Ranking asymmetric auctions

Abstract

We compare the expected revenue in first- and second-price auctions with asymmetric bidders. We consider “close to uniform” distributions with identical supports and show that in the case of identical supports the expected revenue in second-price auctions may exceed that in first-price auctions. We also show that asymmetry over lower valuations has a stronger negative impact on the expected revenue in first-price auctions than in second-price auctions. However, asymmetry over high valuations always increases the revenue in first-price auctions.

Appendix

1.1 A: Auxiliary Lemma

Lemma 2

$$\begin{aligned} F_{i}(v_{i}(b))&= 2b+\varepsilon (V_{i}(b)+H_{i}(2b)+\varepsilon ^{2}(V_{i}(b)H_{i}^{\prime }(2b)+u_{i}(b))+O(\varepsilon ^{3}).\end{aligned}$$

(23)

$$\begin{aligned} f_{i}(v_{i}(b))&= 1+\varepsilon H_{i}^{\prime }(2b)+\varepsilon ^{2}V_{i}(b)H_{i}^{\prime \prime }(2b)+O(\varepsilon ^{3}). \end{aligned}$$

(24)

Proof

Substituting $v_{i}(b)=v_{sym}(b)+\varepsilon V_{i}(b)+\varepsilon ^{2}U_{i}(b)+O(\varepsilon ^{3})$ in $F_{i}(v_{i}(b))$, expanding the series near $v_{sym}(b)$ and collecting the terms in $\varepsilon$ gives

$$\begin{aligned} F_{i}(v_{i}(b))&= v_{i}(b)+\varepsilon H_{i}(v_{i}(b))=\left[ v_{sym}(b)+\varepsilon V_{i}(b)+\varepsilon ^{2}u_{i}(b)+O(\varepsilon ^{3}) \right] \\&\quad +\varepsilon H_{i}\left( (v_{sym}(b)+\varepsilon V_{i}(b)+\varepsilon ^{2}u_{i}(b)+O(\varepsilon ^{3})\right) \\&= v_{sym}(b)+\varepsilon \left[ V_{i}(b)+H_{i}(v_{sym}(b))\right] \\&\quad +\varepsilon ^{2}\left[ V_{i}(b)H_{i}^{\prime }(v_{sym}(b))+u_{i}(b)\right] +O(\varepsilon ^{3}), \end{aligned}$$

where $\varepsilon H_{i}((v_{sym}(b)+\varepsilon V_{i}(b)+\varepsilon ^{2}u_{i}(b)+O(\varepsilon ^{3}))=\varepsilon H_{i}(v_{sym}(b))+\varepsilon ^{2}V_{i}(b)H_{i}^{\prime }(v_{sym}$ $(b))+O(\varepsilon ^{3})$. Substituting $v_{sym}(b)=2b$ (14) completes the proof. A similar calculation proves (24).$\square$

1.2 B: Proof of Theorem 1

The following notations are used later on in this proof. Let

$$\begin{aligned} V(b)={y(b)-H(2b)}, \end{aligned}$$

(25)

where $V(b)$ is given by (16) and

$$\begin{aligned} y{{(b)}=}16b^{3}\int \limits _{2b}^{1}\frac{{H(x)}}{x^{4}}dx. \end{aligned}$$

(26)

Substituting (12, 14, 23) in $R^{first}$ (given by 3) we get

$$\begin{aligned} R^{first}&= \overline{b}-\int \limits _{0}^{\overline{b}}4b^{2}+\varepsilon ^{2}\left[ 2b(V(b)H^{\prime }(2b)+u_{1}(b))-(V(b)+H(2b))^{2}\right] \\&+\,\,\varepsilon ^{2}\left[ 2b(u_{2}(b)+V(b)H^{\prime }(2b))\right] db+O(\varepsilon ^{3}) \\&= \overline{b}-\int \limits _{0}^{\overline{b}}4b^{2}-\varepsilon ^{2}\int \limits _{0}^{\overline{b}_{sym}+\varepsilon ^{2}\overline{b}_{2}}\left[ 2b(V(b)H^{\prime }(2b)+u_{1}(b))-(V(b)+H(2b))^{2}\right] \\&+\,\left[ 2b(u_{2}(b)+V(b)H^{\prime }(2b))\right] db+O(\varepsilon ^{3}) \end{aligned}$$

(Observe that we substitute $\overline{b}=\overline{b}_{sym}+\varepsilon ^{2} \overline{b}_{2}+O(\varepsilon ^{3})$ only in the second integral’s upper limit where the missing $O(\varepsilon ^{3})$ in the integral’s limit is part of the $O(\varepsilon ^{3})$ at the end of the equation). Substituting $\overline{b}-\int \nolimits _{0}^{\overline{b} }4b^{2}db=R_{sym}^{first}+O(\varepsilon ^{3})$ given by Lemma 3 below, $\overline{b}_{sym}=0.5$ (see Eq. 4) gives (Observe that the terms with the $\overline{b}_{2}$ are $O(\varepsilon ^{3})$ )

$$\begin{aligned} R^{first}&= R_{sym}^{first}-\varepsilon ^{2}\int \limits _{0}^{0.5}{\huge [} 2b(V(b)H^{\prime }(2b)+u_{1}(b))-(V(b)+H(2b))^{2} \\&+2b(u_{2}(b)+V(b)H^{\prime }(2b))){\huge ]}db+O(\varepsilon ^{3}) \end{aligned}$$

Denote $u(b)=u_{1}(b)+u_{2}(b)$. Rearranging yields

$$\begin{aligned} R^{first}&= R_{sym}^{first}-\varepsilon ^{2}\int \limits _{0}^{0.5}{ 2b[u(b))+2V(b)H^{\prime }(2b)]db} \\&+\,\varepsilon ^{2}\int \limits _{0}^{0.5}{(V(b)+H(2b))}^{2}{db+} O(\varepsilon ^{3}). \end{aligned}$$

Substituting $2\displaystyle \int \nolimits _{0}^{0.5}{bu(b)db}$ given by Lemma 5 below gives

$$\begin{aligned} R^{first}&= R_{sym}^{first}-\varepsilon ^{2}\int \limits _{0}^{0.5}(1-2b) \left[ \frac{{2H(2b)V(b)+6V^{2}(b)}}{{b}}+4{H^{\prime }(2b)V(b)}\right] {db} \nonumber \\&+\,\varepsilon ^{2}\int \limits _{0}^{0.5}{(V(b)+H(2b))}^{2}{db+} O(\varepsilon ^{3}). \end{aligned}$$

(27)

Substituting $V(b)$ given by ( 25) in (27) we get

$$\begin{aligned} R^{first}&= R_{sym}^{first}-\varepsilon ^{2}\left\{ \int \limits _{0}^{0.5}(1-2b)\left[ \frac{{6y^{2}{{(b)}-10H(2b)}}y{{(b)+4}H^{2}(2b)}}{{b} }\right] db\right. \nonumber \\&\quad \left. -{\int \limits _{0}^{0.5}}(1-2b)[4{H^{\prime }(2b)y(b)}-4{H^{\prime }(2b){H(2b)]}}-{y^{2}{(b)db}}\right\} {+O(}\varepsilon ^{3}), \nonumber \\ \end{aligned}$$

(28)

where $y(b)$ is given by (26). Observe that

$$\begin{aligned} \int \limits _{0}^{0.5}{b{H(2b)}H^{\prime }(2b)db=-}\frac{1}{4}\int \limits _{0}^{0.5}{H}^{2}{(2b)db} \end{aligned}$$

(29)

and

$$\begin{aligned} \int \limits _{0}^{0.5}{{H(2b)}H^{\prime }(2b)db=0.} \end{aligned}$$

(30)

Substituting (29, 30) in (28) gives

$$\begin{aligned} R^{first}&= R_{sym}^{first}-\varepsilon ^{2}\left\{ \int \limits _{0}^{0.5}(1-2b) \left[ \frac{{6y^{2}{{(b)}-10H(2b)}}y{{(b)+4}H^{2}(2b)}}{{b} }\right] {db}\right. \nonumber \\&\quad \left. -{\int \limits _{0}^{0.5}(1-2b)4{H^{\prime }(2b)y(b)} -2H^{2}(2b)-y^{2}{(b)db}}\right\} {+O(}\varepsilon ^{3}). \end{aligned}$$

(31)

Integration by parts of $\int \nolimits _{0}^{0.5}(1-2b){H^{\prime }(2b)y(b)db}$ gives

$$\begin{aligned} \int \limits _{0}^{0.5}(1-2b){H^{\prime }(2b)y(b)db}=-\int \limits _{0}^{0.5} \frac{{H(2b)((1-2b)y}^{\prime }{(b)}-2y{(b))}}{2}db. \end{aligned}$$

(32)

Substituting (32) in (31) yields

$$\begin{aligned} R^{first}&= R_{sym}^{first}-\varepsilon ^{2}\left\{ \int \limits _{0}^{0.5}(1\!-\!2b)\left[ \frac{{6y^{2}{{(b)}\!-\!10H(2b)}}y{{(b)\!+\!4}H^{2}(2b) }}{{b}}\right] {db}\right. \nonumber \\&\left. -{\int \limits _{0}^{0.5}\!-\!{2H(2b)((1\!-\!2b){y}^{\prime }(b)\!-\!2y(b))-} 2H^{2}(2b)\!-\!y^{2}{(b)db}}\right\} {+O(}\varepsilon ^{3}). \nonumber \\ \end{aligned}$$

(33)

Simple calculation shows that

$$\begin{aligned} {y}^{\prime }{(b)=}48b^{2}\int \limits _{2b}^{1}\frac{{H(x)}}{x^{4}}dx- \frac{2{H(2b)}}{b}={\frac{3y{(b)-2H(2b)}}{b}} \end{aligned}$$

(34)

and thus

$$\begin{aligned} {H(2b)}=\frac{3y{(b)-{y}^{\prime }(b)b}}{2}. \end{aligned}$$

(35)

Substituting (34) and (35) in (33) we get

$$\begin{aligned} R^{first}&= R_{sym}^{first}-\varepsilon ^{2}\left\{ \int \limits _{0}^{0.5}(1-2b)(-y{{{{{(b)y}^{\prime }(b)+}}}}\left( {{y}^{\prime }(b)}\right) ^{2}{b}){)db}\right. \nonumber \\&\qquad \qquad \qquad -{\int \limits _{0}^{0.5}-(3y{(b)}-{y}^{\prime }(b)b)\cdot {((1-2b){y }^{\prime }(b)-2y(b))}}\nonumber \\&\qquad \qquad \qquad \left. -\frac{{{(}3y{(b)-{y}^{\prime }(b)b)}^{2}}}{2}{-y^{2}{ (b)db}}\right\} {+O(}\varepsilon ^{3}). \end{aligned}$$

(36)

By integration by parts (observe that $y(1)=0$) we have

$$\begin{aligned} \int \limits _{0}^{0.5}{{b{y}^{\prime }{(b)}}}y{(b)=-}\int \limits _{0}^{0.5} \frac{{y^{2}{(b)}}}{2}db \end{aligned}$$

(37)

and

$$\begin{aligned} \int \limits _{0}^{0.5}{y}^{\prime }{(b)}y{(b)=0.} \end{aligned}$$

(38)

Substituting (37) and (38) in (36) gives

$$\begin{aligned} R^{first}=R_{sym}^{first}+\varepsilon ^{2}\left\{ \int \limits _{0}^{0.5}{ 4y^{2}{(b)}}-\left( 2b-\frac{{9}}{2}{b}^{2}\right) \left( {y}^{\prime }{(b)}\right) ^{2}{ db}\right\} {+O(}\varepsilon ^{3}). \end{aligned}$$

Transformation of $w=2b$ and substituting $z(w)=y(w/2)$ given by (19 ) yields the result. $\square$

Lemma 3

$\overline{b}-\int \nolimits _{0}^{\overline{b} }4b^{2}db=R_{sym}^{first}+O(\varepsilon ^{3}).$

Proof

Since $\bar{b}_{1}=0$ and $\overline{b}_{sym}=0.5$ we have $\overline{b}=\overline{b}_{sym}+\varepsilon \bar{b}_{1}+\varepsilon ^{2} \overline{b}_{2}+O(\varepsilon ^{3})=0.5+\varepsilon ^{2}\overline{b} _{2}+O(\varepsilon ^{3})$. Thus,

$$\begin{aligned} \overline{b}-\int \limits _{0}^{\overline{b}}4b^{2}db&= \overline{b}-\frac{4 }{3}\overline{b}^{3}=\frac{1}{2}+\varepsilon ^{2}\overline{b}_{2}-\frac{4}{3} \left( \frac{1}{2}+\varepsilon ^{2}\overline{b}_{2}\right) ^{3}+O(\varepsilon ^{3}) \nonumber \\&= \frac{1}{3}+O(\varepsilon ^{3})=R_{sym}^{first}+O(\varepsilon ^{3}). \end{aligned}$$

(39)

$\square$

Lemma 4

$$\begin{aligned} u(b)&= \frac{1}{b}\left\{ \int \limits _{0}^{b}\frac{{2H(2s)V(s)+6V^{2}(s)} }{{s}}-{4H^{\prime }(2s)V(s)-2H(2s)H^{\prime }(2s)ds}\right. \nonumber \\&\qquad \qquad \left. +\int \limits _{0}^{b}4s\left( H^{\prime }({2s})\right) ^{2}-{ 4sV(s)H^{\prime \prime }(2s)ds}\right\} , \end{aligned}$$

(40)

where $u(b)=u_{1}(b)+u_{2}(b).$

Proof

The equilibrium equations are given by

$$\begin{aligned} v_{i}^{\prime }(b)=\frac{{F_{i}(v_{i}(b))}}{{f_{i}(v_{i}(b))}}\frac{1}{{ v_{j}(b)-b}},\quad i\ne j. \end{aligned}$$

(41)

Substituting (13, 24, 23), $v_{sym}(b)=2b$ and $v_{i}^{\prime }(b)$ (obtained by differentiating 13) in (41) we get

$$\begin{aligned}&v_{sym}^{^{\prime }}(b)+\varepsilon V_{i}^{^{\prime }}(b)+\varepsilon ^{2}u_{i}^{^{\prime }}(b)+O(\varepsilon ^{3}) \nonumber \\&= \frac{2b+\varepsilon \left[ V_{i}(b)+H_{i}(2b)\right] +\varepsilon ^{2} \left[ V_{i}(b)H_{i}^{\prime }({2b})+u_{i}(b)\right] }{{1+\varepsilon H_{i}^{\prime }(2b)+\varepsilon ^{2}V_{i}(b)H_{i}^{\prime \prime }(2b)+O(\varepsilon ^{3})}} \nonumber \\&\quad \times \frac{1}{{2b}+\varepsilon V_{j}(b)+\varepsilon ^{2}u_{j}(b){-b}} +O(\varepsilon ^{3}),\quad i\ne j. \end{aligned}$$

(42)

To simplify the calculation we define

$$\begin{aligned} A&= \frac{2b+\varepsilon \left[ V_{i}(b)+H_{i}(2b)\right] +\varepsilon ^{2} \left[ V_{i}(b)H_{i}^{\prime }(2b)+u_{i}(b)\right] }{{1+\varepsilon H_{i}^{\prime }(2b)+\varepsilon ^{2}V_{i}(b)H_{i}^{\prime \prime }(2b)+O(\varepsilon ^{3})}}+O(\varepsilon ^{3}); \\ B&= \frac{1}{b+\varepsilon V_{j}(b)+\varepsilon ^{2}u_{j}(b)}+O(\varepsilon ^{3}) \end{aligned}$$

when $\varepsilon$ is sufficiently small, $\varepsilon H_{i}^{\prime }(2b)+\varepsilon ^{2}V_{i}(b)H_{i}^{\prime \prime }(2b)<1$. Then, we can use Taylor’s expansion $\frac{1}{1+\varepsilon C}=1-\varepsilon C+(\varepsilon C)^{2}+O(\varepsilon ^{3})$ in $A$ and $B$ above. The expansion for A is given by

$$\begin{aligned} A&= \left[ 2b+\varepsilon \left[ V_{i}(b)+H_{i}(2b)\right] +\varepsilon ^{2} \left[ V_{i}(b)H_{i}^{\prime }(2b)+u_{i}(b)\right] \right] \\&\times \left[ 1-\varepsilon H_{i}^{\prime }(2b)-\varepsilon ^{2}V_{i}(b)H_{i}^{\prime \prime }(2b)+{\varepsilon ^{2}}H_{i}^{\prime }{}^{2}(2b)\right] +O(\varepsilon ^{3}). \end{aligned}$$

Rearranging gives

$$\begin{aligned} A&= 2b+\varepsilon \left[ V_{i}(b)+H_{i}(2b)-2bH_{i}^{\prime }(2b)\right] \\&\quad +\varepsilon ^{2}\left[ 2bH_{i}^{\prime }{}^{2}(2b) -2bV_{i}(b)H_{i}^{\prime \prime }(2b)-V_{i}(b)H_{i}^{\prime }(2b)\right. \\&\quad \left. -H_{i}(2b)H_{i}^{\prime }(2b)+u_{i}(b)+V_{i}(b)H_{i}^{\prime }(2b) \right] +O(\varepsilon ^{3}). \end{aligned}$$

Since $0\le {\frac{{\varepsilon V_{j}(b)+\varepsilon ^{2}u_{j}(b)}}{{b}}<<1}$ for small $\varepsilon$ ( it easy to see that $\frac{{ V_{j}(b)}}{{b}}$ is bounded for small $b$ ), the Taylor expansion of B is given by

$$\begin{aligned} B&= \frac{1}{{b+\varepsilon V_{j}(b)+\varepsilon ^{2}u_{j}(b)}}=\frac{1}{{b} }\frac{1}{{1+\frac{{\varepsilon V_{j}(b)+\varepsilon ^{2}u_{j}(b)}}{{b}}}}\\&= \frac{1}{{b}}\left[ 1-\frac{{\varepsilon V_{j}(b)}}{{b}}+{\varepsilon ^{2} }\left( -\frac{{u_{j}(b)}}{{b}}+\frac{1}{{b^{2}}}(V_{j}(b)^{2}\right) \right] +O(\varepsilon ^{3}). \end{aligned}$$

Multiplying A and B we get

$$\begin{aligned} A\times B&= 2\left[ 1-\frac{{\varepsilon V_{j}(b)}}{{b}}+{\varepsilon ^{2}}\left( - \frac{{u_{j}(b)}}{{b}}+\frac{1}{{b^{2}}}(V_{j}(b)^{2}\right) \right] \\&\quad +\,\frac{\varepsilon \left[ V_{i}(b)+H_{i}(2b)-2bH_{i}^{\prime }(2b)\right] }{ {b}}\times \left( 1-\frac{{\varepsilon V_{j}(b)}}{{b}}\right) \\&\quad +\,\varepsilon ^{2}\frac{2bH_{i}^{\prime }{}^{2}(2b)-2bV_{i}(b)H_{i}^{\prime \prime }(2b)-H_{i}(2b)H_{i}^{\prime }(2b)+u_{i}(b)}{{b}}+O(\varepsilon ^{3}). \end{aligned}$$

Rearranging (42), comparing the $O\left( \varepsilon ^{2}\right)$ order coefficient ($O(1)$ and $O(\varepsilon )$ are known as shown in Sect. 3) and substituting $V_{1}=-V_{2}$ (15) we get

$$\begin{aligned} u_{i}^{^{\prime }}(b)&= -\left( \frac{2{u_{j}(b)}}{{b}}\right) -\left( \frac{{2H^{\prime }(2b)V(b)}}{{b}}\right) -2{V(b)H^{\prime \prime }(2b)} \\&+2H^{\prime }{}^{2}({2b})+\frac{{V}^{2}{(b)}}{{b^{2}}}+\frac{{u_{i}(b)}}{{b }} \\&+\frac{{H(2b)V(b)}}{{b^{2}}}-\frac{{H(2b)H^{\prime }(2b)}}{{b}}+\frac{2{V} ^{2}{(b)}}{{b^{2}}}. \end{aligned}$$

Let $u(b)=u_{1}(b)+u_{2}(b)$. Then

$$\begin{aligned} u^{\prime }+\frac{{u(b)}}{b}&= \frac{{2H(2b)V(b)+6V^{2}(b)}}{{b^{2}}}-\frac{ {4H^{\prime }(2b)V(b)+2H(2b)H^{\prime }(2b)}}{b} \\&\quad +4H^{\prime 2}({2b})-{4V(b)H^{\prime \prime }(2b).} \nonumber \end{aligned}$$

(43)

Equation (43) is a first order ODE, $u^{\prime }+P(b)u=g(b)$. By (2) the boundary condition is $u(0)=0$. The solution of (43) is given by :

$$\begin{aligned} u(b)=\left( m-\int \limits _{b}^{0.5}{g(s)\cdot e}^{{-\int \nolimits _{s}^{0.5}{P(x)dx}}}{ds}\right) \cdot {{e}^{{\int \nolimits _{b}^{0.5}{ P(s)ds}}},} \end{aligned}$$

where ${P(s)=}\frac{1}{s}$ ,

$$\begin{aligned} g(s)&= \frac{{2H(2b)V(b)+6V^{2}(b)}}{{b^{2}}}-\frac{{4H^{\prime }(2b)V(b)+2H(2b)H^{\prime }(2b)}}{b} \\&\quad +4H^{\prime 2}({2b})-{4V(b)H^{\prime \prime }(2b).} \nonumber \end{aligned}$$

(44)

and $m$ is a constant. Since

$$\begin{aligned} {\exp }\left( \int \limits _{b}^{0.5}{P(s)ds}\right) {=}\frac{1}{2b}. \end{aligned}$$

we have

$$\begin{aligned} u(b)=\frac{m}{2b}-\frac{1}{b}\int \limits _{b}^{0.5}{g(s)s{ds.}} \end{aligned}$$

By the boundary conditions $u(0)=0$ it follows that

$$\begin{aligned} m=2\int \limits _{0}^{0.5}{g(s)s{ds.}} \end{aligned}$$

Substituting $m$ in $u$ gives

$$\begin{aligned} u(b)=\frac{1}{b}\int \limits _{0}^{b}{g(s)s{ds.}} \end{aligned}$$

(45)

Substituting (44) in (45) yields the result. $\square$

Lemma 5

$$\begin{aligned} 2\int _{0}^{0.5}bu(b)&= \int \limits _{0}^{0.5}\left( \int \limits _{0}^{b}{ g(s)s{ds}}\right) db \\&= \int \limits _{0}^{0.5}\left\{ (1-2b)\left[ \frac{{2H(2b)V(b)+6V^{2}(b)}}{ {b}}+4{H^{\prime }(2b)V(b)}\right] \right. \\&\quad \left. -4b{V(b)H^{\prime }(2b)db}\right\} \end{aligned}$$

Proof

Let us find $\int _{0}^{0.5}2bu(b)db$ where $u(b)$ is given by Lemma 4.

$$\begin{aligned} \int _{0}^{0.5}bu(b)db&= \int _{0}^{0.5}\left[ \int \limits _{0}^{b}\Big (\frac{{ 2H(2s)V(s)+6V^{2}(s)}}{{s}}-{4H^{\prime }(2s)V(s)-2H(2s)H^{\prime }(2s)} \right. \\&\quad \left. +4sH^{\prime 2}({2s})-{4sV(s)H^{\prime \prime }(2s)\Big )ds}\right] {db.} \end{aligned}$$

Integrating by parts, multiplying by 2 and using the boundary condition $u(0)=0$ we get

$$\begin{aligned} 2\int _{0}^{0.5}bu(b)&= \int \limits _{0}^{0.5}(1-2b)g(b)bdb \\&= \int \limits _{0}^{0.5}(1-2b)\left[ \frac{{2H(2b)V(b)+6V^{2}(b)}}{{b}}\!-\!{ 4H^{\prime }(2b)V(b)\!-\!2H(2b)H^{\prime }(2b)}\right. \\&\quad \left. +4bH^{\prime 2}({2b})-{4bV(b)H^{\prime \prime }(2b)}\right] {db.} \end{aligned}$$

Since ${V(b)H^{\prime }(2b)}+{2bH^{\prime \prime }(2b)V(b)=V(b)(H}^{\prime }{ (2b)b)}^{\prime }$, we have

$$\begin{aligned} 2\int _{0}^{0.5}bu(b)&= \int \limits _{0}^{0.5}(1-2b)\Bigg [ \frac{{ 2H(2b)V(b)+6V^{2}(b)}}{{b}}-{2H^{\prime }(2b)V(b)}\\&-{2H(2b)H^{\prime }(2b)}+4bH^{\prime 2}({2b})-{2V(b){(H^{\prime }(2b)b)} ^{\prime }{\Bigg ]}db.} \nonumber \end{aligned}$$

(46)

Substituting $V^{^{\prime }}(b)={\frac{3{V(b)+H(2b)}}{b}}-2{H^{^{\prime }}(2b)}$ (obtained by differentiating $V(b)$ defined in Eq. 16) and integrating by parts gives

$$\begin{aligned} \int \limits _{0}^{0.5}2(1-2b){V(b){(H^{\prime }(2b)b)}^{\prime }db} \!&= \!\int \limits _{0}^{0.5}4{V(b)H^{\prime }(2b)bdb} -{\int \limits _{0}^{0.5}(1-2b)}(6V(b)H^{\prime }(2b)\nonumber \\&\quad +2H(2b))H^{\prime }(2b) -4H^{\prime 2}(2b)b)db. \end{aligned}$$

(47)

The substitution of (47) in (46) completes the proof. $\square$