Resource conservation across generations in a Ramsey–Chichilnisky model

Abstract

The Chichilnisky criterion is an explicit social welfare function that satisfies compelling conditions of intergenerational equity. However, it is time inconsistent and has no optimal solution in the Ramsey model. By investigating stationary Markov equilibria in the game that generations with Chichilnisky preferences play, this paper shows how, nevertheless, this criterion can be practically implemented in the Ramsey model, leading to attractive consequences. The time-discounted utilitarian optimum is the unique equilibrium path with a high-productive initial stock, implying that the weight on the infinite future in the Chichilnisky criterion plays no role. However, this part of the Chichilnisky criterion may lead to more stock conservation than the time-discounted utilitarian optimum with a low-productive initial stock. Based on the notion of von Neumann–Morgenstern abstract stability, we obtain uniqueness by assuming that each generation coordinates on an almost best equilibrium and takes into account that future generations will do as well.

Appendix: Proofs

Proof

(Proof of Lemma 1) It is easily checked that the set of all I-feasible consumption paths is convex. Since u is strictly concave, the TDU criterion is strictly concave, and the maximum, if it exists, is unique. $\square$

Proof

(Proof of Lemma 2) Assume that the I-feasible pair $(k^{*}(t), c^{*}(t))$ is I-competitive and satisfies the CVT condition. For any other I-feasible pair (k(t), c(t)) we have, using the definition of $p^{*}(t)$ and the concavity of u:

$$\begin{aligned} \int _{0}^{T}e^{-\delta t}\left( u(c^{*}(t))-u(c(t))\right) \hbox {d}t \ge \int _{0}^{T}p^{*}(t)\left( c^{*}(t)-c(t)\right) \hbox {d}t. \end{aligned}$$

Hence, by Eq. (1) and the property that (3) holds for almost all t:

$$\begin{aligned}&\int _{0}^{T}e^{-\delta t}\left( u(c^{*}(t))-u(c(t))\right) \hbox {d}t\\&\quad \ge \int _{0}^{T} \left( p^{*}(t) \big ( \dot{k}(t)-\dot{k}^{*}(t) \big )+\dot{p}^{*}(t)\left( k(t)-k^{*}(t)\right) \right) \hbox {d}t. \end{aligned}$$

Integrating by parts the right-hand side, and using the fact that $k\left( 0\right) =k_{0}=k^{*}\left( 0\right)$:

$$\begin{aligned} \int _{0}^{T}e^{-\delta t} u(c^{*}(t)) \hbox {d}t - \int _{0}^{T}e^{-\delta t} u(c(t) \hbox {d}t \; \ge \; p^{*} (T) k(T) - p^{*}(T) k^{*}(T) . \end{aligned}$$

(10)

Letting $T\rightarrow \infty$ (keeping in mind that limits exist) and using the CVT condition:

$$\begin{aligned} \int _0^\infty e^{-\delta t} u(c^{*}(t)) \hbox {d}t \; \ge \; \int _0^\infty e^{-\delta t} u(c(t) \hbox {d}t. \end{aligned}$$

Hence, $(k^{*}(t), c^{*}(t))$ is I-optimal. $\square$

Proof

(Proof of Lemma 4) Assume that the pair $(I, \sigma )$ is a one-attractor stationary Markov equilibrium and fix an arbitrary $k_0 \in I$, leading to a unique and absolutely continuous capital path, $\kappa (t;k_0)$ converging to $k_\infty$.

Step 1: $\sigma$ is continuous. Let $\tau _\infty (k_0)$ be the finite or infinite time at which $k_\infty$ is reached. Since $\sigma$ is Markovian, the capital path from $k_0 \in I$ is constant if $k_0 = k_\infty$, increasing on $[k_0, k_\infty )$ if $k_0 < k_\infty$ and decreasing on $(k_\infty , k_0]$ if $k_0 > k_\infty$.

Assume $k_0 \ne k_\infty$. Let $\tau (\cdot ;k_0)$ denote the inverse function of $\kappa (\cdot ;k_0)$, defined on $[k_0, k_\infty )$ if $k_0 < k_\infty$ and on $(k_\infty , k_0]$ if $k_0 > k_\infty$. For fixed $(I, \sigma )$, write $J(k_0, I, \sigma ) = (1-\alpha ) V(k_0) + \alpha \lim _{\rho \rightarrow 0^+} \left( \rho \int _0^{\infty } e^{-\rho t} u(\sigma (\kappa (t;k_0))) \hbox {d}t \right)$, so that $V(k_0)$ is the value of the TDU part of the C-criterion when $\sigma$ is followed from $k_0$.

$$\begin{aligned} V(k_0)&= \delta \int _0^\infty u\left( \sigma (\kappa (t;k_0))\right) e^{-\delta t}\hbox {d}t \\&= \delta \int _{k_0}^{k_\infty } \frac{u\left( \sigma (k)\right) }{f(k) - \sigma (k)} e^{-\delta \tau (k;k_0)} \hbox {d}k \; + \; u(f(k_\infty )) e^{-\delta \tau _\infty (k_0)} \end{aligned}$$

if $k_0 < k_\infty$ and

$$\begin{aligned} V(k_0)&= \delta \int _0^\infty u\left( \sigma (\kappa (t;k_0))\right) e^{-\delta t}\hbox {d}t \\&= \delta \int _{k_\infty }^{k_0} -\frac{u\left( \sigma (k)\right) }{f(k) - \sigma (k)} e^{-\delta \tau (k;k_0)} \hbox {d}k \; + \; u(f(k_\infty )) e^{-\delta \tau _\infty (k_0)} \end{aligned}$$

if $k_0 > k_\infty$. By means of this change of variable, we will show that $\sigma$ must be continuous for all values of $k_0$ if $\sigma$ is an equilibrium strategy.

For values of $k_0$ for which $\sigma$ is continuous, we have that

$$\begin{aligned} \frac{\partial \tau (k;k_0)}{\partial k_0} = - \frac{1}{f(k_0)-\sigma (k_0)} \hbox { for all } k, \hbox { and } \frac{d \tau _\infty (k_0)}{d k_0} = - \frac{1}{f(k_0)-\sigma (k_0)}. \end{aligned}$$

Therefore, for values of $k_0$ for which $\sigma$ is continuous, it follows that

$$\begin{aligned} V'(k_0) = \delta \cdot \frac{V(k_0)-u\left( \sigma (k_0)\right) }{f(k_0)-\sigma (k_0)}, \end{aligned}$$

(11)

independently of whether $k_0 < k_\infty$ or $k_0 > k_\infty$.

Since $\sigma$ is an equilibrium strategy, it follows that for all values of $k_0$ for which $\sigma$ is continuous,

$$\begin{aligned} \sigma (k_0) \hbox { maximizes } \delta u(c) + V'(k_0)\left( f(k_0)- c \right) \hbox {over all } c \end{aligned}$$

(12)

by considering a deviation from $\sigma$ in a sufficiently short time interval $(0, \varDelta )$. Since $\sigma$ is an equilibrium strategy, it follows also that for all values of $k_0$ for which $\sigma$ is continuous,

$$\begin{aligned} V(k_0) \ge u(f(k_0)), \end{aligned}$$

(13)

since otherwise staying put at $k_0$ for $t \in (0, \varDelta )$ by choosing $c(t) = f(k_\infty )$ would be a profitable deviation. Furthermore, Eqs. (11) and (12) imply that $V(k_0) - u(\sigma (k_0)) > 0$ if $k_0 < k_\infty$ and $V(k_0) - u(\sigma (k_0)) < 0$ if $k_0 > k_\infty$.

Suppose that $\tilde{k}$ is a point of discontinuity of $\sigma$. Write $c^- = \lim _{k_0 \uparrow \tilde{k}}\sigma (k_0)$ and $c^+ = \lim _{k_0 \downarrow \tilde{k}}\sigma (k_0)$. It follows from Eqs. (11) and (12) that

$$\begin{aligned} c^- \text { maximizes }&u(c) + \frac{V(\tilde{k})-u\left( c^-\right) }{f(\tilde{k})-c^-}\cdot \left( f(\tilde{k})-c \right) \hbox { over all } c, \end{aligned}$$

(14)

$$\begin{aligned} c^+ \hbox { maximizes }&u(c) + \frac{V(\tilde{k})-u\left( c^+\right) }{f(\tilde{k})-c^+}\cdot \left( f(\tilde{k})-c \right) \hbox { over all } c. \end{aligned}$$

(15)

Since $\tilde{k}$ is a point of discontinuity of $\sigma$, we have that $c^- \ne c^+$. However, if Eq. (13) is satisfied with strict inequality,⁸ then the strict concavity of u implies that the equation

$$\begin{aligned} u'(\tilde{c}) = \frac{V(\tilde{k})-u\left( \tilde{c}\right) }{f(\tilde{k})-\tilde{c}} \end{aligned}$$

(16)

is solved by a unique $\tilde{c}' < f(\tilde{k})$, corresponding to the case where $\tilde{k} < k_\infty$ and a unique $\tilde{c}'' > f(\tilde{k})$, corresponding to the case where $\tilde{k} > k_\infty$. This contradicts that both (14) and (15) can be satisfied and proves that $\sigma$ must be continuous for $k_0 \ne k_\infty$.

It remains to be shown that $\sigma$ is continuous at $k_\infty$. With $k_0 = k_\infty$, Eq. (13) is satisfied with equality. Furthermore, V is continuous at $k_\infty$ as (i) $\sigma (\kappa (t;k_0)) \rightarrow f(k_\infty )$ for all t as $k_0 \rightarrow k_\infty$ in the case where $\tau _\infty (k_0) = \infty$ for $k_0 \ne k_\infty$, and (ii) $\tau _\infty (k_0) \rightarrow 0^+$ as $k_0 \rightarrow k_\infty$ otherwise. It follows from the strict concavity of u and the property that $\sigma (k_0)$ solves (16) for $\tilde{k} = k_0$ that $\sigma (k_0)$ approaches $f(k_\infty )$ continuously also in case (ii).

Step 2: $(\kappa (t;k_0), \sigma (\kappa (t;k_0)))$ satisfies Eq. (3) for almost all $t \in [0, \infty )$. Since $(I, \sigma )$ is a one-attractor stationary Markov equilibrium, then there exists $\varDelta > 0$ such that

$$\begin{aligned} \int _0^\varDelta e^{-rt} \left( u(\sigma (\kappa (t,k_0)) - u(c_{k_0,\varDelta }(t)) \right) \hbox {d}t \ge 0 \end{aligned}$$

(17)

for every I-admissible choice $c_{k_0,\varDelta }$ where the solution $k : [0, \infty ) \rightarrow I$ to $k(0) = k_0$ and $\dot{k} = f(k) - c_{k_0, \varDelta }(t)$ for $t \in [0, \infty )$ satisfies that $k(\varDelta ) = k_1 := \tau (\varDelta ,k_0)$, since then $\sigma (\kappa (t,k_0)) = c_{k_0, \varDelta }(t)$ for $t \in (\varDelta , \infty )$.

There exists a pair $\hat{k} : [0, \varDelta ] \rightarrow I$ and $\hat{c} : [0, \varDelta ] \rightarrow \mathbb {R}_{+}$ such that (i) $\hat{c}$ is absolutely continuous, so that the associated present-value price path $\hat{p}(t)$ defined $\hat{p}(t) = e^{-\delta t}u'(\hat{c}(t))$ is differentiable almost everywhere, (ii) Eq. (3) is satisfied for almost all $t \in [0, \varDelta ]$, and (iii) $\hat{k}(0)=k_0$ and $\hat{k}(\varDelta )=k_1$. By (10) and the strict concavity of the TDU criterion, we have that

$$\begin{aligned} \int _0^\varDelta e^{-rt} \left( u(\sigma (\kappa (t,k_0)) - u(\hat{c}(t)) \right) \hbox {d}t < 0 \end{aligned}$$

if $(\kappa (t,k_0), \sigma (\kappa (t,k_0)))$ does not coincide with $(\hat{k}(t),\hat{c}(t))$ for $t \in [0,\varDelta ]$, contradicting (17). Hence, the pair $(\kappa (t,k_0), \sigma (\kappa (t,k_0)))$ satisfies Eq. (3) for almost all $t \in [0, \varDelta ]$. Since $k_0 \in I$ is arbitrary, it follows that $(\kappa (t;k_0), \sigma (\kappa (t;k_0)))$ satisfies Eq. (3) for almost all $t \in [0, \infty )$.

Step 3: $(\kappa (t;k_0), \sigma (\kappa (t;k_0))$ is I-competitive and satisfies the CVT condition. By Step 1, $\sigma : I \rightarrow \mathbb {R}_{++}$ is continuous, so that, for every $k_0 \in I$, the solution $\kappa (t;k_0)$ has the properties that $\sigma (\kappa (t;k_0))$ is an absolutely continuous function of t and, by Step 2, $(\kappa (t;k_0), \sigma (\kappa (t;k_0))$ satisfies Eq. (3) for almost all $t \in [0, \infty )$. Moreover, by Definition 1, $(\kappa (t;k_0), \sigma (\kappa (t;k_0))$ is I-feasible, so that $(\kappa (t;k_0), \sigma (\kappa (t;k_0))$ is I-competitive. Finally, by Definition 1, $\kappa (t;k_0)$ converges, implying that $(\kappa (t;k_0), \sigma (\kappa (t;k_0))$ satisfies the CVT condition. $\square$

Proof

(Proof of Corollary 1) Assume that the pair $(I, \sigma )$ is an equilibrium, and let $k_0 \in \hbox {int}I$. Let $V(k_0)$ be defined as in the proof of Lemma 4. We need to show that $V'(k_0)$ exists and equals $\delta u'(\sigma (k_0))$.

Case 1: $k_0 \ne k_\infty$. The result follows from (12) of the proof of Lemma 4.

Case 2: $k_0 = k_\infty$ Since $k_0 \in \hbox {int}I$, we are in case (a) of the cases considered in the text preceding the corollary, so that $k_0 = \underline{k}$. The result follows since

$$\begin{aligned} V(k_0) = \delta \int _0^\infty u\left( \sigma (\kappa (t;k_0))\right) e^{-\delta t}\hbox {d}t \end{aligned}$$

is differentiable as a function of $k_0$ at $k_0 = \underline{k}$ and $V(k_0) \ge u(f(k_0))$ for all $k_0 \in I$.

$\square$

Proof

(Proof of Lemma 5) Included in the main text. $\square$

Proof

(Proof of Lemma 6) Let $\sigma = \{(I_1, \sigma _1), \ldots , (I_n, \sigma _n)\}$ be a multiple-attractor equilibrium strategy.

Step 1: At a point of discontinuity $k'$ of $V_{\sigma }, V_{\sigma }(k') = u(f(k'))$. By Propositions 2 and 4 and the observation that $V_{\sigma }$ cannot be discontinuous at $\underline{k}$, if $k'$ is a point of discontinuity, then there is $i \in \{1, \ldots , n\}$ such that either $I_i \subset (0, \underline{k})$ and $k' = \max I_i$ or $I_i \subset (\underline{k}, \infty )$ and $k' = \min I_i$. In both cases,

$$\begin{aligned} V_{\sigma }(k') = (1 - \alpha ) \delta \int _0^{\infty } e^{- \delta t}u(f(k')) \hbox {d}t + \alpha u(f(k')) = u(f(k')) \end{aligned}$$

since $\sigma |_{I_i}(k') = f(k')$ and $\delta \int _0^{\infty } e^{- \delta t} \hbox {d}t = 1$.

Step 2: For all $k' \in (0, \underline{k}), V_{\sigma }(k') \ge u(f(k'))$. By Propositions 2 and 4, if $k' \in (0, \underline{k})$, then there is $i \in \{1, \ldots , n\}$ such that $k' \in I_i$, where the solution $\kappa (\cdot , k') : [0, \infty ) \rightarrow I_i$ to $k(0) = k'$ and $\dot{k} = f(k) - \sigma |_{I_i}(k)$ for $t \in [0, \infty )$ has the property that $k_{\infty } = \lim _{t \rightarrow \infty }\kappa (t,k')$ satisfies $k_{\infty } = \max I_i < \underline{k}$ or $k_{\infty } = \underline{k}$. In either case,

$$\begin{aligned} u(f(k_{\infty })) \ge u(f(k')) \end{aligned}$$

(18)

since $k' \le k_{\infty }$, and both f and u are strictly increasing. Moreover, since the pair (k(t), c(t)) with $k(t) = k'$ and $c(t) = f(k')$ for all $t \in [0, \infty )$ is $I_i$-feasible, it follows from the $I_i$-optimality of $(\kappa (t,k'), \sigma _i(\kappa (t,k')))$ that

$$\begin{aligned} \delta \int _0^{\infty } e^{- \delta t}u(\sigma _i(\kappa (t,k'))) \hbox {d}t \ge \delta \int _0^{\infty } e^{- \delta t}u(f(k')) \hbox {d}t = u(f(k')) \, . \end{aligned}$$

(19)

Hence, by (18) and (19), $V_{\sigma }(k') \ge (1 - \alpha )u(f(k')) + \alpha u(f(k')) = u(f(k'))$.

The two steps imply that discontinuity of the value function $V_{\sigma }$ for $k \in (0, \underline{k})$ is inconsistent with $V_{\sigma }$ being upper semi-continuous. Combined with Lemma 5, this establishes the result since, as argued in footnote 5, $V_{\sigma }$ cannot be discontinuous at $\underline{k}$. $\square$

Proof

(Proof of Lemma 7) It follows from Corollary 1 and the definition of $v_{k_\infty }(k_0)$ that

$$\begin{aligned} \tfrac{\partial }{\partial k_\infty } \upsilon '_{k_\infty }(k_0) \!=\! \tfrac{\partial }{\partial k_\infty }\left[ (1 - \alpha )\delta u'(\sigma _{k_\infty }(k_0)) \right] = (1 - \alpha )\delta u''(\sigma _{k_\infty }(k_0)) \tfrac{\partial }{\partial k_\infty }\left( \sigma _{k_\infty }(k_0) \right) \!>\! 0 \end{aligned}$$

for $(k_\infty , k_0) \in [\underline{k}, \infty ) \times (k_\infty , \infty )$, by the strict concavity of u and the properties of I-competitive paths in the phase diagram in (k, c)-space. $\square$

Proof

(Proof of Lemma 9) Consider the $\varepsilon$-system $(D, \succ _{\varepsilon })$ for the Ramsey–Chichilnisky game, where $\varepsilon$ is some positive number. Suppose that $\succ _{\varepsilon }$ is not strictly acyclic, that is, there exists an infinite sequence $\{(k_1,\sigma _1), (k_2,\sigma _2), \ldots , (k_j,\sigma _j) , \ldots \}$ of elements in D such that, for all $j \in \mathbb {N}, (k_{j+1},\sigma _{j+1}) \succ _{\varepsilon } (k_j,\sigma _j)$. In each of two exhaustive cases, this leads to a contradiction.

Case 1. There exists $j \in \mathbb {N}$ such that $k_{j} \in (0, \underline{k}]$. If $k \in (0, \underline{k}\,]$, then $(k, \sigma ) \in D$ only if $\sigma = \{(I_1, \sigma |_{I_1}), \dots , (I_n, \sigma |_{I_n})\}$ satisfies that there is $\kappa \in (\underline{k}, \infty ) \cup \{\infty \}$ such that $I_1 = (0,\kappa )$ and $\sigma |_{I_1} = \sigma _{\underline{k}}|_{(0,\kappa )}$. Furthermore, by Lemma 8(i), $k'$ is reachable from $k \in (0, \underline{k}]$ if and only if $k' \in (0, \underline{k}]$. Therefore, for any $\varepsilon > 0$, if $k_j \in (0, \underline{k}\,]$, then there is no $(k_{j+1}, \sigma _{j+1}) \in D$ such that $(k_{j+1}, \sigma _{j+1}) \succ _{\varepsilon } (k_j, \sigma _j)$. This establishes the contraction in this case.

Case 2. There does not exist $j \in \mathbb {N}$ such that $k_{j} \in (0, \underline{k}]$. By Lemma 8(ii), $\{k_j\}_{j \in \mathbb {N}}$ is a non-increasing sequence; hence, there is $\hat{k} \in [\underline{k}, \infty )$ such that $\hat{k} = \lim _{j \rightarrow \infty }k_j$. The set $\{V_\sigma (\hat{k}) : \sigma \in \varSigma \}$ is bounded, with—as argued in Sect. 6—$\upsilon _{\underline{k}}(\hat{k})$ being the greatest lower bound, and $(1-\alpha )\sup _{A\left( \hat{k}\right) } \delta \int _{0}^{\infty }e^{-\delta t}u(c(t))\hbox {d}t + \alpha u(f(\hat{k}))$ being an upper bound.⁹ For any $\zeta > 0$ and any $\sigma \in \varSigma , V_\sigma (k)$ is Lipschitz continuous on $[\hat{k}, \hat{k} + \zeta ]$. Hence, for any $\varepsilon > 0$, the existence of an infinite sequence $\{(k_j,\sigma _j), (k_{j+1},\sigma _{j+1}), \ldots , (k_{j+\ell }, \sigma _{j+\ell }), \dots ) \}$ with $\hat{k} \le \cdots \le k_{j+\ell } \le \cdots \le k_{j+1} \le k_j \le \hat{k} + \zeta$ such that, for all $\ell \in \mathbb {N}, (k_{j+\ell +1},\sigma _{j+\ell +1}) \succ _{\varepsilon } (k_{j+\ell },\sigma _{j+\ell })$ contradicts that the set $\{V_\sigma (\hat{k}) : \sigma \in \varSigma \}$ is bounded, for $\zeta > 0$ chosen sufficiently small. $\square$

Proof

(Proof of Lemma 10) If $k \in (0, \underline{k}\,]$, then $(k, \sigma ) \in D$ if $\sigma = \{(\mathbb {R_{++}}, \sigma _{\underline{k}})\}$ and $(k, \sigma ) \in D$ only if $\sigma = \{(I_1, \sigma |_{I_1}), \ldots , (I_n, \sigma |_{I_n})\}$ satisfies that there is $\kappa \in (\underline{k}, \infty ) \cup \{\infty \}$ such that $I_1 = (0,\kappa )$ and $\sigma |_{I_1} = \sigma _{\underline{k}}|_{(0,\kappa )}$. Furthermore, $k'$ is reachable from $k \in (0, \underline{k}]$ if and only if $k' \in (0, \underline{k}]$. Therefore, for any $\varepsilon > 0$, if $k \in (0, \underline{k}\,]$ and $(k, \sigma ) \in D$, then there is no $(k', \sigma ') \in D$ such that $(k', \sigma ') \succ _{\varepsilon } (k, \sigma )$. This establishes the lemma. $\square$

Proof

(Proof of Lemma 11) Using Proposition 6, Lemma 7 and expressions (8) and (9), it follows that, for all $\kappa \in (\underline{k}, \bar{k} \, ], u(f(\kappa )) = \max _{\sigma \in \varSigma }V_\sigma (\kappa )$. Fix $k \in (\underline{k}, \bar{k}\,]$. If $\sigma = \{(I_1, \sigma |_{I_1}), \ldots , (I_n, \sigma |_{I_n})\} \in \varSigma$, then $(k, \sigma ) \in G_{\epsilon }$ if $u(f(k')) - V_\sigma (k') \le \varepsilon$ for all $k' \in (\underline{k}, k\,]$, since then there is no $(k', \sigma ') \in D$ such that $(k', \sigma ') \succ _{\varepsilon } (k, \sigma )$, using the observation in the proof of Lemma 10 that no domination is possible for $k' \in (0, \underline{k})$.

It follows that, for all $k' \in (\underline{k}, k\,], (k', \sigma ') \in G_\varepsilon$, provided that $\sigma ' = \{(I'_1, \sigma '|_{I'_1}), \ldots , (I'_n, \sigma '|_{I'_n})\} \in \varSigma$ satisfies that $u(f(k'')) - V_{\sigma '}(k'') \le \varepsilon$ for all $k'' \in (\underline{k}, k'\,]$. This is consistent with $I'_n = [k' , \infty )$ and $\sigma '|_{I'_n} = \sigma _{k'}|_{I'_n}$ so that $V_{\sigma '}(k') = u(f(k'))$. This shows that $(k, \sigma ) \in G_{\epsilon }$ only if $u(f(k')) - V_\sigma (k') \le \varepsilon$ for all $k' \in (\underline{k}, k\,]$, since otherwise there is $(k', \sigma ') \in G_\varepsilon$ such that $(k', \sigma ') \succ _{\varepsilon } (k, \sigma )$.

Proof

(Proof of Lemma 12) Suppose $k \in (\bar{k}, \infty ), (k, \sigma ) \in G_\varepsilon$ and $v_{\bar{k}}(k') - V_\sigma (k') > \varepsilon$ for some $k' \in (\underline{k}, k]$. By internal $\varepsilon$-stability, there does not exist $(k', \sigma ') \in G_\varepsilon$ where $\sigma ' = \{(I_1, \sigma '|_{I_1}), \ldots , (I_n, \sigma '|_{I_n})\}$ satisfies that $I_n = [\bar{k}, \infty )$. It follows from Lemmas 10 and 11 that, by choosing sufficiently many points of discontinuity in $(\underline{k}, \bar{k}\,]$, there exists $\sigma ' = \{(I_1, \sigma '|_{I_1}), \ldots , (I_n, \sigma '|_{I_n})\}$ such that, for all $k'' \in [0, \bar{k}\,], (k'', \sigma ') \in G_\varepsilon$. Hence, by external $\varepsilon$-stability, there exists $k'' \in (\bar{k}, k'\,]$ and $(k'', \sigma '') \in G_\varepsilon$ such that $V_{\sigma ''}(k'') - v_{\bar{k}}(k'') > \varepsilon$. However, by Lemmas 5 and 7, $v_{\bar{k}}(k'') > V_\sigma (k'')$ for $k'' \in (\bar{k}, k'\,]$ so that $V_{\sigma ''}(k'') - V_{\sigma }(k'') > \varepsilon$ and $(k'', \sigma '') \succ _{\varepsilon } (k, \sigma )$. Since $(k, \sigma ), (k'', \sigma '') \in G_\varepsilon$, this contradicts internal $\varepsilon$-stability and shows that $v_{\bar{k}}(k') - V_\sigma (k') \le \varepsilon$ for all $k' \in (\underline{k}, k]$ if $k \in (\bar{k}, \infty )$ and $(k, \sigma ) \in G_\varepsilon$. $\square$