omega _2,\ldots ;\zeta _1,\zeta _2,\ldots )$. We apply Itô's formula within the interval $[\omega ^{+}_i\wedge \tau ^{\theta }, \omega _{i+1}\wedge \tau ^{\theta })$, then we arrive at

\begin{align*} & e^{-\delta (\omega_{i+1}\wedge\tau^{\theta})} h_n(X^{\theta}_{\omega_{i+1}\wedge\tau^{\theta}} ) - e^{-\delta( \omega^{+}_i\wedge\tau^{\theta})} h_n (X^{\theta}_{\omega^{+}_i\wedge\tau^{\theta}} ) \\ & \quad =c\int_{\omega^{+}_i\wedge\tau^{\theta}}^{\omega_{i+1}\wedge\tau^{\theta}}e^{-\delta s}h^{\prime}_n(X^{\theta}_{s^{-}}) \,{{\rm d}}s - \int_{\omega^{+}_i\wedge\tau^{\theta}}^{\omega_{i+1}\wedge\tau^{\theta}}e^{-\delta s}h^{\prime}_n(X^{\theta}_{s^{-}}) \,{{\rm d}} \tilde{L}^{d}_s -\delta\int_{\omega^{+}_i\wedge\tau^{\theta}}^{\omega_{i+1}\wedge\tau^{\theta}}e^{-\delta s}h_n(X^{\theta}_{s^{-}})\,{{\rm d}}s \\ & \qquad + \sum_{\substack{X^{\theta}_{s}\neq X^{\theta}_{s^{-}}\\ s\in (\omega^{+}_i\wedge\tau^{\theta}, \omega_{i+1}\wedge\tau^{\theta}]}}e^{-\delta s}(h_n(X^{\theta}_{s}) - h_n(X^{\theta}_{s^{-}}) ) \\ & \qquad + \sum_{\substack{X^{\theta}_{s^{+}}\neq X^{\theta}_{s}\\ s\in [\omega^{+}_i\wedge\tau^{\theta}, \omega_{i+1}\wedge\tau^{\theta})}}e^{-\delta s}(h_n(X^{\theta}_{s^{+}}) - h_n(X^{\theta}_{s}) ). \end{align*}

Note that within the interval $[\omega ^{+}_i\wedge \tau ^{\theta }, \omega _{i+1}\wedge \tau ^{\theta })$, the jumps of $X^{\theta }_{s^{+}}-X^{\theta }_{s}$ is equal to the jumps of $L^{d}_s$, that is $X^{\theta }_{s^{+}} - X^{\theta }_{s} = -(L^{d}_{s^{+}} - L^{d}_{s})$, then with the fact that $h_n^{\prime }(\cdot )\ge 1$, we have

(4.1) \begin{align} & -\int_{\omega^{+}_i\wedge\tau^{\theta}}^{\omega_{i+1}\wedge\tau^{\theta}}e^{-\delta s}h^{\prime}_n(X^{\theta}_{s^{-}}) \,{{\rm d}} \tilde{L}^{d}_s +\sum_{\substack{X^{\theta}_{s+}\neq X^{\theta}_{s}\\ s\in [\omega^{+}_i\wedge\tau^{\theta}, \omega_{i+1}\wedge\tau^{\theta})}}e^{-\delta s}(h_n(X^{\theta}_{s+}) - h_n(X^{\theta}_{s}) ) \nonumber\\ & \quad ={-} \int_{\omega^{+}_i\wedge\tau^{\theta}}^{\omega_{i+1}\wedge\tau^{\theta}}e^{-\delta s} h^{\prime}_n(X^{\theta}_{s^{-}})\,{{\rm d}} \tilde{L}^{d}_s - \sum_{\substack{L^{u}_{s+}\neq L^{d}_{s}\\ s\in [\omega^{+}_i\wedge\tau^{\theta}, \omega_{i+1}\wedge\tau^{\theta})}}e^{-\delta s}\left(\int_{0}^{L^{d}_{s+} - L^{d}_s}h^{\prime}_n(X^{\theta}_{s^{-}} - y )\,{{\rm d}}y \right) \nonumber\\ & \quad \le - \int_{\omega^{+}_i\wedge\tau^{\theta}}^{\omega_{i+1}\wedge\tau^{\theta}}e^{-\delta s} \,{{\rm d}} \tilde{L}^{d}_s - \sum_{\substack{L^{d}_{s+}\neq L^{d}_{s}\\ s\in [\omega^{+}_i\wedge\tau^{\theta}, \omega_{i+1}\wedge\tau^{\theta})}}e^{-\delta s}(L^{d}_{s+} - L^{d}_s ) \nonumber\\ & \quad ={-}\int_{\omega^{+}_i\wedge\tau^{\theta}}^{\omega_{i+1}\wedge\tau^{\theta}}e^{-\delta s} \,{{\rm d}}L^{d}_s . \end{align}

On the other hand, for the jumps $X^{\theta }_{s} - X^{\theta }_{s-}$ which only related to the arrival of claims, we define

\begin{align*} M_t & = \sum_{\substack{X^{\theta}_{s}\neq X^{\theta}_{s-}\\ s\le t}}e^{-\delta s}(h_n(X^{\theta}_{s}) - h_n(X^{\theta}_{s-}) )\\ & \quad - \lambda \int_{0}^{t}e^{-\delta s} \int_{0}^{\infty}(h_n(X^{\theta}_{s-} - y) - h_n(X^{\theta}_s))\,{{\rm d}}F(y) \,{{\rm d}}s, \end{align*}

which is obviously a zero mean martingale; then, one can obtain that

\begin{align*} & e^{-\delta (\omega_{i+1}\wedge\tau^{\theta})} h_n(X^{\theta}_{\omega_{i+1}\wedge\tau^{\theta}} ) - e^{-\delta( \omega^{+}_i\wedge\tau^{\theta})} h_n (X^{\theta}_{\omega^{+}_i\wedge\tau^{\theta}} )\\ & \quad \le-\int_{\omega^{+}_i\wedge\tau^{\theta}}^{\omega_{i+1}\wedge\tau^{\theta}}e^{-\delta s} \,{{\rm d}}L^{d}_s + \int_{\omega^{+}_i\wedge\tau^{\theta}}^{\omega_{i+1}\wedge\tau^{\theta}}e^{-\delta s}(\mathcal{A} - \delta )h_n(X^{\theta}_{s-})\,{{\rm d}}s + (M_{\omega_{i+1}\wedge\tau^{\theta}} - M_{\omega^{+}_i\wedge\tau^{\theta}}), \end{align*}

where

\begin{align*} (\mathcal{A}-\delta)h_n(x) & = ch^{\prime}_n(x) - (\lambda +\delta)h_n(x)+ \lambda\int_{0}^{\infty}h_n(x-y)\,{{\rm d}}F(y)\\ & = ch^{\prime}_n(x) - (\lambda +\delta)h_n(x)+ \lambda\int_{0}^{x}h_n(x-y)\,{{\rm d}}F(y)+\lambda\int_{x}^{\infty}\pi(x-y)\,{{\rm d}}F(y)\\ & = (\mathcal{A}_\pi-\delta)h_n(x). \end{align*}

By taking expectation on both sides of the above inequality, one arrives at

(4.2) \begin{align} & \mathbb{E}_x[ e^{-\delta (\omega_{i+1}\wedge\tau^{\theta})}h_n(X^{\theta}_{\omega_{i+1}\wedge\tau^{\theta}} )] - \mathbb{E}_x[e^{-\delta( \omega^{+}_i\wedge\tau^{\theta})} h_n (X^{\theta}_{\omega^{+}_i\wedge\tau^{\theta}} )] \nonumber\\ & \quad \le -\mathbb{E}_x\left[\int_{\omega^{+}_i\wedge\tau^{\theta}}^{\omega_{i+1}\wedge\tau^{\theta}}e^{-\delta s} \,{{\rm d}}L^{d}_s \right]+ \mathbb{E}_x\left[\int_{\omega^{+}_i\wedge\tau^{\theta}}^{\omega_{i+1}\wedge\tau^{\theta}}e^{-\delta s}(\mathcal{A}_\pi - \delta )h_n(X^{\theta}_{s-})\,{{\rm d}}s \right]. \end{align}

Summing both sides of (4.2) from $i=0$ to $i=m$, it follows that

(4.3) \begin{align} & h_n(x) + \sum_{i=1}^{m} \mathbb{E}_{x}[e^{-\delta (\omega_i\wedge\tau^{\theta})}(h_n (X^{\theta}_{\omega^{+}_i\wedge\tau^{\theta}} ) - h_n(X^{\theta}_{\omega_{i}\wedge\tau^{\theta}} ) )] - \mathbb{E}_{x}[e^{-\delta (\omega_{m+1}\wedge\tau^{\theta})}h_n (X^{\theta}_{\omega_{m+1}\wedge\tau^{\theta}} ) ] \nonumber\\ & \quad \ge \mathbb{E}_{x}\left[\int_{0}^{\omega_{m+1}\wedge\tau^{\theta}}e^{-\delta s} \,{{\rm d}}L^{d}_s\right] - \mathbb{E}_x\left[\int_{0}^{\omega_{m+1}\wedge\tau^{\theta}}e^{-\delta s}(\mathcal{A}_\pi - \delta )h_n(X^{\theta}_{s-})\,{{\rm d}}s \right]. \end{align}

Note that when there is a capital injection before $\tau ^{\theta }$, the following equation holds

(4.4) \begin{equation} X^{\theta}_{\omega^{+}_i} = X^{\theta}_{\omega_i} + \zeta_i. \end{equation}

Hence when $\omega _i < \tau ^{\theta }$, from (4.4) and (2.6) one has

$$h_n (X^{\theta}_{\omega^{+}_i} ) = h_n (X^{\theta}_{\omega_i} + \zeta_i ) \le \mathcal{M}h_n\left(X^{\theta}_{\omega_i}\right) + k+\phi\zeta_i,$$

which yields

$$h_n(X^{\theta}_{\omega^{+}_i}) - h_n(X^{\theta}_{\omega_{i}}) \le \mathcal{M}h_n(X^{\theta}_{\omega_i}) - h_n(X^{\theta}_{\omega_{i}}) + k + \phi \zeta_i.$$

But, when $\omega _i \ge \tau ^{\theta }$, we obtain $h_n(X^{\theta }_{\omega ^{+}_i\wedge \tau ^{\theta }} ) = h_n (X^{\theta }_{\omega _{i}\wedge \tau ^{\theta }}) = \pi (X^{\theta }_{\omega _{i}\wedge \tau ^{\theta }})$. Hence, it follows that (4.3) may be expressed as

(4.5) \begin{align} & h_n(x) + \sum_{i=1}^{m}\mathbb{E}_{x} [e^{-\delta \omega_i}(\mathcal{M}h_n (X^{\theta}_{\omega_i} ) - h_n(X^{\theta}_{\omega_{i}}) ) {1}_{\{\omega_i <\tau^{\theta} \}}]\nonumber\\ & \quad \ge \mathbb{E}_x\left[\int_{0}^{\omega_{m+1}\wedge\tau^{\theta}}e^{-\delta s}\,{{\rm d}}L^{d}_s + e^{-\delta (\omega_{m+1}\wedge\tau^{\theta})}h_n (X^{\theta}_{\omega_{m+1}\wedge\tau^{\theta}} ) - \sum_{i=1}^{m}e^{-\delta \omega_i }(k+ \phi \zeta_i){1}_{\{\omega_i <\tau^{\theta} \}} \right] \nonumber\\ & \qquad - \mathbb{E}_x\left[\int_{0}^{\omega_{m+1}\wedge\tau^{\theta}}e^{-\delta s}(\mathcal{A}_\pi - \delta )h_n(X^{\theta}_{s-})\,{{\rm d}}s \right]. \end{align}

Next, we show that

(4.6) \begin{align} & \lim_{m,n \to \infty} \mathbb{E}_x\left[\int_{0}^{\omega_{m+1}\wedge\tau^{\theta}}e^{-\delta s}(\mathcal{A}_\pi - \delta )h_n(X^{\theta}_{s-})\,{{\rm d}}s \right] \nonumber\\ & \quad =\mathbb{E}_x\left[\int_{0}^{\tau^{\theta}}e^{-\delta s}(\mathcal{A}_\pi - \delta )\bar{h}(X^{\theta}_{s-})\,{{\rm d}}s \right] \le 0, \end{align}

where the second inequality holds true since $\bar {h}$ is a viscosity supersolution of (3.3) for $x\ge 0$; note that we also have $\bar {h}^{\prime }(x)\ge 1\ {\rm a.e.}$ In addition, since $h_n$ converges to $\bar {h}$ uniformly on compact sets and $h_n^{\prime }(x)$ converges to $\bar {h}^{\prime }(x)\ {\rm a.e.}$; then, we have

$$e^{-\delta s}(\mathcal{A}_\pi - \delta )h_n(X^{\theta}_{s-}) \stackrel{n\to \infty}{\longrightarrow} e^{-\delta s}(\mathcal{A}_\pi - \delta )\bar{h}(X^{\theta}_{s-}) \quad {\rm a.e.}$$

In addition, with a similar analysis as in Azcue and Muler [Reference Azcue and Muler2] Lemma A.2, we have for $x\ge 0$,

$$1\le \bar{h}^{\prime}(x) \le \frac{\lambda +\delta }{c}\bar{h}(x) - \frac{\lambda}{c}\int_{0}^{x}\bar{h}(x-y)\,{{\rm d}}F(y) - \frac{\lambda}{c}M(x) \le \frac{\lambda \bar{F}(x) + \delta}{c}\bar{h}(x) - \frac{\lambda}{c}M(x) \quad {\rm a.e.},$$

where

$$M(x) = \int_{x}^{\infty}\Pi(x-y)\,{{\rm d}}F(y)< 0,$$

and

$$1\le h_n^{\prime}(x) \le \frac{\lambda \bar{F}(x) + \delta}{c}h_n(x) - \frac{\lambda}{c}M(x).$$

Hence,

\begin{align*} & e^{-\delta s}\vert(\mathcal{A}_\pi - \delta )\bar{h}(X^{\theta}_{s-}) - (\mathcal{A}_\pi - \delta )h_n(X^{\theta}_{s-})\vert\\ & \quad \le e^{-\delta s} \left(c\bar{h}^{\prime}(X^{\theta}_{s-}) + (\lambda+ \delta)\bar{h}(X^{\theta}_{s-}) + \lambda\int_{0}^{X^{\theta}_{s-}}\bar{h}(X^{\theta}_{s-}-y)\,{{\rm d}}F(y) + \lambda |M(X^{\theta}_{s-})|\right)\\ & \qquad + e^{-\delta s} \left(ch_n^{\prime}(X^{\theta}_{s-}) + (\lambda+ \delta)h_n(X^{\theta}_{s-}) + \lambda\int_{0}^{X^{\theta}_{s-}}h_n(X^{\theta}_{s-}-y)\,{{\rm d}}F(y) + \lambda |M(X^{\theta}_{s-})| \right)\\ & \quad \le e^{-\delta s}(\lambda + 2\delta)(\bar{h}(X^{\theta}_{s-}) + h_n(X^{\theta}_{s-})) + 4\lambda e^{-\delta s}|M(X^{\theta}_{s-})|\\ & \quad \le 2 e^{-\delta s}(\lambda + 2\delta)(x+ cs + N) + 4 \lambda e^{-\delta s} M \end{align*}

for sufficiently large $N$ and $M$. Therefore, $e^{-\delta s}\vert (\mathcal {A}_\pi - \delta )\bar {h}(X^{\theta }_{s-}) - (\mathcal {A}_\pi - \delta )h_n(X^{\theta }_{s-})\vert$ is bounded by a positive integrable function as shown above, then (4.6) holds true by dominated convergence theorem. Finally, with the help of monotone and bounded convergence theorem, we let $m, n \to \infty$ on both sides of (4.5), and utilize the uniformly convergence of $h_n$ to $\bar {h}$ and $\bar {h}(x) = \pi (x)$ for $x<0$, and (4.6), we arrive at

(4.7) \begin{align} & \bar{h}(x) + \sum_{i=1}^{\infty}\mathbb{E}_{x} [e^{-\delta \omega_i}(\mathcal{M}\bar{h} (X^{\theta}_{\omega_i} ) - \bar{h}(X^{\theta}_{\omega_{i}}) ) {1}_{\{\omega_i <\tau^{\theta} \}} ]\nonumber\\ & \quad \ge \mathbb{E}_x\left[\int_{0}^{\tau^{\theta}}e^{-\delta s}\,{{\rm d}}L^{d}_s + e^{-\delta \tau^{\theta}}\pi(X^{\theta}_{\tau^{\theta}} ){1}_{\{\tau^{\theta} <\infty\}} - \sum_{i=1}^{\infty}e^{-\delta \omega_i }(k+ \phi \zeta_i){1}_{\{\omega_i <\tau^{\theta} \}} \right]. \end{align}

In addition, since $\mathcal {M}\bar {h}(x) - \bar {h}(x)\le 0$ for all $x\in [0,\infty )$ and the strategy $\theta$ is arbitrary, we get

$$\bar{h}(x)\ge V(x).$$

Proof. Since $\xi$ is a subsolution of (3.3), then according to Definition 3.1, for any continuously differentiable function $h$ with $h\ge \xi$ and $h(x) = \xi (x)$, we have

$$\max\{(\mathcal{A}_\pi-\delta)h(x), 1-h^{\prime}(x), \mathcal{M}h(x) -h(x) \}\ge 0.$$

Then, we construct the continuously differentiable function $h_m = (1+{1}/{m})h - ({1}//{m})w$ such that $h_m\ge \xi _m$ and at $x$ where $h_m(x) = \xi _m(x)$, with the help of Lemma 2.3 we must have

\begin{align*} & \max\{(\mathcal{A}_\pi-\delta)h_m(x), 1-h_m^{\prime}(x), \mathcal{M}h_m(x) -h_m(x)\}\\ & \quad =\max \left\{\left(1+\frac{1}{m}\right)(\mathcal{A}_\pi-\delta)h(x) - \frac{1}{m}(\mathcal{A}_\pi-\delta)w(x), \right.\\ & \qquad \left.\left(1+\frac{1}{m}\right)(1-h^{\prime}(x)) - \frac{1}{m}(1-w^{\prime}(x)), \mathcal{M}h_m(x) -h_m(x)\right\} \\ & \quad \ge \max\left\{\left(1+\frac{1}{m}\right)(\mathcal{A}_\pi-\delta)h(x) - \frac{1}{m}(\mathcal{A}_\pi-\delta)w(x), \right.\\ & \qquad \left(1+\frac{1}{m}\right)(1-h^{\prime}(x)) - \frac{1}{m}(1-w^{\prime}(x)),\\ & \qquad \left.\left(1 +\frac{1}{m}\right)(\mathcal{M}h(x) -h(x)) - \frac{1}{m}(\mathcal{M}w(x) -w(x))\right\} \ge \frac{\kappa(x)}{m}. \end{align*}

The proof for $\eta _m$ is similar, we omit the detail here.

Proof. The proof follows the method used in Albrecher and Thonhauser [Reference Albrecher and Thonhauser1], see also Azcue and Muler [Reference Azcue and Muler2]; however, difficulties raised from the capital injection part in the HJBQVI, which is resolved by utilizing the method discussed in Seydel [Reference Seydel15]. Let $\eta _m$ for $m\in \mathbb {N}$ as defined in Lemma 4.1. Then, it is sufficient to show that $\xi \le \eta _m$ for all $m$ large. For any fixed $m\in \mathbb {N}$, let $0< M :=\sup _{x\ge 0}\{\xi (x)- \eta _m(x)\} <\infty$, and $x^{*}:= \arg\!\max _{x\ge 0}\{\xi (x)- \eta _m(x)\}$. Since $\xi (x)$ is linearly bounded and $\eta _m(x)$ is increasing as polynomial function with degree $p>1$, then we can find a sufficient large $B$ such that $\xi (x)-\eta _m(x)\le 0$ for $x>B$. Furthermore, since $\xi$ and $\eta _m$ are locally Lipschitz continuous, there exists a constant $n>0$ such that

(4.10) \begin{equation} \frac{\xi(y) - \xi(x)}{y-x} \le n, \quad \frac{\eta_m(y) - \eta_m(x)}{y-x}\le n, \quad \text{for } 0\le x\le y \le B. \end{equation}

Then, we consider a set $A$ as

$$A = \{(x,y)\,|\, 0\le x\le y \le B\},$$

we define an auxiliary function

$$H_\epsilon(x,y) := \xi(x)-\eta_m(y) - \frac{\epsilon}{2}(x-y)^{2} - \frac{2n}{\epsilon^{2}(y-x) + \epsilon},$$

and let $M_\epsilon := \sup _{(x,y)\in A}H_{\epsilon }(x,y)$ with the maximizer $(x_{\epsilon }, y_{\epsilon })$. Then, it is obvious that

$$M_{\epsilon} \ge H_{\epsilon}(x^{*},x^{*}) = M - \frac{2n}{\epsilon},$$

which is positive for sufficient large $\epsilon$, then we arrive at

$$\liminf_{\epsilon\to \infty}M_{\epsilon} \ge M >0.$$

Note that we shall prove that the maximizer $(x_{\epsilon }, y_{\epsilon })$ is not on the boundary of set $A$ in order to retain the differentiability at $x_{\epsilon }$ and $y_{\epsilon }$. We postpone the proof to Lemma A.2 in the Appendix. Next, we introduce the other two auxiliary functions,

(4.11) \begin{align} u(x) & = \eta_m(y_\epsilon) + \frac{\epsilon}{2}(x-y_\epsilon)^{2} + \frac{2n}{\epsilon^{2}(y_\epsilon -x) +\epsilon} + H_{\epsilon}(x_\epsilon, y_{\epsilon}), \end{align}

(4.12) \begin{align} v(y) & = \xi(x_\epsilon) -\frac{\epsilon}{2}(x_\epsilon-y)^{2} - \frac{2n}{\epsilon^{2}(y -x_\epsilon) +\epsilon} - H_{\epsilon}(x_\epsilon, y_{\epsilon}). \end{align}

Note that $u$ and $v$ are continuously differentiable, and $\xi (x) - u(x) = H_{\epsilon }(x,y_\epsilon ) - H_{\epsilon }(x_\epsilon,y_\epsilon ) \le 0,$ which reaches the maximum 0 at $x_\epsilon$, i.e. $\xi (x_\epsilon ) = u(x_\epsilon )$. Similarly, $\eta _m(y) - v(y) = H_{\epsilon }(x_\epsilon,y_\epsilon ) - H_{\epsilon }(x_\epsilon,y) \ge 0$ and reaches the minimum at $y_\epsilon$, i.e. $\eta _m(y_\epsilon ) = v(y_\epsilon )$. Since $\xi$ is a subsolution of (3.3) and $\eta _m$ is a supersolution of (4.9), we have at the points $x_\epsilon$ and $y_\epsilon$

\begin{align*} & \max\{(\mathcal{A}_\pi-\delta)(\xi, u)(x_\epsilon), 1-u^{\prime}(x_\epsilon), \mathcal{M}\xi(x_\epsilon) -\xi(x_\epsilon) \}\ge 0 ,\\ & \max\{(\mathcal{A}_\pi-\delta)(\eta_m, v)(y_\epsilon), 1-v^{\prime}(y_\epsilon), \mathcal{M}\eta_m(y_\epsilon) -\eta_m(y_\epsilon) \}\le -\frac{\kappa}{m}, \end{align*}

where $\kappa = \kappa (y_\epsilon )>0$, $\kappa (\cdot )$ is the positive function introduced in Lemma 4.1, and

$$(\mathcal{A}_\pi-\delta)(\xi, u)(x_\epsilon) = cu^{\prime}(x_\epsilon) - (\lambda +\delta)\xi(x_\epsilon) + \lambda\int_{0}^{x_\epsilon}\xi(x_\epsilon-y)\,{{\rm d}}F(y)+ \lambda\int_{x_\epsilon}^{\infty}\pi(x_\epsilon-y)\,{{\rm d}}F(y),$$

and

$$(\mathcal{A}_\pi-\delta)(\eta_m, v)(y_\epsilon) = cv^{\prime}(y_\epsilon) - (\lambda +\delta)\eta_m(y_\epsilon) + \lambda\int_{0}^{y_\epsilon}\eta_m(y_\epsilon-z)\,{{\rm d}}F(z)+ \lambda\int_{y_\epsilon}^{\infty}\pi(y_\epsilon-z)\,{{\rm d}}F(z),$$

which are the operators used in an equivalent formulation of viscosity solution comparing to Definition 3.1, (see e.g. [Reference Azcue and Muler2] Remark 3.3).

By the definition of $u$ and $v$, we have

$$u^{\prime}(x_\epsilon) = v^{\prime}(y_\epsilon) = \epsilon(x_\epsilon - y_\epsilon) + \frac{2 n}{(\epsilon(y_\epsilon - x_\epsilon) + 1)^{2}}.$$

On the other hand, since

$$H_{\epsilon}(x_\epsilon,x_\epsilon)+ H_{\epsilon}(y_\epsilon,y_\epsilon)\le 2H_{\epsilon}(x_\epsilon,y_\epsilon),$$

one has

\begin{align*} & \xi(x_\epsilon)-\eta_m(x_\epsilon) + \xi(y_\epsilon)-\eta_m(y_\epsilon) - \frac{4n}{ \epsilon}\\ & \quad \le 2\left(\xi(x_\epsilon)-\eta_m(y_\epsilon) - \frac{\epsilon}{2}(x_\epsilon-y_\epsilon)^{2} - \frac{2n}{\epsilon^{2}(y_\epsilon-x_\epsilon) + \epsilon} \right). \end{align*}

Rearranging the above inequality and using (4.10), one arrives at

\begin{align*} & \epsilon (x_\epsilon-y_\epsilon)^{2} \le \xi(x_\epsilon) - \xi(y_\epsilon) +\eta_m(x_\epsilon) - \eta_m(y_\epsilon) + \frac{4n (y_\epsilon - x_\epsilon)}{\epsilon(y_\epsilon-x_\epsilon) + 1} \\ \Rightarrow\quad & \epsilon (x_\epsilon-y_\epsilon)^{2} \le 2n|x_\epsilon-y_\epsilon|+ 4n (y_\epsilon - x_\epsilon)\\ \Rightarrow\quad & |x_\epsilon-y_\epsilon|\left(1 - \frac{4n}{\epsilon}\right) \le \frac{2n}{\epsilon}. \end{align*}

Then, we have for $\epsilon$ sufficiently large such that ${4n}/{\epsilon }<1$,

(4.13) \begin{equation} 0\le |x_\epsilon - y_\epsilon| \left(1 - \frac{4n}{\epsilon}\right) \le \frac{2n}{\epsilon}. \end{equation}

Hence, let $(\epsilon _n)_{n\ge 1}$ be an increasing sequence such that $(x_{\epsilon _n},y_{\epsilon _n})\to (\tilde {x},\tilde {y})$ when $\epsilon _n \to \infty$, then according to (4.13), we must have $\tilde {x} = \tilde {y}$.

Case 1: Assume $\mathcal {M}\xi (x_\epsilon ) -\xi (x_\epsilon )\ge 0$. Since $\mathcal {M}\eta _m(y_\epsilon ) -\eta _m(y_\epsilon ) \le -{\kappa }/{m}$, select $\nu >0$ and $\hat {y}\ge 0$ such that $\xi (\tilde {x} +\hat {y}) - k - \phi \hat {y}+ \nu > \mathcal {M}\xi (\tilde {x})$, then we have

\begin{align*} M & \le \liminf_{\epsilon\to \infty}M_{\epsilon}\\ & = \liminf_{\epsilon\to \infty} \left( \xi(x_\epsilon)-\eta_m(y_\epsilon) - \frac{\epsilon}{2}(x_\epsilon-y_\epsilon)^{2} - \frac{2n}{\epsilon^{2}(y_\epsilon-x_\epsilon) + \epsilon} \right)\\ & \le \liminf_{\epsilon \to \infty}\left(\mathcal{M}\xi(x_\epsilon) - \mathcal{M}\eta_m(y_\epsilon) - \frac{\kappa}{m}- \frac{\epsilon}{2}(x_\epsilon-y_\epsilon)^{2} - \frac{2n}{\epsilon^{2}(y_\epsilon-x_\epsilon) + \epsilon}\right)\\ & = \mathcal{M}\xi(\tilde{x}) - \mathcal{M}\eta_m(\tilde{x}) - \frac{\kappa}{m}\\ & <\xi(\tilde{x} +\hat{y}) - k - \phi\hat{y}+ \nu - \eta_m(\tilde{x}+ \hat{y}) +k +\phi\hat{y} - \frac{\kappa}{m} \\ & = \xi(\tilde{x}+ \hat{y}) - \eta_m(\tilde{x}+ \hat{y}) + \nu - \frac{\kappa}{m}\\ & \le M +\nu -\frac{\kappa}{m}, \end{align*}

which is a contradiction when $\nu$ is sufficiently small.

Case 2: Assume $\mathcal {M}\xi (x_\epsilon ) -\xi (x_\epsilon ) <0$, we must have

\begin{equation*}\begin{cases} (\mathcal{A}_\pi-\delta)(\xi, u)(x_\epsilon) \ge 0,\\ (\mathcal{A}_\pi-\delta)(\eta_m, v)(y_\epsilon) \le - \dfrac{\kappa}{m}. \end{cases}\end{equation*}

Then in the following, we derive contradiction from inequality

\begin{equation*}(\mathcal{A}_\pi-\delta)(\xi, u)(x_\epsilon) > (\mathcal{A}_\pi-\delta)(\eta_m, v)(y_\epsilon).\end{equation*}

By noting that $u^{\prime }(x_\epsilon ) = v^{\prime }(y_\epsilon )$, one has

\begin{align*} (\lambda +\delta)(\xi(x_\epsilon) -\eta_m(y_\epsilon) )& < \lambda\int_{0}^{x_\epsilon}\xi(x_\epsilon - z)\,{{\rm d}}F(z) - \lambda\int_{0}^{y_\epsilon}\eta_m(y_\epsilon - z)\,{{\rm d}}F(z)\\ & \quad + \lambda\int_{x_\epsilon}^{\infty}\pi(x_\epsilon - z)\,{{\rm d}}F(z) - \lambda\int_{y_\epsilon}^{\infty}\pi(y_\epsilon - z)\,{{\rm d}}F(z). \end{align*}

Similarly, consider the sequence $(\epsilon _n)_{n\ge 1}$ such that $(x_{\epsilon _n},y_{\epsilon _n})\to (\tilde {x},\tilde {x})$ as $\epsilon _n \to \infty$, then one arrives at

\begin{align*} & (\lambda +\delta)(\xi(\tilde{x}) - \eta_m(\tilde{x})) < \lambda\int_{0}^{\tilde{x}}[\xi(\tilde{x}- z) - \eta_m(\tilde{x} - z) ]\,{{\rm d}}F(z)\le \lambda M\int_{0}^{\tilde{x}}\,{{\rm d}}F(z), \end{align*}

then,

$$M\le \liminf_{\epsilon\to \infty}M_{\epsilon}\le \lim_{n\to \infty}M_{\epsilon_n} = \xi(\tilde{x}) - \eta_m(\tilde{x}) < \frac{\lambda}{\lambda+\delta} M,$$

which is a contradiction. Finally, we complete the proof by noting that the above derivations are still valid when $m\to \infty$.

Proof. Let $h \in \mathcal {LB}^{\pi }(\mathbb {R})$ and $g\in \mathcal {LB}^{\pi }(\mathbb {R})$ be two viscosity solutions of (3.3) with smallest value at zero. On the one hand, let $h$ be the subsolution and $g$ be the supersolution of (3.3) with $h(0)\le g(0)$, then according to Proposition 4.2 and Definition 3.1, we have $h\le g$. On the other hand, let $g$ be the subsolution, $h$ be the supersolution and $g(0)\le h(0)$, then we arrive at $g\le h$. Hence, we have $h=g$.

Proof. According to Proposition 4.1, we have $V_\theta \ge V$; however, since $\theta$ is an admissible strategy, by definition of the value function, $V_\theta \le V$. Hence, we have $V=V_\theta$.

Proof. The proof can refer to Proposition 5.7 and Theorem 5.8 in Azcue and Muler [Reference Azcue and Muler2], where our capital injection and penalty payment at ruin make no difference in the analysis.

Proof. We follow the proof in Proposition 5.10 of Azcue and Muler [Reference Azcue and Muler2]. First, we show that $U_y(x)\ge V(x)$ for $x\in [0,\tilde {x}]$. According to the definition of $U_y(x)$ in (5.2) and Lemma 5.1, we only need to show that $U_y$ is a viscosity supersolution of (3.3) at $y< \tilde {x}$. Note that, $U_y^{\prime }(y^{+}) = 1$, and according to Azcue and Muler [Reference Azcue and Muler4] Definition 3.2 (see also [Reference Azcue and Muler2] Remark 3.5), and the fact that $V^{\prime }(x)\ge 1$, then there exists a test function (say $\varphi$) such that $U_y$ is a viscosity supersolution only when $U_y^{\prime }(y^{-}) = V^{\prime }(y^{-})= 1$, then $\varphi ^{\prime }(y)=1$, and

\begin{align*} & (\mathcal{A}_\pi -\delta)(\varphi,U_y)(y) \\ & \quad = c - (\lambda+\delta)U_y(y) + \lambda\int_{0}^{y}U_y(y-z)\,{{\rm d}}F(z) + \lambda\int_{y}^{\infty}\pi(y-z)\,{{\rm d}}F(z)\\ & \quad = c -(\lambda+\delta)V(y) + \lambda\int_{0}^{y}V(y-z)\,{{\rm d}}F(z) + \lambda\int_{y}^{\infty}\pi(y-z)\,{{\rm d}}F(z) \le 0, \end{align*}

since $V$ is a viscosity supersolution of (3.3) at $y$. And

\begin{align*} & \mathcal{M}U_{y}(x) - U_{y}(x) \\ & \quad = \sup_{z\ge x}\{U_{y}(z) - k-\phi(z-x) \}- U_{y}(x) \\ & \quad =\sup_{z\ge x}\{z-y+ V(y) - k-\phi(z-x) \}- U_{y}(x)\\ & \quad = x -y+ V(y) - k - V(y) - x+y ={-}k<0. \end{align*}

Hence, $U_y$ is a viscosity supersolution of (3.3) at $y$. Next, we show that $U_y(x)\le V(x)$ for $x\in (y,\tilde {x}]$. Consider any $\epsilon >0$, and an $\epsilon$-optimal strategy $\theta$ such that $V(y)\le V_\theta (y)+\epsilon$. Then, for initial surplus $x> y$, consider another strategy $\theta _x$, where the amount of $x-y$ is payout as dividend immediately and follow strategy $\theta$ thereafter; hence, $\theta _x$ is an admissible strategy as well. Then, we have for any $\epsilon >0$ and $x>y$,

$$U_y(x)-\epsilon = V(y) +x-y -\epsilon \le V_\theta(y) +x-y = V_{\theta_x}(x) \le V(x),$$

by letting $\epsilon \to 0$, we arrive at $U_y(x)\le V(x)$ for $x>y$. Hence, $U_y(x) = V(x)$ for $x\in [0,\tilde {x}]$.

Proof. The proof is similar to the proof of Lemma 5.2, we omit the detail here.

Proof. The proof follows the similar steps in Azcue and Muler [Reference Azcue and Muler2], Albrecher and Thonhauser [Reference Albrecher and Thonhauser1] and Xu and Woo [Reference Xu and Woo18].

1. (i) Given that the claim size distribution $F$ is continuous, $(\mathcal {A}^{*}_\pi -\delta )V(\cdot )$ is also continuous, therefore $\mathscr {D}_2$ is closed.

2. (ii) Consider any $\hat {x}\in \mathscr {D}_1$, then we have $(\mathcal {A}^{*}_\pi -\delta )V(\hat {x})<0$ and $V^{\prime }(\hat {x})=1$. Let us consider the auxiliary function defined in (5.2) $U_{\hat {x}-h}$ for each small $h>0$, then we have for any $x\in (\hat {x}-h ,\hat {x})$

   \begin{align*} & (\mathcal{A}^{*}_\pi-\delta)U_{\hat{x}-h}(x)\\ & \quad \le c -(\lambda+\delta)U_{\hat{x}-h}(\hat{x}-h)+ \lambda\int_{0}^{\hat{x}}U_{\hat{x}-h}(\hat{x}-y)\,{{\rm d}}F(y)+ \lambda\int_{\hat{x}}^{\infty}\pi(\hat{x}-y)\,{{\rm d}}F(y) \\ & \quad \le c -(\lambda+\delta)V(\hat{x}-h)+ \lambda\int_{0}^{\hat{x}}V(x-y)\,{{\rm d}}F(y)+ \lambda\int_{\hat{x}}^{\infty}\pi(x-y)\,{{\rm d}}F(y)\\ & \quad = (\mathcal{A}^{*}_\pi-\delta)V(\hat{x}) + (\lambda+\delta)(V(\hat{x})- V(\hat{x}-h)), \end{align*}
   where the second last inequality holds true since $V(\hat {x}-y) - V(\hat {x}-h)\ge (\hat {x}-y) - (\hat {x}-h)$ for $y\in (0,h)$ from Lemma 2.1. Then, since $V$ is continuous, there must exist a sufficient small $\hat {h}>0$ such that $(\mathcal {A}^{*}_\pi -\delta )U_{\hat {x}-\hat {h}}(x)<0$ for all $x\in (\hat {x}-\hat {h},\hat {x})$. In addition,
   \begin{align*} & \mathcal{M}U_{\hat{x}-\hat{h}}(x) - U_{\hat{x}-\hat{x}}(x) \\ & \quad = \sup_{y\ge x}\{U_{\hat{x}-\hat{h}}(y) - k-\phi(y-x) \}- U_{\hat{x}-\hat{h}}(x) \\ & \quad = \sup_{y\ge x}\{y-(\hat{x}-\hat{h})+ V(\hat{x}-\hat{h}) - k-\phi(y-x) \}- U_{\hat{x}-\hat{h}}(x)\\ & \quad = x -(\hat{x}-\hat{h})+ V(\hat{x}-\hat{h}) - k - V(\hat{x}-\hat{h}) - x+(\hat{x}-\hat{h}) ={-}k<0, \end{align*}
   for all $x\in (\hat {x}-\hat {h},\hat {x})$. Therefore, $U_{\hat {x}-\hat {h}}$ is a viscosity supersolution of (3.3) in $(\hat {x}-\hat {h},\hat {x})$. Hence, according to Lemma 5.2, we have $U_{\hat {x}-\hat {h}} = V$ in $[0,\hat {x})$, therefore $(\hat {x}-\hat {h},\hat {x})\in \mathscr {D}_1$, i.e. $\mathscr {D}_1$ is left-open.

   To prove that the lower limit of any connected components of $\mathscr {D}_1$ is in $\mathscr {D}_2$, one need to show for any sufficiently small $h>0$ such that $(\hat {x},\hat {x}+h)\subset \mathscr {D}_1$ and $\hat {x}\notin \mathscr {D}_1$, then $\hat {x}\in \mathscr {D}_2$. Note that if $\hat {x}\notin \mathscr {D}_2$, it must in $\mathscr {N}$. However, since $V(x)$ is a viscosity supersolution of (3.3), if $\hat {x}\in \mathscr {N}$, we have $(\mathcal {A}^{*}_\pi -\delta )V(\hat {x})<0$ and $V^{\prime }(\hat {x})>1$ given the derivative exists. Then, assume that there exists a sequence $x_n\uparrow \hat {x}$ such that $V^{\prime }(x_n)$ exists; we can show that the sequence is not in $\mathscr {D}_2$ since it is closed; and the sequence is also not in $\mathscr {D}_1$, since if so, according to Lemma A.3 we must have $(\hat {x}-h_0,\hat {x})\subset \mathscr {D}_1$ for some $h_0>0$ and in turn $V^{\prime }(x)=1$ for all $x\in (\hat {x}-h_0,\hat {x})$ which is a contradiction. Therefore, the sequence $x_n$ must in $\mathscr {N}$ as well and there exists $h^{\prime }>0$ such that $(\hat {x}-h^{\prime },\hat {x})\subset \mathscr {N}$. In other words, one obtain that $V^{\prime }(\hat {x}^{-}) >1$ and $V^{\prime }(\hat {x}^{+})=1$. Since $V$ is a viscosity subsolution of (3.3) and $\hat {x}\notin \mathscr {C}$, then the test function say $\varphi$ with $\varphi ^{\prime }(\hat {x})=1$ should gives the following inequality

   \begin{align*} & c\varphi^{\prime}(\hat{x}) - (\lambda + \delta)V(\hat{x})+ \lambda\int_{0}^{x}V(x-y)\,{{\rm d}}F(y) + \lambda\int_{x}^{\infty}\pi(x-y)\,{{\rm d}}F(y)\\ & \quad = (\mathcal{A}^{*}_\pi -\delta)V(\hat{x}) \ge 0, \end{align*}
   which is a contradiction. Hence, one arrive at $\hat {x}\in \mathscr {D}_2$.

   Finally, we show that there exists $x^{*}$ large enough such that $(x^{*}, \infty )\subset \mathscr {D}_1$. According to Lemma 5.3, we only need to show that for sufficiently large $x^{*}>0$, the auxiliary function $U_{x^{*}}(x)$ is a viscosity supersolution of (3.3) for $x\in (x^{*},\infty )$. Note that $U^{\prime }_{x^{*}}(x) = 1$ for $x\in (x^{*},\infty )$ by definition and $\mathcal {M}U_{x^{*}}(x)\le U_{x^{*}}(x)$ holds true obviously, then we only to need show that $(\mathcal {A}_\pi -\delta )U_{x^{*}}(x) \le 0$ for $x>x^{*}$. Since $U^{\prime }_{x^{*}}(x^{*})=1$, then

   \begin{align*} & (\mathcal{A}_\pi -\delta)U_{x^{*}}(x)\\ & \quad = c - (\lambda +\delta)U_{x^{*}}(x) + \lambda\int_{0}^{x}U_{x^{*}}(x-z)\,{{\rm d}}F(z) + \lambda\int_{x}^{\infty}\pi(x-z)\,{{\rm d}}F(z)\\ & \quad \le c - (\lambda+\delta )(x-x^{*} +V(x^{*})) + \lambda\int_{0}^{x}(x-z -x^{*} +V({x^{*}}) )\,{{\rm d}}F(z) \\ & \quad \le c -\delta (x-x^{*} +V(x^{*}) ), \end{align*}
   by noting that for each $x^{*}\ge 0$, $c -\delta (x-x^{*} +V(x^{*}))$ is a decreasing function of $x$; hence, when $V(x^{*})\ge c/\delta$, we arrive at $(\mathcal {A}_\pi -\delta )U_{x^{*}}(x)\le 0$ for $x>x^{*}$. Then, from (2.5), we have the lower bound $V(x^{*})\ge x^{*}+{(c+ \lambda (K - \Phi \mu ))}/{(\lambda + \delta )}$, therefore, the result follows by choosing $x^{*} = {c}/{\delta } - {(c+ \lambda (K - \Phi \mu ))}/{(\lambda + \delta )}$.

3. (iii) From Lemmas 2.1 and 2.2, we have that $V(x) -\mathcal {M}V(x)$ is continuous, hence $\mathscr {C}$ is closed.

4. (iv) Consider $x_0\in \mathscr {N}$ and a sequence $x_n\downarrow x_0$, since $\mathscr {D}_2$ and $\mathscr {C}$ are closed sets, the sequence is not in $\mathscr {D}_2$ and $\mathscr {C}$. Assume that $x_n\in \mathscr {D}_1$, since the lower limit of any connected component of $\mathscr {D}_1$ belongs to $\mathscr {D}_2$, there must exist a subsequence $x^{\prime }_n\in \mathscr {D}_2$ such that $x^{\prime }_n \downarrow x_0$, which is a contradiction. Hence, $x_n\in \mathscr {N}$ and $\mathscr {N}$ is right-open. The connected components of $\mathscr {N}$ is bounded follows obviously from the fact that there is no $\hat {x}$ sufficient large such that $[\hat {x},\infty )\subset \mathscr {N}$. On the other hand, since $\mathscr {D}_1$ is left-open, then the upper limit of any connected component of $\mathscr {N}$ is in $\mathscr {D}_2$.

Proof. We denote such band-type strategy as $\theta _b$, hence, we want to show that $V(x)=V_{\theta _b}(x)$ for all $x\ge 0$. The proof follows the fixed point argument. Let's consider a complete metric space $\mathbb {M}$ of continuous functions $f:[0,\infty ) \to \mathbb {R}$ satisfying that

$$f(x) = x-x^{*} + f(x^{*}),\quad \text{for } x\ge x^{*},$$

with the metric being the supremum norm $d(\cdot,\cdot )$ defined as

$$d(f_1,f_2) = \sup_{x\ge 0} |f_1(x)-f_2(x)|.$$

We define an operator $\mathcal {T}: \mathbb {M} \to \mathbb {M}$ as

$$\mathcal{T}(f)(x) = \mathbb{E}_x\left[\int_{0}^{T_1}e^{-\delta t}\,{{\rm d}}L^{\theta_b}_t - \sum_{i=1}^{\infty}e^{-\delta \omega^{\theta_b}}(k+\phi \zeta^{\theta_b}_i){1}_{\{\omega^{\theta_b}< T_1\}} + e^{-\delta T_1}f(X^{\theta_b}_{T_1}) \right],$$

where $T_1$ is the arrival time of the first claim. Note that, according to Proposition 5.1(ii), there exists $x^{*}\in \mathscr {D}_2$ such that $(x^{*},\infty )\subset \mathscr {D}_1$. Since $V^{\prime }(x) = 1$ for $x\in \mathscr {D}_1$, then we have $V(x) = x-x^{*} + V(x^{*})$ for $x\in (x^{*}, \infty )$, hence $V\in \mathbb {M}$. It is also obvious that $|\mathcal {T}f_1(x) - \mathcal {T}f_2(x)|\le ({\lambda }/{(\lambda +\delta )})d(f_1,f_2)$, hence, $\mathcal {T}$ is a contraction mapping with modulus less than 1, i.e. $\mathcal {T}$ admits a unique fixed point. According to Definition 5.1, $\theta _b$ is a stationary strategy, then we have $\mathcal {T}V_{\theta _b} = V_{\theta _b}$. Finally, we show that the value function $V$ is also a fixed point of $\mathcal {T}$, i.e. $\mathcal {T}V=V$. For $x\in \mathscr {D}_2$, we have

(5.3) \begin{align} \mathcal{T}V(x) & = \mathbb{E}_x\left[\int_{0}^{T_1}e^{-\delta t}c\,{{\rm d}}t + e^{-\delta T_1}V(X^{\theta_b}_{T_1}) \right] \nonumber\\ & = \frac{c}{\lambda +\delta} + \int_{0}^{\infty}\lambda e^{-(\lambda+\delta)t}\left\{ \int_{0}^{x}V(x-z)\,{{\rm d}}F(z) + \int_{x}^{\infty}\pi(x-z)\,{{\rm d}}F(z)\right\}{{\rm d}}t \nonumber\\ & = \frac{c}{\lambda+\delta} + \frac{\lambda}{\lambda+\delta}\left\{ \int_{0}^{x}V(x-z)\,{{\rm d}}F(z) +\int_{x}^{\infty}\pi(x-z)\,{{\rm d}}F(z)\right\} \nonumber\\ & = V(x), \end{align}

where the last equation holds since we have $(\mathcal {A}_\pi ^{*}-\delta )V(x) = 0$ for $x\in \mathscr {D}_2$.

For $x\in \mathscr {D}_1$, let $\hat {x}=\inf \{y:(y,x]\subset \mathscr {D}_1\}$, then $\hat {x}\in \mathscr {D}_2$, and from 5.3 we have

$$\mathcal{T}V(x) = x-\hat{x} +\mathcal{T}V(\hat{x}) = V(x).$$

For $x\in \mathscr {N}$, let us consider $\bar {x}= \min \{y>x, y\notin \mathscr {N}\}$, then $\bar {x}\in \mathscr {D}_2$. Let $s= (\bar {x} - x)/c$ and $\bar {y}(t) = x+ct$, one has $\bar {y}(t) \in \mathscr {N}$ for $t< s$. Then, we have

\begin{align*} \mathcal{T}V(x)& = \mathbb{E}_x[e^{-\delta s }V(\bar{x}){1}_{\{T_1>s \}}] + \mathbb{E}_x[e^{-\delta T_1}V(\bar{y}(T_1) - Y_1){1}_{\{T_1\le s\}} ]\\ & =e^{-(\lambda+\delta)s}V(\bar{x}) + \int_{0}^{s}\lambda e^{-(\lambda+\delta)t}\left\{\int_{0}^{\bar{y}(t)}V(\bar{y}(t)-z)\,{{\rm d}}F(z) + \int_{\bar{y}(t)}^{\infty}\pi(\bar{y}(t) - z)\,{{\rm d}}F(z) \right\} {{\rm d}}t\\ & =e^{-(\lambda+\delta)s}V(\bar{x}) + \int_{0}^{s}(- e^{-(\lambda+\delta)t}V(\bar{y}(t)))^{\prime} \,{{\rm d}}t \\ & = V(x), \end{align*}

where the second last equation holds true since $V(x)$ (as an viscosity solution) is an ${\rm a.e.}$ (or weak) solution of

$$ch^{\prime}(x) - (\lambda+\delta)h(x) +\lambda\int_{0}^{x}h(x-z)\,{{\rm d}}F(z) + \lambda \int_{x}^{\infty}\pi(x-z)\,{{\rm d}}F(z)=0,$$

for $x\in \mathscr {N}$, see e.g. Xu and Woo [Reference Xu and Woo18].

Finally, for $x\in \mathscr {C}$, we have immediate capital injection that bring the surplus to $\mathscr {N}$, then

$$\mathcal{T}V(x) = \mathcal{T}V(y^{*}) -k-\phi(y^{*} - x) = V(y^{*}) - k-\phi(y^{*}-x) = \mathcal{M}V(x) = V(x),$$

where $y^{*}\in \mathscr {N}$ satisfying $y^{*} = \arg\!\max _y\{y\ge x| V(y) -k-\phi (y-x) \}$ and for $x\in \mathscr {C}$ we have $\mathcal {M}V(x)=V(x)$.

Then, we have $V = V_{\theta _b}$, and $\theta _b$ is the optimal strategy.