Does counterfeiting benefit genuine manufacturer? The role of production costs

Abstract

A monopolist sells a luxury genuine product which can be illegally copied and sold by a competitive fringe of counterfeiters. Fines imposed on caught counterfeiters are pocketed by the genuine firm. We prove that if production costs are low, then the genuine manufacturer would lobbying for high penalties so that counterfeiters should be thrown out of the market. In this case, the presence of counterfeiters does not provide any benefit to the producer of the original product. Whenever the production cost is neither too high nor too low, the optimal fine guarantees a positive demand for the genuine product as well as for the fake; the genuine producer is better off than in a world without counterfeiters. If production costs are too high, the genuine firm has no more incentive to produce. Its remaining goal is to collect penalty money from counterfeiters. Again, the presence of counterfeiters provides a benefit to the genuine manufacturer. Finally, a comparison between full protection and null protection policies is performed.

Appendix

Proof of Proposition 1

The consumer indexed by $\theta =p_{g1}=\dfrac{p_c}{\alpha }$ has utility zero buying a fake. The consumer indexed by $\theta =p_g$ has utility zero buying a genuine item. The consumer indexed by $\hat{\theta }_g=\dfrac{p_g-p_c}{1-\alpha }$ is indifferent between buying the counterfeit product or the original one. Remember that $p_c=\dfrac{\phi f}{1-\phi }$.

Note that it is: (1) $p_{g1}<p_g<\dfrac{p_g-p_c}{1-\alpha }$ or (2) $\dfrac{p_g-p_c}{1-\alpha }<p_g<p_{g1}.$

Let us consider the case (1).

If $\dfrac{p_g-p_c}{1-\alpha }<1$ (that is $p_g<p_{g2}$) then consumer indexed between $\theta =\hat{\theta }_g$ and $\theta =1$ buy the original item so that $D_g=1-\hat{\theta }_g= \dfrac{1-\alpha - p_g+\dfrac{\phi f}{1-\phi }}{1-\alpha }$. Consumers indexed between $\theta =p_{g1}$ and $\theta =\hat{\theta }_g$ buy the fake and then $D_c=\hat{\theta }_g-p_{g1}= \dfrac{\alpha p_g-\dfrac{\phi f}{1-\phi }}{\alpha (1-\alpha )}.$

If $1\le \dfrac{p_g-p_c}{1-\alpha }$ (that is $p_g\ge p_{g2}$) and $p_{g1}<1$ then consumers indexed between $\theta =p_{g1}$ and $\theta =1$ buy the fake so that $D_c=1-p_{g1}=1-\dfrac{\phi f}{\alpha (1-\phi )}$ and nobody buys the original product.

If $p_{g1}\ge 1$ then nothing is bought, that is $D_c=D_g=0$.

Let us consider the case (2).

If $p_g<1$ then consumer indexed between $\theta =p_g$ and $\theta =1$ buy the original item so that $D_g=1-p_g$ and nobody buys the fake.

If $1\le p_g$ then nothing is bought, that is $D_c=D_g=0$. $\square$

Proof of Proposition 2

It is $\hat{f}<\dfrac{\tilde{f}}{2}$. Note that $\pi _g$ is constant with respect to $p_g$ when $p_g\ge p_{g2}$.

• Let $0\le f< \hat{f}$.

  If $0\le p_g\le p_{g1}$ then $\pi _g=p_g(1-p_g)$ and it is increasing (being $p_g<\dfrac{1}{2}$).

  If $p_{g1}<p_g<p_{g2}$ then

  $$\pi _g=p_g\left( \dfrac{1-\alpha - p_g+\dfrac{\phi f}{1-\phi }}{1-\alpha }\right) +\phi f \left( \dfrac{\alpha p_g-\dfrac{\phi f}{1-\phi }}{\alpha (1-\alpha )}\right).$$

  The graph of $\pi _g$ is a concave parabola. Its derivative with respect to $p_g$ is

  $$\dfrac{d\pi _g}{dp_g}=\dfrac{f\phi (2-\phi )+(1-\phi )(2p_g-(1-\alpha ))}{(1-\phi )(1-\alpha )}.$$

  It is

  $$\dfrac{d\pi _g(p_{g1})}{dp_g}= \dfrac{\alpha (1-\alpha )(1-\phi )-f\phi (2-2\alpha +\alpha \phi )}{\alpha (1-\phi )(1-\alpha )}.$$

  $\dfrac{d\pi _g(p_{g1})}{dp_g}>0$ iff $f<\hat{f}$. It is

  $$\dfrac{d\pi _g(p_{g2})}{dp_g}= -\dfrac{(1-\alpha )(1-\phi )+f\phi ^2}{(1-\phi )(1-\alpha )}<0.$$

  It follows that $\pi _g$ is increasing for $p_{g1}\le p_g\le p_g^*$ and it is decreasing for $p_g^*\le p_g\le p_{g2}$. We conclude that when $0\le f<\hat{f}, \pi _g$ reaches it maximum at $p_g=p_g^*$.

• Let $\hat{f}\le f\le \dfrac{\tilde{f}}{2}$.

  If $0\le p_g\le p_{g1}$ then $\pi _g=p_g(1-p_g)$ and it is increasing (being $p_g<\dfrac{1}{2}$).

  If $p_{g1}<p_g<p_{g2}$ then the graph of $\pi _g$ is a concave parabola and it is

  $$\dfrac{d\pi _g(p_{g1})}{dp_g}<0;\quad \dfrac{d\pi _g(p_{g2})}{dp_g}<0$$

  It follows that $\pi _g$ is decreasing for $p_{g1}\le p_g\le p_{g2}$. We conclude that when $\hat{f}\le f\le \dfrac{\tilde{f}}{2}, \pi _g$ reaches it maximum at $p_g=p_{g1}$.

• Let $\dfrac{\tilde{f}}{2}< f\le \tilde{f}$.

  If $0\le p_g\le p_{g1}$ then $\pi _g=p_g(1-p_g)$ and (being $p_{g1}>\dfrac{1}{2}$) it is increasing with respect to $p_g$ for $0\le p_g\le \dfrac{1}{2}$, decreasing with respect to $p_g$ for $\dfrac{1}{2}\le p_g\le p_{g1}.$

  If $p_{g1}<p_g<p_{g2}$ then the graph of $\pi _g$ is a concave parabola and it is

  $$\dfrac{d\pi _g(p_{g1})}{dp_g}<0;\quad \dfrac{d\pi _g(p_{g2})}{dp_g}<0.$$

  It follows that $\pi _g$ is decreasing for $p_{g1}\le p_g\le p_{g2}$. We conclude that when $\hat{f}\le f\le \dfrac{\tilde{f}}{2}, \pi _g$ reaches its maximum at $p_g=\dfrac{1}{2}$.$\square$

Proof of Proposition 3

The graph of $\pi _g^*$ with respect to the fine f is a concave parabola if $\alpha <\dfrac{4(1-\phi )}{(2-\phi )^2}=\alpha ^*$, a convex parabola if $\alpha >\alpha ^*$. $\pi _g^*$ is a linear function with respect to f in the particular case $\alpha =\alpha ^*$.

• Let $\alpha =\alpha ^*$.

  In this case, it is

  $$\begin{aligned}\pi _g^*&=\dfrac{\phi (2(2-\phi )^3f+\phi (1-\phi ))}{4(1-\phi )(2-\phi )^2}\\ \dfrac{d\pi _g^*}{df}&=\dfrac{\phi (2-\phi )}{2(1-\phi )}>0 \end{aligned}$$

  It is

  $$\begin{aligned}\pi _{g1}&=\dfrac{\phi f(2-\phi )^2(4(1-\phi )^2-\phi f (2-\phi )^2)}{16(1-\phi )^4}\\\dfrac{d\pi _{g1}}{df}&= \dfrac{\phi (2-\phi )^2(2(1-\phi )^2-\phi f (2-\phi )^2)}{8(1-\phi )^4} \end{aligned}$$

  It is $\dfrac{d\pi _{g1}}{df}>0$ if $f<\dfrac{2(1-\phi )^2}{\phi (2-\phi )^2}=\dfrac{\tilde{f}}{2}.$

  Hence if $\alpha =\alpha ^*$ the optimal profit $\pi _{g,max}$ is increasing with respect to f and its maximum value $\dfrac{1}{4}$ is obtained for any $f\ge \dfrac{\tilde{f}}{2}=\dfrac{2(1-\phi )^2}{\phi (2-\phi )^2}.$

• Let $\alpha \ne \alpha ^*$.

  The graph of $\pi _g^*$ with respect to the fine f is a parabola.

  It is

  $$\begin{aligned}\dfrac{d\pi _g^*}{df}&=\dfrac{\phi \left( \alpha (1-\alpha )(1-\phi )(2-\phi )+\left( \alpha (2-\phi )^2-4(1-\phi )\right) \phi f\right) }{2\alpha (1-\alpha )(1-\phi )^2}\\ \dfrac{d\pi _g^*}{df}(0)&=\dfrac{\phi (2-\phi )}{2(1-\phi )}>0\\&\dfrac{d\pi _g^*}{df}(\hat{f})=\dfrac{\phi ^2}{(1-\phi )(2-2\alpha +\alpha \phi )}>0 \end{aligned}$$

  Hence $\pi _g^*$ is increasing with respect to f for $f\in [0,\hat{f}].$

  $$\begin{aligned}&\pi _{g1}=\dfrac{\phi f}{\alpha (1-\phi )} \left( 1-\dfrac{\phi f}{\alpha (1-\phi )}\right) <\dfrac{1}{4}.\\&\dfrac{d\pi _{g1}}{df}= \dfrac{\phi (\alpha (1-\phi )-2\phi f)}{\alpha ^2(1-\phi )^2} \end{aligned}$$

  It is $\dfrac{d\pi _{g1}}{df}>0$ if $f<\dfrac{\tilde{f}}{2}.$

  Hence if $\alpha \ne \alpha ^*$ the optimal profit $\pi _{g,max}$ is increasing with respect to f and its maximum value $\dfrac{1}{4}$ is obtained for any $f\ge \dfrac{\tilde{f}}{2}.\;\quad\square$

In order to prove Proposition 4, we need the following preliminary Lemmas.

Lemma 1

Let $0\le p_g\le p_{g1}.$ Then $\pi _g=(p_g-c)(1-p_g).$ Consequently

• If $0\le f\le f_2$ then $\pi _g$ is increasing with respect to $p_g\in [0,p_{g1}]$ and reaches its maximum $\pi _{g1}$ at $p_g=p_{g1}.$

• If $f_2\le f\le \tilde{f}$ then $\pi _g$ is increasing with respect to $p_g\in [0,\hat{p}_g]$ and decreasing in $[\hat{p}_g,p_{g1}]. \pi _g$ reaches its maximum $\hat{\pi }_g$ at $p_g-\hat{p}_g.$

Proof

It follows from the first order conditions.

Lemma 2

Let $p_{g1}\le p_g\le p_{g2}.$ Then

$$\pi _g=(p_g-c)\left( \dfrac{1-\alpha - p_g+\dfrac{\phi f}{1-\phi }}{1-\alpha }\right) +\phi f \left( \dfrac{\alpha p_g-\dfrac{\phi f}{1-\phi }}{\alpha (1-\alpha )}\right).$$

Consequently

• $\dfrac{\partial \pi _g}{\partial p_g}=0$ if $p_g=p_g^*.$

• $\dfrac{\partial \pi _g}{\partial p_g}\left( p_g=p_{g1}\right) >0$ if $f<f_1.$

• $\dfrac{\partial \pi _g}{\partial p_g}\left( p_g=p_{g2}\right) >0$ if $f<f_4.$

Proof

It is trivial.

Lemma 3

Let

$$\omega _1=\dfrac{2-2\alpha +\alpha \phi }{2-\alpha \phi };\quad \omega _2=1-\alpha +\alpha \phi .$$

• If $0< \alpha <\tilde{\alpha }$ then $0<c1<c2<\omega _1<\omega _2<1;$

• if $\tilde{\alpha }< \alpha <1$ then $0<c2<c1<\omega _1<\omega _2<1.$

Proof

It is trivial.

Lemma 4

• Let $0< \alpha <\tilde{\alpha }.$

  • If $0<c<c_1$ then $f_4<0<f_1<f_2<\tilde{f};$

  • if $c_1<c<c_2$ then $f_4<0<f_2<f_1<\tilde{f};$

  • if $c_2<c<\omega _1$ then $0<f_4<f_2<f_1<\tilde{f};$

  • if $\omega _1<c<\omega _2$ then $0<f_2<f_4<f_1<\tilde{f};$

  • if $\omega _2<c\le 1$ then $0<f_2<\tilde{f}<f_1<f_4;$

• Let $\tilde{\alpha }< \alpha <1.$

  • If $0<c<c_2$ then $f_4<0<f_1<f_2<\tilde{f};$

  • if $c_2<c<c_1$ then $0<f_4<f_1<f_2<\tilde{f};$

  • if $c_1<c<\omega _1$ then $0<f_4<f_2<f_1<\tilde{f};$

  • if $\omega _1<c<\omega _2$ then $0<f_2<f_4<f_1<\tilde{f};$

  • if $\omega _2<c\le 1$ then $0<f_2<\tilde{f}<f_1<f_4;$

Proof

It is trivial.

Lemma 5

Note that the profit of the genuine producer is constant with respect to $p_g$ for $p_g\ge p_{g2}$; consequently, in order to find the optimal price of the genuine item, it is sufficient to examine the behaviour of $\pi _g$ with respect to $p_g$ for $0\le p_g\le p_{g2}.$

• 1. $0<\alpha <\tilde{\alpha }.$

  • Let $0\le c<c_1.$

    Then

    • If $0\le f\le f_1,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_g^*$ and decreasing for $p_g^*\le p_g\le p_{g2}$. We conclude that $\pi _g$ reaches its maximum at $p_g=p_g^*$.

    • If $f_1\le f\le f_2$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le p_{g1}, \pi _g$ is decreasing for $p_{g1}\le p_g\le p_{g2}$. We conclude that $\pi _g$ reaches its maximum at $p_g=p_{g1}$.

    • If $f_2\le f<\tilde{f}$ then then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is decreasing for $p_{g1}\le p_g\le p_{g2}$. We conclude that $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$.

  • Let $c_1\le c< c_2.$

    • If $0\le f\le f_2,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_g^*$ and decreasing for $p_g^*\le p_g\le p_{g2}$. We conclude that when $0\le f\le f_3, \pi _g$ reaches its maximum at $p_g=p_g^*$.

    • If $f_2\le f\le f_1,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_g^*$ and decreasing for $p_g^*\le p_g\le p_{g2}$. Therefore in this case $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$ or at $p_g=p_g^*$ and it is necessary to compare $\pi _g^*$ and $\hat{\pi }_g$.

    • If $f_1\le f\le \tilde{f}$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is decreasing with respect to $p_g$ for $p_{g1}\le p_g\le p_{g2}$. We conclude that $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$.

  • Let $c_2\le c< \omega _1.$

    • If $0\le f\le f_4,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_{g2}.$ We conclude that $\pi _g$ reaches its maximum at any $p_g\ge p_{g2}$.

    • If $f_4\le f\le f_2,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_g^*$ and decreasing for $p_g^*\le p_g\le p_{g2}$. We conclude that $\pi _g$ reaches its maximum at $p_g=p_g^*$.

    • If $f_2\le f\le f_1,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_g^*$ and decreasing for $p_g^*\le p_g\le p_{g2}$. Therefore in this case $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$ or at $p_g=p_g^*$ and it is necessary to compare $\pi _g^*$ and $\hat{\pi }_g$.

    • If $f_1\le f\le \tilde{f}$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is decreasing with respect to $p_g$ for $p_{g1}\le p_g\le p_{g2}$. We conclude that $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$.

  • Let $\omega _1\le c< \omega _2.$

    • If $0\le f\le f_2,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_{g2}.$ We conclude that when $0\le f\le f_1, \pi _g$ reaches its maximum at any $p_g\ge p_{g2}$.

    • If $f_2\le f\le f_4,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_{g2}.$ Therefore in this case $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$ or at any $p_g\ge p_{g2}$ and it is necessary to compare $\pi _{g,2}$ and $\hat{\pi }_g$.

    • If $f_4\le f\le f_1,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_g^*$ and decreasing for $p_g^*\le p_g\le p_{g2}$. Therefore in this case $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$ or at $p_g=p_g^*$ and it is necessary to compare $\pi _g^*$ and $\hat{\pi }_g$.

    • If $f_1\le f\le \tilde{f}$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is decreasing with respect to $p_g$ for $p_{g1}\le p_g\le p_{g2}$. We conclude that $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$.

  • Let $\omega _2\le c\le 1.$

    • If $0\le f\le f_2,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_{g2}.$ We conclude that $\pi _g$ reaches its maximum at any $p_g\ge p_{g2}$.

    • If $f_2\le f\le \tilde{f},$ then then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_{g2}.$ Therefore in this case $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$ or at any $p_g\ge p_{g2}$ and it is necessary to compare $\pi _{g,2}$ and $\hat{\pi }_g$.

• 2. $\tilde{\alpha }<\alpha <1.$

  • Let $0\le c<c_2.$

    • If $0\le f\le f_1,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_g^*$ and decreasing for $p_g^*\le p_g\le p_{g2}$. We conclude that $\pi _g$ reaches its maximum at $p_g=p_g^*$.

    • If $f_1\le f\le f_2$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le p_{g1}, \pi _g$ is decreasing for $p_{g1}\le p_g\le p_{g2}$. We conclude that $\pi _g$ reaches its maximum at $p_g=p_{g1}$.

    • If $f_2\le f<\tilde{f}$ then then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is decreasing for $p_{g1}\le p_g\le p_{g2}$. We conclude that $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$.

  • Let $c_2\le c<c_1.$

    • If $0\le f\le f_4,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_{g2}$. We conclude that $\pi _g$ reaches its maximum at any $p_g\ge p_{g2}$.

    • If $f_4\le f\le f_1,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_g^*$ and decreasing for $p_g^*\le p_g\le p_{g2}$. We conclude that $\pi _g$ reaches its maximum at $p_g=p_g^*$.

    • If $f_1\le f\le f_2$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le p_{g1}, \pi _g$ is decreasing for $p_{g1}\le p_g\le p_{g2}$. We conclude that $\pi _g$ reaches its maximum at $p_g=p_{g1}$.

    • If $f_2\le f<\tilde{f}$ then then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is decreasing for $p_{g1}\le p_g\le p_{g2}$. We conclude that $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$.

  • Let $c_1\le c< \omega _1.$

    • If $0\le f\le f_4,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_{g2}.$ We conclude that $\pi _g$ reaches its maximum at any $p_g\ge p_{g2}$.

    • If $f_4\le f\le f_2,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_g^*$ and decreasing for $p_g^*\le p_g\le p_{g2}$. We conclude that $\pi _g$ reaches its maximum at $p_g=p_g^*$.

    • If $f_2\le f\le f_1,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_g^*$ and decreasing for $p_g^*\le p_g\le p_{g2}$. Therefore in this case $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$ or at $p_g=p_g^*$ and it is necessary to compare $\pi _g^*$ and $\hat{\pi }_g$.

    • If $f_1\le f\le \tilde{f}$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is decreasing with respect to $p_g$ for $p_{g1}\le p_g\le p_{g2}$. We conclude that $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$.

  • Let $\omega _1\le c< \omega _2.$

    • If $0\le f\le f_2,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_{g2}.$ We conclude that when $0\le f\le f_1, \pi _g$ reaches its maximum at any $p_g\ge p_{g2}$.

    • If $f_2\le f\le f_4,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_{g2}.$ Therefore in this case $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$ or at any $p_g\ge p_{g2}$ and it is necessary to compare $\pi _{g,2}$ and $\hat{\pi }_g$.

    • If $f_4\le f\le f_1,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_g^*$ and decreasing for $p_g^*\le p_g\le p_{g2}$. Therefore in this case $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$ or at $p_g=p_g^*$ and it is necessary to compare $\pi _g^*$ and $\hat{\pi }_g$.

    • If $f_1\le f\le \tilde{f}$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is decreasing with respect to $p_g$ for $p_{g1}\le p_g\le p_{g2}$. We conclude that $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$.

  • Let $\omega _2\le c< 1.$

    • If $0\le f\le f_2,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_{g2}.$ We conclude that $\pi _g$ reaches its maximum at any $p_g\ge p_{g2}$.

    • If $f_2\le f\le \tilde{f},$ then then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_{g2}.$ Therefore in this case $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$ or at any $p_g\ge p_{g2}$ and it is necessary to compare $\pi _{g,2}$ and $\hat{\pi }_g$.

Proof

It is a consequence of Lemmas 1, 2, 3 and 4.

Lemma 6

Let

$$\begin{aligned}&\lambda _1=\dfrac{\alpha (1-\phi )-\sqrt{\alpha (1-\phi )}\sqrt{\alpha (1-\phi )-(1-c)^2}}{2\phi }.\\&\lambda _2=\dfrac{\alpha (1-\phi )+\sqrt{\alpha (1-\phi )}\sqrt{\alpha (1-\phi )-(1-c)^2}}{2\phi }. \end{aligned}$$

• If $0<c<c_4$ then $\pi _{g,2}<\hat{\pi }_g$ for any value of f;

• if $c_4<c$ then $\pi _{g,2}>\hat{\pi }_g$ for $\lambda _1<f<\lambda _2$ while $\pi _{g,2}<\hat{\pi }_g$ for $f<\lambda _1$ or $f>\lambda _2.$

Proof

Note that $\pi _{g,2}>\hat{\pi }_g$ if

$$4\phi ^2f^2-4\alpha \phi (1-\phi )f+ \alpha (1-\phi )(1-c)^2<0.$$

From Lemma 5 we need to compare $\hat{\pi }_g$ and $\pi _{g,2}$ in the cases

• $\omega _1<c<\omega _2$ and $f_2<f<f_4$;

• $\omega _2<c<1$ and $f_2<f<\tilde{f}.$

Lemma 7

• 1. $c_4<\omega _1$ for any $\alpha , \phi \in [0,1].$

• 2. $\lambda _1<f_2$ for any $0<\alpha <1, 0<\phi <1, c>0$.

• 3. $\lambda _2>f_2$ for any $0<\alpha <1, 0<\phi <1, c>\omega _1$.

• 4. $\lambda _2>f_4$ if $\omega _1<c<c_5, \lambda _2<f_4$ if $c_5<c<1.$

• 5. $\lambda _2<\tilde{f}$ for any $\alpha , \phi \in ]0,1[.$

Proof

• 1. $c_4<\omega _1$ if $4\alpha (1-\phi )-(2-\alpha \phi )^2<0$, that is $-4(1-\alpha )-\alpha ^2\phi ^2<0.$

• 2. $\lambda _1<f_2$ if $\sqrt{1-\phi }\sqrt{\alpha (1-\phi )-(1-c)^2}+c\sqrt{\alpha }(1-\phi )>0.$

• 3. $\lambda _2>f_2$ if

  $$\sqrt{1-\phi }\sqrt{\alpha (1-\phi )-(1-c)^2}> c\sqrt{\alpha }(1-\phi )$$

  or

  $$\left[ 1+\alpha (1-\phi )\right] c^2-2c+ \left[ 1-\alpha +\alpha \phi \right] <0$$

  that is if

  $$\dfrac{1-\alpha +\alpha \phi }{1+\alpha +\alpha \phi }<c<1.$$

  Since $\dfrac{1-\alpha +\alpha \phi }{1+\alpha +\alpha \phi }<\omega _1$ for any $\alpha , \phi \in ]0,1[$ we conclude that $\lambda _2>f_2$ for any $c>\omega _1.$

• 4. $\lambda _2<f_4$ if

  $$\sqrt{\alpha }\phi \sqrt{1-\phi } \sqrt{\alpha (1-\phi )-(1-c)^2}< (1-\phi )\left[ 2c+2\alpha -\alpha \phi -2\right]$$

  that is

  $$\left\{ \begin{array}{l} c>c_3\\ \alpha \phi ^2(1-\phi )\left[ \alpha (1-\phi )-(1-c)^2\right] < (1-\phi )^2\left[ 2c+2\alpha -\alpha \phi -2\right] ^2 \end{array} \right.$$

  or

  $$\begin{aligned} \left\{ \begin{array}{lcl} c>c_3&{}&{}\\ c<y_1&{}or&{}c>y_2 \end{array} \right. \end{aligned}$$

  where

  $$\begin{aligned}&y_1= \dfrac{-2\alpha \phi (1-\phi )\sqrt{1-\alpha }-\alpha (\phi ^2-6\phi +4)+4(1-\phi )}{\alpha \phi ^2+4(1-\phi )}.\\&y_2= \dfrac{2\alpha \phi (1-\phi )\sqrt{1-\alpha }-\alpha (\phi ^2-6\phi +4)+4(1-\phi )}{\alpha \phi ^2+4(1-\phi )}; \end{aligned}$$

  It is $y_1<\omega _1<y_2<\omega _2$ and $c_3<\omega _1$ for any $\alpha , \phi \in ]0,1][.$ Observing that $c_5=y_2$ completes the proof.

• 5. $\lambda _2<\tilde{f}$ if $\sqrt{1-\phi }\sqrt{\alpha (1-\phi )-(1-c)^2}<\sqrt{\alpha }(1-\phi )$ or $\alpha (1-\phi )-(1-c)^2<\alpha (1-\phi )$ or $-(1-c)^2<0.$

Lemma 8

• 1. Let $\omega _1<c<c_5.$ Then $\pi _{g,2}>\hat{\pi }_g$ for any $f\in ]f_2,f_4[.$

• 2. Let $c_5<c<\omega _2.$ Then

  • $\pi _{g,2}>\hat{\pi }_g$ if $f\in ]f_2,f_5[;$

  • $\pi _{g,2}<\hat{\pi }_g$ if $f\in ]f_5,f_4[;$

• 3. Let $\omega _2<c<1.$ Then

  • $\pi _{g,2}>\hat{\pi }_g$ if $f\in ]f_2,f_5[;$

  • $\pi _{g,2}<\hat{\pi }_g$ if $f\in ]f_5,\tilde{f}[.$

Proof

It is a consequence of Lemmas 6 and 7. Note that $f_5=\lambda _2$.

Lemma 9

Let $c>c_1$ and

$$\begin{aligned}&z_1=\dfrac{\alpha (1-\phi )\left\{ c\phi -(1-\alpha )(2-\phi )-\sqrt{1-\alpha }\left[ 2c-c\phi -\phi \right] \right\} }{\phi \left[ \alpha (2-\phi )^2-4(1-\phi )\right] };\\&z_2=\dfrac{\alpha (1-\phi )\left\{ c\phi -(1-\alpha )(2-\phi )+\sqrt{1-\alpha }\left[ 2c-c\phi -\phi \right] \right\} }{\phi \left[ \alpha (2-\phi )^2-4(1-\phi )\right] }. \end{aligned}$$

• If $0<\alpha <\alpha ^*$ then $z_2<z_1, \pi _g^*>\hat{\pi }_g$ for $z_2<f<z_1$ while $\pi _g^*<\hat{\pi }_g$ for $z<z_2$ or $z>z_1;$

• if $\alpha ^*<\alpha <1$ then $z_1<z_2, \pi _g^*<\hat{\pi }_g$ for $z_1<f<z_2$ while $\pi _g^*>\hat{\pi }_g$ for $z<z_1$ or $z>z_2.$

Proof

Note that $\pi _g^*<\hat{\pi }_g$ if

$$\phi ^2\left[ \alpha (2-\phi )^2-4(1-\phi )\right] f^2-2\alpha \phi (1-\phi )\left[ c\phi -(1-\alpha )(2-\phi )\right] f +\alpha ^2(1-\phi )^2\left[ c^2-(1-\alpha )\right] <0.$$

Solving the equation

$$\phi ^2\left[ \alpha (2-\phi )^2-4(1-\phi )\right] f^2-2\alpha \phi (1-\phi )\left[ c\phi -(1-\alpha )(2-\phi )\right] f +\alpha ^2(1-\phi )^2\left[ c^2-(1-\alpha )\right] =0$$

with respect to c we obtain the solutions $z_1$ and $z_2$. It is $z1<z_2$ if $\alpha >\alpha ^*$ and $z1>z_2$ if $\alpha <\alpha ^*.$

From Lemma 5 we need to compare $\hat{\pi }_g$ and $\pi _{g}^*$ in the cases

• $c_1<c<\omega _1$ and $f_2<f<f_1$;

• $\omega _1<c<\omega _2$ and $f_4<f<f_1.$

Lemma 10

Let $c>c_1$.

• 1. Let

  $$\sigma _1= \dfrac{2(1-\phi )(1-\alpha )[2-2\alpha +\alpha \phi ]-\alpha \phi ^2\sqrt{1-\alpha }}{4(1-\phi )(1-\alpha )-\alpha \phi (2-\phi )\sqrt{1-\alpha }}.$$

  It is $z_1<f_4$ for any $0<\alpha <1, 0<\phi <1, c>\sigma _1$.

• 2. $f_2<z_1$ for any $0<\alpha <1, 0<\phi <1, c>c_1$.

• 3. $z_1<f_1$ for any $0<\alpha <1, 0<\phi <1, c>c_1$.

• 4. Let

  $$\sigma _2= \dfrac{\alpha \phi ^2\sqrt{1-\alpha } +2(1-\alpha )(1-\phi )[\alpha \phi -2\alpha +2]}{\alpha \phi \sqrt{1-\alpha }(2-\phi )+4(1-\phi )(1-\alpha )}.$$

  It is $z_2<f_4$ if

  • $0<\alpha <\alpha ^*, 0<\phi <1, c>\sigma _2$;

  • $\alpha ^*<\alpha <1, 0<\phi <1, c<\sigma _2.$

• 5. It is $z_2<f_2$ if $\alpha <\alpha ^*$.

• 6. It is $z_2<f_1$ if $\alpha <\alpha ^*$.

Proof

• 1. Let

  $$\rho _1=\dfrac{4(1-\alpha )(1-\phi )-\alpha \phi (2-\phi )\sqrt{1-\alpha }}{\alpha (2-\phi )^2-4(1-\phi )}$$

  It is $z_1<f_4$ if ($\rho _1<0$ and $c>\sigma _1$) or ($\rho _1>0$ and $c<\sigma _1$).

  Since it is $\rho _1<0$ for any $\alpha , \phi \in ]0,1[$, it follows that $z_1<f_4$ if $c>\sigma _1$.

• 2. $f_2<z_1$ if

  $$\dfrac{\left[ 2\sqrt{1-\alpha }+2\alpha -\alpha \phi -2\right] \left[ c\phi -2c+\phi \right] }{\alpha (2-\phi )^2-4(1-\phi )}>0$$

  which is satisfied for any $0<\alpha <1, 0<\phi <1, c>c_1$.

• 3. $z_1<f_1$ if

  $$\dfrac{\left[ c\phi -2c+\phi \right] \left[ \sqrt{1-\alpha }(2-2\alpha +\alpha \phi )-2(1-\alpha )\right] }{\alpha (2-\phi )^2-4(1-\phi )}<0$$

  which is satisfied for any $0<\alpha <1, 0<\phi <1, c>c_1$.

• 4. Let

  $$\rho _2=\dfrac{-4(1-\alpha )(1-\phi )-\alpha \phi (2-\phi )\sqrt{1-\alpha }}{\alpha (2-\phi )^2-4(1-\phi )}$$

  It is $z_2<f_4$ if ($\rho _2<0$, and $c<\sigma _2$) or ($\rho _2>0$ and $c>\sigma _2$).

  Since it is $\rho _2<0$ if $\alpha >\alpha ^*$, it follows that $z_2<f_4$ if $c<\sigma _2$ and $\alpha >\alpha ^*$ or if $c>\sigma _2$ and $\alpha <\alpha ^*$.

• 5. It is trivial.

• 6. It is trivial.

Lemma 11

It is

$$\sigma _2<\omega _1<\sigma _2<\omega _1$$

for any $0<\alpha <1, 0<\phi <1$.

Proof

• 1. $\sigma _2<\omega _1$ if

  $$-\dfrac{2(1-\alpha )^{3/2}+(1-\alpha )[2-2\alpha +\alpha \phi ]}{\alpha \phi \sqrt{1-\alpha }(2-\phi )+4(1-\phi )(1-\alpha )}<0.$$

• 2. $\omega _1<\sigma _1$ if

  $$\dfrac{2\alpha \phi (1-\phi )(1-\alpha ) \left[ 2-2\alpha +\alpha \phi -2\sqrt{1-\alpha }\right] }{(2-\alpha \phi )\left[ \alpha \phi \sqrt{1-\alpha }(2-\phi )-4(1-\alpha )(1-\phi )\right] }>0.$$

  Note that the numerator and the denominator of the above fraction are both positive if $\alpha >\alpha ^*$ and both negative if $\alpha <\alpha ^*$.

• 3. $\sigma _1<\omega _2$ if

  $$\dfrac{-\left\{ \alpha \phi (1-\phi )\left[ \sqrt{1-\alpha }(2-2\alpha +\alpha \phi )-2(1-\alpha \right] \right\} }{4(1-\alpha )(1-\phi )-\alpha \phi \sqrt{1-\alpha }(2-\phi )}>0.$$

  Note that the numerator and the denominator of the above fraction are both positive if $\alpha <\alpha ^*$ and both negative if $\alpha >\alpha ^*$.

Lemma 12

• 1. Let $c_1<c<\omega _1.$ Then

  • $\pi _g^*>\hat{\pi }_g$ if $f\in ]f_2,z_1[;$

  • $\pi _g^*<\hat{\pi }_g$ if $f\in ]z_1,f_1[;$

• 2. Let $\omega _1<c<\sigma _1.$ Then

  • $\pi _g^*>\hat{\pi }_g$ if $f\in ]f_4,z_1[;$

  • $\pi _g^*<\hat{\pi }_g$ if $f\in ]z_1,f_1[;$

• 3. Let $\sigma _1<c<\omega _2.$ Then $\pi _g^*<\hat{\pi }_g$ if $f\in ]f_4,f_1[.$

Proof

It is a consequence of Lemmas 9, 10 and 11.

Now we are ready to prove Proposition 4.

Proof of Proposition 4

Since $f_3=z_1$ and since it can be shown that $\sigma _1=c_5$, from Lemma 5, 8 and 12 we obtain the following results.

Note that the profit of the genuine producer is constant with respect to $p_g$ for $p_g\ge p_{g2}$; consequently, in order to find the optimal price of the genuine item, it is sufficient to examine the behaviour of $\pi _g$ with respect to $p_g$ for $0\le p_g\le p_{g2}.$

• 1. $0<\alpha <\tilde{\alpha }.$

  • Let $0\le c<c_1.$

    Then

    • If $0\le f\le f_1,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_g^*$ and decreasing for $p_g^*\le p_g\le p_{g2}$. We conclude that $\pi _g$ reaches its maximum at $p_g=p_g^*$.

    • If $f_1\le f\le f_2$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le p_{g1}, \pi _g$ is decreasing for $p_{g1}\le p_g\le p_{g2}$. We conclude that $\pi _g$ reaches its maximum at $p_g=p_{g1}$.

    • If $f_2\le f<\tilde{f}$ then then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is decreasing for $p_{g1}\le p_g\le p_{g2}$. We conclude that $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$.

  • Let $c_1\le c< c_2.$

    • If $0\le f\le f_2,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_g^*$ and decreasing for $p_g^*\le p_g\le p_{g2}$. We conclude that when $0\le f\le f_3, \pi _g$ reaches its maximum at $p_g=p_g^*$.

    • If $f_2\le f\le f_3,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_g^*$ and decreasing for $p_g^*\le p_g\le p_{g2}$. Therefore in this case $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$ or at $p_g=p_g^*$. Comparing $\pi _g^*$ and $\hat{\pi }_g$ we conclude that $\pi _g$ reaches its maximum at $p_g=p_g^*.$

    • If $f_3\le f\le f_1,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_g^*$ and decreasing for $p_g^*\le p_g\le p_{g2}$. Therefore in this case $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$ or at $p_g=p_g^*$. Comparing $\pi _g^*$ and $\hat{\pi }_g$ we conclude that $\pi _g$ reaches its maximum at $p_g=\hat{p}_g.$

    • If $f_1\le f\le \tilde{f}$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is decreasing with respect to $p_g$ for $p_{g1}\le p_g\le p_{g2}$. We conclude that $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$.

  • Let $c_2\le c< \omega _1.$

    • If $0\le f\le f_4,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_{g2}.$ We conclude that $\pi _g$ reaches its maximum at any $p_g\ge p_{g2}$.

    • If $f_4\le f\le f_2,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_g^*$ and decreasing for $p_g^*\le p_g\le p_{g2}$. We conclude that $\pi _g$ reaches its maximum at $p_g=p_g^*$.

    • If $f_2\le f\le f_3,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_g^*$ and decreasing for $p_g^*\le p_g\le p_{g2}$. Therefore in this case $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$ or at $p_g=p_g^*$. Comparing $\pi _g^*$ and $\hat{\pi }_g$ we conclude that $\pi _g$ reaches its maximum at $p_g=p_g^*.$

    • If $f_3\le f\le f_1,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_g^*$ and decreasing for $p_g^*\le p_g\le p_{g2}$. Therefore in this case $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$ or at $p_g=p_g^*$. Comparing $\pi _g^*$ and $\hat{\pi }_g$ we conclude that $\pi _g$ reaches its maximum at $p_g=\hat{p}_g.$

    • If $f_1\le f\le \tilde{f}$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is decreasing with respect to $p_g$ for $p_{g1}\le p_g\le p_{g2}$. We conclude that $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$.

  • Let $\omega _1\le c< c_5.$

    • If $0\le f\le f_2,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_{g2}.$ We conclude that when $0\le f\le f_1, \pi _g$ reaches its maximum at any $p_g\ge p_{g2}$.

    • If $f_2\le f\le f_4,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_{g2}.$ Therefore in this case $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$ or at any $p_g\ge p_{g2}.$ Comparing $\hat{\pi }_g$ and $\pi _{g,2}$ we conclude that $\pi _g$ reaches its maximum at any $p_g\ge p_{g2}.$

    • If $f_4\le f\le f_3,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_g^*$ and decreasing for $p_g^*\le p_g\le p_{g2}$. Therefore in this case $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$ or at $p_g=p_g^*$. Comparing $\pi _g^*$ and $\hat{\pi }_g$ we conclude that $\pi _g$ reaches its maximum at $p_g=p_g^*.$

    • If $f_3\le f\le f_1,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_g^*$ and decreasing for $p_g^*\le p_g\le p_{g2}$. Therefore in this case $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$ or at $p_g=p_g^*$. Comparing $\pi _g^*$ and $\hat{\pi }_g$ we conclude that $\pi _g$ reaches its maximum at $p_g=\hat{p}_g.$

    • If $f_1\le f\le \tilde{f}$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is decreasing with respect to $p_g$ for $p_{g1}\le p_g\le p_{g2}$. We conclude that $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$.

  • Let $c_5\le c<\omega _2.$

    • If $0\le f\le f_2,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_{g2}.$ We conclude that when $0\le f\le f_1, \pi _g$ reaches its maximum at any $p_g\ge p_{g2}$.

    • If $f_2\le f\le f_5,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_{g2}.$ Therefore in this case $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$ or at any $p_g\ge p_{g2}.$ Comparing $\hat{\pi }_g$ and $\pi _{g,2}$ we conclude that $\pi _g$ reaches its maximum at any $p_g\ge p_{g2}.$

    • If $f_5\le f\le f_4,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_{g2}.$ Therefore in this case $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$ or at any $p_g\ge p_{g2}.$ Comparing $\pi _g^*$ and $\hat{\pi }_g$ we conclude that $\pi _g$ reaches its maximum at $p_g=\hat{p}_g.$

    • If $f_4\le f\le f_3,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_g^*$ and decreasing for $p_g^*\le p_g\le p_{g2}$. Therefore in this case $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$ or at $p_g=p_g^*$. Comparing $\pi _g^*$ and $\hat{\pi }_g$ we conclude that $\pi _g$ reaches its maximum at $p_g=\hat{p}_g.$

    • If $f_3\le f\le f_1,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_g^*$ and decreasing for $p_g^*\le p_g\le p_{g2}$. Therefore in this case $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$ or at $p_g=p_g^*$. Comparing $\pi _g^*$ and $\hat{\pi }_g$ we conclude that $\pi _g$ reaches its maximum at $p_g=\hat{p}_g.$

    • If $f_1\le f\le \tilde{f}$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is decreasing with respect to $p_g$ for $p_{g1}\le p_g\le p_{g2}$. We conclude that $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$.

  • Let $\omega _2\le c< 1.$

    • If $0\le f\le f_2,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_{g2}.$ We conclude that $\pi _g$ reaches its maximum at any $p_g\ge p_{g2}$.

    • If $f_2\le f\le f_5,$ then then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_{g2}.$ Therefore in this case $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$ or at any $p_g\ge p_{g2}.$ Comparing $\hat{\pi }_g$ and $\pi _{g,2}$ we conclude that $\pi _g$ reaches its maximum at any $p_g\ge p_{g2}.$

    • If $f_5\le f\le \tilde{f},$ then $\pi _g=$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_{g1}$. Therefore in this case $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$ or at any $p_g\ge p_{g2}.$ Comparing $\hat{\pi }_g$ and $\pi _{g,2}$ we conclude that $\pi _g$ reaches its maximum at at $p_g=\hat{p}_g$.

  • $\tilde{\alpha }<\alpha <1.$

    • Let $0\le c<c_2.$

      • If $0\le f\le f_1,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_g^*$ and decreasing for $p_g^*\le p_g\le p_{g2}$. We conclude that $\pi _g$ reaches its maximum at $p_g=p_g^*$.

      • If $f_1\le f\le f_2$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le p_{g1}, \pi _g$ is decreasing for $p_{g1}\le p_g\le p_{g2}$. We conclude that $\pi _g$ reaches its maximum at $p_g=p_{g1}$.

      • If $f_2\le f<\tilde{f}$ then then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is decreasing for $p_{g1}\le p_g\le p_{g2}$. We conclude that $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$.

    • Let $c_2\le c<c_1.$

      • If $0\le f\le f_4,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_{g2}$. We conclude that $\pi _g$ reaches its maximum at any $p_g\ge p_{g2}$.

      • If $f_4\le f\le f_1,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_g^*$ and decreasing for $p_g^*\le p_g\le p_{g2}$. We conclude that $\pi _g$ reaches its maximum at $p_g=p_g^*$.

      • If $f_1\le f\le f_2$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le p_{g1}, \pi _g$ is decreasing for $p_{g1}\le p_g\le p_{g2}$. We conclude that $\pi _g$ reaches its maximum at $p_g=p_{g1}$.

      • If $f_2\le f<\tilde{f}$ then then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is decreasing for $p_{g1}\le p_g\le p_{g2}$. We conclude that $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$.

    • Let $c_1\le c< \omega _1.$

      • If $0\le f\le f_4,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_{g2}.$ We conclude that $\pi _g$ reaches its maximum at any $p_g\ge p_{g2}$.

      • If $f_4\le f\le f_2,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_g^*$ and decreasing for $p_g^*\le p_g\le p_{g2}$. We conclude that $\pi _g$ reaches its maximum at $p_g=p_g^*$.

      • If $f_2\le f\le f_3,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_g^*$ and decreasing for $p_g^*\le p_g\le p_{g2}$. Therefore in this case $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$ or at $p_g=p_g^*$. Comparing $\pi _g^*$ and $\hat{\pi }_g$ we conclude that $\pi _g$ reaches its maximum at $p_g=p_g^*.$

      • If $f_3\le f\le f_1,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_g^*$ and decreasing for $p_g^*\le p_g\le p_{g2}$. Therefore in this case $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$ or at $p_g=p_g^*$. Comparing $\pi _g^*$ and $\hat{\pi }_g$ we conclude that $\pi _g$ reaches its maximum at $p_g=\hat{p}_g.$

      • If $f_1\le f\le \tilde{f}$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is decreasing with respect to $p_g$ for $p_{g1}\le p_g\le p_{g2}$. We conclude that $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$.

    • Let $\omega _1\le c< c_5.$

      • If $0\le f\le f_2,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_{g2}.$ We conclude that when $0\le f\le f_1, \pi _g$ reaches its maximum at any $p_g\ge p_{g2}$.

      • If $f_2\le f\le f_4,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_{g2}.$ Therefore in this case $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$ or at any $p_g\ge p_{g2}.$ Comparing $\hat{\pi }_g$ and $\pi _{g,2}$ we conclude that $\pi _g$ reaches its maximum at any $p_g\ge p_{g2}.$

      • If $f_4\le f\le f_3,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_g^*$ and decreasing for $p_g^*\le p_g\le p_{g2}$. Therefore in this case $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$ or at $p_g=p_g^*$. Comparing $\pi _g^*$ and $\hat{\pi }_g$ we conclude that $\pi _g$ reaches its maximum at $p_g=p_g^*.$

      • If $f_3\le f\le f_1,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_g^*$ and decreasing for $p_g^*\le p_g\le p_{g2}$. Therefore in this case $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$ or at $p_g=p_g^*$. Comparing $\pi _g^*$ and $\hat{\pi }_g$ we conclude that $\pi _g$ reaches its maximum at $p_g=\hat{p}_g.$

      • If $f_1\le f\le \tilde{f}$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is decreasing with respect to $p_g$ for $p_{g1}\le p_g\le p_{g2}$. We conclude that $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$.

    • Let $c_5\le c<\omega _2.$

      • If $0\le f\le f_2,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_{g2}.$ We conclude that when $0\le f\le f_1, \pi _g$ reaches its maximum at any $p_g\ge p_{g2}$.

      • If $f_2\le f\le f_5,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_{g2}.$ Therefore in this case $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$ or at any $p_g\ge p_{g2}.$ Comparing $\hat{\pi }_g$ and $\pi _{g,2}$ we conclude that $\pi _g$ reaches its maximum at any $p_g\ge p_{g2}.$

      • If $f_5\le f\le f_4,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_{g2}.$ Therefore in this case $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$ or at any $p_g\ge p_{g2}.$ Comparing $\pi _g^*$ and $\hat{\pi }_g$ we conclude that $\pi _g$ reaches its maximum at $p_g=\hat{p}_g.$

      • If $f_4\le f\le f_1,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_g^*$ and decreasing for $p_g^*\le p_g\le p_{g2}$. Therefore in this case $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$ or at $p_g=p_g^*$. Comparing $\pi _g^*$ and $\hat{\pi }_g$ we conclude that $\pi _g$ reaches its maximum at $p_g=\hat{p}_g.$

      • If $f_1\le f\le \tilde{f}$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is decreasing with respect to $p_g$ for $p_{g1}\le p_g\le p_{g2}$. We conclude that $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$.

    • Let $\omega _2\le c< 1.$

      • If $0\le f\le f_2,$ then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_{g2}.$ We conclude that $\pi _g$ reaches its maximum at any $p_g\ge p_{g2}$.

      • If $f_2\le f\le f_5,$ then then $\pi _g$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_{g2}.$ Therefore in this case $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$ or at any $p_g\ge p_{g2}.$ Comparing $\hat{\pi }_g$ and $\pi _{g,2}$ we conclude that $\pi _g$ reaches its maximum at any $p_g\ge p_{g2}.$

      • If $f_5\le f\le \tilde{f},$ then $\pi _g=$ is increasing with respect to $p_g$ for $0\le p_g\le \hat{p}_g$ and decreasing for $\hat{p}_g\le p_g\le p_{g1}, \pi _g$ is increasing with respect to $p_g$ for $p_{g1}\le p_g\le p_{g1}$. Therefore in this case $\pi _g$ reaches its maximum at $p_g=\hat{p}_g$ or at any $p_g\ge p_{g2}.$ Comparing $\hat{\pi }_g$ and $\pi _{g,2}$ we conclude that $\pi _g$ reaches its maximum at at $p_g=\hat{p}_g$.

In order to prove Proposition 5, we need the following preliminary Lemmas.

Lemma 13

It is

• 1.
  $$\dfrac{\partial \pi _g^*}{\partial f}>0\iff \left\{ \begin{array}{lcl} f<f_6&{} {and} &{}0<\alpha <\alpha ^*;\\ f>f_6&{} {and} &{}\alpha ^*<\alpha <1.\\ \end{array} \right.$$
• 2.
  $$\dfrac{\partial \pi _{g,2}}{\partial f}>0\iff f<\dfrac{\tilde{f}}{2}.$$
• 3.
  $$\dfrac{\partial \pi _{g1}}{\partial f}>0\iff f<f_2.$$

Proof

It is trivial.

Lemma 14

Let

$$\omega _3=\dfrac{(1-\alpha )(2-\phi )}{\phi }.$$

It is

• 1. $c_2<c_3<c_5$ for any $\alpha , \phi \in ]0,1[;$

• 2. $c_1<c_3\iff \alpha <\alpha ^*;$

• 3. $c_2<\omega _3$ for any $\alpha , \phi \in ]0,1[;$

• 4. $c_1<\omega _3\iff \alpha <\alpha ^*;$

• 5. $c_4<c_3\iff \alpha <\alpha ^*;$

• 6. $c_1<c_4\iff \alpha <\alpha ^*.$

Proof

It is trivial.

Lemma 15

It is $c_4<c_5$ for any $\alpha , \phi \in ]0,1[.$

Proof

$c_5-c_4>0\iff$

$$2\sqrt{\alpha }(1-\phi )(2-\phi )< 2\sqrt{\alpha }\phi \sqrt{1-\alpha }(1-\phi )+ \sqrt{1-\phi }\left[ \alpha \phi ^2+4(1-\phi )\right]$$

or

$$2\sqrt{\alpha }\sqrt{1-\phi }\left[ 2-\phi -\phi \sqrt{1-\alpha }\right] < \alpha \phi ^2+4(1-\phi ).$$

Squaring both sides of the above inequality, after some algebraic manipulations we obtain

$$8\alpha \phi (1-\phi )(2-\phi )\sqrt{1-\alpha }+ \alpha ^2\phi ^2\left[ \phi ^2+4(1-\phi )\right] + 16(1-\phi )^2(1-\alpha )>0$$

which is true for any $\alpha , \phi \in ]0,1[.$

Lemma 16

It is

• 1.
  $$f_6>0\iff \left\{ \begin{array}{lcl} c<\omega _3&{} {and} &{}0<\alpha <\alpha ^*;\\ c>\omega _3&{} {and} &{}\alpha ^*<\alpha <1.\\ \end{array} \right.$$
• 2.
  $$f_6<f_1\iff \left\{ \begin{array}{lcl} c>c_1&{} {and} &{}0<\alpha <\alpha ^*;\\ c<c_1&{} {and} &{}\alpha ^*<\alpha <1.\\ \end{array} \right.$$
• 3.
  $$f_6<f_3\iff \left\{ \begin{array}{lcl} c>c_1&{} {and} &{}0<\alpha <\alpha ^*;\\ c<c_1&{} {and} &{}\alpha ^*<\alpha <1.\\ \end{array} \right.$$
• 4.
  $$f_6<f_4\iff \left\{ \begin{array}{lcl} c>c_3&{} {and} &{}0<\alpha <\alpha ^*;\\ c<c_3&{} {and} &{}\alpha ^*<\alpha <1.\\ \end{array} \right.$$
• 5.
  $$\dfrac{\tilde{f}}{2}<f_4\iff c>c_3.$$
• 6.
  $$\dfrac{\tilde{f}}{2}<f_5\iff c>c_4.$$

Proof

It is trivial.

Now we are ready to prove Proposition 5.

Proof of Proposition 5

From Lemmas 13, 14, 15 and 16 and from Proposition 4 we obtain the following results.

• 1. $0<\alpha <\tilde{\alpha }.$

  • Let $0\le c<c_1.$

    Then

    $$\pi _{g,max}(f)= \left\{ \begin{array}{lcll} \pi _g^*(f)&{}if&{}0<f<f_1;&{} increasing;\\ \pi _{g1}(f)&{}if&{}f_1<f<f_2;&{} increasing;\\ \hat{\pi }_g (f)&{}if&{}f_2<f;&{} constant.\\ \end{array} \right.$$

    Therefore the highest profit for the genuine producer is guaranteed by any $f\ge f_{opt}=f_2$. The ensuing profit is $\pi _{opt}=\hat{\pi }_g.$

  • Let $c_1\le c<c_2.$

    Then

    $$\pi _{g,max}(f)= \left\{ \begin{array}{lcll} \pi _g^*(f)&{}if&{}0<f<f_3;&{} increasing\; in\; [0,f_6] \; and\; decreasing\; in [f_6,f_3];\\ \hat{\pi }_g (f)&{}if&{}f_3<f;&{} constant.\\ \end{array} \right.$$

    Therefore the highest profit for the genuine producer is guaranteed by $f=f_6$. The ensuing profit is $\pi _{opt}=\pi _g^*(f=f_6).$

  • Let $c_2\le c<c_3.$

    Then

    $$\pi _{g,max}(f)= \left\{ \begin{array}{lcll} \pi _{g,2}(f)&{}if&{}0<f<f_4;&{} increasing;\\ \pi _g^*(f)&{}if&{}f_4<f<f_3;&{} increasing\; in\; [f_4,f_6] \; and\; decreasing\; in [f_6,f_3];\\ \hat{\pi }_g (f)&{}if&{}f_3<f;&{} constant.\\ \end{array} \right.$$

    Therefore the highest profit for the genuine producer is guaranteed by $f=f_6$. The ensuing profit is $\pi _{opt}=\pi _g^*(f=f_6).$

  • Let $c_3\le c<c_5.$

    Then

    $$\pi _{g,max}(f)=\left\{ \begin{array}{lcll} \pi _{g,2}(f)&{}if&{}0<f<f_4;&{}increasing\; in\; \left[ 0,\,\dfrac{\tilde{f}}{2}\right] \; and\; decreasing\; in \left[ \dfrac{\tilde{f}}{2},f_4\right] ;\\ \pi _g^*(f)&{}if&{}f_4<f<f_3;&{} decreasing;\\ \hat{\pi }_g (f)&{}if&{}f_3<f;&{} constant.\\ \end{array} \right.$$

    Therefore the highest profit for the genuine producer is guaranteed by $f=\dfrac{\tilde{f}}{2}$. The ensuing profit is $\pi _{opt}=\pi _{g,2}\left( f=\dfrac{\tilde{f}}{2}\right) .$

  • Let $c_5\le c\le 1.$

    Then

    $$\pi _{g,max}(f)= \left\{ \begin{array}{lcll} \pi _{g,2}(f)&{}if&{}0<f<f_5;&{}increasing\; in\; \left[ 0,\,\dfrac{\tilde{f}}{2}\right] \; and\; decreasing\; in \left[ \dfrac{\tilde{f}}{2},f_5\right] ;\\ \hat{\pi }_g (f)&{}if&{}f_5<f;&{} constant.\\ \end{array} \right.$$

    Therefore the highest profit for the genuine producer is guaranteed by $f=\dfrac{\tilde{f}}{2}$. The ensuing profit is $\pi _{opt}=\pi _{g,2}\left( f=\dfrac{\tilde{f}}{2}\right) .$

• 2. $\tilde{\alpha }<\alpha <\alpha ^*.$

  • Let $0\le c<c_2.$

    Then

    $$\pi _{g,max}(f)= \left\{ \begin{array}{lcll} \pi _g^*(f)&{}if&{}0<f<f_1;&{} increasing;\\ \pi _{g1}(f)&{}if&{}f_1<f<f_2;&{} increasing;\\ \hat{\pi }_g (f)&{}if&{}f_2<f;&{} constant.\\ \end{array} \right.$$

    Therefore the highest profit for the genuine producer is guaranteed by any $f\ge f_{opt}=f_2$. The ensuing profit is $\pi _{opt}=\hat{\pi }_g.$

  • Let $c_2\le c<c_1.$

    Then

    $$\pi _{g,max}(f)= \left\{ \begin{array}{lcll} \pi _{g,2}(f)&{}if&{}0<f<f_4;&{} increasing;\\ \pi _g^*(f)&{}if&{}f_4<f<f_1;&{} increasing;\\ \pi _{g1}(f)&{}if&{}f_1<f<f_2;&{} increasing;\\ \hat{\pi }_g (f)&{}if&{}f_2<f;&{} constant.\\ \end{array} \right.$$

    Therefore the highest profit for the genuine producer is guaranteed by any $f\ge f_{opt}=f_2$. The ensuing profit is $\pi _{opt}=\hat{\pi }_g.$

  • Let $c_1\le c<c_3.$

    Then

    $$\pi _{g,max}(f)= \left\{ \begin{array}{lcll} \pi _{g,2}(f)&{}if&{}0<f<f_4;&{} increasing;\\ \pi _g^*(f)&{}if&{}f_4<f<f_3;&{} increasing\; in\; [f_4,f_6] \; and\; decreasing\; in [f_6,f_3];\\ \hat{\pi }_g (f)&{}if&{}f_3<f;&{} constant.\\ \end{array} \right.$$

    Therefore the highest profit for the genuine producer is guaranteed by $f=f_6$. The ensuing profit is $\pi _{opt}=\pi _g^*(f=f_6).$

  • Let $c_3\le c<c_5.$

    Then

    $$\pi _{g,max}(f)= \left\{ \begin{array}{lcll} \pi _{g,2}(f)&{}if&{}0<f<f_4;&{}increasing\; in\; \left[0,\,\dfrac{\tilde{f}}{2}\right] \; and\; decreasing\; in \left[ \dfrac{\tilde{f}}{2},f_4\right] ;\\ \pi _g^*(f)&{}if&{}f_4<f<f_3;&{} decreasing;\\ \hat{\pi }_g (f)&{}if&{}f_3<f;&{} constant.\\ \end{array} \right.$$

    Therefore the highest profit for the genuine producer is guaranteed by $f=\dfrac{\tilde{f}}{2}$. The ensuing profit is $\pi _{opt}=\pi _{g,2}\left( f=\dfrac{\tilde{f}}{2}\right) .$

  • Let $c_5\le c\le 1.$

    Then

    $$\pi _{g,max}(f)= \left\{ \begin{array}{lcll} \pi _{g,2}(f)&{}if&{}0<f<f_5;&{}increasing\; in\; \left[ 0,\,\dfrac{\tilde{f}}{2}\right] \; and\; decreasing\; in \left[ \dfrac{\tilde{f}}{2},f_5\right] ;\\ \hat{\pi }_g (f)&{}if&{}f_5<f;&{} constant.\\ \end{array} \right.$$

    Therefore the highest profit for the genuine producer is guaranteed by $f=\dfrac{\tilde{f}}{2}$. The ensuing profit is $\pi _{opt}=\pi _{g,2}\left( f=\dfrac{\tilde{f}}{2}\right) .$

• 3. $\alpha ^*<\alpha <1.$

  • Let $0\le c<c_2.$

    Then

    $$\pi _{g,max}(f)= \left\{ \begin{array}{lcll} \pi _g^*(f)&{}if&{}0<f<f_1;&{} decreasing\; in\; [0,f_6] \; and\; increasing\; in [f_6,f_1];;\\ \pi _{g1}(f)&{}if&{}f_1<f<f_2;&{} increasing;\\ \hat{\pi }_g (f)&{}if&{}f_2<f;&{} constant.\\ \end{array} \right.$$

    Therefore the highest profit for the genuine producer is guaranteed by any $f\ge f_{opt}=f_2$. The ensuing profit is $\pi _{opt}=\hat{\pi }_g.$

  • Let $c_2\le c<c_3.$

    Then

    $$\pi _{g,max}(f)= \left\{ \begin{array}{lcll} \pi _{g,2}(f)&{}if&{}0<f<f_4;&{} increasing;\\ \pi _g^*(f)&{}if&{}f_4<f<f_1;&{} increasing;\\ \pi _{g1}(f)&{}if&{}f_1<f<f_2;&{} increasing;\\ \hat{\pi }_g (f)&{}if&{}f_2<f;&{} constant.\\ \end{array} \right.$$

    Therefore the highest profit for the genuine producer is guaranteed by any $f\ge f_{opt}=f_2$. The ensuing profit is $\pi _{opt}=\hat{\pi }_g.$

  • Let $c_3\le c<c_4.$

    Then

    $$\pi _{g,max}(f)= \left\{ \begin{array}{lcll} \pi _{g,2}(f)&{}if&{}0<f<f_4;&{} increasing\; in\; \left[ 0,\,\dfrac{\tilde{f}}{2}\right] \; and\; decreasing\; in \left[ \dfrac{\tilde{f}}{2},f_4\right] ;\\ \pi _g^*(f)&{}if&{}f_4<f<f_1;&{} decreasing\; in\; [f_4,f_6] \; and\; increasing\; in [f_6,f_1];\\ \pi _{g1}(f)&{}if&{}f_1<f<f_2;&{} increasing;\\ \hat{\pi }_g (f)&{}if&{}f_2<f;&{} constant.\\ \end{array} \right.$$

    Comparing $\pi _{g,2}\left( f=\dfrac{\tilde{f}}{2}\right)$ and $\hat{\pi }_g$ we obtain that the highest profit for the genuine producer is guaranteed by any $f\ge f_{opt}=f_2$. The ensuing profit is $\pi _{opt}=\hat{\pi }_g.$

  • Let $c_4\le c<c_1.$

    Then

    $$\pi _{g,max}(f)= \left\{ \begin{array}{lcll} \pi _{g,2}(f)&{}if&{}0<f<f_4;&{} increasing\; in\; \left[ 0,\,\dfrac{\tilde{f}}{2}\right] \; and\; decreasing\; in \left[ \dfrac{\tilde{f}}{2},f_4\right] ;\\ \pi _g^*(f)&{}if&{}f_4<f<f_1;&{} decreasing\; in\; [f_4,f_6] \; and\; increasing\; in [f_6,f_1];\\ \pi _{g1}(f)&{}if&{}f_1<f<f_2;&{} increasing;\\ \hat{\pi }_g (f)&{}if&{}f_2<f;&{} constant.\\ \end{array} \right.$$

    Comparing $\pi _{g,2}\left( f=\dfrac{\tilde{f}}{2}\right)$ and $\hat{\pi }_g$ we obtain that the highest profit for the genuine producer is guaranteed by $f=\dfrac{\tilde{f}}{2}$. The ensuing profit is $\pi _{g,2}\left( f=\dfrac{\tilde{f}}{2}\right) .$

  • Let $c_1\le c<c_5.$

    Then

    $$\pi _{g,max}(f)= \left\{ \begin{array}{lcll} \pi _{g,2}(f)&{}if&{}0<f<f_4;&{} increasing\; in\; \left[ 0,\,\dfrac{\tilde{f}}{2}\right] \; and\; decreasing\; in \left[ \dfrac{\tilde{f}}{2},f_4\right] ;\\ \pi _g^*(f)&{}if&{}f_4<f<f_3;&{} decreasing\; in [f_4,f_3];\\ \hat{\pi }_g (f)&{}if&{}f_3<f;&{} constant.\\ \end{array} \right.$$

    Therefore the highest profit for the genuine producer is guaranteed by $f=\dfrac{\tilde{f}}{2}$. The ensuing profit is $\pi _{g,2}\left( f=\dfrac{\tilde{f}}{2}\right) .$

  • Let $c_5\le c\le 1.$

    Then

    $$\pi _{g,max}(f)= \left\{ \begin{array}{lcll} \pi _{g,2}(f)&{}if&{}0<f<f_5;&{}increasing\; in\; [0,\,\dfrac{\tilde{f}}{2}] \; and\; decreasing\; in [\dfrac{\tilde{f}}{2},f_5];\\ \hat{\pi }_g (f)&{}if&{}f_5<f;&{} constant.\\ \end{array} \right.$$

    Therefore the highest profit for the genuine producer is guaranteed by $f=\dfrac{\tilde{f}}{2}$. The ensuing profit is $\pi _{opt}=\pi _{g,2}\left( f=\dfrac{\tilde{f}}{2}\right) .$

Proof of Proposition 6

• Let $0\le c<1-\alpha .$

  From (4) and (5) we have that $W(\phi =0)>W(\phi =1)$ iff $M>M^*$, where

  $$M^*= \dfrac{\alpha \left( (1-\alpha )(2-5q)-(2-q)c^2\right) }{8(1-\alpha )}$$

  It is $\dfrac{\partial M^*}{\partial q}= \dfrac{\alpha (c^2-5(1-\alpha ))}{8(1-\alpha )}<0$ for any $c\in [0,1-\alpha ]$. Therefore $M^*$ is a decreasing function of q. It is $M^*>0$ iff $q<q^*$ where

  $$q^*=\dfrac{2(1-\alpha -c^2)}{ 5(1-\alpha )-c^2}$$

  It is $0<q^*<1$ for any $c\in [0,1-\alpha [.$

• Let $1-\alpha \le c\le 1.$

  From (4) and (6) we have that $W(\phi =0)>W(\phi =1)$ iff $M>M^*$, where

  $$M^*= \dfrac{(2-q)c^2-2(2-q)c+2-q-4\alpha q}{8}$$

  It is $\dfrac{\partial M^*}{\partial q}=- \dfrac{4\alpha +(1-c)^2}{8}<0$ for any $c\in [1-\alpha ,1]$. Therefore $M^*$ is a decreasing function of q. It is $M^*>0$ iff $q<q^*$ where

  $$q^*=\dfrac{2(1-c)^2}{(1-c)^2+4\alpha }$$

  It is $0<q^*<1$ for any $c\in [1-\alpha ,1].$