Industry equilibrium and welfare in monopolistic competition under uncertainty

Abstract

This paper offers a new theory that describes the influence of uncertainty on economic fundamentals. This theory posits that uncertainty can improve social welfare. We argue that in an economy, where spending of the customers for the differentiated good correlates with larger substitutability of its varieties, the equilibrium output decreases and the prices increase when uncertainty appears. Alternatively, if such spending and substitutability anti-correlate, the predictions for the price and output changes are reversed. The arguments are based on general equilibrium modeling with the monopolistic competition of firms which produce varieties of the differentiated good under limited information regarding the consumer demand. The impact of uncertainty on the equilibrium is assessed by using the relationship between the weighted elasticity of substitution between varieties, the elasticity of the consumer utility, and the income share spent on the differentiated good.

Appendix

Proof of Proposition 1

Let $s<Q$. Then, according to (10), the profit $\pi$ is an increasing function of p. We will prove that in the opposite case, $s\ge Q$, the profit $\pi$ is a decreasing function of p. Then the choice $s=Q$ maximizes the profit $\pi$.

Indeed, let $s\ge Q$. Differentiating (11) we have

$$\begin{aligned} \frac{\partial \pi }{\partial p}=Q(p)+p \frac{\partial Q}{\partial p}. \end{aligned}$$

(46)

Substituting Eq. (14) into (46) we get $\frac{\partial \pi }{\partial p}= Q\big (1-{\mathfrak{S}}(q,q_a)\big )$. Since $\sigma >1$ it follows that ${\mathfrak{S}}>1$ and the profit $\pi$ is a decreasing function of Q. $\square$

Proof of Proposition 2

The proof is split into Lemmas 1–7. Lemma 1 relates the average price to the weighted elasticity of the substitution found with a specific value of the random variable $\zeta$.

Lemma 1

Assume that the optimal prices, which solve maximization problem (17) for any feasible value$z\in Z$of$\zeta$, are continuous with respect toz. Then, first, there exists a value$z^*$of the random variable$\zeta$such that the mean${\bar{p}}$of the optimal prices is given by the equation

$$\begin{aligned} \bar{p}=\frac{mw{\mathfrak{S}}^*}{{\mathfrak{S}}^*-1}, \end{aligned}$$

(47)

where${\mathfrak{S}}^*={\mathfrak{S}}\big (q(z^*),q_a(z^*)\big )$, and the demands$q(z^*)$and$q_a(z^*)$satisfy equation

$$\begin{aligned} \frac{\bar{p}}{{\mathfrak{S}}(q(z^*),q_a(z^*))}= \int \frac{p(z)}{{\mathfrak{S}}(q(z),q_a(z))}f(z)\,dz \end{aligned}$$

(48)

as well as (4), (6), (7), and (8) (with$z=z^*$). Second, Eqs. (27) and (28) hold.

Proof

Differentiating (17) by s we obtain

$$\begin{aligned} \frac{\partial {{\bar{\pi }}}}{\partial s}= {\bar{p}} + s\int \limits _0^{\infty }\frac{\partial p(z)}{\partial s}f(z)dz -mw. \end{aligned}$$

The derivative

$$\begin{aligned} \frac{\partial p}{\partial s}=-\frac{p}{s{\mathfrak{S}}} \end{aligned}$$

(49)

can be found from (14) with $Q=s$. Then the first order condition $\partial {{\bar{\pi }}}/\partial s=0$ becomes

$$\begin{aligned} \bar{p}-s\int \frac{p(z)}{s{\mathfrak{S}}\big (q(z),q_a(z)\big )}f(z)dz -mw=0. \end{aligned}$$

By the mean value theorem there exists a value $z^*$ such that ${\mathfrak{S}}\big (q(z^*),q_a(z^*)\big )$ can be taken off the last integral

$$\begin{aligned} \bar{p}\left( 1- \frac{1}{{\mathfrak{S}}\big (q(z^*),q_a(z^*)\big )}\right) =mw \end{aligned}$$

justifying (47). Substituting (47) into (17) and using free entry condition ${{\bar{\pi }}}=0$ we get (27). Then the substitution of (27) into (18) leads to (28). $\square$

Lemma 2

Letzbe a revealed value of the exponent$\zeta$. Then the equilibrium prices, wages, and individual demands are given by Eqs.(29), (30), and (31) respectively.

Proof

Income (19) of the skilled workers consists of the wages and firm profits (or losses), whereas the income of the workers employed in the traditional sector consists only of their wages $y_a=w_a=1$. Then the balance of money given by Eqs. (6) and (8) can be written as:

$$\begin{aligned}&pqN=\left( w+ \frac{\pi N}{L}\right) {z}, \end{aligned}$$

(50)

$$\begin{aligned}&pq_aN={z}. \end{aligned}$$

(51)

Multiplying (50) by L and (51) by $L_a$ and summing up the results, we get

$$\begin{aligned} pQN=z\left( wL+ \pi N +L_a\right) . \end{aligned}$$

From (11) and (18) it follows that

$$\begin{aligned} wL+ \pi N = wL+pQN-w(ms+\varphi )N = pQN, \end{aligned}$$

(52)

so that

$$\begin{aligned} p=\frac{zL_a}{(1-z)QN}. \end{aligned}$$

(53)

At last, from (27) and (28) we conclude that $QN=\left( {\mathfrak{S}}^*-1\right) L/\left( m{\mathfrak{S}}^*\right)$ proving (29).

Taking the expected value of the both sides of (29) and evaluating ${\bar{p}}$ in the left-hand side with (47), we obtain (30).

Finally, from (50), (51), and (53) it follows that

$$\begin{aligned} \frac{q}{q_a}=\frac{pQN}{L}=\frac{zL_a}{(1-z)L}. \end{aligned}$$

(54)

Then (31) follows from linear system of Eq. (54) on q and $q_a$ and the definition of aggregate demand (12). $\square$

Substituting formulas (29), (47), (30) and (31) for prices, average prices, wages and individual demands into (48), we get (32).

By this moment, we have proved that the optimal supply (together with ${\mathfrak{S}}^*$ which is yet unknown) satisfies the system of Eqs. (27) and (32). Now we intend to establish the existence and uniqueness of the optimal supply. The exclusion of ${\mathfrak{S}}^*$ from this system results in the equation

$$\begin{aligned} {\mathbf{E}}\left( \frac{\zeta }{1-\zeta }\right) =\left( \frac{ms}{\varphi }+1\right) \int _Z\frac{z}{1-z} \frac{f(z)}{z\sigma \left( \frac{zs}{L} \right) + (1-z)\sigma \left( \frac{(1-z)s}{L_a} \right) }dz. \end{aligned}$$

(55)

Lemma 3

We suppose that Assumption 1holds. Then Eq. (55) has a unique solution with respect tos.

Proof

The left hand side $\mathrm{lhs}$ of (55) does not depend on s. We are going to prove that its right hand side as a monotonic function of s crosses once the level given by the left hand side. The right hand side $\mathrm{rhs}(0)=\int _Z\frac{z}{1-z}\frac{f(z)}{\sigma (0)}dz= \frac{1}{\sigma (0)}{\mathbf{E}}\big (\frac{\zeta }{1-\zeta }\big )<\mathrm{lhs}$. However, as $s\rightarrow \infty$,

$$\begin{aligned} z\sigma \left( \frac{zs}{L} \right) + (1-z)\sigma \left( \frac{(1-z)s}{L_a} \right) < \max \left\{ \sigma \left( \frac{zs}{L} \right) , \sigma \left( \frac{(1-z)s}{L_a}\right) \right\} . \end{aligned}$$

If the function $\sigma (\varkappa )$ is increasing, we continue

$$\begin{aligned} z\sigma \left( \frac{zs}{L} \right) + (1-z)\sigma \left( \frac{(1-z)s}{L_a} \right) < \max \left\{ \sigma \left( \frac{s}{L} \right) , \sigma \left( \frac{s}{L_a}\right) \right\} . \end{aligned}$$

By the Lagrange formula, there is $\kappa$ such that

$$\begin{aligned} \frac{\mathrm{lhs}(\infty )}{\mathrm{rhs}(\infty )}< \frac{\sigma (0)+\sigma '(\kappa )\frac{s}{\min \{L,L_a\}}}{\frac{ms}{\varphi }+1}. \end{aligned}$$

If s is sufficiently large, then $\sigma (0)/\big (\frac{ms}{\varphi }+1\big )<1/2$. By Assumption 1, we have $\sigma '(\kappa )\frac{s}{\min \{L,L_a\}}<\frac{ms}{2\varphi }$. Eventually, we have $\mathrm{lhs}(\infty )<\mathrm{rhs}(\infty )$. Regarding the monotonicity of the right hand side, we put

$$\begin{aligned} h(s)=\frac{z\sigma \left( \frac{zs}{L}\right) + (1-z)\left( \frac{(1-z)s}{L_a}\right) }{\frac{ms}{\varphi }+1} \end{aligned}$$

(56)

and establish that $h'(s)<0$. The explicit computation yields that the inequality $h'(s)<0$ is equivalent to

$$\begin{aligned}&\left( \frac{z^2}{L}\sigma '\left( \frac{zs}{L} \right) + \frac{(1-z)^2}{L_a}\sigma '\left( \frac{(1-z)s}{L_a} \right) \right) \left( \frac{ms}{\varphi }+1\right) \\&\quad <\left( z\sigma \left( \frac{zs}{L} \right) + (1-z)\sigma \left( \frac{(1-z)s}{L_a} \right) \right) \frac{m}{\varphi }. \end{aligned}$$

(57)

Initially, we are going to establish that

$$\begin{aligned} \frac{z}{L}\sigma '\left( \frac{s}{L}\right) \left( \frac{ms}{\varphi }+1 \right) < \sigma \left( \frac{zs}{L}\right) \frac{m}{\varphi }. \end{aligned}$$

(58)

Case A$ms/\varphi \le 1$. Then the inequality $\sigma '(\varkappa )<\frac{m}{2\varphi }\sigma (\varkappa )$ followed from (23) and used with $\varkappa =zs/L$ implies that

$$\begin{aligned} \frac{z}{L}\sigma '\left( \frac{s}{L}\right) \left( \frac{ms}{\varphi }+1 \right) \le \frac{2z}{L}\sigma '\left( \frac{s}{L}\right) < \sigma \left( \frac{zs}{L}\right) \frac{m}{\varphi }, \end{aligned}$$

i. e. inequality (58) holds.

Case B$ms/\varphi >1$. Then inequality ${\mathcal{E}}_{\sigma }(\varkappa )<1/2$ which is a part of (23) used with $\varkappa =zs/L$ implies that

$$\begin{aligned} \frac{z}{L}\sigma '\left( \frac{s}{L}\right) \left( \frac{ms}{\varphi }+1 \right) \le \frac{z}{L}\sigma '\left( \frac{s}{L}\right) \frac{2ms}{\varphi }< \sigma \left( \frac{zs}{L}\right) \frac{m}{\varphi }, \end{aligned}$$

i. e. inequality (58) holds once again. The inequality

$$\begin{aligned} \frac{1-z}{L}\sigma '\left( \frac{s}{L}\right) \left( \frac{ms}{\varphi }+1 \right) < \sigma \left( \frac{(1-z)s}{L}\right) \frac{m}{\varphi } \end{aligned}$$

(59)

is proved in the same manner. Inequalities (58) and (59) result in (57).

Let the function $\sigma (\varkappa )$ be decreasing. Then we have

$$\begin{aligned} z\sigma \left( \frac{zs}{L} \right) + (1-z)\sigma \left( \frac{(1-z)s}{L_a} \right) < \sigma (0) \end{aligned}$$

and $\mathrm{lhs}(\infty )<\mathrm{rhs}(\infty )$. The function h(s) decreases because the numerator in (56) decreases in s, whereas the denominator increases. Thus, the monotonically decreasing right hand side is larger than the left hand side at zero but smaller at infinity. Therefore, a unique solution of Eq. (55) exists. $\square$

Now we turn to the second order conditions. The proof is prefaced by the following technical lemma.

Lemma 4

For anyx, y, and$\delta \ge 0$such that$1+\delta \le y\le x\le 2y$, the inequality

$$\begin{aligned} x^2-4xy+y^2+(1+\delta )x+(1+\delta )y\le 0 \end{aligned}$$

(60)

holds.

Proof

We fix an arbitrary $y\ge 1+\delta$ and find the roots $x_{\pm }$ of the left hand side (lhs) of Eq. (60) considered as a function of x:

$$\begin{aligned} x_{\pm }=\frac{4y-(1+\delta )\pm \sqrt{12y^2-12y(1+\delta )+(1+\delta )^2}}{2}. \end{aligned}$$

The lhs, as a quadratic function with respect to x, is negative, if x is located between the roots: $x_-<x<x_+$. The direct computation gives evidence that the lesser root $x_{-}$ is less that y. The inequality $2y\le x_{+}$ is equivalent to $-3y^2+3(1+\delta )y\le 0$ and $1+\delta \le y$. Therefore, from the inequality $y<x<2y$ it follows that $x_-<x<x_+$ and inequality (60) holds. $\square$

Lemma 5 exposes the second order condition for the firm optimization problem.

Lemma 5

We suppose that Eq. (17) defines the expected profits${\bar{\pi }}$and Assumptions 1and2hold. Then$\partial ^2{\bar{\pi }}/\partial s^2<0$.

Proof

A direct computation of the partial derivatives of the profits (17) together with the equality $s=Q$ leads to the equation

$$\begin{aligned} \frac{\partial ^2{\bar{\pi }}}{\partial s^2}= \int _0^{\infty }\left( 2\frac{\partial p(z)}{\partial Q}+ Q\frac{\partial ^2 p(z)}{\partial Q^2}\right) f(z)dz. \end{aligned}$$

(61)

We are going to show that for any z the expression in the brackets of the last integral is negative:

$$\begin{aligned} 2\frac{\partial p}{\partial Q}+ Q\frac{\partial ^2 p}{\partial Q^2}<0. \end{aligned}$$

(62)

The algebraic routine based on (49) gives

$$\begin{aligned} \frac{\partial ^2 p}{\partial Q^2}= \frac{\partial }{\partial Q} \left( \frac{\partial p}{\partial Q}\right) = \frac{\partial }{\partial Q} \left( -\frac{ p}{{\mathfrak{S}}Q}\right) = \frac{\partial }{\partial p} \left( -\frac{ p}{{\mathfrak{S}}Q}\right) \frac{\partial p}{\partial Q}, \end{aligned}$$

justifying that inequality (62) is equivalent to $-\frac{{\mathfrak{S}}-1}{p}Q-\frac{Q}{{\mathfrak{S}}} \frac{\partial {\mathfrak{S}}}{\partial p}<0$. To find the derivative of ${\mathfrak{S}}=\left( qL\sigma (q)+q_aL_a\sigma (q_a)\right) /Q$ with respect to p we perform an auxiliary computation based on (4) and (7):

$$\begin{aligned} \frac{\partial }{\partial p}\left( \frac{qL}{Q}\right) = \frac{qq_aLL_a}{pQ^2}\left( \sigma (q_a)-\sigma (q)\right) , \, \frac{\partial }{\partial p}\left( \frac{q_aL_a}{Q}\right) = \frac{qq_aLL_a}{pQ^2}\left( \sigma (q)-\sigma (q_a)\right) . \end{aligned}$$

Then

$$\begin{aligned} \frac{\partial {\mathfrak{S}}}{\partial p} = -\sigma '(q)\sigma (q)\frac{q^2 L}{pQ}- \sigma '(q_a) \sigma (q_a)\frac{q_a^2 L_a}{pQ}- \frac{qq_aLL_a}{pQ^2}\left( \sigma (q)- \sigma (q_a)\right) ^2. \end{aligned}$$

Eventually, (62) is equivalent to the inequality

$$\begin{aligned} {\mathfrak{S}}Q^2+Q(qL(\sigma (q)+\sigma '(q)q)+ q_aL_a(\sigma (q_a)+\sigma '(q_a)q_a)) -2{\mathfrak{S}}^2Q^2<0. \end{aligned}$$

(63)

Writing ${\mathfrak{S}}$ and Q as the corresponding sums, we get

$$\begin{aligned}&q^2\sigma (q)(\sigma (q)+\sigma '(q)q+1)L^2+ q_a^2\sigma (q_a)(\sigma (q_a)+\sigma '(q_a)q_a+1)L_a^2\\&\quad +qq_a(\sigma (q)(\sigma (q)+\sigma '(q)q+1)+ \sigma (q_a)(\sigma (q_a)+\sigma '(q_a)q_a+1))LL_a\\&\quad -2( q^2\sigma ^2(q)L^2+ q_a^2\sigma ^2(q_a)L_a^2+ 2q\sigma (q)q_a\sigma (q_a)LL_a)<0. \end{aligned}$$

We have to prove that

$$\begin{aligned} \sigma (q)(\sigma (q)+\sigma '(q)q+1)+ \sigma (q_a)(\sigma (q_a)+\sigma '(q_a)q_a+1) < 4\sigma (q)\sigma (q_a). \end{aligned}$$

This inequality coincides with (60) established in Lemma 4 and used with $x=\max \{\sigma (q),\sigma (q_a)\}$ and $y=\min \{\sigma (q),\sigma (q_a)\}$. If $\sigma '<0$, Lemma 4 is applied with $\delta =0$. If $\sigma '>0$, Lemma 4 is applied with $\delta =\max \{\sigma '(q)q,\sigma '(q_a)q_a\}$. The applicability of Lemma 4 is provided by Assumption 2. $\square$

In the next two Lemmas, we discuss the relation between Assumptions 2 and 2’.

Lemma 6

Assumption 2’implies that the individual demandsqand$q_a$satisfy the inequality

$$\begin{aligned} \frac{1}{2}<\frac{q}{q_a}<2 \end{aligned}$$

(64)

at the stationary point of the firm optimization problem.

Proof

From the condition (25) we obtain $\frac{L_a}{L_a+2L}<1-z<\frac{2L_a}{2L_a+L}$. Therefore, $\frac{L}{2L_a}<\frac{z}{1-z}<\frac{2L}{L_a}$ and $\frac{1}{2}<\frac{zL_a}{(1-z)L}<2$. Then the statement of the Lemma follows from (31). $\square$

Lemma 7

Assumption 2’and the condition$\sigma '(\varkappa ) < (\sigma (\varkappa )-1)/\varkappa$of Assumption 1imply the double inequality (24) of Assumption 2.

Proof

Case A: $\sigma (q)$ is increasing. Then (23) yields that the function $\sigma (q)/q$ is decreasing. If, additionally, $q\ge q_a$, inequality (64) implies

$$\begin{aligned} 1\le \frac{\sigma (q)}{\sigma (q_a)} = \frac{\sigma (q)}{\sigma (q_a)}\cdot \frac{q_a}{q}\cdot \frac{q}{q_a} =\frac{\sigma (q)/q}{\sigma (q_a)/q_a}\cdot \frac{q}{q_a}<2. \end{aligned}$$

If $q<q_a$, then $1>\frac{\sigma (q)}{\sigma (q_a)}= \frac{\sigma (q)/q}{\sigma (q_a)/q_a}\cdot \frac{q}{q_a}>1/2$.

Case B: $\sigma (q)$ is decreasing. Let $q\ge q_a$. From Assumption 2’ it follows that

$$\begin{aligned} 1\ge \frac{\sigma (q)}{\sigma (q_a)} = \frac{\sigma (q)q}{\sigma (q_a)q_a}\cdot \frac{q_a}{q} >1/2. \end{aligned}$$

If $q<q_a$, then $1<\frac{\sigma (q)}{\sigma (q_a)}=\frac{\sigma (q)q}{\sigma (q_a)q_a}\cdot \frac{q_a}{q}<2$. $\square$

Concluding the Proof of Proposition 2, we note that the existence and uniqueness of the optimal supply is provided by Lemma 3. The equilibrium relationships are proved in Lemma 2. Finally, Lemma 5 yields the second order conditions. $\square$

To prove Proposition 3 we formulate three Lemmas. The first of them describes the sign of the covariance between two functions of a random variable X (see also Schmidt 2014).

Lemma 8

Letf(x) andg(x) be increasing functions that are differentiable on an interval [a, b]. Then for any random variableXdefined on [a, b] such that the expected values off(X) andg(X) exist, the inequality$\mathbf{Cov}(f(X),g(X))>0$holds. If one of these functions is decreasing and the other is increasing on [a, b], then$\mathbf{Cov}(f(X),g(X))<0$.

Proof

Let the expected values of f(X) and g(X) be attained by the corresponding functions at points $x_f$ and $x_g$ respectively: $f(x_f)={\mathbf{E}}(f(X))$, $g(x_g)={\mathbf{E}}(g(X))$. Without loss of generality we assume that $x_f\le x_g$. Then the product $(f(x)-f(x_f))(g(x)-g(x_g))$ is positive on $[a,b]\setminus [x_f,x_g]$ and negative on $(x_f,x_g)$. Let ${\tilde{x}}_g\in (a,x_f)$ be a point such that $\int _{{\tilde{x}}_g}^{x_g} f(x)d\mu =f(x_f)$, where the measure $d\mu (x)$ defines the random variable X. We put $\varDelta f(x,x_f)=f(x)-f(x_f)$. Then, by monotonicity of g, we have:

$$\begin{aligned}&\int _{{\tilde{x}}_g}^{x_f} \varDelta f(x,x_f)(g(x)-g(x_g))d\mu> \int _{{\tilde{x}}_g}^{x_f} \varDelta f(x,x_f)(g(x_f)-g(x_g))d\mu \\&\quad> \int _{x_f}^{x_g} \varDelta f(x,x_f)(g(x_g)-g(x_f))d\mu >- \int _{x_f}^{x_g} \varDelta f(x,x_f)(g(x)-g(x_g))d\mu . \end{aligned}$$

We conclude that the integral $\int (f(x)-f(x_f))(g(x)-g(x_g))d\mu$ is positive over the interval $[{\tilde{x}}_g,x_g]$. It is positive outside this interval because the integrated function is positive. In the same way we establish that if one of the functions is decreasing and the other function is increasing, then $\mathbf{Cov}(f(X),g(X))<0$. $\square$

The following Lemma allows us to find the sign of the derivative of ${\mathfrak{S}}(q,q_a)$ considered as a function of $v(\varkappa )$.

Lemma 9

Letq(z) and$q_a(z)$be the equilibrium demands of the skilled and unskilled workers for high-tech goods corresponding to a revealed valuezof the random variable$\zeta$. Then for some$\alpha \in [0,1]$

$$\begin{aligned} \frac{d{\mathfrak{S}}}{dz}(q,q_a)=v'(\alpha q+(1-\alpha )q_a)(q-q_a). \end{aligned}$$

(65)

Proof

In this Lemma and further on, we explore the weighted elasticity of substitution ${\mathfrak{S}}(q(z),q_a(z))$ as a function of z. It is still denoted by symbol ${\mathfrak{S}}$, now as ${\mathfrak{S}}(z)$.

Due to (31), the function ${\mathfrak{S}}(q(z),q_a(z))$ has the following form

$$\begin{aligned} {\mathfrak{S}}(z)= z\sigma \left( \frac{z s}{L}\right) + (1-z)\sigma \left( \frac{(1-z) s}{L_a}\right) . \end{aligned}$$

Therefore,

$$\begin{aligned} \frac{d{\mathfrak{S}}(z)}{dz}=\sigma \left( \frac{z s}{L}\right) +\frac{z s}{L}\sigma '\left( \frac{z s}{L}\right) - \sigma \left( \frac{(1-z) s}{L_a}\right) -\frac{(1-z) s}{L_a}\sigma '\left( \frac{(1-z) s}{L_a}\right) . \end{aligned}$$

With the definition of v, we re-write the obtained equation as

$$\begin{aligned} \frac{d{\mathfrak{S}}(z)}{dz}=v(q)-v(q_a), \end{aligned}$$

(66)

and (65) follows from the mean-value theorem. $\square$

The following Lemma relates the convexity of the function $1/{\mathfrak{S}}(z)$, as a function of z, to the monotonicity of v.

Lemma 10

Let Assumptions 1–3be satisfied. If$v(\varkappa )$is a non-increasing function, then the function$1/{\mathfrak{S}}(z)$is convex. On the contrary, if$v(\varkappa )$is a non-decreasing function and, additionally, inequalities (36) and (38) hold, then the function$1/{\mathfrak{S}}(z)$is concave.

Proof

Clearly, $\frac{d}{dz}\left( \frac{1}{{\mathfrak{S}}(z)} \right) = -{\mathfrak{S}}^{-2}(z){\mathfrak{S}}'(z)$ and $\frac{d^2}{dz^2}\left( \frac{1}{{\mathfrak{S}}(z)} \right) = \frac{-{\mathfrak{S}}''{\mathfrak{S}}+2{\mathfrak{S}}'^2}{{\mathfrak{S}}^3}$, where all the primes are the derivatives with respect to z. Combining the last equation with $\frac{d}{dz}\left( v(q) \right) = v'(q)\frac{dq}{dz}$ and (66), we have

$$\begin{aligned} \frac{d^2}{dz^2}\left( \frac{1}{{\mathfrak{S}}(z)} \right) =-\frac{v'(q)s/L+v'(q_a)s/L_a}{{\mathfrak{S}}^2}+ 2\frac{\big (v(q)-v(q_a)\big )^2}{{\mathfrak{S}}^3}. \end{aligned}$$

(67)

If the function v is non-increasing then $v' \le 0$ and the convexity of $1/{\mathfrak{S}}(z)$ follows from (67).

Assume that $v(\varkappa )$ is a non-decreasing function. By (67), the second derivative of $1/{\mathfrak{S}}(z)$ is negative if and only if

$$\begin{aligned} \left( v'(q)\frac{s}{L}+v'(q_a)\frac{s}{L_a}\right) {\mathfrak{S}}(z) >\big (v(q)-v(q_a)\big )^2, \end{aligned}$$

or $\left( v'(q)\frac{s}{L}+v'(q_a)\frac{s}{L_a}\right) \big ( z\sigma (q)+(1-z)\sigma (q_a) \big )> 2v'(\xi )(q-q_a)\big (v(q)-v(q_a)\big ),$ where $\xi \in [q_a,q]$. As condition (36) guarantees that $q>q_a$, it is enough to prove that

$$\begin{aligned} v'(q_a)\frac{sz}{L_a}\sigma (q)> 2v'(\xi )q v(q), \quad \text{or} \quad v'(q_a)\frac{qL}{L_a}\sigma (q) > 2v'(\xi )q \left( \sigma (q)+q\sigma '(q)\right) . \end{aligned}$$

Since $v'(\varkappa )$ is a non-increasing function it follows that $v'(\varkappa )\ge v'(\xi )$. Then the last inequality follows from (38). $\square$

Proof of Proposition 3

We establish inequalities (37) for the decreasing function v(q). The first of them is equivalent to $\frac{\varphi ({\mathfrak{S}}^*-1)}{m}< \frac{\varphi \big ({\mathfrak{S}}(z_0)-1\big )}{m}$ or to

$$\begin{aligned} {\mathfrak{S}}^*<{\mathfrak{S}}(z_0). \end{aligned}$$

(68)

The second inequality (37) is the direct consequence of the first one. Proving the third inequality, we combine (29) and (30) to obtain that

$$\begin{aligned} \frac{m {\mathfrak{S}}^*}{{\mathfrak{S}}^*-1 } < \frac{m {\mathfrak{S}}\left( z_0\right) }{{\mathfrak{S}}\left( z_0\right) -1 }. \end{aligned}$$

The latter is equivalent to (68). We are going to prove that for any decreasing function v(q) inequality (68) holds. From the definition (32) of ${\mathfrak{S}}^*$ it follows that (68) is equivalent to ${\mathbf{E}}\left( \frac{\zeta }{1-\zeta }\frac{1}{{\mathfrak{S}}(\zeta )} \right) {\mathfrak{S}}(z_0) > {\mathbf{E}}\left( \frac{\zeta }{1-\zeta }\right)$ and

$$\begin{aligned} {\mathbf{E}}\left( \frac{1}{{\mathfrak{S}}\left( \zeta \right) }\right) + \frac{ \mathbf{Cov}\left( \frac{\zeta }{1-\zeta },\frac{1}{{\mathfrak{S}}(\zeta )}\right) }{{\mathbf{E}}\left( \frac{\zeta }{1-\zeta }\right) } > \frac{1}{{\mathfrak{S}}(z_0)}, \end{aligned}$$

(69)

We intend to establish that the covariance is positive in the last inequality. Introducing the functions $\chi _1(z)=z/(1-z)$ and $\chi _2(z)={\mathfrak{S}}^{-1}(z)$, we obtain that $\chi _1'(z)>0$ and $\chi _2'(z)=-{\mathfrak{S}}^{-2}(z){\mathfrak{S}}'(z)$. Equation (65) and the inequality $q>q_a$ provided by (36) implies that ${\mathfrak{S}}'(z)<0$ and $\chi _2'(z)>0$. Therefore, the derivatives of $\chi _1$ and $\chi _2$ have the same sign. According to Lemma 8, $\mathbf{Cov}\displaystyle \left( \frac{\zeta }{1-\zeta },\frac{1}{{\mathfrak{S}}(\zeta )}\right) >0.$ Jensens’ inequality and the convexity of $1/{\mathfrak{S}}(z)$ (proved in Lemma 10 for decreasing functions v) result in the inequality ${\mathbf{E}}\left( 1/{\mathfrak{S}}\left( \zeta \right) \right) > 1/{\mathfrak{S}}\left( {\mathbf{E}}(\zeta )\right)$. From the last two inequalities it follows that (69) holds. The case of an increasing function v(q) is considered in the same way. $\square$

We turn to the proof of Proposition 4, presenting three technical lemmas.

Lemma 11

Let Assumptions 1, 2 (or 2’), and 4hold. We put

$$\begin{aligned} \nu ^* = \ln ^2 \left( \left( \frac{L}{\varphi \sigma (0)} \right) ^{1/\gamma } \frac{1}{L_a} \right) , \quad \gamma = (\sigma (0) - 1) / \sigma (0), \end{aligned}$$

(70)

and additionally assume that$\nu ^*$is positive and the function$v(\varkappa )$non-increases. Then the inequality

$$\begin{aligned} \frac{W(z_0)}{W(z^*)} \le \frac{1}{1+{(\nu ^*)^2 D/4}}, \end{aligned}$$

(71)

where$D = \mathrm{Var}\zeta$is the dispersion of the random variable$\zeta$, holds if the upper and lower-tier utilities are related to each other by the equation$U=u^{-1}$.

Proof

The direct calculations provide that the utilities ${\hat{\mathbf{U}}}$ and ${\hat{\mathbf{U}}}_a$ of skilled and unskilled workers are

$$\begin{aligned} {\hat{\mathbf{U}}}= \big (U(Nu(q))\big )^z\left( (1-z)\frac{z}{1-z}\frac{L_a}{L}\right) ^{1-z}, {\hat{\mathbf{U}}}_a=\big (U(Nu(q_a))\big )^z\left( (1-z)\right) ^{1-z}. \end{aligned}$$

Then

$$\begin{aligned} W=L_a\left( \frac{Q}{L_a} \right) ^z \left( \left( \frac{U(Nu(q))L}{Qz}\right) ^zz+ \left( \frac{U(Nu(q_a))L_a}{Q(1-z)}\right) ^z(1-z) \right) . \end{aligned}$$

Put, $F(\varkappa ,N)=\frac{Qu^{-1}(Nu(\varkappa ))}{L_a\varkappa }$, $F=F(q,N)$, $F_a=F(q_a,N)$, where q, $q_a$, and N are the equilibrium values. Then substituting $u^{-1}$ for U gives

$$\begin{aligned} W(z)=L_a \big ( F^zz+F_a^z(1-z) \big ). \end{aligned}$$

Simplifying notation, the function $h = W/L_a$ is used below instead of W. We are going to use Jensen’s inequality in a sharpened form ${\mathbf{E}}(h(\zeta )) \ge h(z_0) + \min _{z} h''(z)\mathrm{Var}\zeta / 2$. This approach requires an estimate of $h''$ from below by a positive term. Computing the derivative of h with respect to z, we use the equilibrium variables $q=Qz/L$ and $q_a=Q(1-z)/L_a$ substituted for the first variable $\varkappa$:

$$\begin{aligned} h''= & {}\, F^z\left( \ln F+\frac{\partial F}{\partial q}\frac{q}{F}\right) ^2z+ 2F^z\biggl (\underbrace{\ln F}_A+ \underbrace{\frac{\partial F}{\partial q}\frac{q}{F}}_A \biggr ) + F^z\frac{q}{F}\biggl ( \underbrace{2\frac{\partial F}{\partial q}}_B+ \underbrace{\frac{q\partial ^2 F}{\partial q^2}}_B \\&-\underbrace{\frac{q}{F}\left( \frac{\partial F}{\partial q}\right) ^2 \biggr )}_B+ F_a^z\left( \ln F_a-\frac{\partial F_a}{\partial q_a} \frac{q_a}{F_a}\frac{z}{1-z}\right) ^2(1-z)- 2F_a^z\biggl (\underbrace{\ln F_a}_A \\&-\underbrace{\frac{\partial F_a}{\partial q_a}\frac{Qz}{L_aF_a}}_{B'}\biggr )+ F_a^z\frac{q_a}{F_a} \biggl (\underbrace{-2\frac{\partial F_a}{\partial q_a}}_{A}+ \frac{\partial ^2 F_a}{\partial q_a^2} \underbrace{\frac{Qz}{L_aF_a}}_{B'}-\underbrace{\frac{Qz}{L_aF_a} \left( \frac{\partial F_a}{\partial q_a}\right) ^2}_{B'} \biggr ). \end{aligned}$$

(72)

The objective is to estimate the right hand side of (72) from below by the first term:

$$\begin{aligned} h'' \ge F^z\left( \ln F+\frac{\partial F}{\partial q}\frac{q}{F}\right) ^2z. \end{aligned}$$

(73)

To this end, we evaluate (72), transforming the terms marked by A to $\ln F+{\mathcal{E}}_F$, and extracting the derivative of $(\ln F+{\mathcal{E}}_F)$ in the terms marked by B and $B'$:

$$\begin{aligned} h''\ge & {} F^z\left( \ln F+\frac{\partial F}{\partial q}\frac{q}{F}\right) ^2z+ 2f^z(\underbrace{\ln F + {\mathcal{E}}_F}_A) - 2F_a^z(\underbrace{\ln F_a + {\mathcal{E}}_{F_a}}_A) \\&+F^z\frac{q}{F} \underbrace{\frac{\partial }{\partial q}(\ln F + {\mathcal{E}}_F)}_{B} +F_a^z\frac{Qz}{F_aL_a} \underbrace{\frac{\partial }{\partial q_a} (\ln F_a + {\mathcal{E}}_{F_a})}_{B'} \end{aligned}$$

(74)

We intend to derive that

$$\begin{aligned} \frac{\partial }{\partial \varkappa }(\ln F(\varkappa ,N) + {\mathcal{E}}_F(\varkappa ,N)) \ge 0, \end{aligned}$$

(75)

for any $N > 1$ and, therefore, the function $\ln F(\varkappa ,N) + {\mathcal{E}}_F(\varkappa ,N)$ is monotone increasing with respect to $\varkappa$. Since $q>q_a$ in the equilibrium, this monotonicity together with the monotonicity of F implies that the sum of the terms marked by A in (72) and (74) is non-negative. The positiveness of the terms marked by B and $B'$ follows from (75) directly. Therefore, (75) yields (74) and (73).

We focus on the proof of (75). The dependence of N is omitted and the prime is used instead of the partial derivative with respect to $\varkappa$ to simplify notation. With $g(\varkappa ,N)=u^{-1}(Nu(\varkappa ))=F(\varkappa ,N)\varkappa L_a/Q$, we have ${\mathcal{E}}_F(\varkappa )={\mathcal{E}}_g(\varkappa )-1$. Therefore, (75) is transformed into

$$\begin{aligned} {\mathcal{E}}_g+{\mathcal{E}}_g'\varkappa \ge 1. \end{aligned}$$

(76)

The straightforward computations of the derivative of g (omitted here), which involve the notation $\eta =U(N u(\varkappa ))$, $\eta '=Nu'(\varkappa )/u'(\eta )$, and ${\mathcal{E}}_g(\varkappa )= \frac{Nu'(\varkappa )\varkappa }{\eta u'(\eta )}$, result in

$$\begin{aligned} {\mathcal{E}}_{g}(\varkappa )= {} \frac{{\mathcal{E}}_u(\varkappa )}{{\mathcal{E}}_u(u^{-1}(Nu(\varkappa )))}, \end{aligned}$$

(77)

$$\begin{aligned} {\mathcal{E}}_g'(\varkappa )\varkappa= {} \frac{{\mathcal{E}}_u^2(\varkappa )}{{\mathcal{E}}_u(\eta )} \left( \frac{(\sigma (\varkappa )- 1)}{\sigma (\varkappa ){\mathcal{E}}_u(\varkappa )} - \frac{(\sigma (\eta )-1)}{\sigma (\eta ){\mathcal{E}}_u(\eta )} \right) . \end{aligned}$$

(78)

Since $N>1$ and ${\mathcal{E}}_u$ is monotone non-increasing, Eq. (77) yields that ${\mathcal{E}}_g\ge 1$ and ${\mathcal{E}}_F \ge 0$. Further, the function $(\sigma (\cdot )-1)/\big (\sigma (\cdot ){\mathcal{E}}_u(\cdot )\big )$ is non-increasing (Assumption 4) and $\eta \ge \varkappa$. Therefore, ${\mathcal{E}}_g'(\varkappa )\varkappa \ge 0$, and (77), (78) result in (76). The latter leads to (75) and, eventually, to (73).

The monotone increasing function F (with respect to $\varkappa$) attains its global minimum at zero (understood as the limit). In a positive neighborhood of zero, the utility $u(\varkappa )$ is at most $C\varkappa ^{\gamma }$ up to smaller terms, where $\gamma$ is the Hölder exponent; $\gamma = (\sigma (0) - 1) / \sigma (0)$. Therefore, $\ln F \ge \sqrt{\nu ^*}$, where $\nu ^*$ is defined in (70). Since ${\mathcal{E}}_F \ge 0$ and $\nu ^*>0$, inequality (73) yields the estimate $h''\ge F^z {\nu ^*} z$. With Jensen’s inequality,

$$\begin{aligned} \frac{W(z_0)}{W(z^*)} \le \left( 1 - \frac{\nu ^*}{2W(z^*)} L_a\int _{Z} F^z z (z-z_0)^2 f(z)\,dz \right) . \end{aligned}$$

We note that $L_aF^{z}z \ge L_a\big ( F^{z}z + F_a^{z}(1-z)\big ) / 2 =\frac{W(z)}{2}$. The function W(z), as we have already proved, is convex. Applying Jensen’s inequality once again, we have $\int W(z) (z-z_0)^2 f(z)\,dz \ge W(z_0) D$. The last three estimates lead to the inequality

$$\begin{aligned} \frac{W(z_0)}{W(z^*)} \le \left( 1 - \frac{W(z_0)\nu ^* D}{4W(z^*)} \right) , \end{aligned}$$

which is equivalent to (71). $\square$

Lemma 12

From assumptions of Proposition 4it follows that the inequality

$$\begin{aligned} \frac{{\mathfrak{S}}_0}{{\mathfrak{S}}^*} - 1 \le \frac{3 (-\sigma '(0))\varphi B_3}{2m\min \{L,L_a\}}D, \end{aligned}$$

(79)

where$B_3 = (2-{\mathcal{E}}_{\sigma '}(0))(\sigma (0)-1)$and$D = \mathrm{Var}\zeta$, holds.

Proof

The definition (32) of the ${\mathfrak{S}}^*$ is written as

$$\begin{aligned} {\mathfrak{S}}^* = E\left( \frac{\zeta }{1-\zeta } \right) \Big / E\left( \frac{\zeta }{1-\zeta } \frac{1}{{\mathfrak{S}}(\zeta )} \right) . \end{aligned}$$

in the terms of expected values. Then the denominator is

$$\begin{aligned} E\left( \frac{\zeta }{1-\zeta } \frac{1}{{\mathfrak{S}}(\zeta )} \right) = E\left( \frac{\zeta }{1-\zeta } \right) E\left( \frac{1}{{\mathfrak{S}}(\zeta )} \right) + \rho \sqrt{\mathrm{Var}\left( \frac{\zeta }{1-\zeta } \right) \mathrm{Var} \left( \frac{1}{{\mathfrak{S}}(\zeta )} \right) }, \end{aligned}$$

(80)

where $\rho$ is the correlation coefficient between the corresponding random variables. The expansion of ${\mathfrak{S}}^{-1}(z)$ into the Taylor series at $z = z_0 = E\zeta$ leads to

$$\begin{aligned} \frac{1}{{\mathfrak{S}}(z)} = \frac{1}{{\mathfrak{S}}(z_0)} + \left( \frac{1}{{\mathfrak{S}}} \right) '(z_0)(z-z_0) + \frac{1}{2} \left( \frac{1}{{\mathfrak{S}}} \right) ''({\hat{z}})(z-z_0)^2, \end{aligned}$$

(81)

where ${\hat{z}}$ is located between z and $z_0$. The substitution of (81) into the expected value results in

$$\begin{aligned} E\left( \frac{\zeta }{1-\zeta } \right) E\left( \frac{1}{\zeta } \right) \le E\left( \frac{\zeta }{1-\zeta } \right) \left( \frac{1}{{\mathfrak{S}}_0} + \frac{1}{2} \max \left\{ \left( \frac{1}{{\mathfrak{S}}} \right) ''({\hat{z}}) \right\} D. \right) \end{aligned}$$

Estimating the variance of ${\mathfrak{S}}^{-1}$ with the Taylor expansion given by (81), reduced to the first order term, we get

$$\begin{aligned} \mathrm{Var} \left( \frac{1}{{\mathfrak{S}}(\zeta )} \right) \le \max \left\{ \left| \left( \frac{1}{{\mathfrak{S}}} \right) '\right| ({\hat{z}})\right\} ^2 D. \end{aligned}$$

In the same manner, $\mathrm{Var}\big ( (1-\zeta )^{-1} \big ) \le \max \big \{ (1-z)^{-2} \big \} D$. Substituting the derived variances into (80), we have

$$\begin{aligned}&E\left( \frac{\zeta }{1-\zeta } \frac{1}{{\mathfrak{S}}(\zeta )} \right) \le E\left( \frac{\zeta }{1-\zeta } \right) \left( \frac{1}{{\mathfrak{S}}_0}+ \max \left\{ \left( \frac{1}{{\mathfrak{S}}} \right) ''D \right\} \right) \\&\qquad + \rho \max \left\{ \frac{1}{1-z} \right\} \max \left( \frac{1}{{\mathfrak{S}}} \right) ' D. \end{aligned}$$

Then

$$\begin{aligned} \frac{{\mathfrak{S}}_0}{{\mathfrak{S}}^*} - 1 \le \left( \max \left\{ \left( \frac{1}{{\mathfrak{S}}} \right) '' \right\} + \frac{\max \left\{ \frac{1}{1-z} \right\} \max \left| \left( \frac{1}{{\mathfrak{S}}} \right) '\right| }{E\left( \frac{\zeta }{1-\zeta } \right) } \right) {\mathfrak{S}}_0 D. \end{aligned}$$

(82)

Using (65), we estimate $|(1/{\mathfrak{S}})'|$. By the assumption of Proposition 4, $-v'({\tilde{q}}) \le -v'(0) = -\sigma '(0)(2-{\mathcal{E}}_{\sigma '}(0))$ for any ${\tilde{q}}$. Further, $q - q_a \le q = \frac{Qz}{L} \le \frac{\varphi ({\mathfrak{S}}(z)-1) z}{mL}$. Since ${\mathfrak{S}}(z) = z\sigma (q_{z}) + (1-z)\sigma (q_{a,z}) \le \sigma (0)$ and $z < 2L / (2L + L_a)$ it follows that

$$\begin{aligned} \left| \left( \frac{1}{{\mathfrak{S}}} \right) '\right| \le \frac{-2\sigma '(0) \varphi B_3}{m(2L+L_a)}, \end{aligned}$$

where $B_3 = (2-{\mathcal{E}}_{\sigma '}(0))(\sigma (0)-1)$. Substituting just obtained estimates of $v'(q)$ and $v(q_a) - v(q)$ into (67) and using (44), we have:

$$\begin{aligned} \left( \frac{1}{{\mathfrak{S}}(z)} \right) '' \le \frac{-3\sigma '(0) \varphi B_3}{2m \min \{L,L_a\}}. \end{aligned}$$

Finally, the estimate $(1-z)^{-1} / E\big ( \zeta / (1-\zeta ) \big ) \le 2(L+L_a)/L$ is obtained by the substitution of $2L / (2L+L_a)$ for z in the numerator and $L / (L_a + L)$ in the denominator. Substituting the auxiliary estimates into (82), we get (79). $\square$

Lemma 13

Let condition (43) hold. We additionally assume that the functions$v(\varkappa )$, $-v'(\varkappa )$, $\sigma (\varkappa )$, $-\sigma '(\varkappa )$are monotone non-increasing. Then

$$\begin{aligned} U(N_0u(q_0)) \le U(Nu(q)) \left( 1 - B_1B_2 \frac{{\mathfrak{S}}_0 - {\mathfrak{S}}^*}{{\mathfrak{S}}_0} \right) ^{-1}, \end{aligned}$$

(83)

where$B_1$and$B_2$are defined in Assumption 5, and index 0 indicates that the corresponding variable is related to${\mathfrak{S}}_0$.

Proof

The mean value theorem written with the elasticities yields the following equation:

$$\begin{aligned} \frac{U(N_0u(q_0) - U(Nu(q))}{\big (U(Nu(q))\big )({\hat{{\mathfrak{S}}}})} = \frac{\big (U(Nu(q))\big )'_{{\mathfrak{S}}^*}({\hat{{\mathfrak{S}}}}){\hat{{\mathfrak{S}}}}}{\big (U(Nu(q))\big )({\hat{{\mathfrak{S}}}})}\cdot \frac{{\mathfrak{S}}_0 - {\mathfrak{S}}^*}{{\hat{{\mathfrak{S}}}}}, \end{aligned}$$

(84)

where ${\hat{{\mathfrak{S}}}}$ is located between ${\mathfrak{S}}^*$ and ${\mathfrak{S}}_0$. We are going to estimate the right hand side of (84). Let ${\mathfrak{S}}$ be located between ${\mathfrak{S}}^*$ and ${\mathfrak{S}}_0$, and q is related to ${\mathfrak{S}}$ by Eq. (31); z is fixed. The direct computation of the derivatives leads to

$$\begin{aligned} {\mathcal{E}}_{Nu(q)}({\mathfrak{S}}) = \frac{{\mathcal{E}}_u(q){\mathfrak{S}}}{{\mathfrak{S}}-1} - 1. \end{aligned}$$

(85)

The right hand side of (85) is monotone increasing (Assumption 4) and zero at $q=0$ (u is Hölder continuous). By Proposition 3, under the assumptions of Proposition 4, ${\mathfrak{S}}^* \le {\mathfrak{S}}_0$. Therefore, $Nu(q) \le N_0u(q_0)$. Returning to (84) and changing ${\hat{{\mathfrak{S}}}}$ to either ${\mathfrak{S}}^*$ or ${\mathfrak{S}}_0$ in an appropriate way, we have

$$\begin{aligned} \frac{U(N_0u(q_0)) - U(Nu(q))}{U(N_0u(q_0))} \le {\mathcal{E}}_{U(Nu(q))}({\mathfrak{S}}_0) \frac{{\mathfrak{S}}_0 - {\mathfrak{S}}^*}{{\mathfrak{S}}^*}. \end{aligned}$$

(86)

Since the elasticity of the composition is the product of the elasticities, we continue:

$$\begin{aligned} \frac{U(N_0u(q_0)) - U(Nu(q))}{U(N_0u(q_0))} \le {\mathcal{E}}_U(Nu(q)) \left( \frac{{\mathcal{E}}_u(q){\mathfrak{S}}_0}{{\mathfrak{S}}_0-1} - 1 \right) \frac{{\mathfrak{S}}_0 - {\mathfrak{S}}^*}{{\mathfrak{S}}^*}. \end{aligned}$$

Then (83) follows from (43). $\square$

Proof of Proposition 4

We estimate the bracket in the right hand side of (83) by using the elementary inequality $(1-x)^{-1} < 1 + 2x$, which holds if $x < 1/2$. The latter follows from (44). Then (83) is re-written as

$$\begin{aligned} U(N_0u(q_0)) \le U(Nu(q)) \big ( 1 + 2B_1B_2D ({\mathfrak{S}}_0 - {\mathfrak{S}}^*) / {\mathfrak{S}}^* \big ). \end{aligned}$$

Using (79), we change the last inequality to $U(N_0u(q_0)) \le U(N^*u(q^*)) ( 1 + \nu _2D )$, where

$$\begin{aligned} \nu _2 = \frac{3\varphi (-\sigma '(0)))}{2m\min \{L,L_a\}}B_1B_2B_3 \end{aligned}$$

(87)

and the variables with stars are computed with ${\mathfrak{S}}^*$.

In the same way, $U(N_0u(q_{a,0})) \le U(N^*u(q^*_a)) ( 1 + \nu _2 D)$. Taking into account that $z > 1-z$, we estimate the welfare:

$$\begin{aligned} W_0(z_0)\le & {} L_a^{1-z} (1+\nu _2D)^z \left( \left( \frac{U(N^*u(q^*))L}{z} \right) ^z z \right. \\&+\left. \left( \frac{U(N^*u(q^*_a))L_a}{1-z} \right) ^z (1-z) \right) \le (1+\nu _2D)^z W(z_0). \end{aligned}$$

Combining the last inequality with (71), we have

$$\begin{aligned} \frac{W_0(z_0)}{W(z^*)} \le \frac{(1+\nu _2D)^z}{1+\nu ^* D/4}. \end{aligned}$$

(88)

Therefore, the inequality $W_0(z_0) < W(z^*)$ follows from $4z\nu _2 < \nu ^*$. With definitions (70) and (87) of $\nu ^*$ and $\nu _2$ the last inequality becomes

$$\begin{aligned} \frac{6L\varphi (-\sigma '(0))}{m\min \{L,L_a\}^2}B_1B_2B_3 < \ln ^2 \left( \left( \frac{L}{\varphi \sigma (0)} \right) ^ {\sigma (0)/(\sigma (0)-1)} \frac{1}{L_a} \right) . \end{aligned}$$

This inequality and the third inequality in (43) yields (44). $\square$