Security bid auctions for agency contracts

Abstract

A principal uses security bid auctions to award an incentive contract to one among several agents in the presence of hidden action and hidden information. Securities range from cash to equity and call options. “Steeper” securities are better surplus extractors, yet reduce effort incentives. In view of this trade-off, a hybrid share auction that includes a cash reward to the winner, a minimum share, and an option to call a fixed wage contract, tends to outperform all other auctions, although it is not an optimal mechanism. However, by adding output targets a hybrid share auction can (arbitrary closely) implement the optimal mechanism.

Appendix

In this appendix we fully explain and prove Proposition 7 in a sequence of lemmas.

Because the game is symmetric (players’ productivity parameters are i.i.d. random variables), we restrict attention to symmetric mechanisms with respect to the permutation of type profiles.

For convenience we define (and omit the subscripts in $T_i$, $Q_i$):

$$\begin{aligned} t(x_i) :=E_{\mathrm{X}_{-i}}\big (T(x_i,\mathrm{X}_{-i})\big ) , \quad q(x_i): = E_{\mathrm{X}_{-i}} \big (Q(x_i,\mathrm{X}_{-i})\big ). \end{aligned}$$

(34)

Define $\gamma (x_i,z_i)$ as the winner’s cost of fulfilling his output requirement when the agent reports type $z_i$ while his true type is $x_i$, i.e.,

$$\begin{aligned} \gamma (x_i,z_i)&= {\left\{ \begin{array}{ll} c_L &{} \quad \text {if} \, \psi (z_i) \le x_i + e_L \\ c_H &{} \quad \text {if} \, x_i + e_L < \psi (z_i) \le x_i + e_H \\ \infty &{} \quad \text {otherwise}. \end{array}\right. } \end{aligned}$$

(35)

We say that it is feasible for type $x_i$ to report $z_i$ if $\gamma (x_i,z_i)<\infty$. For convenience define

$$\begin{aligned} U(x_i,z_i):=t(z_i)-q(z_i)\gamma (x_i,z_i). \end{aligned}$$

(36)

A mechanism is truthfully implementable if it satisfies the output restriction (OR), is incentive compatible (IC), and satisfies individual rationality (IR):

$$\begin{aligned} \psi (x_i)&\le x_i + e_H \quad \, \text {for all } x_i \in [0,1] \end{aligned}$$

(OR)

$$\begin{aligned} U(x_i,x_i)&\ge U(x_i,z_i) \quad \! \text {for all }x_i,z_i \in [0,1] \end{aligned}$$

(IC)

$$\begin{aligned} U(x_i,x_i)&\ge 0\qquad \quad \, \text {for all }x_i \in [0,1]. \end{aligned}$$

(IR)

Given two truthfully implementable mechanisms $(T,Q,\psi )$ and $(\hat{T},Q,\hat{\psi })$, we say that $(\hat{T},Q,\hat{\psi })$ improves upon $(T,Q,\psi )$ if $q(x)\hat{\psi }(x)-\hat{t}(x) \ge q(x)\psi (x)-t(x)$ for all $x \in [0,1]$ and $q(x)\hat{\psi }(x)-\hat{t}(x)>q(x)\psi (x)-t(x)$ for some $x \in [0,1]$. The binary relation “improves upon” defined on the set of truthfully implementable mechanisms defines a partial ordering. Because we are looking for the optimal mechanism, we will focus on the maximal truthfully implementable mechanisms. In particular, maximal truthfully implementable mechanisms have the property that $\psi (x)$ is equal to either $x+e_L$ or $x+e_H$. The following two lemmas show that there is no loss of generality when we restrict our search for the optimal mechanism to the ones where $\psi (x)$ is equal to either $x+e_L$ or $x+e_H$.

Lemma 3

Suppose that $(T,Q,\psi )$ is a truthfully implementable mechanism and $\psi (x')<x'+e_L$ for some $x'$. Define $\hat{\psi }$ by $\hat{\psi }(x)=\psi (x)$ for $x\ne x'$ and $\hat{\psi }(x')=x'+e_L$. Then $(T,Q,\hat{\psi })$ is truthfully implementable.

Proof

We only need to check (IC):

$$\begin{aligned} \hat{U}(x',x')&=t(x')-q(x')c_L=U(x',x')\\&\ge U(x',x)=t(x)-q(x)\gamma (x',x)=\hat{U}(x',x),\\ \hat{U}(x,x)&=t(x)-q(x)\gamma (x,x)=U(x,x)\\&\ge U(x,x')=t(x')-q(x')\gamma (x,x')\ge \hat{U}(x,x'). \end{aligned}$$

The last inequality holds because $\hat{\gamma }(x,x')\ge \gamma (x,x')$, where $\hat{\gamma }(x,z)$ is the cost function (35) with $\psi$ replaced by $\hat{\psi }$. $\square$

Lemma 4

Suppose that $(T,Q,\psi )$ is a truthfully implementable mechanism and $x'+e_L<\psi (x')<x'+e_H$ for some $x'$. Define $\hat{\psi }$ by $\hat{\psi }(x)=\psi (x)$ for $x\ne x'$ and $\hat{\psi }(x')=x'+e_H$. Then $(T,Q,\hat{\psi })$ is truthfully implementable.

The proof is the same as that of the previous lemma. Hence we will focus on truthfully implementable mechanisms where $\psi (x)$ is equal to either $x+e_L$ or $x+e_H$. Define $X_H :=\{x~|~\psi (x)=x+e_H\}$ and $X_L :=\{x~|~\psi (x)=x+e_L\}$.

Lemma 5

A maximal truthfully implementable mechanism $(T,Q,\psi )$ satisfies $\psi (x)=x+e_H$ for all $x\in (1-\Delta e,1]$.

Proof

Suppose to the contrary that $\psi (x')=x'+e_L$ for some $x'\in (1-\Delta e,1]$. Define $\hat{\psi }$ by $\hat{\psi }(x)=\psi (x)$ for $x\ne x'$ and $\hat{\psi }(x')=x'+e_H$ and $\hat{T}$ by $\hat{T}_i(x_i,x_{-i})=T_i(x_i,x_{-i})$ for $x_i\ne x'$ and $\hat{T}_i(x',x_{-i})=T_i(x',x_{-i})+q(x')\Delta c$. Then $\hat{t}(x')=t(x')+q(x')\Delta c$ and $(\hat{T},Q,\hat{\psi })$ is truthfully implementable as shown below. This contradicts the fact that $(T,Q,\psi )$ is a maximal truthfully implementable mechanism, because

$$\begin{aligned} q(x)\hat{\psi }(x)-\hat{t}(x) =q(x)(\psi (x)+\Delta e)-(t(x)+q(x)\Delta c) >q(x)\psi (x)-t(x). \end{aligned}$$

We now show that $(\hat{T},Q,\hat{\psi })$ is truthfully implementable. It is obvious that (IR) holds for $(\hat{T},Q,\hat{\psi })$. In checking the condition (IC), it is sufficient to check the condition for the types above $x'$, because the types below $x'$ cannot fulfill the output requirement.

$$\begin{aligned} \hat{U}(x,x')&=t(x')+q(x')\Delta c-q(x')c_H\\&=t(x')-q(x')c_L=U(x,x')\le U(x,x)=\hat{U}(x,x). \end{aligned}$$

$\square$

Lemma 6

A mechanism $(T,Q,\psi )$ satisfies (IC) if and only if the following conditions hold:

• (i) U(x,x) is non-decreasing in $x$

• (ii) $x\in X_H \text { and } x'\ge x+\Delta e \Rightarrow U(x',x')\ge U(x,x)+q(x)\Delta c$

• (iii) $x'\in X_L \text { and } x'\in (x,x+\Delta e] \Rightarrow U(x',x')\le U(x,x)+q(x')\Delta c$.

Proof

(1) Necessity:

• (i) Due to the transitivity of the inequality, it is sufficient to show $U(x',x')\ge U(x,x)$ for $x,x'$ with $x'\in (x,x+\Delta e]$. If $x\in X_H$, then $U(x',x')\ge U(x',x)=t(x)-q(x)c_H=U(x,x)$. If $x\in X_L$, then $U(x',x')\ge U(x',x)=t(x)-q(x)c_L=U(x,x)$.

• (ii) Suppose $x\in X_H$ and $x'\ge x+\Delta e$. Then (IC) implies

  $$\begin{aligned} U(x',x')\ge U(x',x)=t(x)-q(x)c_L=U(x,x)+q(x)\Delta c. \end{aligned}$$

• (iii) Suppose $x'\in X_L$ and $x'\in (x,x+\Delta e]$. Then (IC) implies

  $$\begin{aligned} U(x,x)\ge U(x,x')=t(x')-q(x')c_H=U(x',x')-q(x')\Delta c. \end{aligned}$$

(2) Sufficiency: If $z\in X_H$ and $z\le x-\Delta e$, then

$$\begin{aligned} U(x,z)=t(z)-q(z)c_L=U(z,z)+q(z)\Delta c\le U(x,x) \text { by (ii)}. \end{aligned}$$

If $z\in X_H$ and $z\in (x-\Delta e,x)$, then

$$\begin{aligned} U(x,z)=t(z)-q(z)c_H=U(z,z)\le U(x,x) \text { by (i)}. \end{aligned}$$

If $z\in X_L$ and $z<x$, then

$$\begin{aligned} U(x,z)=t(z)-q(z)c_L=U(z,z)\le U(x,x) \text { by (i)}. \end{aligned}$$

If $z\in X_L$ and $z\in (x,x+\Delta e]$, then

$$\begin{aligned} U(x,z)=t(z)-q(z)c_H=U(z,z)-q(z)\Delta c\le U(x,x) \text { by (iii)}. \end{aligned}$$

If either $z\in X_L$ and $z>x+\Delta e$ or $z\in X_H$ and $z>x$, then $U(x,z)=-\infty <U(x,x)$. $\square$

As long as the principal can extract the entire surplus, he prefers to assign high output, $x+e_H$, because $e_H-c_H\ge e_L-c_L$. Lemma 5 shows that $(1-\Delta e,1]\subset X_H$. The following lemma shows that the principal can fully extract the surplus when $X_H=(1-\Delta e,1]$.

Lemma 7

The principal can extract the entire surplus if the mechanism prescribes $\psi (x)=x+e_H$ for $x\in (1-\Delta e,1]$ and $\psi (x)=x+e_L$ for $x\in [0,1-\Delta e]$.

Proof

Consider the mechanism (16)–(18) for the particular choice of $\hat{x}=1-\Delta e$. Under this mechanism, $q(x)=G(x)$ and $t(x)=q(x)\gamma (x,x)$. Hence $U(x,x)=t(x)-q(x)\gamma (x,x)=0$ for all $x$, and thus (IR) is satisfied. Furthermore, (IC) is satisfied by Lemma 6. Because it always selects the highest type, it maximizes the revenue among all truthful mechanisms with $X_H=(1-\Delta e,1]$. $\square$

Note that the principal can extract the entire surplus with $X_H=[0,1]$ if $\Delta e > 1$. Given $X_H=(1-\Delta e,1]$, the optimal selection rule is $Q$, because the output $\psi (x)$ is increasing in winner’s type $x$, whereas the payment is already determined when agents report their types.

While Lemma 7 shows that full surplus extraction is achieved if the mechanism prescribes high effort for all $x$ drawn from an interval $(1-\Delta e,1]$, in a final step we need to show that the optimal mechanism must exhibit $X_{H}=[\hat{x},1]$ for some $\hat{x}$.

To prepare the proof, we introduce the following auxiliary definition: For an arbitrary $X_H$ define $\hat{x} :=\inf \{ x~|~x\in X_H \}.$ Define a function $u:[0,1]\rightarrow \mathbb R$ recursively as follows:²⁴

$$\begin{aligned} u(x)&:={\left\{ \begin{array}{ll} 0 &{} \quad \text {if} \, x < \hat{x}+\Delta e \\ \sup \{u(y)+q(y)\triangle c\mathbb {1}_{X_{H}}~|~ y\le x-\Delta e,~y\in X_H\} &{} \quad \text {otherwise}. \end{array}\right. } \end{aligned}$$

(37)

Also let

$$\begin{aligned} t(x) = u(x)+G(x)\gamma (x,x). \end{aligned}$$

(38)

Then, $U(x,x)=u(x)$ and the conditions in Lemma 6 are satisfied. Furthermore, the payment is minimized for the given $X_H$.

Because $f(x)$ is continuous and positive everywhere $f$ has a minimum, $m$, and a maximum, $M$, on $[0,1]$. Define $\mu :={M}/{m}$.

The optimal mechanism chooses the right balance between high effort and information rent. When $X_H=[1-\Delta e,1]$, the information rent is 0. In the final step of the proof of Proposition 7 we show that when we increase the probability of high effort, the high effort region should be increased from the top in order to minimize the information rent. Hence the optimal mechanism must have the high effort region $X_H$ in the form of an interval, $[\hat{x},1]$, for some $\hat{x}$. The proof assumes that $F$ is not the uniform distribution; the case of the uniform distribution is trivial and hence omitted.

Proof of Proposition 7

Suppose $X_H=(x_0,1]$ and consider increasing the high effort region $X_H$ by a small interval, $[x',x'+ \delta (x')]$, so that the probability of high effort is increased by $\varepsilon > 0$. In other words consider choosing $x'$ and $\delta (x')$ so that $F(x'+\delta (x'))^n-F(x')^n=\varepsilon$ and the high effort region becomes $X_H^1:=[x',x'+\delta (x')]\cup X_H$. Also let $x^*$ be the particular value of $x$ for which $F(x^*)^n=F(x_0)^n-\varepsilon$ so that $x^*+\delta (x^*)=x_0$ and let $X_H^*:=[x^*,1]$. Denote the $u$-function defined for the effort prescriptions $X_H^1$ by $u^1(x)$ and the $u$-function defined for $X_H^*$ by $u^*(x)$.

Because the principal’s expected revenue is $\Pi =n\int _0^1 \big (G(x)\psi (x)-t(x)\big )f(x)dx$, the change from $X_H$ to $X_H^1$ changes the principal’s expected revenue by:

$$\begin{aligned} \Delta \Pi&= n\int _{x'}^{x'+\delta (x')}G(x)(\Delta e-\Delta c)f(x)dx-n\int _0^1 \Delta u(x)f(x)dx\\&= \varepsilon (\Delta e-\Delta c) - n\int _0^1 \Delta u(x)f(x)dx, \end{aligned}$$

where the second equation follows from the fact that $n\int _{x'}^{x'+\delta (x')}G(t)f(t)dt=F(x'+\delta (x'))^n-F(x')^n=\varepsilon$ and $\Delta u(x):=u^1(x)-u(x)$ for the change from $X_H$ to $X_H^1$. If one replaces $\Delta u(x)$ by $\Delta u^*(x):=u^*(x)-u(x)$, the above equation applies to the change from $X_H$ to $X_H^*$.

Hence, $x'$ and $\delta (x')$ should be chosen to minimize $\int _0^1 \Delta u(x)f(x)dx$, the increase in information rent. A particular choice of $x'$ [and $\delta (x')$] is drawn in Fig. 2. There, $x'$ is chosen so that $x'+\delta (x')$ is slightly below $x_0$.

Fig. 2

Increase in payment due to increase in $X_H$

The graph of $\Delta u(x)$ is flat on the interval $(x'+\delta (x')+\Delta e,x_0+\Delta e]$, because $(x'+\delta (x'),x_0]\subset X_L$ [hence $u^1(x)$ is flat and $u(x)=0$].

Because an increase in the probability of high effort can be divided into many small steps, we can choose $\varepsilon >0$ arbitrarily small without loss of generality.

Consider the function $\xi (x):=G(x)\Delta c(x_0-x)$. Evidently, $\xi (0)=\xi (x_0)=0$, $\xi (x)>0$ for $x\in (0,x_0)$, and $\xi '(x)=G'(x)\Delta c(x_0-x)-G(x)\Delta c$ is positive near 0 and negative near $x_0$. Hence there exist $a>0$ and $x_a,x_b\in (0,x_0)$ such that $\xi (x_a)=\xi (x_b)=a$, $\xi (x)\ge a$ for $x\in [x_a,x_b]$, $\xi '(x)>0$ for $x\in [0,x_a]$, and $\xi '(x)<0$ for $x\in [x_b,x_0]$.

Choose $\varepsilon$ so small that $x_b<x_0-{\mu ^2}/{(\mu -1)}(x_0-x^*)$ and $G(x_0)-G(x_0-{\mu ^2}/{(\mu -1)}(x_0-x^*))< {G\left( x_0-{\mu ^2}/{(\mu -1)}(x_0-x^*)\right) }/{\mu ^2}$. We distinguish four cases:

(1) Suppose $x'+\delta (x')\in [x_a,x_0-{\mu ^2}/{(\mu -1)}(x_0-x^*)]$. Then,

$$\begin{aligned} \int _0^1&\Delta u(x)f(x)dx - \int _0^1 \Delta u^*(x)f(x)dx \\&>\sum _{j=1}^{k'-1} \left( \int _{x'+\delta (x')+j\Delta e}^{x_0+j\Delta e} \Delta u(x)f(x)dx - \int _{x^*+j\Delta e}^{x_0+j\Delta e} \Delta u^*(x)f(x)dx \right) , \end{aligned}$$

where $k'$ is defined as the smallest integer $k$ for which $x'+ k \Delta e >1$.²⁵ We show that each term of the sum is positive [the second last inequality follows from the fact that $\xi '(x)$ is negative for all $x \ge x_b$]:

$$\begin{aligned}&\int _{x'+\delta (x')+j\Delta e}^{x_0+j\Delta e} \Delta u(x)f(x)dx - \int _{x^*+j\Delta e}^{x_0+j\Delta e} \Delta u^*(x)f(x)dx \\&\quad = G(x'+\delta (x'))\Delta c\int _{x'+\delta (x')+j\Delta e}^{x_0+j\Delta e}f(x)dx - \int _{x^*+j\Delta e}^{x_0+j\Delta e} G(x-j\Delta e)\Delta c f(x)dx \\&\quad \ge \Delta c\left( G(x'+\delta (x'))(x_0-(x'+\delta (x')))m - G(x_0)(x_0-x^*)M \right) \\&\quad \ge \Delta c\left( G\left( x_0-\frac{\mu ^2}{\mu -1}(x_0-x^*)\right) \frac{\mu ^2}{\mu -1}(x_0-x^*)m - G(x_0)(x_0-x^*)M \right) \\&\quad =\Delta c(x_0-x^*)M\left( \frac{\mu }{\mu -1}G\left( x_0-\frac{\mu ^2}{\mu -1}(x_0-x^*)\right) - G(x_0) \right) >0. \end{aligned}$$

(2) Suppose $x'+\delta (x')\in (x_0-{\mu ^2}/{(\mu -1)}(x_0-x^*), x^*]$. First note

$$\begin{aligned} \varepsilon =F(x_0)^n-F(x^*)^n=n\int _{x^*}^{x_0}G(x)f(x)dx\ge n(x_0-x^*)G(x^*)m, \end{aligned}$$

hence $x_0-x^*\le {\varepsilon }/{(nG(x^*)m)}$. Also $\delta (x')\ge {\varepsilon }/{(nG(x'+\delta (x'))M)}$, because

$$\begin{aligned} \varepsilon =n\int _{x'}^{x'+\delta (x')}G(x)f(x)dx\le n\delta (x')G(x'+\delta (x'))M. \end{aligned}$$

We have:

$$\begin{aligned} \int _0^1&\Delta u(x)f(x)dx - \int _0^1 \Delta u^*(x)f(x)dx \\&=\sum _{j=1}^{k'-1} \left( \int _{x'+j\Delta e}^{x_0+j\Delta e} \Delta u(x)f(x)dx - \int _{x^*+j\Delta e}^{x_0+j\Delta e} \Delta u^*(x)f(x)dx \right) . \end{aligned}$$

Each term of the sum is positive because:

$$\begin{aligned}&\int _{x'+j\Delta e}^{x_0+j\Delta e} \Delta u(x)f(x)dx - \int _{x^*+j\Delta e}^{x_0+j\Delta e} \Delta u^*(x)f(x)dx\\&\quad =\Delta c\left( \int _{x'+j\Delta e}^{x'+\delta (x')+j\Delta e} G(x-j\Delta e) f(x)dx \right. \\&\quad \quad \quad \quad \, \left. + G(x'+\delta (x'))\int _{x'+\delta (x')+j\Delta e}^{x_0+j\Delta e} f(x)dx - \int _{x^*+j\Delta e}^{x_0+j\Delta e} G(x-j\Delta e)f(x)dx \right) \\&\quad >\Delta c\left( \max \left\{ G(x'),G\left( x_0-\frac{\mu ^2}{\mu -1}(x_0-x^*)\right) \right\} m\delta (x') \right. \\&\quad \quad \quad \qquad \left. -\big ( G(x_0)-G(x'+\delta (x'))\big )M(x_0-x^*) \right) \\&\quad \ge \Delta c \left( G(x_0-\frac{\mu ^2}{\mu -1}(x_0-x^*))\frac{m\varepsilon }{nG(x'+\delta (x'))M}\right. \\&\quad \quad \quad \qquad \left. -\big ( G(x_0)-G(x'+\delta (x'))\big )\frac{M\varepsilon }{nG(x^*)m} \right) \\&\quad \ge \frac{\Delta c \varepsilon }{n} \left( G\left( x_0-\frac{\mu ^2}{\mu -1}(x_0-x^*)\right) \frac{m}{G(x'+\delta (x'))M}\right. \\&\qquad \qquad \quad \quad \left. -\left( G(x_0)-G(x_0-\frac{\mu ^2}{\mu -1}(x_0-x^*))\right) \frac{M}{G(x^*)m} \right) \\&\quad \ge \frac{\Delta c \varepsilon G\left( x_0-\frac{\mu ^2}{\mu -1}(x_0-x^*)\right) }{n} \left( \frac{m}{G(x'+\delta (x'))M}-\frac{M}{\mu ^2 G(x^*)m} \right) >0. \end{aligned}$$

(3) Suppose $x'+\delta (x')\in (x^*, x_0)$. In this case, $\Delta u(x)$ and $\Delta u^*(x)$ coincide on $[x^*+j\Delta e,x'+\delta (x')+j\Delta e]$, and we need to compare the length of the intervals $[x',x^*]$ and $[x'+\delta (x'),x_0]$, which we denote by $\ell$ and $\ell ^*$, respectively. It is easy to show that $\ell \ge {\eta }/{(nG(x')m)}$ and $\ell ^*\le {\eta }/{(nG(x_0)M)}$, where $\eta =\varepsilon -\big (F(x'+\delta (x'))^n-F(x^*)^n \big )$. Hence, we have

$$\begin{aligned}&\int _{x'+j\Delta e}^{x_0+j\Delta e} \Delta u(x)f(x)dx - \int _{x^*+j\Delta e}^{x_0+j\Delta e} \Delta u^*(x)f(x)dx\\&\quad =\Delta c\left( \int _{x'+j\Delta e}^{x'+\delta (x')+j\Delta e} G(x-j\Delta e) f(x)dx \right. \\&\quad \quad \quad \quad \left. + G(x'+\delta (x'))\int _{x'+\delta (x')+j\Delta e}^{x_0+j\Delta e} f(x)dx - \int _{x^*+j\Delta e}^{x_0+j\Delta e} G(x-j\Delta e)f(x)dx \right) \\&\quad =\Delta c\left( \int _{x'+j\Delta e}^{x^*+j\Delta e} G(x-j\Delta e) f(x)dx \right. \\&\quad \quad \quad \qquad \left. - \int _{x'+\delta (x')+j\Delta e}^{x_0+j\Delta e} \big (G(x-j\Delta e)-G(x'+\delta (x')) \big )f(x)dx \right) \\&\quad >\Delta c\Big (G(x')m\ell -\big ( G(x_0)-G(x'+\delta (x'))\big )M\ell ^* \Big )\\&\quad \ge \Delta c \left( G(x_0-\frac{\mu ^2}{\mu -1}(x_0-x^*))\frac{m\eta }{nG(x')M}\right. \\&\quad \qquad \quad \quad \left. -\big ( G(x_0)-G(x'+\delta (x'))\big )\frac{M\eta }{nG(x_0)m} \right) \\&\quad \ge \frac{\Delta c \eta }{n} \left( G\left( x_0-\frac{\mu ^2}{\mu -1}(x_0-x^*)\right) \frac{m}{G(x')M}\right. \\&\quad \quad \left. -\left( G(x_0)-G\left( x_0-\frac{\mu ^2}{\mu -1}(x_0-x^*)\right) \right) \frac{M}{G(x_0)m} \right) \\&\quad \ge \frac{\Delta c \eta G\left( x_0-\frac{\mu ^2}{\mu -1}(x_0-x^*)\right) }{n} \Big ( \frac{m}{G(x')M}-\frac{M}{\mu ^2 G(x_0)m} \Big ) >0. \end{aligned}$$

(4) Finally, suppose $x'+\delta (x')<x_a$. Then,

$$\begin{aligned}&\int _{x'+j\Delta e}^{x_0+j\Delta e} \Delta u(x)f(x)dx - \int _{x^*+j\Delta e}^{x_0+j\Delta e} \Delta u^*(x)f(x)dx \\&\quad >\int _{x'+\delta (x')+j\Delta e}^{x_0+j\Delta e} \Delta u(x)f(x)dx - \int _{x^*+j\Delta e}^{x_0+j\Delta e} \Delta u^*(x)f(x)dx \\&\quad = \Delta c\left( G(x'+\delta (x'))\int _{x'+\delta (x')+j\Delta e}^{x_0+j\Delta e}f(x)dx - \int _{x^*+j\Delta e}^{x_0+j\Delta e} G(x-j\Delta e)f(x)dx \right) \\&\quad \ge \Delta c\Big ( G(x'+\delta (x'))(x_0-(x'+\delta (x')))m - G(x_0)(x_0-x^*)M \Big )\\&\quad \ge \Delta c\Big ( G(\delta (0))(x_0-\delta (0))m - G(x_0)(x_0-x^*)M \Big ) \quad (\because \xi '(x)>0)\\&\quad \ge \Delta c \varepsilon \left( \varepsilon ^{-\frac{1}{n}} (x_0-\delta (0))m - \frac{G(x_0)M}{nG(x^*)m} \right) , \end{aligned}$$

which is positive when $\varepsilon$ is small.

Hence, $\int _0^1 \Delta u(x)f(x)dx$ is minimized when $x'=x^*$. Therefore, the set $X_H$ is the interval $[\hat{x}, 1]$, and the function $u(x)$ can be stated explicitly as in Eq. (19), which completes the proof.