Global convergence in an overlapping generations model with two-sided altruism

Abstract

We study the evolutionary dynamics of an overlapping generations economy with two-sided altruism, where each generation cares about the utilities of its parental generation, its offspring, and its own. We assume that the young and old generations in each period act jointly, as a central planner. We then introduce the concepts of a two-sided altruistic path and a two-sided altruistic stationary path. Then, we go on to prove that the former always converges to the latter, in the two-sided altruistic model. This stationary capital level is higher than the steady-state level in the model with a constant discount factor.

Appendix

A. Proof of Lemma 2: By definition, w ^τ(c) = max _{c ≥ x ≥ 0}[u(x) + τu(c − x)]. Its solution x is denoted as a function of c by x(c). Because Assumption U3 assures that x is an interior of [0, c], the first order equation is u ^′(x) − τu ^′(c − x) = 0. Totally differentiating it gives [u ^″(x) + τu ^″(c − x)]dx − τu ^″(c − x)dc = 0. Therefore, x is smooth in c by the implicit function theorem, and $\frac{dx}{dc}=\frac{dx}{dc}$. Due to strict concavity u ^″(⋅) < 0, $0<\frac{dx}{dc}<1$ holds. Thus, ${w}_{\tau}^{\prime }(c)=\tau {u}^{\prime}\left( c- x(c)\right)+\left[{u}^{\prime}\left( x(c)\right)-\tau {u}^{\prime}\left( c- x(c)\right)\right]\frac{dx}{dc}=\tau {u}^{\prime}\left( c- x(c)\right)>0$, and${w}_{\tau}^{{\prime\prime} }(c)=\tau {u}^{{\prime\prime}}\left( c- x(c)\right)-\tau {u}^{{\prime\prime}}\left( c- x(c)\right)\frac{dx}{dc}=\tau {u}^{{\prime\prime}}\left( c- x(c)\right)\left(1-\frac{dx}{dc}\right)<0$. Both prove that w _τ (c) is strictly increasing and strictly concave in c. ${ \lim}_{c\to 0}{w}_{\tau}^{\prime }(c)=\infty$ follows from Assumption U3.

Next, again by definition, w _1/β (c) = u(x(c)) + β ⁻¹ u(c − x(c)), where x(c) is already defined as argmax _{c ≥ x ≥ 0}[u(x) + τu(c − x)].

Then w _1/β (c) = u(x(c)) + τu(c − x(c)) + (1/β − τ)u(c − x(c))=w _τ (c) + (1/β − τ)u(c − x(c)). Differentiating w _1/β (c) with c gives ${w}_{1/\beta}^{\prime }(c)={w}_{\tau}^{\prime }(c)+\left(1/\beta -\tau \right){u}^{\prime}\left( c- x(c)\right)\left(1-\frac{dx}{dc}\right)$. Since 1/β − τ > 0, u ^′(⋅) > 0, and $1-\frac{dx}{dc}>0$, we have ${w}_{1/\beta}^{\prime }(c)>{w}_{\tau}^{\prime }(c)>0$. Hence, w _1/β (c) is strictly increasing, and ${ \lim}_{c\to 0}{w}_{1/\beta}^{\prime }(c)=\infty$ follows from Assumption U3.

Again, differentiating ${w}_{1/\beta}^{\prime }(c)$ with respect to c gives ${w}_{1/\beta}^{{\prime\prime} }(c)={w}_{\tau}^{{\prime\prime} }(c)+\left(1/\beta -\tau \right)\left[{u}^{{\prime\prime}}\left( c- x(c)\right){\left(1-\frac{dx}{dc}\right)}^2-{u}^{\prime}\left( c- x(c)\right)\frac{dx}{dc}\right]$.

Thus, the sufficient condition for the strict concavity of w _1/β (c), ${w}_{\beta}^{{\prime\prime} }(c)<0$ is $\frac{d^2 x}{{d c}^2}\ge 0$. Denoting Δ = {u ^″(x(c)) + τu ^″(c − x(c))},

$$ \begin{array}{l}\frac{d^2 x}{{d c}^2}=\frac{\tau}{\Delta^2}\left[\begin{array}{c}\hfill {u}^{{\prime\prime\prime}}\left( c- x(c)\right)\left(1-\frac{ d x}{ d c}\right)\left\{{u}^{{\prime\prime}}\left( x(c)\right)+\tau {u}^{{\prime\prime}}\left( c- x(c)\right)\right\}\hfill \\ {}\hfill -{u}^{{\prime\prime}}\left( c- x(c)\right)\left\{{u}^{{\prime\prime\prime}}\left( x(c)\right)\frac{ d x}{ d c}+\tau {u}^{{\prime\prime\prime}}\left( c- x(c)\right)\left(1-\frac{ d x}{ d c}\right)\right\}\hfill \end{array}\right]\\ {}=\frac{\tau}{\Delta^2}\left[{u}^{{\prime\prime\prime}}\left( c- x(c)\right){u}^{{\prime\prime}}\left( x(c)\right)\left(1-\frac{ d x}{ d c}\right)-{u}^{{\prime\prime}}\left( c- x(c)\right){u}^{{\prime\prime\prime}}\left( x(c)\right)\frac{ d x}{ d c}\right]\\ {}=\frac{\tau}{\Delta^3}\left[{u}^{{\prime\prime\prime}}\left( c- x(c)\right){\left\{{u}^{{\prime\prime}}\left( x(c)\right)\right\}}^2-\tau {\left\{{u}^{{\prime\prime}}\left( c- x(c)\right)\right\}}^2{u}^{{\prime\prime\prime}}\left( x(c)\right)\right].\end{array} $$

(20)

Because Δ < 0, $\frac{d^2 x}{{d c}^2}\ge 0$ is equivalent tou ^‴(c − x(c)){u ^″(x(c))}² − τ{u ^″(c − x(c))}² u ^‴(x(c)) ≤ 0, that is, $\frac{u^{{\prime\prime\prime}}\left( c- x(c)\right)}{\tau {\left\{{u}^{{\prime\prime}}\left( c- x(c)\right)\right\}}^2}\le \frac{u^{{\prime\prime\prime}}\left( c- x(c)\right)}{\tau {\left\{{u}^{{\prime\prime}}\left( c- x(c)\right)\right\}}^2}$. This is satisfied from Lemma 1 and Assumption U4 because u ^′(x(c)) − τu ^′(c − x(c)) = 0 and x(c) > c − x(c) > 0. □.

B. Proof of Lemma 3: Set k ₀ > 0. Suppose k ₁ = 0. Therefore, k _t  = 0 (all t ≥ 1). Then W(k ₀) = w _τ (f(k ₀) − 0). Consider an alternative solution ${k}_1^{\prime }=\varepsilon >0$ and ${k}_2^{\prime }=0$, where the corresponding utility is w _τ (f (k ₀) − ε) + βw _1/β  (f (ε) − 0). Because ${\left\{{k}_t\right\}}_{t=0}^{\infty }$ is an optimal solution,

$$ {w}_{\tau}\left( f\left({k}_0\right)\right)\ge {w}_{\tau}\left( f\left({k}_0\right)-\varepsilon \right)+\beta {w}_{1/\beta}\left( f\left(\varepsilon \right)\right). $$

(21)

Arranging and dividing both sides by ε gives

$$ \frac{w_{\tau}\left( f\left({k}_0\right)\right)-{w}_{\tau}\left( f\left({k}_0\right)-\varepsilon \right)}{\varepsilon}\ge \beta \frac{w_{1/\beta}\left( f\left(\varepsilon \right)\right)-{w}_{1/\beta}\left( f(0)\right)}{\varepsilon}. $$

(22)

Due to Assumption F1 and Lemma 2, f(0) = 0 and w _1/β (0) = 0. By taking the limits as ε → 0, the left-hand side converges to ${w}_{\tau}^{\prime}\left( f\left({k}_0\right)\right)$, while the right-hand side diverges to ${ \lim}_{\varepsilon \to 0}\beta {w}_{1/\beta}^{\prime}\left( f\left(\varepsilon \right)\right){f}^{\prime}\left(\varepsilon \right)=\infty$ due to Assumption F4 and Lemma 2. The above inequality leads to a contradiction because ε can be arbitrarily small. Therefore, k ₁ > 0. Similarly k _t + 1 > 0 (t ≥ 1) can be confirmed by replacing w _τ by w _1/β .

Next we shall prove that c _t  = f(k _t ) − k _t + 1 > 0 for all t ≥ 0. Suppose k ₀ > 0, but c ₀ = 0 (i.e., f(k ₀) = k ₁) and c ₁ > 0 (i.e., f(k ₁) > k ₂). Then

$$ W\left({k}_0\right)={w}_{\tau}(0)+\beta {w}_{1/\beta}\left( f\left({k}_1\right)-{k}_2\right)+{\beta}^2\overline{W}\left({k}_2\right). $$

(23)

As k ₁ > 0 and c ₁ = f(k ₁) − k ₂ > 0, an alternative solution of ${k}_1^{\prime }={k}_1-\varepsilon >0$ and ${k}_2^{\prime }={k}_2$ is feasible for sufficiently small ε > 0, where the corresponding utility is ${w}_{\tau}\left(\varepsilon \right)+\beta {w}_{1/\beta}\left( f\left({k}_1-\varepsilon \right)-{k}_2\right)+{\beta}^2\overline{W}\left({k}_2\right)$. Because ${\left\{{k}_t\right\}}_{t=0}^{\infty }$ is the optimal solution,

$$ {w}_{\tau}(0)+\beta {w}_{1/\beta}\left( f\left({k}_1\right)-{k}_2\right)+{\beta}^2\overline{W}\left({k}_2\right)\ge {w}_{\tau}\left(\varepsilon \right)+\beta {w}_{1/\beta}\left( f\left({k}_1-\varepsilon \right)-{k}_2\right)+{\beta}^2\overline{W}\left({k}_2\right). $$

(24)

Arranging and dividing both sides by ε gives

$$ \beta \frac{w_{1/\beta}\left( f\left({k}_1\right)-{k}_2\right)-{w}_{1/\beta}\left( f\left({k}_1-\varepsilon \right)-{k}_2\right)}{\varepsilon}\ge \frac{w_{\tau}\left(\varepsilon \right)-{w}_{\tau}(0)}{\varepsilon}. $$

(25)

By taking the limits as ε → 0, the left-hand side converges to $\beta {w}_{1/\beta}^{\prime}\left( f\left({k}_1\right)-{k}_2\right){f}^{\prime}\left({k}_1\right)$, whereas the right-hand side diverges to ${ \lim}_{\varepsilon \to 0}{w}_{\tau}^{\prime}\left(\varepsilon \right)=\infty$ due to Lemma 2. The above inequality leads to a contradiction because ε can be arbitrarily small. Therefore, c ₀ = c ₁ = 0, or c ₀ > 0.

Suppose that c ₀ = c ₁ = 0 and T is the smallest in t ≥ 1 such that c _T  = 0 and c _T + 1 > 0. If such a T does not exist, then c _t  = 0 for all t ≥ 1 must be true. Clearly, this is not optimal. Therefore, such a T exists. Then, by the same reasoning as above, it leads to a contradiction. Therefore, c _T + 1 > 0 must imply c _T  > 0. Due to recursive induction, c _t  > 0 for all t ≤ T + 1. Hence, c ₀ > 0 and c ₁ > 0 must be true. Because c ₀ > 0 and c ₁ > 0, either c _t  > 0 for all t ≥ 0 or c _t  = 0 for some t ≥ 2. Suppose that T is the smallest in t ≥ 2 with c _t  = 0 and c _t + 1 > 0. Then the same reasoning as above leads to a contradiction. Hence, c _t  > 0 for all t ≥ 0. □

C. Proof of Lemma 4: Let ${\left\{{k}_t^{\prime}\right\}}_{t=0}^{\infty }$ and ${\left\{{k}_t\right\}}_{t=0}^{\infty }$ be the optimal solution from ${k}_0^{\prime }$ and k ₀, respectively.

$$ \begin{array}{l}\frac{1}{2}\left[ W\left({k}_0^{\prime}\right)+ W\left({k}_0\right)\right]=\frac{1}{2}\sum_{t=0}^{\infty }{\beta}^t\left[{\tilde{w}}_t\left( f\left({k}_t^{\prime}\right)-{k}_{t+1}^{\prime}\right)+{\tilde{w}}_t\left( f\left({k}_t\right)-{k}_{t+1}\right)\right]\\ {}\le \sum_{t=0}^{\infty }{\beta}^t{\tilde{w}}_t\left(\frac{f\left({k}_t^{\prime}\right)+ f\left({k}_t\right)}{2}-\frac{k_{t+1}^{\prime }+{k}_{t+1}}{2}\right)\\ {}\le \sum_{t=0}^{\infty }{\beta}^t\left[{\tilde{w}}_t\left( f\left(\frac{k_t^{\prime }+{k}_t}{2}\right)-\frac{k_{t+1}^{\prime }+{k}_{t+1}}{2}\right)\right]\\ {}\le W\left(\frac{k_0^{\prime }+{k}_0}{2}\right),\end{array} $$

(26)

where ${\tilde{w}}_t(c)={w}_{\tau}(c)$ for t = 0, and ${\tilde{w}}_t(c)={w}_{1/\beta}(c)$ for t ≥ 1. The second inequality strictly holds for ${k}_0^{\prime}\ne {k}_0$. Thus, for ${k}_0^{\prime}\ne {k}_0$, $\frac{1}{2}\left[ W\left({k}_0^{\prime}\right)+ W\left({k}_0\right)\right]< W\left(\frac{1}{2}\right)$ holds, implying that W(⋅) is strictly concave.

Next, let ${k}_0^{\prime }={k}_0$ and suppose that two optimal solutions, ${\left\{{k}_t^{\prime}\right\}}_{t=0}^{\infty }$ and ${\left\{{k}_t\right\}}_{t=0}^{\infty }$, exist. Let’s assume that T is the smallest, such that $f\left({k}_T^{\prime}\right)-{k}_{T+1}^{\prime}\ne f\left({k}_T\right)-{k}_{T+1}$. Then, the first inequality in (26) becomes strict. Since ${\left\{\frac{k_t^{\prime }+{k}_t}{2}\right\}}_{t=0}^{\infty }$ gives a higher utility than ${\left\{{k}_t\right\}}_{t=0}^{\infty }$, there is a contradiction to the optimality of ${\left\{{k}_t^{\prime}\right\}}_{t=0}^{\infty }$ and ${\left\{{k}_t\right\}}_{t=0}^{\infty }$. If there is no T such that $f\left({k}_T^{\prime}\right)-{k}_{T+1}^{\prime}\ne f\left({k}_T\right)-{k}_{T+1}$, then ${\left\{{k}_t^{\prime}\right\}}_{t=0}^{\infty }$ and ${\left\{{k}_t\right\}}_{t=0}^{\infty }$ coincide. Thus, ${k}_t^{\prime }={k}_t$ for all t ≥ 0. □

D. Proof of Theorem 1: Suppose that ${k}_0^{\prime }>{k}_0$ and ${k}_1^{\prime }<{k}_1$. Then $\left({k}_0,{k}_1^{\prime}\right)$ and $\left({k}_0^{\prime },{k}_1\right)$ are feasible because

$$ f\left({k}_0\right)-{k}_1^{\prime }> f\left({k}_0\right)-{k}_1>0\ \mathrm{and}\ f\left({k}_0^{\prime}\right)-{k}_1^{\prime }> f\left({k}_0^{\prime}\right)-{k}_1>0 $$

(27)

are true. By the uniqueness of an optimal solution (Lemma 4),

$$ {w}_{\tau}\left( f\left({k}_0\right)-{k}_1\right)+\beta \overline{W}\left({k}_1\right)>{w}_{\tau}\left( f\left({k}_0\right)-{k}_1^{\prime}\right)+\beta \overline{W}\left({k}_1^{\prime}\right) $$

(28)

and

$$ {w}_{\tau}\left( f\left({k}_0^{\prime}\right)-{k}_1^{\prime}\right)+\beta \overline{W}\left({k}_1^{\prime}\right)>{w}_{\tau}\left( f\left({k}_0^{\prime}\right)-{k}_1\right)+\beta \overline{W}\left({k}_1\right) $$

(29)

must be true. Combining (28) and (29) gives

$$ {w}_{\tau}\left( f\left({k}_0\right)-{k}_1\right)+{w}_{\tau}\left( f\left({k}_0^{\prime}\right)-{k}_1^{\prime}\right)>{w}_{\tau}\left( f\left({k}_0\right)-{k}_1^{\prime}\right)+{w}_{\tau}\left( f\left({k}_0^{\prime}\right)-{k}_1\right). $$

(30)

This contradicts the strict concavity of w _τ (⋅) because $f\left({k}_0^{\prime}\right)-{k}_1$ and $f\left({k}_0\right)-{k}_1^{\prime }$ lie in $\Big( f\left({k}_0^{\prime}\right)-{k}_1^{\prime },$ f(k ₀) − k ₁). Therefore, ${k}_1^{\prime}\ge {k}_1$ must be true. Suppose that ${k}_1^{\prime }={k}_1$. By the Euler equation,

$$ -{w}_{\tau}^{\prime}\left( f\left({k}_0\right)-{k}_1\right)+\beta {w}_{1/\beta}^{\prime}\left( f\left({k}_1\right)-{k}_2\right){f}^{\prime}\left({k}_1\right)=0. $$

(31)

Since ${k}_0^{\prime }>{k}_0$ and ${k}_1^{\prime }={k}_1$, then both ${\left\{{k}_t\right\}}_{t=1}^{\infty }$ and ${\left\{{k}_t^{\prime}\right\}}_{t=1}^{\infty }$ are optimal solutions of Problem (II) from ${k}_1^{\prime }={k}_1$. Since the optimal solution of Problem (II) is unique, ${k}_2^{\prime }={k}_2$ holds, and $-{w}_{\tau}^{\prime }(x)+\beta {w}_{1/\beta}^{\prime}\left( f\left({k}_1\right)-{k}_2\right){f}^{\prime}\left({k}_1\right)=0$ has two distinct solutions, x = f(k ₀) − k ₁ and $x= f\left({k}_0^{\prime}\right)-{k}_1$. This contradicts the strictly decreasing property of ${w}_{\tau}^{\prime }(x)$. Hence, ${k}_1^{\prime }>{k}_1$.

Next, ${\left\{{k}_t\right\}}_{t=1}^{\infty }$ and ${\left\{{k}_t^{\prime}\right\}}_{t=1}^{\infty }$ are the optimal solutions of Problem (II), given k ₁ and ${k}_t^{\prime }$, respectively. By the principle of optimality, $\overline{W}\left({k}_1\right)={w}_{1/\beta}\left( f\left({k}_1\right)-{k}_2\right)+\beta \overline{W}\left({k}_2\right)$ and $\overline{W}\left({k}_1^{\prime}\right)={w}_{1/\beta}\left( f\left({k}_1^{\prime}\right)-{k}_2^{\prime}\right)+\beta \overline{W}\left({k}_2^{\prime}\right)$. Therefore, the same argument as above may be applied, and ${k}_1^{\prime }>{k}_1$ implies ${k}_2^{\prime }>{k}_2$. Finally, it is proved that ${k}_0^{\prime }>{k}_0$ implies ${k}_t^{\prime }>{k}_t$ for t ≥ 1. □.

E. Proof of Theorem 2: The Euler equation for t ≥ 1 is

$$ -{w}_{1/\beta}^{\prime}\left({c}_t\right)+\beta {w}_{1/\beta}^{\prime}\left({c}_{t+1}\right){f}^{\prime}\left({k}_{t+1}\right)=0,\mathrm{where}\ {c}_t= f\left({k}_t\right)-{k}_{t+1}. $$

(32)

Rewriting this gives $\beta {f}^{\prime}\left({k}_{t+1}\right)=\frac{w_{1/\beta}^{\prime}\left({c}_t\right)}{w_{1/\beta}^{\prime}\left({c}_{t+1}\right)}$. This equation implies that, if k ^∗ > k _t + 1, then $\frac{w_{1/\beta}^{\prime}\left({c}_t\right)}{w_{1/\beta}^{\prime}\left({c}_{t+1}\right)}>1$ and c _t  < c _t + 1. If k _t + 1 > k ^∗, then c _t  > c _t + 1. Assuming k ^∗ > k ₁ > 0, k ^∗ > k _t + 1 > 0 for t ≥ 1 by Theorem 1, assuring that c _t  < c _t + 1 for all t ≥ 1. That is, c _t is increasing, which is impossible if k _t converges to 0. Since ${\left\{{k}_t\right\}}_{t=1}^{\infty }$ is a monotone interior solution, it must monotonically increase and converge to some interior point, say $\tilde{k}$. Then $\tilde{k}$ satisfies

$$ -{w}_{1/\beta}^{\prime}\left( f\left(\tilde{k}\right)-\tilde{k}\right)+\beta {w}_{1/\beta}^{\prime}\left( f\left(\tilde{k}\right)-\tilde{k}\right){f}^{\prime}\left(\tilde{k}\right)=0. $$

(33)

Therefore, $\tilde{k}$ must coincide with k ^∗.

Assume k ₁ > k ^∗. Then k _t  > k ^∗ for all t ≥ 1. If k _t converges to $\overline{k}$ where $f\left(\overline{k}\right)=\overline{k}>0$, then c _t converges to zero. Clearly, accumulating capital without consuming is not optimal. Therefore, k _t does not converge to $\overline{k}$. Because ${\left\{{k}_t\right\}}_{t=1}^{\infty }$ is a monotone interior solution, it must converge to k ^∗. □.

F. Proof of Lemma 6: Consider a function of e, ${w}_e\left( f\left({k}_0\right)-{k}_1\right)+\beta \overline{W}\left({k}_1\right)$. Let k ₀ be given. Furthermore, both ${\left\{{k}_t\right\}}_{t=1}^{\infty }$ and ${\left\{{k}_t^{\prime}\right\}}_{t=1}^{\infty }$ are unique optimal solutions for e = τ and 1/β, respectively. Suppose that ${k}_1^{\hbox{'}}>{k}_1$. Then, by the uniqueness of an optimal solution,

$$ {w}_{\tau}\left( f\left({k}_0\right)-{k}_1\right)+\beta \overline{W}\left({k}_1\right)>{w}_{\tau}\left( f\left({k}_0\right)-{k}_1^{\prime}\right)+\beta \overline{W}\left({k}_1^{\prime}\right) $$

(34)

and

$$ \kern0.5em {w}_{1/\beta}\left( f\left({k}_0\right)-{k}_1^{\prime}\right)+\beta \overline{W}\left({k}_1^{\prime}\right)>{w}_{1/\beta}\left( f\left({k}_0\right)-{k}_1\right)+\beta \overline{W}\left({k}_1\right). $$

(35)

Combining (28) and (29),

$$ {w}_{\tau}\left( f\left({k}_0\right)-{k}_1\right)+{w}_{1/\beta}\left( f\left({k}_0\right)-{k}_1^{\prime}\right)>{w}_{\tau}\left( f\left({k}_0\right)-{k}_1^{\prime}\right)+{w}_{1/\beta}\left( f\left({k}_0\right)-{k}_1\right). $$

(36)

From definitions (12) and (13),

$$ \left(1/\beta -\tau \right)\left[ u\left({c}^o\left( f\left({k}_0\right)-{k}_1^{\prime}\right)\right)- u\left({c}^o\left( f\left({k}_0\right)-{k}_1\right)\right)\right]>0, $$

(37)

where u(c ^o(c)) is, by the proof of Lemma 2, increasing in c. Therefore, 1/β > τ implies that $f\left({k}_0\right)-{k}_1^{\prime }> f\left({k}_0\right)-{k}_1$. Hence, ${k}_1>{k}_1^{\prime }$, which is a contradiction. Consequently, ${k}_1\ge {k}_1^{\prime }$.

Suppose that ${k}_1={k}_1^{\prime }$. Then the Euler equation,

$$ -{w}_e^{\prime}\left( f\left({k}_0\right)-{k}_1\right)+\beta {w}_{1/\beta}^{\prime}\left( f\left({k}_1\right)-{k}_2\right){f}^{\prime}\left({k}_1\right)=0, $$

(38)

must hold for e = τ , 1/β. However, by 1/β > τ and Property W3, it holds that ${w}_{1/\beta}^{\prime}\left( f\left({k}_0\right)-{k}_1\right)>{w}_{\tau}^{\prime}\left( f\left({k}_0\right)-{k}_1\right)$. This contradicts the Euler equation above. Hence, ${k}_1>{k}_1^{\prime }$ must hold. □.

G. Proof of Theorem 3: Let $\widehat{K}$ be the set of positive initial capital stocks of two-sided altruistic stationary paths. Since an optimal solution ${\left\{{\overline{k}}_t\right\}}_{t=0}^{\infty }$ from ${\overline{k}}_0=\overline{k}$ where $f\left(\overline{k}\right)=\overline{k}>0$ satisfies ${\overline{k}}_1<\overline{k}$ by Lemma 3, $\overline{x}\equiv \sup \left\{\widehat{k}|\widehat{k}\in \widehat{K}\right\}<\overline{k}$ must be true.

Let ${\left\{{k}_t\right\}}_{t=0}^{\infty }$ and ${\left\{{k}_t^{\prime}\right\}}_{t=0}^{\infty }$ with ${k}_0={k}_0^{\hbox{'}}$ be solutions of

$$ \underset{f\left({k}_0\right)\ge {k}_1\ge 0}{ \max}\left[{w}_{\tau}\left( f\left({k}_0\right)-{k}_1\right)+\beta \overline{W}\left({k}_1\right)\right] $$

(39)

and

$$ \underset{f\left({k}_0\right)\ge {k}_1\ge 0}{ \max}\left[{w}_{1/\beta}\left( f\left({k}_0\right)-{k}_1\right)+\beta \overline{W}\left({k}_1\right)\right], $$

(40)

respectively. If $0<{k}_0\left(={k}_0^{\prime}\right)<{k}^{\ast }$, ${\left\{{k}_t^{\prime}\right\}}_{t=0}^{\infty }$ is monotonically increasing and converges to k ^∗ by Theorem 2. Thus, ${k}_0<{k}_1^{\prime }$ is true. Since ${k}_1>{k}_1^{\prime }$ by Lemma 6, k ₁ > k ₀ holds. Hence, ${k}_0< \inf \left\{\widehat{k}|\widehat{k}\in \widehat{K}\right\}\equiv \underset{\bar{\mkern6mu}}{x}$.

Consider a two-sided altruistic path ${\left\{{x}_0^n\right\}}_{n=0}^{\infty }$ from ${x}_0^0\in \left(0,\overline{k}\right)$. For each generation n, the optimal solution, ${\left\{{x}_t^n\right\}}_{t=0}^{\infty }$, satisfies the Euler equation

$$ -{w}_{\tau}^{\prime}\left( f\left({x}_0^n\right)-{x}_1^n\right)+\beta {w}_{1/\beta}^{\prime}\left( f\left({x}_1^n\right)-{x}_2^n\right){f}^{\hbox{'}}\left({x}_1^n\right)=0. $$

(41)

Since ${\left\{{x}_0^n\right\}}_{n=0}^{\infty }$ is a monotone sequence by Lemma 8 that lies in $\left(0,\overline{k}\right)$, ${x}_0^n$ converges to some $\widehat{k}\in \left(0,\overline{k}\right)$. ${x}_1^n\left(={x}_0^{n+1}\right)$ also converges to $\widehat{k}$. Since $\left({x}_0^n,{x}_1^n,{x}_2^n\right)$ satisfies the Euler equation, the limit as n → ∞ should also satisfy it. Therefore, ${x}_2^n$ converges to some value ${\widehat{k}}_2$ satisfying

$$ -{w}_{\tau}^{\prime}\left( f\left(\widehat{k}\right)-\widehat{k}\right)+\beta {w}_{1/\beta}^{\prime}\left( f\left(\widehat{k}\right)-{\widehat{k}}_2\right){f}^{\hbox{'}}\left(\widehat{k}\right)=0. $$

(42)

This implies ${\widehat{k}}_2=\widehat{k}$. Consequently, the two-sided altruistic path ${\left\{{x}_0^n\right\}}_{n=0}^{\infty }$ converges to a two-sided altruistic stationary path $\Big\{\widehat{k},$ $\widehat{k},$ $\widehat{k},\cdots \Big\}$. □.