Will Technological Change Save the World? The Rebound Effect in International Transfers of Technology

Abstract

Technological change and its transfer to developing countries is often portrayed by policy-makers as a critical part of the solution to a resource problem such as climate change, based on the assumption that the transfer of resource-conserving technologies to developing countries will result in reduced use of natural capital by those countries. We demonstrate here, in a capital conversion based model of development, that the free transfer of resource-conserving technologies to developing countries will increase the options available to those countries, but that the way that they expend these options need not be in the direction of conserving resources. This is another example of the potential for a rebound effect to determine ultimate outcomes, here in the context of international technology transfer policy. The transfer of technologies is as likely to simply move developing countries more rapidly down the same development path as it is to alter the choices they make along that path. For this reason, the transfer of resource-conserving technologies, without incentives provided to alter development priorities, may not result in any resource-conservation at all.

Appendix

1.1 Proof Proposition 1

The Hamiltonian of this problem $\mathcal {H}=U(c,N)+\lambda \left[ f(k)-c-m\right] +\mu \left[ \Psi (m)\right. \left. -\,\phi f(k)\right]$ yields the following first order conditions:

$$\begin{aligned}&U_{c}=\lambda \end{aligned}$$

(13)

$$\begin{aligned}&-\lambda +\mu \Psi ^{\prime }(m)\le 0\;\quad \text{ with } \text{ equality } \text{ if } m>0\nonumber \\&m = {\left\{ \begin{array}{ll} 0 &{} \text{ if } \lambda \ge \mu \\ m^{*}=\Psi ^{\prime -1}\left( \lambda /\mu \right) >0 &{} \text{ if } \lambda =\mu \Psi ^{\prime }(m) \end{array}\right. } \end{aligned}$$

(14)

$$\begin{aligned}&\frac{\dot{\lambda }}{\lambda }={\left\{ \begin{array}{ll} {\displaystyle \rho -\left( 1-\frac{\mu }{\lambda }\phi \right) f^{\prime }(k)} &{} \text{ if } \lambda \ge \mu \\ {\displaystyle \rho -\left( 1-\frac{\phi }{\Psi ^{\prime }(m)}\right) f^{\prime }(k)} &{} \text{ if } \lambda =\mu \Psi ^{\prime }(m) \end{array}\right. } \end{aligned}$$

(15)

$$\begin{aligned}&\frac{\dot{\mu }}{\mu }={\left\{ \begin{array}{ll} {\displaystyle \rho -\frac{U_{N}\left( c,N\right) }{\mu }} &{} \text{ if } \lambda \ge \mu \\ {\displaystyle \rho -\frac{U_{N}\left( c,N\right) }{U_{c}}\Psi ^{\prime }(m)} &{} \text{ if } \lambda =\mu \Psi ^{\prime }(m) \end{array}\right. } \end{aligned}$$

(16)

The dynamics of consumption and environmental management obtain following time differentiation of (13) and $\lambda =\mu \Psi ^{\prime }(m)$ for $m>0$:

$$\begin{aligned}&{\displaystyle \frac{\dot{c}}{c}={\left\{ \begin{array}{ll} \frac{1}{\sigma }{\displaystyle \left[ \left( 1-\frac{\mu }{\lambda }\phi \right) f^{\prime }(k)-\rho \right] } &{} \text{ if } \lambda \ge \mu \\ \frac{1}{\sigma }{\displaystyle \left[ \left( 1-\frac{\phi }{\Psi ^{\prime }(m)}\right) f^{\prime }(k)-\rho \right] } &{} \text{ if } \lambda =\mu \Psi ^{\prime }(m) \end{array}\right. }} \end{aligned}$$

(17)

$$\begin{aligned}&\dot{m}={\left\{ \begin{array}{ll} 0 &{} \text{ if } \lambda \ge \mu \\ {\displaystyle \frac{\Psi ^{\prime }(m)}{\Psi ^{\prime \prime }(m)}\left\{ \frac{U_{N}}{U_{c}}\Psi ^{\prime }(m)-\sigma \frac{\dot{c}}{c}-\rho \right\} } &{} \text{ if } \lambda =\mu \Psi ^{\prime }(m) \end{array}\right. } \end{aligned}$$

(18)

Part (a) Consider a developing country that is initially endowed with low physical capital stock $k_{0}$ and large natural capital stock $N_{0}$. So long as the shadow value of physical capital is high relative to the shadow value of natural capital so that $\lambda \ge \mu$, the social planner will optimally choose not to manage the environment and set $m^{*}=0$. To see that there is an optimal time at which environmental management is undertaken, we resort to the same argument as Stokey (1998). Since $\lambda$ is decreasing while $\mu$ is increasing , so that the ratio $\lambda /\mu$ declines monotonically, there exists a finite time period $\tilde{t}$ (i.e. $\tilde{t}<\infty$) that satisfies $\lambda (\tilde{t})=\mu (\tilde{t})$ (with $\dot{\lambda }(\tilde{t})<0$ and $\dot{\mu }(\tilde{t})>0$) so that for $t<\tilde{t}$, $\lambda (t)\ge \mu (t)$ and therefore $m\left( t\right) =0$.

Now for $t\le \tilde{t}$, $\dot{\lambda }(t)<0$ if and only if ${\displaystyle f^{\prime }\left( k\left( t\right) \right) >\frac{\rho }{1-{\displaystyle \frac{\mu \left( t\right) }{\lambda \left( t\right) }}\phi }}$ i.e. $\displaystyle k\left( t\right) <f^{\prime -1}\left( \frac{\rho }{1-{\displaystyle \frac{\mu \left( t\right) }{\lambda \left( t\right) }}\phi }\right)$. In particular for $t=0$, we have ${\displaystyle k_{0}<f^{\prime -1}\left( \frac{\rho }{1-{\displaystyle \frac{\mu _{0}}{\lambda _{0}}}\phi }\right) }$. In addition, from (16), for $t\le \tilde{t}$, $\dot{\mu }>0$ if and only if $U_{N}<\rho \mu \le \rho \lambda =\rho U_{c}$ or equivalently ${\displaystyle \frac{U_{c}}{U_{N}}>\frac{1}{\rho }}$ , i.e. if the marginal rate of N for c is large enough. In other words if N is large enough. In particular for $t=0$, we have ${\displaystyle \frac{U_{c}\left( c_{0},N_{0}\right) }{U_{N}\left( c_{0},N_{0}\right) }>\frac{1}{\rho }}$.

Part (b) For $t\le \tilde{t}$, it is straightforward to see that $\dot{N}=-\phi f\left( k\right) <0$ and $\dot{m}=0$. By definition of $\tilde{t}$, when $t\le \tilde{t}$, $\dot{\lambda }<0$ and $\dot{\mu }>0$. By the dynamics of c (Eq. 17), $\dot{c}>0$ if and only if $\dot{\lambda }<0$. This in turn implies that $\dot{k}=f(k)-c>0$.

Part (c) Part (a) implies that the developing country will use management $m\left( t\right) >0$ for $t>\tilde{t}$. In the transition phase towards $\tilde{t}$, we know that $\lambda \ge \mu \Psi ^{\prime }(m)$, which implies that ${\displaystyle 1-\frac{\mu }{\lambda }\phi \ge 1-\frac{\phi }{\Psi ^{\prime }(m)}}$. In addition, since $f^{\prime }\left( k\left( t\le \tilde{t}\right) \right)>f^{\prime }\left( k\left( t>\tilde{t}\right) \right)$, it follows that:

$$\begin{aligned} {\displaystyle \frac{1}{\sigma }\left[ \left( 1-\frac{\mu }{\lambda }\phi \right) f^{\prime }\left( k\left( t\le \tilde{t}\right) \right) -\rho \right] }> & {} \frac{1}{\sigma }\left[ \left( 1-\frac{\phi }{\Psi ^{\prime }(m)}\right) f^{\prime }\left( k\left( t>\tilde{t}\right) \right) -\rho \right] \nonumber \\ \frac{\dot{c}\left( t\le \tilde{t}\right) }{c\left( t\le \tilde{t}\right) }> & {} {\displaystyle \frac{\dot{c}\left( t>\tilde{t}\right) }{c\left( t>\tilde{t}\right) }} \end{aligned}$$

(19)

That is, consumption growth slows down beyond the optimal date $\tilde{t}$. In addition, multiplying (19) by $-\sigma$ yields:

$$\begin{aligned} {\displaystyle \rho -\left( 1-\frac{\mu }{\lambda }\phi \right) f^{\prime }\left( k\left( t\le \tilde{t}\right) \right) }< & {} \rho -\left( 1-\frac{\phi }{\Psi ^{\prime }(m)}\right) f^{\prime }\left( k\left( t>\tilde{t}\right) \right) \nonumber \\ \frac{\dot{\lambda }\left( t\le \tilde{t}\right) }{\lambda \left( t\le \tilde{t}\right) }< & {} {\displaystyle \frac{\dot{\lambda }\left( t>\tilde{t}\right) }{\lambda \left( t>\tilde{t}\right) }} \end{aligned}$$

(20)

Inequality (20) suggests that the shadow value of physical capital decelerates at a slower rate beyond $\tilde{t}$. It must be the case that the growth in the stock of physical capital slows down relative to the earlier phase before $\tilde{t}$.⁴⁰

In addition, since for $t>\tilde{t}$, $\dot{N}=\Psi (m)-\phi f(k)$, the depletion of natural capital decelerates due to the investment in environmental management. Note that for sufficiently small $\phi <\Psi \left( m\right) /f\left( k\right)$, $\dot{N}>0$, i.e. the natural capital is regenerated.

Part (d) We established earlier that at $\tilde{t}$, $\lambda (\tilde{t})=\mu (\tilde{t})$. This is implies that:

$$\begin{aligned}&\displaystyle \int _{\tilde{t}}^{\infty }\dot{\lambda }(s)ds = \int _{\tilde{t}}^{\infty }\dot{\mu }(s)ds\nonumber \\&\displaystyle \int _{\tilde{t}}^{\infty }\left[ \left( \rho -f^{\prime }\left( k\left( s\right) \right) \right) \lambda (s)+\phi \mu (s)f^{\prime }\left( k\left( s\right) \right) \right] ds = \int _{\tilde{t}}^{\infty }\left[ \rho \mu (s)-U_{N}\left( c\left( s\right) ,N\left( s\right) \right) \right] ds\nonumber \\&\displaystyle \int _{\tilde{t}}^{\infty }\left[ \left( \rho -f^{\prime }\left( k\left( s\right) \right) \right) \lambda (s)+\phi \mu (s)f^{\prime }\left( k\left( s\right) \right) \right] -\left[ \rho \mu (s)-U_{N}\left( c\left( s\right) ,N\left( s\right) \right) \right] ds = 0\nonumber \\ \end{aligned}$$

(21)

Consider ${\displaystyle J(\beta )=\int _{a}^{b}h(x,\beta )\,\mathrm {d}x}$, Leibniz rule states that $\displaystyle \frac{\mathrm {d}J}{\mathrm {d}\beta }=\int _{a}^{b}h_{\beta }(x,\beta )\,\mathrm {d}x+h(b,\beta )\frac{\mathrm {d}b}{\mathrm {d}\beta }-h(a,\beta )\frac{\mathrm {d}a}{\mathrm {d}\beta }$. Applying Leibniz rule to Eq. (21), we obtain:

$$\begin{aligned}&\displaystyle \left. {\left\{ \begin{array}{ll} \int _{\tilde{t}}^{\infty }\mu (s)f^{\prime }\left( k\left( s\right) \right) ds\\ -\left[ \underbrace{\left( \rho -f^{\prime }\left( k\left( \tilde{t}\right) \right) \right) \lambda (\tilde{t})+\phi \mu (\tilde{t})f^{\prime }\left( k\left( \tilde{t}\right) \right) }_{\dot{\lambda }\left( \tilde{t}\right) }-\underbrace{\left( \rho \mu (\tilde{t})-U_{N}\left( c\left( \tilde{t}\right) ,N\left( \tilde{t}\right) \right) \right) }_{\dot{\mu }\left( \tilde{t}\right) }\right] {\displaystyle \frac{d\tilde{t}}{d\phi }} \end{array}\right. }\right\} = 0 \nonumber \\\end{aligned}$$

(22)

$$\begin{aligned}&\displaystyle {\displaystyle \frac{d\tilde{t}}{d\phi }} = \frac{{\displaystyle \int _{\tilde{t}}^{\infty }\mu (s)f^{\prime }\left( k\left( s\right) \right) ds}}{\dot{\lambda }\left( \tilde{t}\right) -{\displaystyle \dot{\mu }\left( \tilde{t}\right) }}<0 \end{aligned}$$

(23)

This is true because $\dot{\lambda }\left( \tilde{t}\right) <0$ and $\dot{\mu }\left( \tilde{t}\right) >0$ by Part (a) of this proof (see above). $\square$

1.2 Proof Proposition 2

The system of Eqs. (9–12) that defines the steady state after total differentiation can be written in matrix form as:

where the matrix is denoted $\Xi$; ${\displaystyle a_{1}\equiv \frac{\Psi ^{\prime \prime }\left( m\right) }{\Psi ^{\prime }\left( m\right) ^{2}}\phi }f^{\prime }\left( k\right) <0$; ${\displaystyle a_{2}\equiv \left( 1-\frac{\phi }{\Psi ^{\prime }\left( m\right) }\right) }f^{\prime \prime }\left( k\right) <0$; $a_{3}\equiv -\rho U_{cc}>0$; $a_{4}\equiv U_{N}\Psi ^{\prime \prime }\left( m\right) <0$; $a_{5}\equiv U_{NN}\Psi ^{\prime }\left( m\right) <0$.

Define $\Xi =\left( \begin{array}{cccc} 0 &{} a_{1} &{} a_{2} &{} 0\\ a_{3} &{} a_{4} &{} 0 &{} a_{5}\\ -1 &{} -1 &{} f^{\prime }\left( k\right) &{} 0\\ 0 &{} \Psi ^{\prime }\left( m\right) &{} -\phi f^{\prime }\left( k\right) &{} 0 \end{array}\right)$; $M_{c}=\left( \begin{array}{cccc} \frac{f^{\prime }\left( k\right) }{\Psi ^{\prime }\left( m\right) } &{} a_{1} &{} a_{2} &{} 0\\ 0 &{} a_{4} &{} 0 &{} a_{5}\\ 0 &{} -1 &{} f^{\prime }\left( k\right) &{} 0\\ f\left( k\right) &{} \Psi ^{\prime }\left( m\right) &{} -\phi f^{\prime }\left( k\right) &{} 0 \end{array}\right)$;

$M_{m}=\left( \begin{array}{cccc} 0 &{} \frac{f^{\prime }\left( k\right) }{\Psi ^{\prime }\left( m\right) } &{} a_{2} &{} 0\\ a_{3} &{} 0 &{} 0 &{} a_{5}\\ -1 &{} 0 &{} f^{\prime }\left( k\right) &{} 0\\ 0 &{} f\left( k\right) &{} -\phi f^{\prime }\left( k\right) &{} 0 \end{array}\right)$; $M_{k}=\left( \begin{array}{cccc} 0 &{} a_{1} &{} \frac{f^{\prime }\left( k\right) }{\Psi ^{\prime }\left( m\right) } &{} 0\\ a_{3} &{} a_{4} &{} 0 &{} a_{5}\\ -1 &{} -1 &{} 0 &{} 0\\ 0 &{} \Psi ^{\prime }\left( m\right) &{} f\left( k\right) &{} 0 \end{array}\right)$

$M_{N}=\left( \begin{array}{cccc} 0 &{} a_{1} &{} a_{2} &{} \frac{f^{\prime }\left( k\right) }{\Psi ^{\prime }\left( m\right) }\\ a_{3} &{} a_{4} &{} 0 &{} 0\\ -1 &{} -1 &{} f^{\prime }\left( k\right) &{} 0\\ 0 &{} \Psi ^{\prime }\left( m\right) &{} -\phi f^{\prime }\left( k\right) &{} f\left( k\right) \end{array}\right)$

Using Cramer’s rule, we can derive the comparative statics:

$$\begin{aligned} \frac{dc^{\infty }}{d\phi }=\frac{\det M_{c}}{\det \Xi };\;\;\;\;\frac{dm^{\infty }}{d\phi }=\frac{\det M_{m}}{\det \Xi };\;\;\;\;\frac{dk^{\infty }}{d\phi }=\frac{\det M_{k}}{\det \Xi };\;\;\;\;\frac{dN^{\infty }}{d\phi }=\frac{\det M_{N}}{\det \Xi } \end{aligned}$$

The determinant of $\Xi$ is negative and given by:

$$\begin{aligned} \det \Xi =-a_{5}\left[ a_{1}\phi f^{\prime }\left( k\right) +a_{2}\Psi ^{\prime }\left( m\right) \right] <0 \end{aligned}$$

The direction of the four effects is therefore given by the sign of the determinants $\det M_{c}$, $\det M_{m}$, $\det M_{k}$ and $\det M_{N}$.

Part (a) we have:

$$\begin{aligned} \det M_{c}= & {} \frac{a_{5}\left[ -\left( \Psi ^{\prime }\left( m\right) -\phi \right) f^{\prime }\left( k\right) ^{2}+\left( a_{2}+a_{1}f^{\prime }\left( k\right) \right) f\left( k\right) \Psi ^{\prime }\left( m\right) \right] }{\Psi ^{\prime }\left( m\right) }>0\\ \det M_{k}= & {} a_{5}\left[ a_{1}f(k)-f^{\prime }(k)\right] >0 \end{aligned}$$

It follows that $dc^{\infty }/d\phi<0;\;\;\text{ and } \;\;dk^{\infty }/d\phi <0$.

Part (b) & (c)

$$\begin{aligned} \det M_{m}= & {} -a_{5}\left[ \frac{\phi f^{\prime }\left( k\right) ^{2}}{\Psi ^{\prime }\left( m\right) }+a_{2}f(k)\right] \end{aligned}$$

As a result, ${{\mathrm{sgn}}}\left( \det M_{m}\right) {=}{{\mathrm{sgn}}}\left( {\displaystyle \Delta }\right)$ where $\displaystyle \Delta {\equiv }\frac{\phi f^{\prime }\left( k\right) ^{2}}{\Psi ^{\prime }\left( m\right) }+ a_{2}f(k){=}\bigg [\phi +\left( \Psi ^{\prime }\left( m\right) -\phi \right) \frac{f\left( k\right) f^{\prime \prime }\left( k\right) }{f^{\prime }\left( k\right) ^{2}}\bigg ]\frac{f^{\prime }\left( k\right) ^{2}}{\Psi ^{\prime }\left( m\right) }$.

$\Delta <0$ if and only if ${\displaystyle \Psi ^{\prime }\left( m\right) >\left( 1-\frac{f^{\prime }\left( k\right) ^{2}}{f\left( k\right) f^{\prime \prime }\left( k\right) }\right) \phi }$. Then $\det M_{m}<0$ and consequently we always have $dm^{\infty }/d\phi >0$.

$\Delta >0$ if and only if ${\displaystyle \Psi ^{\prime }\left( m\right) <\left( 1-\frac{f^{\prime }\left( k\right) ^{2}}{f\left( k\right) f^{\prime \prime }\left( k\right) }\right) \phi }$. Then $\det M_{m}>0$ and consequently we always have $dm^{\infty }/d\phi <0$.

In addition, we have:

$$\begin{aligned} \det M_{N}= & {} -\underbrace{\frac{f^{\prime }\left( k\right) ^{2}}{\Psi ^{\prime }\left( m\right) }\rho U_{cc}\Gamma }_{>0}+\underbrace{\Psi ^{\prime \prime }\left( m\right) U_{N}}_{<0}\underbrace{\Delta }_{\lessgtr 0} \end{aligned}$$

where ${\displaystyle \Gamma \equiv \left[ 1-\frac{f\left( k\right) f^{\prime \prime }\left( k\right) }{f^{\prime }\left( k\right) ^{2}}\right] \left( \Psi ^{\prime }\left( m\right) -\phi \right) -\frac{\Psi ^{\prime \prime }\left( m\right) }{\Psi ^{\prime }\left( m\right) }\phi f\left( k\right) }>0$.

If $\Delta <0$ then $\det M_{N}>0$ and consequently we always have $dN^{\infty }/d\phi <0$.

If $\Delta >0$ then the sign of $\det M_{N}>0$ if and only if:

$$\begin{aligned} -\rho U_{cc}> & {} -\Psi ^{\prime \prime }\left( m\right) U_{N}\frac{\Psi ^{\prime }\left( m\right) }{f^{\prime }\left( k\right) ^{2}}\frac{\Delta }{\Gamma } \end{aligned}$$

(24)

Now $\text{ det }M_{c}$ and $\text{ det }M_{m}$ can be written as a function of $\Gamma$ and $\Delta$ respectively so that $\text{ det }M_{c}=-U_{NN}f^{\prime }\left( k\right) ^{2}\Gamma$ and $\text{ det }M_{m}=-U_{NN}\Psi ^{\prime }\left( m\right) \Delta$. As a result, the necessary and sufficient condition (24) can be written:

$$\begin{aligned} -\rho U_{cc}> & {} -\Psi ^{\prime \prime }\left( m\right) U_{N}\frac{\text{ det }M_{m}}{\text{ det }M_{c}} \end{aligned}$$

This is equivalent to

$$\begin{aligned} -\frac{U_{cc}}{U_{c}}> & {} -\Psi ^{\prime \prime }\left( m\right) \frac{U_{N}}{U_{c}}\frac{1}{\rho }\frac{dm^{\infty }/d\phi }{dc^{\infty }/d\phi } \end{aligned}$$

By Eq. (10), we established that ${\displaystyle \frac{U_{N}}{U_{c}}\frac{1}{\rho }=\frac{1}{\Psi ^{\prime }\left( m^{\infty }\right) }}$, i.e., the discounted marginal rate of substitution between natural capital (as a public good) and consumption goods is equal to the marginal cost of providing N. This implies that:

$$\begin{aligned} -\frac{U_{cc}}{U_{c}}> & {} -\frac{\Psi ^{\prime \prime }\left( m\right) }{\Psi ^{\prime }\left( m^{\infty }\right) }\frac{dm^{\infty }/d\phi }{dc^{\infty }/d\phi } \end{aligned}$$

After simple calculation and re-interpretation of the expression in terms of elasticities, we obtain:

$$\begin{aligned} \sigma> & {} \varepsilon \frac{\theta _{m}}{\theta _{c}} \end{aligned}$$

(25)

where ${\displaystyle \sigma =-\frac{U_{cc}}{U_{c}}c}$ and ${\displaystyle \varepsilon =-\frac{\Psi ^{\prime \prime }\left( m\right) }{\Psi ^{\prime }\left( m\right) }m}$ represent the elasticity of marginal utility of consumption and the elasticity of marginal product of management, respectively; ${\displaystyle \theta _{c}=\frac{dc}{d\phi }\frac{\phi }{c}}$ and ${\displaystyle \theta _{m}=\frac{dm}{d\phi }\frac{\phi }{m}}$ denote the elasticity of consumption and management with respect to $\phi$.

Part (d) From the expressions of $\det M_{N}$ and $\det \Xi$, we can write:

$$\begin{aligned}&\displaystyle \frac{dN^{\infty }}{d\phi }=\frac{\det M_{N}}{\det \Xi }\\&\displaystyle \quad =\frac{{\displaystyle -\frac{f^{\prime }\left( k\right) ^{2}}{\Psi ^{\prime }\left( m\right) }}\rho U_{cc}\left\{ {\displaystyle \left[ 1-\frac{f\left( k\right) f^{\prime \prime }\left( k\right) }{f^{\prime }\left( k\right) ^{2}}\right] \left( \Psi ^{\prime }\left( m\right) -\phi \right) -\frac{\Psi ^{\prime \prime }\left( m\right) }{\Psi ^{\prime }\left( m\right) }\phi f\left( k\right) }\right\} +\Psi ^{\prime \prime }\left( m\right) U_{N}\Delta }{{\displaystyle -U_{NN}\Psi ^{\prime }\left( m\right) \left\{ \frac{\Psi ^{\prime \prime }\left( m\right) }{\Psi ^{\prime }\left( m\right) }\phi ^{2}f^{\prime }\left( k\right) ^{2}+\left( 1-\frac{\phi }{\Psi ^{\prime }\left( m\right) }\right) f^{\prime \prime }\left( k\right) \Psi ^{\prime }\left( m\right) \right\} }} \end{aligned}$$

After some tedious though simple computation, we obtain:

$$\begin{aligned} \frac{dN^{\infty }}{d\phi }=\frac{{\displaystyle \rho \sigma c^{-1-\sigma }}\left\{ {\displaystyle \left[ f^{\prime }\left( k\right) ^{2}-f\left( k\right) f^{\prime \prime }\left( k\right) \right] \left( \Psi ^{\prime }\left( m\right) -\phi \right) -\frac{\Psi ^{\prime \prime }\left( m\right) }{\Psi ^{\prime }\left( m\right) }\phi f\left( k\right) f\left( k\right) f^{\prime }\left( k\right) ^{2}}\right\} +\Psi ^{\prime \prime }\left( m\right) U_{N}\Delta }{{\displaystyle \chi \eta N^{-1-\eta }\left\{ \Psi ^{\prime \prime }\left( m\right) \phi ^{2}f^{\prime }\left( k\right) ^{2}+\left( \Psi ^{\prime }\left( m\right) -\phi \right) f^{\prime \prime }\left( k\right) \Psi ^{\prime }\left( m\right) ^{2}\right\} }} \end{aligned}$$

Now $c^{-1-\sigma }=e^{(-1-\sigma )\ln c}$ so that for $g\left( \sigma \right) =\sigma c^{-1-\sigma }=\sigma e^{(-1-\sigma )\ln c}$, we have $g^{\prime }\left( \sigma \right) =c^{-1-\sigma }+\sigma (-1)\ln \left( c\right) e^{(-1-\sigma )\ln c}=c^{-1-\sigma }\left[ 1-\sigma \ln \left( c\right) \right]$,

$$\begin{aligned} \frac{\partial }{\partial \sigma }\left( \frac{dN^{\infty }}{d\phi }\right) =\rho c^{-\sigma -1}\left[ 1-\sigma \ln \left( c\right) \right] \Omega \end{aligned}$$

where ${\displaystyle \Omega \equiv \frac{{\displaystyle \left[ f^{\prime }\left( k\right) ^{2}-f\left( k\right) f^{\prime \prime }\left( k\right) \right] \left( \Psi ^{\prime }\left( m\right) -\phi \right) -\frac{\Psi ^{\prime \prime }\left( m\right) }{\Psi ^{\prime }\left( m\right) }\phi f\left( k\right) f\left( k\right) f^{\prime }\left( k\right) ^{2}}}{{\displaystyle \chi \eta N^{-1-\eta }\left[ \Psi ^{\prime \prime }\left( m\right) \phi ^{2}f^{\prime }\left( k\right) ^{2}+\left( \Psi ^{\prime }\left( m\right) -\phi \right) f^{\prime \prime }\left( k\right) \Psi ^{\prime }\left( m\right) ^{2}\right] }}}<0$ since both the numerator is positive and the denominator is negative.

$$\begin{aligned} \frac{\partial }{\partial \sigma }\left( \frac{dN^{\infty }}{d\phi }\right)<0\iff 1-\sigma \ln \left( c^{\infty }\right) >0\iff c^{\infty }<e^{1/\sigma } \end{aligned}$$